gabor j szekely contests in higher mathematics mikl ¦s schweitzer Part II by Leon Petrakovsky

3.3 THEORY OF FUNCTIONS

237

where the convention ao := 0 and its consequence Da0 = -a1 are used. It follows that the series (2) is convergent: 00

E ak sin kx = E Dk (x) Dak =: f (X) k=1

(7)

k=0

for every x with the possible exception of the case x = 0 (mod 27r). In the following, we make use of an inequality of the Sidon type: for any integer n > 2 and numerical sequence {bk}, it

1/2

2n-1

E bkDk(x) dx < C > kbk

ir/n

k=n

(8)

k=n

where C is a positive constant. To see this, we first apply the CauchySchwarz inequality and then exploit the orthogonality of the system {cos (k +

x}:

2n-1

rir

2n-1

E bkDk(x) dx=1/n k=n 1/2

T'n

2)2 1/2

bkcos l k+ 2 l x k=n

\\\

1/2

2n-1

k=n

cos (k + 2) x dx 2 sin 2

2n-1

Cn- > bk

k=n

(2 sin

(2n_1

kbk

k=n

Let s > 1 be an integer. According to (7),

j-1 > Dk (x)Dak dx

2'-1

(x) dx >

j=1

J'r/(7+1)

2'-1

k=0

E Dk(x)Dak dx := I1 - I2 . 7=1

3T k=7

Using the inequality

<k+1 (0 <x <ir; k=0,1,...), we obtain that 2'-1

I1 >

it/J

9-1

l(7+1) k=0 = I11 - 112 .

Dak X- -

2'-1

9=1

ir/7

7-1

> (k + 1) jakl dx

fit/(j+1) k=0

(9)

3: SOLUTIONS TO THE PROBLEMS

238

Since

1+1)'

ln(1+i J -

i(i

by the foregoing 2' Ill >

7(7a1 + 1)/

It is easy to see that 2°-1

sl0akl,

j=1.7(7+1)

j=1

whence

2°-1

i k=1 lAakl

Ill >

ld l j=1

Similarly, 2"-1j-1 V-1

y(7

112 = 7r j=1

k= =0

Therefore

l0ak1 < 7rE lLak1 < 27r ILakl .

+ 1)

k=0

2°-1

k=1 00

laj l

I1 1 E

_ (1 + 27r) E l0akl .

j=1

(10)

k=1

We turn to estimating 12: 2'-1

I2= E f1=1 j=21-1 2''--1

1=1 a

2'-1

2^-1 \

,r/'

2'-1

, /(7+1)

k=j +n=l}1,

it/j

2'-1

/(j+1)

k=j

Dk(x)Aak

2^-1

7r/j

j=21-1 n=1+1

L,+1

Dk(x)Dak dx =: I21 + 122 )

k=2^-'

Making use of the elementary inequality

sinx>

-x

( 0<x< 2)

we obtain 8

I21 !5

2'-1

7r/j

1=1 j=21-1 >r9 /(+1) 2 sin 2 k=21-1 g

E !=1

2'-1

7r2-1+1 *7r

2-- -

k=z'-1

lAakldx

l0ak l dx 7r 1n2 9

2"-1

1: lAakl k=1

3.3 THEORY OF FUNCTIONS

239

We now apply (8): 2-t+1

122 = E

Dk(x)Aak dx

L_1

1=1 n=1+1

0o n -1

1r 2

2n-1 k=2n-1 2n-1

-1+1

Dk(X)Dak dx

n=21=1 2 i

k=2n-1 2" -1

- n2

k-2n 1

2n-1

n=2

k=2"-1

Dk(x)Lak dx 1/2

E kILakI2

Consequently, ir In 2 12 <-

2째-1

1/2

2n-1

kkakI+C> E klDakI2

(11)

k=2n-1

n=2

k=1

Relying on (9), (10), and (11), we find that 2째-1

ir2--

jaki

E ak sin kx k=1

k=1 o0

2-1

n=2

k=2n-1

-CE

k=1

1/2

kIDakI2

In view of conditions (3), (4), and (5), this already yields the desired conclusion.

Problem F.56. Let h : [0, oc) --> [0, oo) be a measurable, locally integrable function, and write

H(t) :=

t h(s)ds

(t >_ 0).

Prove that if there is afconstant Bfo with H(t) < Bt2 for all t, then

e-H(t)

t eH(u)du

dt = oo.

Solution 1. Interchanging the order of (integration, we obtain oo

e-H(t) J0

t eH(u) du dt =

J J 0o 0

= L째째 Ju f00

H(t)-H(u)] du dt

e[-H(t)-H(u)] dt du

fexp L- J h(s) dsI

3. SOLUTIONS TO THE PROBLEMS

240

It is sufficient to prove the existence of a constant c > 0 such that

ft h(s)ds] dtdu>c I(T):=exp[-J f T

u+1/T

IT,

for large T. We have u+1/T

I(T) >

eXp - u

IT,

u+1/T

j2T

u+1/T

h(s) ds du.

exp

h(s)ds dtdu

Introduce the notation u+1/T

h(s) ds > 10B

u E [T, 2T] : f

Qt :=

and find an upper estimate for the Lebesgue measure µ(QT) of the set QT: T

T+1/T

u+1/T

10Bp(T) < IT, f

-1/T h(s) duds

h(s) ds du < IT,

=Th(s)ds<B+ (2T j 2T+1/T

whence µ(QT) !5

C2T +

T+

Therefore,

1(T)>

e-los du >

e-10B

T J[T,2T]\QT

> e- 10B

(T - µ(QT))

1--) 1=: C>0'

provided that T > 1.

Solution 2. We first show that if g : [1, oo) -* (0, oo) is measurable and locally integrable, further t

f g(s) ds < B, t2

(B1 = constant)

for every t, then

11 ds = oo. 1

g(s)

3.3 THEORY OF FUNCTIONS

241

Indeed, let T > 1 be any real number. By the Bunyakovski-Schwarz inequality

(12T T2 =

(j27' 1 dt

'1g (t)

9(t) dt

(j2Tg(t)dt) (12T1) g(t)

< B1 4T2 fT

g(t)

dt .

So 2T

dt >

g(t)

which yields the assertion. Let g(t)

(T > 0),

4B1 1

eH(t) fot

(t > 1).

eH(u) du

According to the problem, we have to prove that rÂ°Â°

g(t) dt = oo.

To this end, by the above remark, it is sufficient to verify the inequality t

(t > 1)

g(s) ds < Blt2 1

with a suitable B1. This is simple:

ft g(s) = J 1

eH(e)

f0 eH(u) du t

eH(u) du.

= In f eH(") du - In 0

Since H is monotone increasing,

f with a suitable B1.

g(s)

In [teH(t)] = lnt + < Bt2 + In t < B1 t2

3. SOLUTIONS TO THE PROBLEMS

242

Problem F.57. Consider the equation f'(x) = f (x + 1). Prove that (a) each solution f : [0, oo) -> (0, oo) has an exponential order of growth, that is, there exist numbers a > 0, b > 0 satisfying If (x) I < a ebx,

x>0;

(b) there are solutions f : [0, oo) -+ (-oo, oo) of nonexponential order of growth.

Solution. In the statement of the problem, a constant c has been omitted. The statement is correctly, f'(x) = cf (x + 1), where 0 < c < 1/e. Then there exists a A > 0 such that A = ce' and so eAx is a positive solution. On the other hand, the equation f'(x) = f (x + 1) has no positive solution f. If it had one, then f would be strictly increasing, and the Lagrange mean value theorem would give f (1) > f (1) - f (0) = f'(i;) = f (e+1) > f (1), a contradiction. The equation f'(x) = cf (x+l) has a positive solution on [0, oo) if and only if c < 1/e (see T. Krisztin, Exponential bound for positive solutions of functional differential equations, unpublished manuscript). (a) Proof of the exponential order of growth. If f : [0, oo) -> (0, oo) sat-

isfies the equation f'(x) = cf (x + 1), put a(x) = f'(x)/ f (x). Then Ux

f (x) = f (0) exp

a(s) ds

and

a(x) = cexp

ra+1

a(s) ds) > 0;

that is,

Ina(x) =JX+1a(s)ds, x>0. c

Choose k so that k > a(0) and k > ln(k/c). Begin to define the sequence {xn}Â° o in the following way: xo = 0,

x1 = max{x E (0,1]: a(x) < k}.

Since

<lnk <k, J a(s)ds=lna(0) c c o

x1 is well defined. If x0, ... , xn are given, put xn+1 = max{x E (xn,, xn + 11: a(x) < k} . Since

JXn+1 n

a(s)ds=Ina(x") <Ink <k, c

3.3 THEORY OF FUNCTIONS

243

xn+l is well defined. Since a(x) > k on (xn+1, xn + 1], it follows that Xn+2 > xn + 1. Hence 2n

[0,n] C U[xl-1,x1] l=1

Therefore, if x E In - 1, n), then 2n

a(s) ds < j a(s) ds <

a(s) ds 1=1

< 1=1

x,-1

y1_1+1

a(s) as =

x,_

< 2nk < 2xk + k

< 2n In

a(xl-1)

Consequently,

(JX

f (0)e 2k a 2(x > 0).

a(s ) ds

f (x) = f (0) exp

(b) There is a function ¢ E Coo [0, 1], not identically zero, such that 0(n)(0) = 0(n)(1) = 0; n = 0, 1, 2, .... For example, O(x) = exp (1(x(x - 1))) if x E (0,1), and f (0) = f (1) = 0. Then the formula n = 0, 1.... ;

f (x + n) = nC q(n) (x),

x E [0,1] ,

defines a solution of the equation f'(x) = cf (x + 1). Let X E (0, 1) be such that ¢(x) # 0. By Taylor's theorem, there exists an 77 E (0, x) such that

0(l) (0) xl

O(x) 1=0

0(n)

xn =

_(n) x++

Hence

If (n + n) I = C

0(n)

(,) = 10(X) l

(cx)n

which increases faster than any a ebn as n -4 oo. Thus f has a nonexponential order of growth.

3. SOLUTIONS TO THE PROBLEMS

244

3.4 GEOMETRY Problem G.1. Find the minimum possible sum of lengths of edges of a prism all of whose edges are tangent to a unit sphere. Solution. If the edges of a prism are tangent to a sphere, then each face of the prism intersects the sphere in a circle, tangent to the sides of the face. Hence, each face is a circumscribed polygon.

The translation along the generators of the prism that takes one base polygon to the other moves the inscribed circle of the first polygon to the inscribed circle of the second. These circles are congruent and have parallel planes since the bases of the prism are congruent and parallel. Since the translation that moves one of two congruent and parallel circles on the sphere to the other is perpendicular to the plane of the circles, the generators of the prism are perpendicular to the base, thus, the prism in question is a right prism. Sides of a right prism are rectangles. Since they are also circumscribed, and only squares are circumscribed among rectangles, each side of the prism is a square. Consequently, the edges of the bases and the generators have the same length, the bases are regular, and the sum of the lengths of the edges of the prism is three times the perimeter of the base. Let us cut the prism and the sphere by a plane through the center of the sphere, parallel to the base. The prism is cut in a polygon congruent to the base polygons; the sphere is cut in a great circle. The great circle contains the point of contact of every tangents of the sphere orthogonal to the plane of the great circle. Therefore, it contains a point from each generator. We get that the base polygon is inscribed in a unit circle. Since the base polygon has equal sides, it must be a regular polygon. We conclude that the prisms that satisfy the prescribed condition are regular prisms, with the length of the generators equal to the length of the sides of the base polygon scaled in such a way that the circumcircle of the base polygon has radius 1. Conversely, every prism of this type satisfies the condition of the problem, since the center of such a prism is located at distance 1 from each edge. Indeed, the distance to the center from a generator is clearly equal

to 1. On the other hand, the orthogonal projection of the center onto a lateral face, which is a square, is the center of the face. Thus the distances to the center from the sides of this square are equal.

To get the sum of the lengths of the edges of such a prism, we have to multiply by 3 the perimeter of a regular polygon inscribed in a unit circle. It remains to decide which of these regular polygons has the shortest perimeter. The perimeter of a regular n-gon inscribed in a unit circle is equal to 71

sin a

2n sin - = 2Tr

3.4 GEOMETRY

245

where a = it/n. Since sin x is concave on the interval (0, 7r), the slope of the chord connecting the origin to the point (x, sin x) is decreasing as x is increasing. Therefore, sin a/a is minimal when a = 7r/n is maximal, that is, n is minimal. Thus, the minimum is attained by a regular triangle. As for the prisms, the minimum of the sum of the edge lengths is obtained for a regular prism over a triangular base, with lateral faces being squares, and the value of the minimum is 3. 6 sin 60Â° =9i.

Problem G.2. Show that the perimeter of an arbitrary planar section of a tetrahedron is less than the perimeter of one of the faces of the tetrahedron.

Solution. Obviously, planar sections of a tetrahedron are triangles or quadrangles, since each face of the tetrahedron contains at most one edge of the intersection provided that the intersection does not coincide with the face. The case of quadrangles can be reduced to the case of triangles. Namely, we show that if we translate the plane of a quadrangle intersection in both directions until it passes through the nearest vertex, then one of the translated planes intersects the tetrahedron in a triangle (possibly degenerated to an edge), the perimeter of which is not shorter than the perimeter of the quadrangle. If both translated planes intersect the tetrahedron in a triangle, then introduce the notation shown in Figure G.1 and express the sides of the quadrangle with the help of the sides of the two triangles. Since the planes of the three sections are parallel, the ratio of the segments of a straight line cut off by these planes is the same for all straight lines. Set

.A=BP: AB=BQ: BT =FR: OF=GS: AG. Then for the corresponding sides we have p = .dal,

q = bl + (1 - A)(b2 - bl) = .tbl + (1 - .)b2i

r=c2+)1(cl-c2)=Acl+(1-)%)c2i s=(1-.))a2. From these, p + q + r + s = A(al + bl + cl) + (1 - A) (a2 + b2 + C2), that is,

K = AK, + (1 - A)K2, where K = p + q + r + s, Kl = al + bl + cl, and K2 = a2 + b2 + c2. This implies K < max{K1, K2} .

If one of the triangles AAUT, or LBGF, or both degeneratse to the edges AD or BC, respectively, then this inequality remains valid, putting Kl = 2AD or K2 = 2BC, respectively. Then we get that the perimeter of one of

246

3. SOLUTIONS TO THE PROBLEMS

Figure G.1.

the faces of the tetrahedron is greater than the perimeter of the intersection quadrangle. Hence, it is enough to consider the case of triangle intersections. We may suppose that the intersection triangle has a vertex in common with

the tetrahedron. Otherwise we may translate the plane of the triangle to reach this situation (see Figure G.2). The translated plane meets the tetrahedron in a triangle similar to the original one, with ratio of similarity greater than 1 and it passes through that vertex of the nonintersected face of the tetrahedron that is nearest to the intersecting plane.

If the intersecting plane is parallel to the face not met by it, the assertion is trivial. Otherwise, we show that the perimeter of AKBM is smaller than

3.4 GEOMETRY

247

the greater of the perimeters of OKBC and LKBD. For this purpose, consider the smallest ellipsoid of revolution with foci K and B that contains

both C and D. Since M is an internal point of it, we get KM + MB < max{KC + CB, KD + DB} . But then, either kKBM < kKBD < kABD or kKBM < kKBC, where kPQR denotes the perimeter of OPQR. In the first case, the assertion is proved, the second case can be finished by a repeated application of the previous arguments.

Problem G.3. Show that the center of gravity of a convex region in the plane halves at least three chords of the region.

Solution. Let us denote the disc by T and its barycenter by S. If X is a point of the boundary curve G of T, then denote by Y(X) the second intersection point of the straight line XS and the curve G. Let f (X) = )FS - Y(X)S. We have f (X) = -f (Y(X)) for any X E G. Since f (X) changes continuously as X runs over the are XI (X ), it attains any value

between f (X) and -f (X). This implies the existence of a point X1 E G such that X1S = Y(X1)S. If X1S were the only straight line whose intersection with T is halved by S, then f would be positive along one of the arcs Y(X1). Reflecting this arc through S, the reflected are together with the other arc Xfj-Y(XI) of G would bound a domain T1, the barycenter

of which is different from S, since T1 lies in a half-plane bounded by a straight line passing through S. Therefore, the barycenter of the union T2 = T1 U T should be different from S. On the other hand, T2 is centrally symmetric with respect to S, so its barycenter should be invariant under the reflection in S, which means that its barycenter should be S, which is a contradiction. The same arguments show that if f has finite zeros, then it can not be nonnegative on a half-arc of G. Suppose that there are only two straight

lines X1S and X2S for which the segment of the straight line cut off by the figure is halved by S (X1, X2 E G). By the previous remark, f must be negative (or positive) on the arcs '(X1) and 11'(X2), which is a contradiction, since the sign of f is opposite on arcs opposite to one another. We conclude that at least three chords of the figure are halved by S. The existence of four such chords cannot be proved in general, as it is shown by the case of an arbitrary triangle. Problem G.4. Let A1i A2, ... , A,,, be the vertices of a closed convex n-gon K numbered consecutively. Show that at least n - 3 vertices Ai have the property that the reflection of Ai with respect to the midpoint of Ai_1Ai+1 is contained in K. (Indices are meant mod n.)

Solution. We shall call a vertex P of K reflectible with respect to K if its reflection in the midpoint of the segment connecting the two neighbors of P belongs to K.

3. SOLUTIONS TO THE PROBLEMS

248

1. Let us begin with the first nontrivial case, with the case of quadrangles. For both pairs of opposite sides, the sum of angles lying on one of the two sides is at least -7r; let us denote by A a common vertex of two such sides and denote the other vertices in a cyclic order by B, C, D (see Figure G.3). We show that A is reflectible with respect to the quadrangle. We can construct the reflection A' of A by taking a parallel to AB through D, which intersects side BC in a point E because <ADC + <BAD > 7r,

and then taking the parallel to AD through B, which intersects the segment DE at A' because <DAB + <ABC > 7r. Therefore, A' belongs to the quadrangle ABCD.

Figure G.3.

Figure G.4.

2. If we show that among any four vertices of the convex polygon K, at least one is reflectible with respect to K, then the assertion of the problem is proved.

We shall say that a vertex P of the polygon precedes the neighboring

vertex R if going around K starting from P in the direction of R is a movement of positive orientation. Consider four arbitrary vertices of K and apply for them the previous notation, that is, let <ADC+<BAD > 7r, <DAB + <ABC > 7r (see Figure G.4). If the vertex E of K that follows A is not B, then E is an internal point of the intersection S of three halfplanes S1, S2, S3, where S1 is bounded by the straight line AD and contains ABCD, S2 is bounded by BC and contains ABCD, and S3 is bounded by

AB and does not contain ABCD. Similarly, if the vertex F that precedes A is not D, then F is an internal point of the domain T = T1 fl T2 fl T3, where the half-plane T1 is bounded by the straight line AB and contains ABCD, T2 is bounded by DC and contains ABCD, and T3 is bounded

by AD and does not contain ABCD. It is enough to prove that A is reflectible with respect to the quadrangle AECF, since then it is obviously reflectible with respect to K as well. For this purpose, it suffices to show that <EAF + <AFC > 7r and <FAE + <EAC > Tr. It is enough to see

3.4 GEOMETRY

249

the first inequality; the second can be shown in a similar way. But we have

<EAF + <AFC = <EAC + (<CAF + <AFC) = QEAC + 7r - QACF > QBAC + it - QACD = QBAC + (<CAD + QADC) = <BAD + <ADC > 7r, which proves the assertion of the problem. Remarks. 1. The assertion cannot be improved; that is, there exists for all n a convex n-gon having exactly n - 3 reflectible vertices. Indeed, consider a convex quadrangle ABCD with no parallel sides. It is easy to see that such a quadrangle has only one reflectible vertex; denote it by A. Now consider

a convex n-gon K such that B, C, and D are vertices of K while the other vertices of K lie in a neighborhood of radius e of A (obviously, such a convex n-gon exists for any n). It can be easily seen that if a is sufficiently small, then B, C, and D are not reflectible with respect to K. 2. The assertion does not hold for concave polygons. Problem G.S. Is it true that on any surface homeomorphic to an open disc there exist two congruent curves homeomorphic to a circle ?

Solution. The answer is no: There exists a surface homeomorphic to a disc which does not contain two congruent curves homeomorphic to a circle. We show this by giving an example.

We start from a minimal surface, that is, from a surface for which H = 1/2(gl + 92) = 0. (Here H denotes the Minkowski curvature, and gi and 92 are the principal curvatures of the surface.) Although unnecessary, we explain why we are looking for counterexamples among minimal surfaces. If two congruent copies of a surface intersect one another in a closed curve, then by covering one copy with the other, the two positions of the intersection curve give two congruent curves on the surface, which are different in general. If two copies are tangent to one another at an isolated common point, then by a slight movement of one of them, they can be made to intersect one another along a curve. The choice of minimal surfaces is justified by the fact that if H > 0 at a point of a surface, then we can always find a congruent surface so that this point is an isolated common point of them. We can get such a surface by reflecting the original surface in the tangent plane at the point in question and then rotating it 90Â° about the normal of the surface. This follows from Euler's theorem, since by the inequality gl cos2 (p + 92 sin2 (p > -g2 cos2 (p - gl sin2 (p ,

the curvatures of the normal sections of the original surface are always greater than the curvatures of the corresponding normal sections of the

250

3. SOLUTIONS TO THE PROBLEMS

transformed surface, and therefore the first surface is above the second one in a small neighborhood (looking at them from the direction of the normal vector).

Now we show that in a sufficiently small domain of a minimal surface, any two congruent closed curves bound congruent domains of the sur-

For this purpose, take two copies of the surface and move one of them so that the congruent curves cover one another. Suppose that the piece of the minimal surface is small enough to ensure that both surfaces can be obtained as the graph of the single-valued functions z = f 1(x, y) and z = f2 (x, y). The areas of the domains bounded by the closed curve are obtained as integrals of the lengths of the vectors ml(-1,p1,Q1) and

face.

m2 (-1, p2) q2), respectively, over the same domain. (Here pi and qi denote the partial derivatives of fi.)

We use the fact that the variation of the surface area of a minimal surface is 0, that is, if a minimal surface is embedded into a one-parameter

family of surfaces all having the same boundary, then the derivative of the area of these surfaces with respect to the parameter is zero at the minimal surface. Consider the one-parameter family of surfaces defined by z = Af, 1 + (1- A) f2. The area of a member of this family is obtained as the integral of the magnitude of the vector amt + (1 - A)m2. This magnitude is a strictly convex function of the parameter A if m1 54 m2 and does not depend on A if m1 = m2. Consequently, the area, obtained by integrating convex functions of A, is itself a convex function of A. Its derivatives at A = 0 and 1 can vanish only if it is a constant function, and this happens

only if m1 = m2 at any point of the domain, that is, if the two pieces of surfaces coincide.

According to this observation, it is enough to find a minimal surface that does not contain different congruent pieces. Since two irreducible algebraic surfaces that have a domain in common coincide, it is enough to present an algebraic minimal surface that has no or only a finite number of automorphisms different from the identity. (An automorphism of a surface is a bijective mapping of the surface onto itself that can be extended to a congruence of the space.) If the surface has finite automorphisms, then a small neighborhood of a point different from its automorphic images has no automorphisms different from the identity. Actually, any algebraic minimal surface would do, provided it is not a surface of revolution. However, we need to find only one of them. A simple computation shows that for the surface

x = u3 - 3uv2 + 3u,

y=v3-3u2v+3v, z = 6uv,

H = 0 and K = -2(U2 (u2 + v2 + 1)2 (K is the Gauss curvature). The latter attains its minimum only at it = v = 0; thus this point and the Dupin indicatrix at this point must be fixed by any automorphism of the surface.

3.4 GEOMETRY

251

There may be only a finite number of such automorphisms, however, since the Dupin indicatrix is a hyperbola by K < 0.

Problem G.6. The plane is divided into domains by n straight lines in general position, where n > 3. Determine the maximum and minimum possible number of angular domains among them. (We say that n lines are in general position if no two are parallel and no three are concurrent.)

Solution. The minimal number of angular domains is three. Indeed, the convex hull of the intersection points of the straight lines is a convex polygon having at least three vertices. Each vertex is an intersection point of two straight lines and those half-lines of these lines, which go outside the convex hull, bound an angular domain of the considered subdivision, since there are no intersection points on them.

On the other hand, one can always position n _> 3 straight lines in the plane so that the number of angular domains is exactly 3. Such a construction is given by n tangents to a quadrant (see Figure G.5).

Figure G.5.

The maximal number of angular domains is (-1)n+l

-1

that is, n if n is odd and n - 1 if n is even. Indeed, each straight line is divided by the others into two half-lines and some segments. Angular domains can be bounded only by these half-lines, and each half-line bounds at most one angular domain (otherwise there would be a point that lies on three of the straight lines). It follows that 2n half-lines can bound no more than n angular domains.

For odd n, we can present an (essentially unique) construction with n straight lines and n angular domains: Consider the longest diagonals of a regular n-gon. These n straight lines are in general position. There is an

252

3. SOLUTIONS TO THE PROBLEMS

Figure G.6.

angular domain at each vertex of the polygon. Thus the number of angular domains is n (see Figure G.6).

We prove that if n straight lines in general position bound n angular domains, then n is odd. Removing the two half-lines considered above from each straight line gives n segments. Any endpoint of such a segment is shared by exactly two segments. If two of the segments have different endpoints, then they cross one another since the intersection point of their straight lines can lie neither outside nor at the end of the segments. Thus, the union of the segments yields a (number of) self-intersecting closed broken line(s). Fix an orientation on each of these broken lines, and consider three consecutive segments a, b, and c on one of them. Since a and c intersect, they lie on the same side of b. Omitting the endpoints of a and b

(that is, three points), we can couple the remaining vertices, saying that the points P, Q form a pair if P is on the same side of b as the segments a and c, and Q is the neighboring vertex that comes after P according to the fixed orientation on the broken line containing P. Hence, the number, n, of vertices is odd. In the case of even n, we can attain the maximal (n-1) angular domains. This follows directly from Figure G.6; if we remove one straight line, the number of angular domains decreases by 2.

Problem G.7. Let A = A1A2A3A4 be a tetrahedron, and suppose that for each j $ k, [Aj, Ask] is a segment of length p extending from A3 in the direction of Ak. Let pj be the intersection line of the planes and [AkA1A.,,,,].

Show that there are infinitely many straight lines that

intersect the straight lines P1, P2, P3, P4 simultaneously.

Solution. It is natural to assume that we are in the projective space obtained from the Euclidean space by joining ideal elements.

3.4 GEOMETRY

253

Exclude first the singular cases, and assume the tetrahedron has no edges of length p. Let {j, k,1, m} = {1, 2, 3, 4}. Denote the intersection point of the straight lines AkAl and AikA;l by Bj,,,,,. Obviously, Bum, is the intersection point of the plane S,,,, _ [A;AkAI] opposite to A,, and the straight line pj. Consider now the plane S3 (see Figure G.7). The points Bkj, B1 5 B.j are adjacent to a straight line ej, as is easy to see by a multiple application of Menelaos' theorem. This implies that ej intersects P1, P2, P3, P4 simultaneously. (Indeed, the points Bki, Bid, Bmj belong to pk, pi, p,,,,,, respectively, while pj and ej are coplanar.) The line ej is not an edge of the tetrahedron, since it would result, the excluded case. The lines el, e2, e3, e4 are all different since they lie in different planes of the tetrahedron, but none of them coincides with an edge.

Figure G.7. Therefore, the lines pi, P2, P3, P4 are intersected by four different straight lines simultaneously. If two of the lines pi (i = 1, 2, 3, 4) are intersecting, then their point in common must be contained in the plane spanned by the two other straight lines; hence we can easily find infinitely many straight lines intersecting all four of them. If P1, P2, p3, p4 are mutually skew, then take P1, P2, P3. It is known that the straight lines that intersect all of these three lines sweep out a doubly ruled second-order surface. One family of straight lines on this surface is given by the straight lines intersecting pl, P2, P3, while pl, P2, P3 belong to the other family. The line p4 has more than two points in common with the surface, hence it is also a generator of it and intersects every straight line in the first family. Now let us discuss the singular cases. (a) If the three edges starting from one of the vertices Ak have length

p, then Pk is not properly defined, while the three other lines are adjacent to Ak. As Pk is not determined, the statement cannot be applied to this case. If, however, we define Pk as an arbitrary straight

3. SOLUTIONS TO THE PROBLEMS

254

line in the two coinciding planes that should define it, the statement is clearly true.

(b) If there are two edges of length p starting from a vertex Ak, then denote by A, the fourth vertex, not lying on these edges. It is easy to see that in this case pk is the intersection of the planes Sk and S1, pi c S1, while the two other lines, pj, p,,,,, go through Ak. This means that any straight line e such that Ak E e C S1 intersects all the lines A. (c) Assume now that the tetrahedron has exactly one edge of length p. In this case, one of the straight lines ei coincides with this edge, and therefore we can only say that there are three different straight lines

among el, e2, e3, e4. However, to apply the arguments used in the generic case, it suffices to have three straight lines that intersect all of P1, P2, P3, P4. Thus, P1, P2, P3, p4 are intersected by infinitely many straight lines simultaneously.

Figure G.8.

(d) Assume, finally, that the tetrahedron has two edges of length p, opposite to one another, say A1A3 = A2A4 = p = a. Introduce the notation A1A2 = b, A2A3 = c, A3A4 = d, A1A4 = f (see Figure G.8). Applying Menelaos' theorem, we obtain the following divided ratios: (A1A3B24) _ (A2A4B31) _

b-a a-c'

a-d '

(A2A4B13)

a-frn

c -a

a-d'

(A1A3B42)=

-a

Turning to cross ratios,

(AlA3B24B42) _

- a)la - d) = (A2A4B13B31),

(a - c)(f -a)

3.4 GEOMETRY

255

but then (A1A3B24B42) = (B31B13A4A2)

Let us take into consideration that Al and B31 are on p3i A3 and B13 are on pi; B24 and A4 are on p2i B42 and A2 are on p4. The equality we have just obtained means that p1, P2, P3, p4 cut the skew lines A1A3 and A2A4 in two 4-tuples having the same cross ratio. Thus these lines belong to the same family of generators on a doubly ruled second-order surface. This implies the statement of the problem.

Problem G.8. Consider the radii of normal curvature of a surface at one of its points P0 in two conjugate directions (with respect to the Dupin

indicatrix). Show that their sum does not depend on the choice of the conjugate directions. (We exclude the choice of asymptotic directions in the case of a hyperbolic point.)

Solution. If P0 is a parabolic point, then the indicatrix is a couple of parallel lines, and a direction not parallel to them is conjugate only to their (that is, the asymptotic) direction. The radius of normal curvature is infinite in the asymptotic direction and finite in any other direction, so the sum of the radii of normal curvature in two conjugate directions is always infinite.

It is known that the absolute value of the radius of normal curvature in a given direction is the square of the length of the segment from Po to the indicatrix point in the given direction. According to this, we need only to show that the sum (for the elliptic case) or the difference (for the hyperbolic case) of the square of the length of conjugate half-diameters of the Dupin indicatrix does not depend on the choice of the conjugate diameters.

A theorem of Apollonius states that the indicatrix is an ellipse at an elliptic point.

Now suppose that P0 is a hyperbolic point, that is, the indicatrix is a pair of conjugate hyperbolas whose equation is of the form X2

- b2 = Âą1

in a properly chosen coordinate system. Let us denote by Si and s2 the common asymptotes of them (see Figure G.9). Let S be a point on one of the hyperbolas and e1 be the tangent of the hyperbola at S. This tangent intersects s1 at the point L, s2 at the point M. Draw a tangent (different from s2) from M to the other hyperbola. This tangent touches the hyperbola at the point T and crosses s1 at N. It is known that the area of the triangle enclosed by the asymptotes and a tangent of the hyperbolas is equal to ab. Thus, LPOLM and LPOMN are of the same area, which, due to their common altitude MM*, yields POL = PON.

It is also known that S halves LM and T halves MN. Hence,

L POST is the median triangle of ILMN; consequently, POS 11 e2 and

3. SOLUTIONS TO THE PROBLEMS

256

Figure G.9. POT II el.

This means, however, that POS and POT are conjugate half-

diameters. Applying the cosine law for the triangles OPOLM and OPOMN, and using the equality P0L = PON shown above, we get

4 (PoT2 - P-S2) =LM2 - NM2 =LPo2 + PoM2 - 2LPOPoM cos a[NPo2 + PO M2 - 2NPOPOM cos(7r - a)]

=-4LPOPOMcosa=-16

LPo PoM

cosa,

where LPo/2 and POM/2 are the contravariant coordinates of the vector P with respect to a basis consisting of unit vectors pointing in the directions of the asymptotes. The product of them (as it is known) does not depend on the choice of the point S of the hyperbola. In such a way, 2 ( 2 - AO-S) does not depend on S, that is, on the choice of the conjugate pair of diameters.

Problem G.9. Show that a segment of length h can go through or be tangent to at most 2[h/../) + 2 nonoverlapping unit spheres. ([.] is integer part.)

Solution. Let us denote by e the supporting straight line of the given segment of length h, project the centers of the spheres onto this straight line, and consider the open intervals of length on e centered at the projections of the centers. We claim that no point of e belongs to more than two such intervals. If not, there would exist a cylinder of radius 1 and altitude m < f containing three points A, B, and C (the centers of the spheres) that are at least 2 units away from one another. We shall show that this is impossible.

3.4 GEOMETRY

257

Figure G.10.

In the following, we shall use the term "the plane of a point" for the plane that is orthogonal to e and passes through the point. The plane of A intersects the cylinder in a circle centered at K. Let us project B and C onto the plane of A. Denoting the projections by B' and C', respectively (see Figure G.10), we have

m2 +B'A2 > 22, that is,

B'A2>22-m2>2, hence, LB'KA is obtuse angled at K, since the radius of the circle equals 1. It follows that if we move A to the perimeter of the circle along the radius KA, AB' and hence AB increase (and similarly, so do AC' and

hence AC). The same is true for the other points, so we may suppose that A, B, and C are on the surface of the cylinder, at distance 1 from e. for the height of the cylinder, we may also Since we required only m < suppose that one of the points, say A, lies in the upper base of the cylinder and one of the points, say C, lies in the lower base.

Figure G.11.

258

3. SOLUTIONS TO THE PROBLEMS

Consider the plane of B, and let B1 be the reflection of B in e, A' and C' be the projections of A and C onto the plane of B, respectively (see Figure G.11). Since the distances between the planes of A, B and C are smaller

than f , while the distances between the points are not less than 2, the distances between the projections of these points are greater than 0. This implies that the endpoints A' and C' lie on different arcs determined by B and B1. If we move both A' and C' toward B along the arc, and move A and C in a corresponding way, we can manage to have AFB- = CB = 2. But in this case A'B1 is just the distance of the planes of A and B, and C'B1 is just the distance of the planes of C and B. Therefore

A'B1+C'B1<m<V. Consequently,

A'C' < A'B1 + CBl < v f2. Thus,

AC2=m2+A'C'2<2+2,

and so AC < 2.

This is a contradiction that proves the propo-

sition. Let us elongate the segment of length h in both directions by a segment

of length /. Since the segment goes through or is tangent to the spheres, the intervals assigned to the spheres are all contained in this enlarged segment. Since every point of the enlarged segment is covered by at most two intervals, we get that the number of spheres is at most

h+2 2

vf2-

2[VJ+2.

To see this, we put the intervals in an increasing order by their left endpoints (this ordering is not always unique). Then the intervals standing at odd (or even) positions are necessarily disjoint. Thus, the number of all intervals is at most

[h]+,+[h]+,.

It is easy to see that the given estimation is exact.

Problem G.10. Characterize those configurations of n coplanar straight lines for which the sum of angles between all pairs of lines is maximum.

Solution. We may suppose that all straight lines go through one point. Starting from an arbitrary straight line, let us number the straight lines in counterclockwise orientation (the order of coinciding straight lines is

3.4 GEOMETRY

259

arbitrary): el, ... , en. For 1 < i < [n/2], we say that el is the ith neighbor of ek if

l - k + i (mod n) . We shall denote the ith neighbor of ek by ek. Let us denote by (e, f) the angle between the straight lines e and f and denote by < e, f > the angle of the counterclockwise rotation that takes e to f. For i fixed,

=i(<e1,e2>+<e2ie3>+...+<en,e1>)=i7r, and equality holds if and only if < ek, ek >< 2

(1)

for all k. Thus, for odd n, the sum of angles between all pairs of straight lines can be estimated as follows:

n-1

a i7r = n2-17r n-1

[(ei, ei) + ....+ (en, e')] <

(2)

and equality holds if and only if (1) holds for all k = 1, 2, ... , n and i = 1, 2, ... , (n - 1)/2. For this, it is enough to require that ek,ekn-i)/2

>< 2

(3)

hold for k = 1, 2, ... , n. For even n, the sum of angles is 2-1

[(ei, ei) + ... + (en, a ,)] + z=i

n7r <ilr+22

[(el, ei/2) + ....+ (en, e; /2)J

n27r

and equality holds if and only if n/2

holds f o r k = 1, 2, ... , n. For even n, these inequalities imply the requested

characterization, since the inequalities < ek, ek/2 ><

and

< ek+n/2,

can be satisfied simultaneously only if n/2 < ek, ek

>= 2 ,

n/2

>< 2

260

3. SOLUTIONS TO THE PROBLEMS

that is the (n/2)th neighbors are perpendicular to one another. To summarize, if n is even, the maximum is attained if the family of straight lines is composed of n/2 orthogonal pairs and any family of orthogonal pairs of straight lines yields maximal sum of angles. Returning to the case of odd n, let us observe that orthogonal couples of straight lines can be removed from the family, since the omission of an orthogonal pair there decreases the sum of angles by (n - 1) 2 and n2

(n - 2)2

=n-1.

Suppose that after the removal of orthogonal pairs there remain 2m + 1 straight lines. These straight lines must be different, because if e were a double line then the mth neighbor of the mth neighbor f of e would be e, which would mean that e and f are perpendicular. If e is an arbitrary member of the 2m + 1 straight lines, then conditions (3) imply that each quadrant of the plane determined by e and the straight line perpendicular to e must meet exactly m straight lines. Adding to such a family of an odd number of straight lines an arbitrary number of orthogonal pairs, we get all the configurations that yield the maximum, and the maximum is equal to [(n2 - 1)/4] (a/2) by (2). Finally, if we do not suppose that the straight lines go through one point, then taking a configuration characterized above and replacing each straight line by a parallel one, we can get all the requested configurations.

Problem G.11. Let f (n) denote the maximum possible number of right triangles determined by n coplanar points. Show that lim

f n) = oc and

n-.oo n2

lim f () = 0. n-+oo n3

Solution. Lower estimation. We shall show the following sharper result. There exists a constant c > 0 such that f (n) > cn2 log n

(n > 3).

Suppose first that n = (3k + 1)2, where k is a natural number. Consider those points of the lattice of points with integer coordinates, which lie in the square spanned by the vertices (0, 0), (0, 3k), (3k, 3k), (3k, 0). The number of these points is obviously (3k + 1)2. We shall count only those right triangles for which the right-angled vertex (q, r) satisfies k < q < 2k, k < r < 2k and the two other vertices of which lie in the square of sides 2k centered at (q, r). Obviously, all these right triangles are good for us. We have to determine the number of right triangles spanned by the

lattice points of the square (-k, -k), (-k, k), (k, k), (k, -k) so that the right-angled vertex lies in the origin (0, 0).

3.4 GEOMETRY

261

The perpendicular sides of such a right-angled triangle lie on straight lines given by equations of the form ix = jy and jx = -iy, where (i, j) = 1, -k < i, j < k. Obviously, we estimate the number of right triangles from below if we take into consideration only those straight lines that satisfy

0 < i < j as well. In this case, each of the straight lines ix = jy and jx = -iy contains 2 [k/j] lattice points different from (0, 0), so the number of right-angled triangles determined by them is 2

[k]

>.T 2

For a fixed j, we find at least

right triangles (cp denotes Euler's phi function), which gives k2

E 0(j)

j=1

right-angled triangles altogether. Now returning to the original problem, the (3k + 1)2 points we constructed yield at least k

(2k + 1)2k2

P j2

3=1

right triangles. Introducing the function

fi(x) = 1:W(7) j<x

and summing up partially, we get

f ((3k + 1)2) > (2k + 1)2k2 j=1

(2j

+2 j (2) 2U + 1)2

Combining this with the well-known inequality 'D(x) > clx2,

we get

(1)

k-1 > c3k4log k.

f((3k + 1)2) > c2(2k + 1)2k2 j=1 j

3. SOLUTIONS TO THE PROBLEMS

262

Now let n be an arbitrary number. Set k = [(f - 1)/3] > V/n-/4. Then

f(n) > f((3k+1)2) > c3k4logk> cn2logn. Upper estimation. We shall prove a sharper result again: using the "descente infinie" method we show that

f(n) < n2/.

(2)

The statement is trivial for n = 1, 2, 3, 4, 5 (for these values n2 Vjn- > (3) ). If (2) were not true, then there would be a smallest natural number n such

that f(n) > n2V/n.

Let us take n points P1i ... , PP, on the plane that span f (n) right-angled triangles. We claim that there must be a straight line that contains from the n points at least 2/ ones. Consider all possible ordered couples (Pi, Pj), and assign to each of them the number of right triangles PPP3Pk such that the right angle is at Pi. This number is the number of points different from Pi on the straight line perpendicular to PiPj at Pi. The sum of these numbers is twice the number of right triangles and so it is at least 2n2 Vrn-. Since the number of summands is n(n - 1)(< n2), one of the summands is at least 2 f, and thus one of the perpendicular lines contains at least 2Vrn_ points.

Let us remove the points of this straight line and examine how many right triangles are destroyed by this at most. We divide the triangles destroyed into two groups: (A) triangles for which the right-angled vertex is omitted; (B) triangles with right-angled vertex lying off and one of the other vertices lying on the critical straight line. We can give an upper bound for the number of triangles of type A in such a way that we have n(n - 1)/2 choices for the hypotenuse of such a triangle, and a segment PPP may serve as hypotenuse for at most two right triangles of type A (Thales' circle!), so the number of triangles of type A is at most n(n - 1) < n2. In the case of triangles of type B, we have n(n - 1)/2 choices for those vertices that are not required to lie on the critical straight line, and we can choose at most two points on the critical straight line which form a triangle of type B together with a fixed couple of points P2P3 so that the right-angled vertex is either at Pi or at Pj. It follows that the number of triangles of type B is also at most n(n - 1) < n2. In such a way, the removal of the points of the critical straight line destroys at most 2n2 right-angled triangles; consequently, the remaining,

at most, n - 2 points span at least n2

- 2n2 right triangles. By the

induction hypothesis, n2Vn-

- 2n2 < (n - 2 V/n) 2

n-2 n<(n-2/ )2/i,

3.4 GEOMETRY

263

from which

2n2 < 4nj, that is,

n<4,

for which cases we have already seen that the assertion is true. Remarks. 1. We could have avoided the use of (1) using only the elementary fact that cp(p) = p - 1 for prime numbers. However, this would have given us only the estimation f (n) > c4 n2 log log n,

where c4 is a, suitable positive constant. 2. Laszlo Lovasz remarks that there is a positive constant c5 such that the number of right triangles in the lattice construction described in the first part of the solution is at most c5 n2log n.

3. Bela Bollobas proves the statement of the problem in the following more

general form: Let us denote by f (n, k) the maximal number of rightangled triangles spanned by n points of the k(> 2)-dimensional space, where the maximum is taken for all possible configurations of the points. Then dkn3_2'-k , cn2 log n < f (n, k) <

where c and dk are positive constants. The lower bound comes from our previous arguments because of f(n,k) _> f(n,2). We prove the upper bound for k = 3. Denote by R a collection of n points in the three-dimensional space, and denote by p the number of those straight lines that contain more than 3/ points of R. Let el, e2, ... , e, stand for these straight lines. el contains at least 3V,`n points of R, el and e2 together contain at least 3/i(3 / - 1) points, because two different straight lines may have at most one common point, and so on, el, ... , en' cover at least p'-1

E(3/- 1)>p3 ' -P2 i=1

points of R for every p' < p. Consequently, p < V/n. We prove that if "big" straight lines, the number of right triangles decreases at most by 4n512. Let e be a "big" straight line. Those right we remove at most

triangles that have at least one vertex on e can be divided into two classes: i.

Triangles that have two vertices on e. The third vertex of such a triangle can be chosen out of at most n points. This point, together with a point on e, leaves only two possibilities, and so the number of right triangles of this class is at most 2n2.

264

ii.

3. SOLUTIONS TO THE PROBLEMS

Triangles that have only one vertex on e. We have n(n - 1)/2 < n2/2 choices for the two other vertices, which form a right-angled triangle together with at most 4 points on e, thus the number of right triangles of this type is less than 4n2/2 = 2n2. Since we have at most V/n "big" straight lines, their removal causes the vanishing of not more than v 4n2 = 4n5/2 right triangles. After the removal, each straight

line contains at most 3V/'n points. Let us denote by q the number of planes passing through the point P and least 6n3/4 points of the remaining family. By the idea used above, we get q'6n3/4

- gj23v < n

for all q' < q, since two planes have at most 3 f common points. From here, q < n1/4. There are less than n1/43v/n triangles satisfying

that its right-angled vertex is at P, one of its perpendicular sides lies in one of these "big" planes, while the other is perpendicular to the plane. The number of right triangles with right-angled vertex at P, not satisfying these conditions, can be estimated by the sum i-1 fi s:, where l denotes the number of orthogonal pairs of a plane and a straight line passing through P such that both the plane and the straight line contains at least one point different from P and fi, si denote the number of points different from P lying on the plane and the straight line of the ith pair, respectively. But then fi < 6n7/4 . fisi < 6n3/4 Thus, the number of right triangles in R is less than 4n512 + 6n714n < 10n1114

We remark that the case k > 3 requires the consideration of not only straight lines and 1 (1 < k)-dimensional linear subspaces but also that of circles and spheres.

Problem G.12. Suppose that a bounded subset S of the plane is a union of congruent, homothetic, closed triangles. Show that the boundary of S can be covered by a finite number of rectifiable arcs.

Solution 1. Let us denote the boundary of S by H and let P E H be an arbitrary point. Then one can find a sequence of points {P,} converging to P such that each point P,,, lies on the boundary of a triangle A,,, constituting S.

We may suppose that each point P,, lies on the boundary of A at the same position (that is, the translation that takes Auto A,,,, takes P,,, to P,,,.). (This can be shown by our assumptions. Translate the triangles A,,,

to a fixed triangle A; denote by Q the point corresponding to P. The sequence {Q} is bounded, and it thus contains a convergent subsequence

3.4 GEOMETRY

265

{Qnk}, which tends to Q. Translating A back to Ank, Q corresponds to Pnk . Obviously, Pnk -> P and the points P"'k lie at the same position on the boundary of Ank ). Obviously, each triangle An contains a homothetic triangle A of half size such that Pn is a vertex of A lying at the same position for all n. These triangles converge to a homothetic triangle Ap with vertex P such that the interior of A contains no point of H. Let us number the three vertex positions of the triangles homothetic to An. This gives rise to a natural ordering of the vertices of triangles homothetic to An. Let us define the sets Hi C H (i = 1, 2,3) in such a way that a point P E H belongs to Hi if and only if P is the ith vertex of A ,. Obviously, H = H1 U H2 U H3. Let us study H1. Denote by a the angle lying at the first vertex of OP. Let us introduce a coordinate system on the plane so that the y-axis shows in the direction of the bisectrix of the angle a, the direction of the x-axis coincides with the direction of the supplementary angle of a. Let us divide the plane into bands of equal width parallel to the x-axis. If the width of the bands is chosen to be small enough (for example, if it is smaller than

the bisectrix of Ap that passes through P), then the intersection of H1 and one of the closed bands is the graph of a function y = f (x), where f is defined on a bounded subset of the x-axis and satisfies Lipschitz condition

with constant k = cot (a/2), which gives the possibility to extend it to the closure of its original domain. The extension of f is also k-Lipschitz. After this, by extending f to the components of the open complement of its domain in a linear way, we can obtain a k-Lipschitz extension of f, defined on a sufficiently large bounded interval. The graph of this extension is a rectifiable curve that covers the intersection of H1 and the considered band. Since S is bounded, it crosses only a finite number of bands; thus, Hl can be covered by a finite number of rectifiable arcs. We can proceed similarly for H2 and H3, so the theorem is proved. Solution 2. Let 1[8` be the two-dimensional Euclidean vector space, and let H C 1E82 be a nonsingular closed triangle whose angles are the regions A1, A2, A3, that is,

H=A1nA2nA3.

(1)

The statement of the problem can be reformulated as follows.

Statement. If M (C R2) is bounded, than the boundary of M + H can be covered by a finite number of rectifiable arcs. (From now on, we use operations on complexes.)

Since M is bounded, there exists a finite set V C 1[82 such that M C V + 12 H. Setting M(v) = (v + 12 H) n H, we get

M(V);

(2)

hence,

M+H=

(M(v) + H).

(3)

3. SOLUTIONS TO THE PROBLEMS

266

First we show that M(v) + H = xf11(M(v) + Ai)

(4)

for all v c V. It is clear from (1) that M(v) + H C iA1(M(v) + Ai). Suppose now that

p E inl(M(v) + Ai). Then p E M(v) + Ai (i = 1, 2, 3), but

(p - M(v)) n (Ai \ H) C (p-v- 1H)n(Ai\H) =0 must be satisfied at least for one i. Thus

(p-M(v))nH

and this means that

p E M(v) + H. Hence,

M(v) + H D inl(M(v) + Ai) holds as well, and this proves (3). Let us denote by fr (X) the boundary of a set X. Using that V is finite, (3) and (4) yield

fr (M + H) C U U fr (M(v) + Ai) vEV i=1

Now it is enough to show the following proposition.

Proposition. Let A be the angular domain of angle 2a (0 < 2a < 7r), the vertex of which is the origin and the bisectrix of which is the negative y-axis. Let M C R2 be a bounded set; furthermore,

T={(x,y) ER2:0<x<1}. Then

G=fr(M+A)nT is a rectifiable curve.

Proof. To show this, let (t, g(t)) denote for 0 < t < 1 the point of G the whose abscissa is t, that is, set

g(t) = sup{y - x - tj cot a: (x, y) E M}. Since g satisfies the Lipschitz condition

Ig(t) -g(t')I < It - t'Icota, the latter proposition is trivial.

3.4 GEOMETRY

267

Problem G.13. Let F be a surface of nonzero curvature that can be represented around one of its points P by a power series and is symmetric around the normal planes parallel to the principal directions at P. Show that the derivative with respect to the arc length of the curvature of an arbitrary normal section at P vanishes at P. Is it possible to replace the above symmetry condition by a weaker one?

Solution 1. Since the normal planes parallel to the principal directions are perpendicular, the symmetry of F around them implies the symmetry around their intersection, that is, around the normal of F at P. Since the plane of normal sections contains the normal line, normal sections at P are also symmetric around the normal. The analyticity of the surface ensures that the normal sections have arc length and curvature at each point, which is a differentiable function of the are length. Let g(s) be the curvature expressed as a function of the arc length. If P corresponds to the parameter so, then the symmetry of the normal section around the normal at P gives the identity g(so + s) = g(so - s). Differentiating with respect to s at s = 0, we get g'(so) = -g'(so), that is, g'(so) = 0, which proves the proposition. As we see, symmetry around the normal planes parallel to the principal

directions can be substituted by the symmetry around the normal of F at P.

Solution 2. By our assumption, introducing a coordinate system with origin at P, the surface is given by an equation of the form i k

aikx y

i,k=0

where aoo = 0. We may also suppose that the x- and y-axes show in the principal directions of F. In this case, alo = a01 = 0. With such a choice of coordinate system, according to our assumptions, z(x, y) = z(-x, y) =

z(-x, -y) = z(x, -y), which can be the case only if aik = 0 whenever either i or k is odd. Then the equation of the surface has the form 00

a2i,2kx2iy2k

i,k=0

A normal section at P can be parameterized as follows:

r=r(t): x=clt,y=c2t, cc

a2i,2k(Clt)2i i, k=0

(C2t)2k ==

bnt2n

n=1

where the constants c1, c2 (ci +c2 # 0) depend on the plane of the normal section, and the coefficients bn are functions of a2i,2k.

3. SOLUTIONS TO THE PROBLEMS

268

The curvature of the curve r = r(t) can be computed by the formula (rl r")2

9(t) =

Ir/l3/2

By our assumption, g(O) # 0, and instead of needing to prove the equation dg/ds = 0, it is enough to show the relation g'(0) = 0, since

dg 1

dg dt dt ds

dt Ir'I

Taking into consideration that

r ' = (Cl, r" =

0" n=1

t2n-1

2n(2n - 1)

0, 0,

bnt2n-2

n=1

r /l/ =

2n(2n - 1) (2n - 2)

0, 0,

bnt2n-3

n=1

the relation g'(0) = 0 is obtained by direct computation. The above solution shows that in order to weaken the symmetry con-

dition, it is enough to require that the equation z = EÂ°k=o aikxiyk of the surface in the coordinate system attached to the normal and principal directions does not contain terms of order three.

Problem G.14. Let a(Sn, k) denote the sum of the kth powers of the lengths of the sides of the convex n-gon Sn inscribed in a unit circle. Show that for any natural number greater than 2 there exists a real number ko between 1 and 2 such that a(Sn, ko) attains its maximum for the regular n-gon.

Solution. We shall show the following sharper result. If 1 < k < (tan(7r/n))/(ir/n), then a (Sn, k) < a(SS,, k) for any convex n-gon inscribed in a unit circle, where Sam, is the regular n-gon inscribed in a unit circle. P r o o f . T o show this statement, it is enough to consider only those poly-

gons Sn that contain the center of the circle on the boundary or inside. Indeed, let S n be a polygon not satisfying this condition, and let A1, ... , An be the vertices of Sn. Suppose that A1A2 is the closest side of Sn to the center of the circle. Denote by A'2 the antipodal pair of Al, and consider the polygon S'n with vertices Al, A'2, A3, ... , An. Then AlA2 > A1A2, and A2A3 > A2A3 (since the angle opposite to the side A3A2 in DA2A2A3 is obtuse). Fork > 1, obviously U(SS,, k) > Q(Sn, k).

3.4 GEOMETRY

269

Now let Sn be a convex n-gon inscribed in a unit circle, such that S,,, contains the center of the circle on the boundary or inside. Let x1 ,. .. , xn denote half of the central angles corresponding to the sides of Sn. Then n

u(Sn, k) = 2k > sink xi, where 0 < xi < 2 ,

xi = 7r. i=1

i=1

Our proposition is well known for k = 1, so it suffices to deal with the case k > 1. Let us fix the exponent 1 < k < (tan(7r/n)/(ir/n). Since the function (tan x)/x is continuous and strictly increasing in the interval (0, it/2, furthermore, lima-o(tan x)/x = 1, there exists a unique real number a = a(k)

such that 0 < a < ir/n and k = (tan a)/a. Consider the function f (x) = sink x. Since f"(x) = k(k - 1) sink-2 x cos2 x - k sink x = k sink-2 x(k cos2 x - 1) , one can easily check that f (x) is convex on the interval and concave on the interval (arc cos (1//), 7r/2). There(0, arc cos fore, it is concave on the interval (a, -7r/2) because

f" (a) = Set

t a

a sink-2 a

( tan a

cos2 a-1 ) <0. if 0 < x < a, if a < x < 7r/2.

k sink-1 a cos ax, sink x,

Obviously, g is continuous in the interval [0, 7r/2]. One can see easily by a study of the derivative that the function (sink x)lx is strictly increasing on the interval [0, a] and strictly decreasing on the interval [a, 7r/21. One can derive from this that g is convex on [0, 7r/2] and also that f (x) < g(x) holds for x E [0, 7r/2]. These facts, together with the inequality n

Eg(xi) < ng (n)l i=1

obtained by an application of Jensen's inequality, yield n

a(Sn, k) = 2k

f(xi) :5 n2kg (n/l

i=1

Since a < 7r/n, we have f (ir/n) = g(ir/n), and thus a(Sn, k) _< n2k sink The proposition is proved.

= a(S*, k).

3. SOLUTIONS TO THE PROBLEMS

270

Remark. Many contestants observed that if 3 < n' < n and 1 < k < (tan(ir/n))/(ir/n), then we also have the inequality u(Sn,, k) < o(Sn, k). Indeed, considering the derivative of the function y = issink(7r/x), we see that

y' = sink - is

k7r is

k7r

sink-1 - cos - = sink-1 - 1 sin - - - cos -J > 0 is

provided that is > 2 and k < (tan(ir/x))/(ir/x) 7r/n' > it/n and

tan, n'

Using the relations

tan n n

we obtain

v(S,, k) = 2kn sink - > 2kn'sink = a(S,, k) > a(Sn, k). n n'

Problem G.15. Let h be a triangle of perimeter 1, and let H be a triangle of perimeter A homothetic to h. Let hl, h2i ... be translates of h such that, for all i, hi is different from hi+2 and touches H and ht+i (that is, intersects without overlapping). For which values of A can these triangles be chosen so that the sequence hl, h2i ... is periodic? If A > 1 is such a value, then determine the number of different triangles in a periodic chain hl, h2i ... and also the number of times such a chain goes around the triangle H.

Solution. We may restrict ourselves to the case of regular triangles since any triangle can be transformed into a regular one by an affine transformation and affinities preserve homothetic and touching position of triangles and also the ratio of perimeters of homothetic triangles. Let A, B, C be the vertices of the triangle H; let Pi, Qi, Ri be the vertices of the triangle hi; and let a, b, c and pi, qj, ri be the sides of the triangles H and hi opposite to the vertices A, B, C and Pi, Qi, R, respectively. These sides are considered to be half-open segments excluding the endpoint C

from side a, the endpoint Ri from side pi, etc. It is assumed that the vertices A, B, and C correspond homothetically to the vertices Pi, Qi, and Ri, respectively (see Figure G.12). There are two ways a triangle hi can touch the triangle H: either a vertex of H lies on the side of hi opposite to the homothetically corresponding vertex of hi, or a vertex of hi lies on the side of H opposite to the homothetically corresponding vertex. According to our assumptions, we always have exactly one of the two cases. In the first case, we say that hi touches

a vertex of H; in the second, we say that hi touches a side of H. The common point of the two triangles will be called the point of contact. We fix the orientations (Pi, Qi, Ri) and (A, B, C) of the triangles hi and H.

3.4 GEOMETRY

271

Figure G.12.

If hi touches a side of H, then denote by f (hi) the distance between the point of contact and the vertex of H that comes next to the contact point according to the fixed orientation. If hi touches a vertex of H, then denote by f (hi) the distance between the point of contact and the vertex of hi that comes next to the contact point according to the orientation opposite to the fixed one. Obviously, 0 < f (hi) < A/3 in the first case, 0 < f (hi) < 1/3 in the second (see Figure G.13).

Figure G.13.

It is clear that the contact points of hi and hi+i are contained in one (closed) side of H. Thus, the fixed orientation of H gives rise to an ordering of the triangles hi and hi+i We may establish the following facts. Let h' denote an arbitrary translate of h touching H. We want to describe the translates h" of h that touch both h' and H and come after h' (according to the fixed orientation). (See Figures G.14.a-e). (a) If h' touches a side of H and f (h') > 1/3, then h" is unique, it touches

the same side of H, and the distance of contact points is 1/3 (thus f(h") = f(h') - 1/3)) (Figure G.14.a). (b) If h' touches a side of H and f (h') < 1/3, then h" is unique, it touches the vertex of H next to the contact point of h' with respect to the

272

3. SOLUTIONS TO THE PROBLEMS

f(h')>1/3 H

Figure G.14.a

Figure G.14.b

fixed orientation, and f (h") = 1/3 - f (h') (Figure G.14.b).

(c) If h' touches a vertex of H and f (h') < A/3, then h" is unique, it touches the side of H that contains the contact point of h', and f (h") = A/3 - f (h') (Figure G.14.c). (d) If h' touches a vertex of H and f (h') = A/3, then we can choose for h" any triangle that touches the vertex of H next to the contact point of h' satisfying f (h") < (1 - ))/3 (Figure G.14.d). (e) Finally, if h' touches a vertex of H and f (h') > A/3, then h" is unique again, it touches the vertex of H next to the contact point of h', and it satisfies f (h") = (1 - A) /3 (Figure G.14.e). I. Suppose first that A < 1. We show that in this case one can always find a periodic chain of triangles.

Let h1 touch the vertex A of H so that f(h') = 1 - A. Set k = - (-A/(1 - A)). Starting from h1 and advancing according to the fixed orientation, one can construct the sequence h2, ... , h2k+1 uniquely by the

3.4 GEOMETRY

273

f(h')< 3

f(h")= 3 -f(h') Figure G.14.c

Figure G.14. d

previous observations (a)-(e), so that the even triangles h 2 ,-- . , h2k touch a side of H, hl, ... , h2k+1 touch a vertex of H, and f (h2k+1) >- A. Then we can choose h2k+2 according to (d) so that f (h2k+2) = 1 - A is fulfilled. The triangle h2k+2 can be obtained from h1 by a rotation of angle 2iri/3 about the center of H. Therefore, continuing the construction from h2k+2, choosing f (h4k+3) = 1 - A, we can close the chain by setting h6k+4 = hi, which finishes the proof.

274

3. SOLUTIONS TO THE PROBLEMS

Figure G.14.e

II. Now let us examine the case A > 1. Then only the cases (a), (b), and (c) are possible, and each h" is uniquely determined by h'. The same would be true if we changed the orientation; hence, there are exactly two translates of h that touch H and a given h'. Consequently, given hl and h2 that touches hl and H, the sequence hl, h2i ... can be continued uniquely. Suppose, we are given a periodic chain hl, h2i ... with period n(hn+l = hl) winding around H k times. The phrase "the chain winds around H k times" makes sense since the contact points of the triangles hi follow one another according to a fixed orientation of the boundary of H. We may suppose that this orientation coincides with the orientation (A, B, C). Observations (a), (b), and (c) make clear that the contact points of the triangles hi that touch a side of H form a cyclic sequence of n - 3k elements running around H k times such that the peripheral distance between two consecutive contact points is 1/3. Thus,

3(n-3k)=Ak, that is,

A = n 33 k,

n = 3k(A + 1).

(1)

One obtains from this equation that in the case A > 1, the existence of a periodic chain implies the rationality of A. In this case, the number of times

3.4 GEOMETRY

275

this chain goes around the triangle H is the denominator in the reduced simple fraction form of 3A, and the number of triangles contained in the chain is obtained as the product of this winding number and 3(A + 1) (of course, there is a periodic chain for any integer multiple of n and k with period n and winding number k). We still have to show that if (1) holds for some natural numbers k and n, then there exists a periodic chain with period n. Then it will also prove that the rationality of A is sufficient for the existence of a periodic chain. In addition to this proposition, we show that no matter where we put the starting triangle hl, the sequence of his will get closed at the nth step after k rounds around H. We may restrict ourselves to the case when h1 touches a side of H, since either h1 or h2 touches a side of H, and if the chain starting from h2 gets closed after the nth step, then so does the chain starting from h1 because of the unique reconstructability of the sequence in both directions. Let us construct the beginning part hl, ... , h,,,+1 of the sequence until it goes around H k times and the distance between the contact points of h,,,,+1 and h1 is less than 1/3. This situation is surely arrived at sooner or later, because at most A consecutive elements of the chain can touch the same side of H. Then, as above, we have

0< kA -

1 m-3k <3. 3

(2)

Expressing A with the help of (1) and multiplying by 3, we get

0<In-ml<1. The integers n and m can satisfy the latter inequality and the equivalent relation (2) only if equality holds on the left, that is, n = m and hn+1 = h1.

Problem G.16. The traffic rules in a regular triangle allow one to move only along segments parallel to one of the altitudes of the triangle. We define the distance between two points of the triangle to be the length of

the shortest such path between them. Put (n21) points into the triangle in such a way that the minimum distance between pairs of points is maximal. Solution. In the course of the solution, "distance" will refer to the ordinary distance of points while the italic "distance" will mean distance in the new

metric. Let ABC be the given triangle, AB = 1. In order not to lose the idea of the solution among the precise verification of simple geometric facts, we shall omit technical details. (a) Construction for lower bound. Let us divide each side of the triangle into n - 1 equal parts, connect each node of BC to a corresponding node

of CA by a segment parallel to AB, and divide these segments into n 2, n - 3, ... , 2,1 equal parts, respectively, in the order of their distances

276

3. SOLUTIONS TO THE PROBLEMS

from the side AB. We obtain the same set of points if we take through all nodes on the sides AB, BC, and CA straight lines parallel to the two other sides and consider all possible intersections of these straight lines. One may expect that this system of (n21) points is optimal. Let us denote by rn the minimum of distances for this system. (b) Theorem. Let (2) < N < (n21). If we are given N points in LABC, then one can always find two among them with distance < rn. If N = (n2 1) and the distance of any two points is at least rn, then the system of points coincides with the one constructed in (a). (c) For the proof, we need the proposition that a ball of radius r with center R in the metric space described in the problem is a regular hexagon centered at R, the vertices of which are obtained by moving R off to distance r parallel with one of the altitudes. The proof of this proposition is left to the reader. (d) We show here that if the distance of any two points of a certain subset

of the AABC is greater than rn = 2c = //n -1, then the subset contains at most (2) points. Consider triangles of the triangular lattice constructed in (a). Triangles homothetic to LABC with ratio 1/(n - 1) will be called

recumbent, and those homothetic to ABC with ratio -1/(n - 1) will be called standing. There are (2) recumbent triangles. Let us place a ball of radius c that is a regular hexagon of side c around the centers of recumbent triangles. They cover the recumbent triangles and the standing ones as well,

since each standing triangle is surrounded by three recumbent ones and the hexagons put around the centers of the neighboring triangles cover the standing triangle obviously. Thus, the given (2) balls cover AABC, that is, every point of the considered subset belongs to one of these hexagons. Because the diameter of such a hexagon is 2c = rn, a system of points for which the distance between any two points is greater than rn meets each hexagon in at most one point; hence, it contains at most (2) points. This proves the first part of the theorem. (e) We can show the uniqueness for N = (n21) points in the following way. If q < 1 and we shrink AABC from A with ratio q, then we get a triangle AB'C' that contains at most (2) points. If q -> 1, we obtain that side BC contains at least n points. Of course, side BC cannot contain

more than n points, and the points have to divide side BC into n - 1 equal parts. Let AB+C+ be the dilatation of ABC with respect to A with ratio (1 - 1/(n - 1)). The hexagons of side c around the points lying on BC cover the difference of the triangles ABC and AB+C+. Consequently, tAB+C+ contains (2) points, so that the distance of any two points is at

least rn = rn_1 (1 - 1/(n - 1)) = rn_1AB+. Since the theorem is trivial for n = 2, we may complete the proof by induction.

Problem G.17. Let C be a simple arc with monotone curvature such that C is congruent to its evolute. Show that under appropriate differentiability conditions, C is a part of a cycloid or a logarithmic spiral with polar equation r = ae''9.

3.4 GEOMETRY

277

Solution. Let rl = r(s), s E [0j] be the parameterization of C by arc length for which the radius of curvature p(s) is a monotone increasing function. The evolute E is parameterized then by r2 = r(s) + p(s)n(s), where n is the principal normal vector. The length of the arc of the evolute bounded by r2(0) and r2(s) is Q(s) = p(s) - po, where po = p(0). Let 4) be the congruence that superimposes C on E. There are two possibilities: 1. 4) takes r(O) to r2(0);

2. 4) takes r(O) to r2($). The radius of curvature of the evolute at the point r2 (s) is p(p(s) - po) in the first case and p(t - p(s) + po) in the second. Since, by Frenet equations, dr2

d2r2

da - n'

dal

P(s)P'(s)

we have p(p(s) - po) = p(s)p'(s) in the first case and p(B - p(s) + po) _ p(s)p'(s) in the second. We show that s = p(s) - po for all s E [0, .ÂŁ] in the first case. Otherwise, we could find an interval [a, b] by the suitable differentiability conditions

such that a = p(a) - po, b = p(b) - po, and s > p(s) - po or s < p(s) - po holds everywhere inside [a, b]. However, in this case, we would have

b - a = P(b) - P(a) = f b P (s)ds = b P(P(s) - Po) ds j4 b - a a

P(s)

since p(s) - po > s implies p(p(s) - po) > p(s) and p(s) - po < s implies p(p(s) - po) < p(s). We conclude that the solutions belonging to the first case have a natural equation of the form s = p(s) - po. These curves are logarithmic spirals the polar, equation of which has the form r = ae'9, and one easily checks that any arc of such a spiral is a solution. In the second case, .ÂŁ - s = p(t - p(s) + po) holds for all s E [0, f]. To see this, we have to show first that 4) is an orientation-reversing transformation of the plane. We may suppose without loss of generality that

while r(s) is running along the curve C, the unit tangent vector t(s) is rotating counterclockwise. Meanwhile r2(8) is running along the evolute E, and its unit tangent vector n(s) is also rotating counterclockwise. Since 4) takes r(0) to r2(P), the point 4)(r(s)) is running along E in the opposite direction and its unit tangent vector is rotating clockwise. Therefore, 4) reverses orientation. Orientation-reversing isometries of the plane are glide reflections, so the evolute F of the evolute of C is a translate of C. Let r3(s) denote the center of curvature of E at r2(s). The length of the arc of F lying between r3 (f) and r3(s) is equal to p(2 - p(s) + po) - Po Monotonicity of the curvature implies that the direction of the tangent vector uniquely characterizes the point at which the vector is tangent to the curve. As a consequence, we get that the translation above takes r(s) to r3(s).

For this reason, p(f - p(s) + po) = f - s holds, as we wanted to show. From here, we obtain p(s)p'(s) = f-s+po by our previous considerations. Thus, in the second case, the solutions are given by arcs of the curves

3. SOLUTIONS TO THE PROBLEMS

278

with natural equation p2(s) + y - s + po)2 = c. These curves are cycloids given by x = a(u - sin u), y = a(l - cos u), where a = 4 po + (t + po)2. We may conclude that, in the second case, the solutions are those subarcs i'2 of the arc corresponding to the parameter domain u E [0,7r] for which a = p2(Pl) + [iP2 + p(Pl)]2 (for example, the whole arc fits the requirement). Remark. We may pose the more general problem of finding curves that are similar to their evolute. The solution of this problem involves epi- and hypocycloids and logarithmic spirals of the general polar equation r = aeÂ°o as well.

Problem G.18. Given four points A1, A2, A3, A4 in the plane in such a way that A4 is the centroid of the AA1A2A3i find a point A5 in the plane that maximizes the ratio mine<j<j<k<5 T (A1AiAk) max1<i<a <k<5 T (AzAiAk)

(T (ABC) denotes the area of the triangle LXABC.)

Solution. The medians, which go through the centroid of the triangle, divide the plane into six angular regions. We claim that there is exactly one point in each of these regions for which the ratio in question attains its maximal value, namely 1/4. Let F be the midpoint of A2A3, let E denote the reflection of A in F, let R denote the trisecting point of the segment A3E that is closer to A3, and finally, let A5 denote the point on the halfline A4R for which A4A5 = 3/2. A4R. A5 is the point in the angular region A3A4E, that yields the maximum (see Figure G.15).

Figure G.15.

Suppose for a while that A5 is an arbitrary point in the angle A3A4E.

If this point is not in tA3A4E, then using the notation T(A4A5E) =

3.4 GEOMETRY

279

2T(AlA5A4) = 2b, T(A3A5A4) = a, T(A1A2A4) = c, T(A1A2A5) = d, we can express the area of the hatched domain as a + 2b + c = d,

(1)

since this area is the sum of the area of three triangles each having a side of length A1A2 = A3E, and the sum of the corresponding altitudes equals the altitude of LA1A2A5 corresponding to the side AlA2. Formula (1) remains true also when A5 is inside AA3A4E, only we have to take signed areas. In both cases, the areas with no sign satisfy

a+2b+c<d. From this, min (a, b, c) <

4d.

(2)

Therefore, the ratio in question is < 1/4. The necessary and sufficient condition of having equality in (2) is a =

b = c = 1/4. On one hand, this means that A5 is four times farther from the straight line A1A2 than A4 is; on the other hand, it implies that T(A3A4A5) = T(A1A4A5), that is, A4A5 1 A1A3. These two conditions together are satisfied only by the point A5 constructed above. One can show by a direct computation that in this case AA1A2A5 has the largest area among all DAjAjAk and that T(AiA;Ak) > d/4 for all i,7,k. Problem G.19. Let K be a compact convex body in the n-dimensional Euclidean space. Let P1, P2,. .., Pn+1 be the vertices of a simplex having maximal volume among all simplices inscribed in K. Define the points

Pn+2, Pn+3, ... successively so that Pk (k > n + 1) is a point of K for which the volume of the convex hull of Pi,. .. , Pk is maximal. Denote this volume by Vk. Decide, for different values of n, about the truth of the statement "the sequence Vn+l, Vn+2,... is concave."

Solution. The statement makes no sense for n = 0 and is obviously true for n = 1. Consider the case n = 2. It is clear that the vertices of the starting simplex lie on the boundary of K, otherwise we could move the point lying inside K farther from the opposite hyperplane. We show by general induction for the two-dimensional case that further points are also chosen from the boundary.

The base clue has already been proved. Suppose that the first n points lie on the boundary. In this case, they are vertices of a convex n-gon, each side of which cuts off a (possibly trivial) part of the convex figure. The new point is chosen from one of these parts, which results in the polygon "growing" by a triangle. Because of maximality, the new point must come from the boundary.

280

3. SOLUTIONS TO THE PROBLEMS

To prove concavity, we have to show that when taking two consecutive differences of the sequence, the second does not exceed the first. If the new points come from arcs belonging to different sides, then this is obvious, since otherwise we would have to add these points to the sequence in the opposite order. If two consecutive points are chosen from the same are over a side of the convex hull of preceding points, then let A and B

denote the endpoints of this side and C and D be the points added to the sequence. Since C is taken first, the area of AABC is greater than or equal to the area of LABD. Denoting by E the intersections of the diagonals of the quadrangle spanned by the points ABCD, the statement follows from the inequality T(EAB) > T(ECD), since the first difference is T(ABC), while the second is T(ABD) + T(ECD) - T(EAB). Observe that the supporting half-lines of the convex figure taken at A and B in the half-plane bounded by AB containing C (see Figure G.16) are parallel or intersect each other; otherwise the one from A and B that was chosen later into the sequence would not have the maximality property. Let us draw a straight line through C parallel to AB. This intersects the straight line BD at a point D' lying outside the segment BD, so it suffices to prove the inequality with D' instead of D. The triangles ABE' and CD'E' are similar, and by the above observation, the segment AB is not shorter than the segment CD', and thus the statement is implied.

Figure G.16.

Now we are going to prove that the sequence is not necessarily concave for n > 3. For n = 3, the regular octahedron serves as a counterexample. Let us choose four arbitrary, noncoplanar vertices of the octahedron for the vertices of the first inscribed simplex. The volume of this simplex is maximal. Indeed, suppose that ABCD is an inscribed simplex of maximal volume having as many common vertices with the octahedron as possible. If one of the vertices, say A, is not a vertex of the octahedron, then consider

the plane through A parallel to the plane BCD. By the maximality assumption, this plane may not contain an internal point of the octahedron, but it has a nonempty intersection with the octahedron and thus contains at least one vertex of the octahedron, which could have been chosen instead of A. The contradiction proves our proposition. One can similarly

3.4 GEOMETRY

281

show, that the fifth point can also be taken from the vertices of the octahedron. The point is that opposite faces of the octahedron are parallel, so the points lying at maximal distance from two neighboring faces form an edge of the octahedron. The sixth point then must be the sixth vertex of the octahedron. The second difference of this sequence is twice as much as the first one; hence the sequence is not concave. Turning to the general case, consider a prism over the octahedron, that

is, the convex hull of a three-dimensional octahedron and an (n - 3)dimensional simplex lying in a complementary subspace of the octahedron.

We can show that this is a suitable counterexample in the same way as above, since due to maximality, the starting simplex must contain all the vertices not belonging to the octahedron and the further points must be chosen from the three-dimensional plane generated by the octahedron.

Problem G.20. Let us connect consecutive vertices of a regular heptagon inscribed in a unit circle by connected subsets (of the plane of the circle) of diameter less than 1. Show that every continuum (in the plane of the circle) of diameter greater than 4, containing the center of the circle, intersects one of these connected sets.

Solution. Let A1, &..., A7 be the vertices of the heptagon, let 0 be the center of the circumscribed circle, and let H1, H2,. . . , H7 denote the connected sets that connects the pairs A1, A2, A2, A3, ... ,A7A1, respectively. Denote by C a continuum in question, by D the circle of radius 2 centered at O.

By the assumption d(Hi) < 1 (i = 1, 2, ... , 7), Hi is inside D and 0 Â˘ Hi. On the other hand, conditions d(C) > 4 and 0 E C imply that there exists a point Q E C which is outside D. Suppose to the contrary that UiHi and C are disjoint. In this case, the distance 6(x) of a point x E Hi from the closed set C U D not covering x is positive. Consider the open disc U(x, 6(x)) of radius 6(x) centered at x for all x E Hi. The union Fi = UXEH; U(x, 6(x)) of these discs is open and connected, and the

same holds for the union r = UiI'i. The set IF is inside D; furthermore, IF n (C U D) = 0. Since IF is open and connected, one can draw a closed broken line T* inside P that connects the points A1, A2i ... , A7. Let T be the boundary of the connected component of R2 \ T* containing O. T is a simple, closed, broken line inside D such that T and C are disjoint, 0 E C is inside T, and Q E C is outside T. We are going to show that the existence of such a T contradicts the fact that C is a continuum. Denote by B the set of points that lie inside T, by K the set of points lying outside T. By C n D = 0, we have

C= (CnB)U(CnK), where

(a) both C n B and C n K are open-closed in the subspace topology of C; furthermore,

282

3. SOLUTIONS TO THE PROBLEMS

b) none of them is empty because 0 E C fl B and Q E C fl K. However, these two facts together contradict the connectedness of the continuum C. Remark. One can give a counterexample for the proposition of the prob-

lem if C is required only to be connected, dropping the assumption that it is also closed. Indeed, draw two disjoint, infinite, broken lines into a square EFLJ so that one of them starts from E, the other starts from F, and both have closure that cover the segment U. Thus, M1 = T1 U J and M2 = T2 U J are two disjoint connected sets. Let us shrink and move the square and the figures contained in it in such a way that the points E, F that correspond to the vertices E and F get onto the segment A7A1 in the order A7, E, F, Al while L gets inside the heptagon. (We denote the image of a figure under this similarity by putting a_bar over the corresponding sign.) Let Q* be a point on the half-line OF whose distance from 0 is greater than 4. Then the claim of the proposition does not hold for the connected sets Hj = AjAj+1 (j = 1,2,... , 6), H7 = E7E1 U M1 U LA1, and the bounded and connected set C = OJ U M2 U FQ*. (This remark is due to LAszlo Babai.)

Problem G.21. What is the radius of the largest disc that can be covered by a finite number of closed discs of radius 1 in such a way that each disc intersects at most three others?

Solution. Draw a closed disc of radius 1 around each vertex of a square of side V r2-. These four discs cover the disc of radius drawn around the center of the square. We are going to show that a circle of radius greater than i does not have a covering having the prescribed properties. In the following, we shall denote by' the boundary of a set (the boundary of A is A') and "disc" will always mean a closed disc.

Let K be a disc of radius greater than vf2-, and suppose that it has a covering with the prescribed properties. Let us consider a subsystem A1, A2, ... , A,,,, of the covering in such a way that (i) the system A1, A2, ... , A,,,. covers K', (ii) if any of the discs A1, A2i ... , are removed, the remaining discs fail to cover K'. Such a subsystem can be produced obviously by omitting unnecessary discs one by one. There are no single points and empty sets among the intersections K' n Ai, since the discs different from Ai cover a closed subset of K', so if K' n Ai is empty or consists of one point, then Ai can be omitted. We claim that m > 5. Obviously, K' n Ai is an arc with diameter not greater than 2; however, the central angle of such an arc is less than 7r/2 in a circle of radius greater than f . Hence one needs at least five discs to cover K'. Now let X1 and X2 be the endpoints of the arc K'nA1. We may suppose without loss of generality that A2 covers X1 and A3 covers X2 since if both

3.4 GEOMETRY

endpoints of an arc K' n Ai were contained in the disc Aj (i whole arc would lie in Aj and Ai could be omitted.

283

j), then the

Now if we suppose that ((Al n K) \ A2) \ A3 is not empty, then it is an arc of positive length. Observe that this arc must be covered by one disc, otherwise Al would intersect more then three discs. Denote by A this

disc, and consider the arc A3 n K. This arc must be covered by A and Al together. Otherwise, A would also have a point in common with the disc that covers the remaining part, that is, A would intersect at least four other discs. Therefore, A U Al D A3 n K. But then A covers the endpoint of K' n A3 not contained in A1, and since four circles can not cover K', A must intersect at least one more disc, which would be the fourth disc intersected by A, and this is impossible. We conclude that ((A' n K) \ A2) \ A3 = 0, that is, A2 u A3 D A'1 n K. Similarly, we can find further two discs, A4 and A5, such that Al U A4 A3 n K, Al u A5 D A1n K, and A1,... , A5 are different. However, this is a contradiction, since in this case Al would intersect more than four other circles.

Thus, a disc of radius v admits a covering with the prescribed properties, but larger discs do not. On the basis of the above proof, we can also see that the covering of the disc of radius f is "essentially" unique (that is, unique up to rotations).

Problem G.22. Assume that a face of a convex polyhedron P has a common edge with every other face. Show that there exists a simple closed polygon that consists of edges of P and passes through all vertices.

Solution 1. Let S be the distinguished face of P, and suppose that P has n vertices off S. The proof goes by general induction with respect to n. The base clause is obviously true for n = 1; now let n > 1. It is easy to see that by omitting the edges of S from the graph of edges of P, we obtain a tree. (Indeed, imagine that P is a planet and its edges are dams, S is filled with water, while the other faces are empty basins and explode the dams that bound S.) For this reason, P has a vertex B not belonging to S such that the edges starting from B end on S with the exception of one edge. Let us denote by C the endpoint of the exceptional edge and by A1, A2, ... , A,,, the endpoints of the other edges ordered in correspondence with an orientation of S. Finally, let A0 be the vertex of S that precedes the vertex A1, andletAm+1 be the vertex that follows the vertex A,,,, according to the above orientation of S. (Figure G.17 shows a case with m = 3.)

Now if the half-line CB intersects the plane of S in a point A, then construct a polyhedron P' by gluing the prism AA1 ... AmB to P. P' is obviously convex itself and satisfies the conditions of the proposition. However, B is not a vertex of P, and thus P' has only n - 1 vertices off the plane of S. By the induction hypothesis, the edge skeleton of P' contains a simple closed polygon that passes through each vertex of P'. This polygon goes through A in one of the orders AOAA77.+1, AOAC, CAA,,,+1.

284

3. SOLUTIONS TO THE PROBLEMS

Figure G.17.

According to these three cases, replace vertex A of this polygon by the paths A1... A,,,._1BA,,,,, A1A2 ... A,..B, BA1A2 ... A,,,., respectively. The simple closed polygon we obtain this way passes through each vertex of P.

If CB is parallel to the plane of S, then choose a plane -y such that but has empty intersection with P. Now if we take the projective augmentation of the Euclidean space by adding a -y crosses the half-line BB

plane w of points at infinity and apply a projectivity II that takes -y to w, then the polyhedron IIP will completely consist of proper points, and the intersection point of the half-line (IIC) (IIB) and the plane of HS will also be proper; consequently, we may apply the previous consideration. We can proceed similarly if the half-line BAG intersects the plane of S.

Solution 2. Let S be the distinguished face, and let G be the graph obtained by removing the edges of S from the graph formed by the vertices and edges of the polyhedron. G is a tree, and the degree of a vertex of G not belonging to S is at least three. Let us choose an arbitrary face, B1, of P, and color the edges of G that lie on the boundary of B1. Suppose that we have already chosen the faces B1i B2, ... , Br in such a way that (1) Bi and Bj have no vertex in common if i # j, (i, j = 1, 2, ... , r), and

(2) for all i (i = 1, 2, ... , r), there exists an edge of G that connects Bi to one of the faces B1, B2, ... , Bi_1. Using that G is connected, we get that if P has a vertex not belonging to S that is not covered by the faces B1, B2, ... , Br, then there is a vertex x among these that is connected to some Bi (i E {1, 2, ... , r}) by an edge xy. Since the degree of x is at least three, x is covered by a face B,.+1 such that xy is not an edge of B,.+1. If some vertices of B,.+1 were covered by the faces B j (j = 1, 2, ... , r), then one of these vertices, say z, could be connected to x along edges of G without passing through points of the faces Bp On the other hand, according to the induction hypothesis, we

3.4 GEOMETRY

285

can join x and z by a path in G in such a way that all the vertices we go through belong to some Bj, which contradicts the fact that G is a tree. Consequently, the choice of faces B1, B2, ... , B,.+1 meets requirements (1) and (2). Now color the edges of G that lie on the boundary of Br+1. At the end of this process also color the edges of S that do not belong to any of the faces Bj. Thus, we obtain a system of colored edges that cover all vertices of P and form a polygon with the required properties.

Problem G.23. Let D be a convex subset of the n-dimensional space, and suppose that D' is obtained from D by applying a positive central dilatation and then a translation. Suppose also that the sum of the volumes of D and D' is 1, and D fl D' $ 0. Determine the supremum of the volume of the convex hull of DUD' taken for all such pairs of sets D, D'.

Solution. The conditions imply that D and D' are bounded convex sets with nonempty interior. The problem is equivalent to the determination of the supremum of the quantity V(co(D U D'))//(V(D) + V(D')) , where besides, having the properties just mentioned, D and D' have nonempty intersection and D' is obtained from D by a positive central dilatation and a translation. Let the ratio of the dilatation be a fixed positive number A. We show that in the case of a fixed A, the supremum in question is (1 + A + ... + A'')/(1 + An) = f (A), and this supremum is attained. The case n = 1 is trivial; let n > 1 in the following. First, we show that the case A = 1 can be derived from the case A $ 1. Indeed, if A = 1, then stretch D' from a point p E D fl D' with ratio 1 + E. Since D and the obtained D meet each other at p, we have

V(co (D U D')) <V(co(DUD'6 )) < f(1+e)(V(D)+V(D'')). 6 0, the right-hand side tends to f (1)(V(D) + V(D')). Taking the limit Assume now that A # 1. Changing the role of D and D' if necessary, we may also suppose that 0 < A < 1, since f (A) = f (A-1). Then D' can be obtained from D by one central dilatation, the center 0 of which will be chosen for the origin. V (D) = sup V(P), where supremum is taken for (closed) polyhedrons

P c D. It is easy to see that V(P) tends to V (D) if and only if r(P, D) = supXED d(x, P) tends to 0, where d(x, P) denotes the distance between x and P. It is also not difficult to see that if r(P, D) _< E, then r(co (P U P'), co (D U D')) < E (P' denotes the image of P in D'). Suppose that P contains a common point of D and D' and its preimage with respect to the similarity; then P fl P' $ 0. Taking such polyhedrons P, if V(P) -* V (D), then V(co (PUP')) -> V(co (DUD')), and thus to prove that the supremum is < f (A), we may restrict ourselves to polyhedrons.

Now let D be a (closed) polyhedron, D' = AD, p E D fl D'. We may assume that 0 Â˘ D. It is easy to see that co (D U AD) = UA<N,<1yD. Set E = (Uo<1,<1 jD) \ D. Then

U uD=DU(E\AE), a<Âľ<1

3. SOLUTIONS TO THE PROBLEMS

286

and thus,

U AD =V(D)+(1-a")V(E).

a<µ<1

We show that V (E) < V (D) A/(1 - A) . Let F1, F2, .. ., Fr be those (n - 1)-dimensional faces of D whose hyperplane separates 0 strictly from D. Then the segment that connects 0 to p intersects the hyperplane iri at a point pi. Therefore, using the notation q = (1/A) p, we have

d(0,7i) =

epid(4,ii) <

4pi

pd(4,7ri)

_ 1-

d(4,7r )

Since E is the union of pyramids over the faces Fi with vertex 0, and since the pyramids over the faces Fi with vertex q are contained in D, the previous inequality gives

Ed(O,iri)VV_i(Fi)

V(E) =

i=1 A

1-a i=1

>d(4,7i)V _i(Fi)

1-AV(D)

(V,i_1 denotes the (n - 1)-dimensional volume). From this,

V(co (DUD'))

V(D)+V(D')

- V (A<µ<lµD) - V(D) + (1 - a")V(E) V(D)(1+An)

V(D)(1+A' )

1 + (A + A2 + . .. + An)

1+A

.f (A)

Now we show that f (A) is an exact maximum. Let D be a simplex with vertices p1, p2, ... , pn+l; the corresponding vertices of D' are pi, p2, ... , p;+1+

and let p1 = p2. (For A = 1, we choose for D' the translate of D for which p'1 = P2.) In this case, co (D U D') is made up of the simplex D' and a truncated pyramid (or a prism if A = 1), so its volume is AnV(D) + (1 + A + + An-1)V(D) = f (A) (V(D) + V(D')). It remains to determine f (A). We claim that f (A) < f (l) (n + 1)/2, that is, (n+1)(1+A"`)

However, this follows from summing up the inequalities

1+An-A1-An-2=(1-AZ)(1-An-z)>0 for 0<i<n.

3.4 GEOMETRY

287

Problem G.24. Consider the intersection of an ellipsoid with a plane or passing through its center O. On the line through the point 0 perpendicular to a, mark the two points at a distance from 0 equal to the area of the intersection. Determine the loci of the marked points as or runs through all such planes.

Solution. Let B be an arbitrary body, 0 a point of it. Denote by F(a; B, 0) the pair of points P = {P+, P- } that lie on the straight line perpendicular to a at 0 and whose distance from 0 is equal to the area of the intersection of or and B: Area (a fl B). Let 4)(B) stand for the figure formed by the points P as or is varied: 4) (B) = U F(a; B, O) . a

Let E and t be a plane and a straight line intersecting one another orthog-

onally at 0, U the stretching in the direction of t with ratio A, and U* the stretching in the direction of E with ratio A. Obviously, U and U* are affinities.

Lemma.

4)(Lf(B)).

Proof. Set a' = U(a),

P = .P(a; B, O),

P' = F(u';U(B), 0) =.F(U(a);U(B),U(O)) The planes a and a' intersect each other in a straight line m, lying in the

plane E. Thus P' belongs to the plane A spanned by t and P. (Figure G.18 shows the trace of the mentioned figures in the plane A viewed from the direction of m.) Consider the points U E or and U' E a' that lie over a given point N E E n A (that is, UN, U'N I E). They satisfy NU' = ANU, and by elementary properties of affinities,

OP' _ Area (a' fl U (B)) _ OU' QP Area (a fl B) OU This equation shows that AOPP' can be obtained from tODU' by a dilatation and a right-angled rotation about 0 in the plane A. This similarity takes E n A to t, the point N to the intersection M oft and PP', and since similarities preserve angles and the ratio between corresponding segments,

we get that PP' I t and MP' = MP, which means that U* (P) = P', and this proves the lemma. Now let a, b, c be the half-axes of the ellipsoid E (E(a, b, c)). The equation of E takes the canonical form x2

a2+b2+C2=1

3. SOLUTIONS TO THE PROBLEMS

288

Figure G.18.

in a suitable coordinate system. Consider the sphere S2 (a) of radius a around the origin. Then ID(S2(a)) is also a sphere: S2(a27r). Now let 111 be the stretching with ratio A = b/a in the direction of the y-axis. Ul (S2 (a) ) is the ellipsoid El (a, b, a), and by the lemma, -D(El) = D(Ul(S2(a))) U, (.1, (S2(a))) = U (S2(a27r)) = E2(abir, a27r, ab7r) .

Now let U2 be the stretching with ratio c/a in the direction of the z-axis. Then U2 (El) is the given ellipsoid E(a, b, c), and a repeated application of the lemma gives ,D (E) = D(U2(El)) =1d2 (f'(E1)) = U2 (E2) = E3(bc7r, ac7r, ab7r),

which was to be determined. Remarks.

1. We remark that the proposition of the lemma is true also for n-dimensional bodies B and (n - 1)-dimensional hyperplanes. Thus, an application of the lemma gives an immediate answer to the n-dimensional version of the problem as well. 2. We also remark that the lemma, which lies in the base of the solution, concerns an arbitrary body B and an arbitrary point 0 of it. Hence, it is applicable to a much wider class of problems.

Problem G.25. Construct on the real projective plane a continuous curve, consisting of simple points, which is not a straight line and is intersected in a single point by every tangent and every secant of a given conic.

Solution. Removing a straight line from the projective plane we obtain a Euclidean plane. Let us introduce Cartesian coordinates (x, y) on it and

3.4 GEOMETRY

289

denote by K the circle of radius 1 centered at (0, -2). K is a conic of the projective plane. All conics of the projective plane are equivalent, that is, one can, find a projective transformation P for any conic C such that P takes C to K. Therefore, if r is a solution of the problem with respect

to the circle K, then P-1(F) is a solution with respect to C since the projectivity P and its inverse P-1 are bijections that take straight lines into straight lines, and secants and tangents of a conic into a secant or tangent of the image conic, respectively. Thus if F meets the requirements of the problem with the conic K, then so does P-1 (F) with the conic C. First, we construct on the Euclidean plane a continuous curve r consisting solely of simple points that is not a straight line and is intersected in a single point by every secant and tangent of K that is not parallel to the x-axis. Let y = f (x) be a function defined on the whole x-axis such that (a) f is continuous and differentiable, (b) 0 < f(x) < 1,

(c) the absolute value of the derivative of f is less than 1, (d) f is increasing on the left of x = 0 and decreasing on the right, (e) the limits lima-,,,, f (x) and lima-_,,. f (x) exist. We show that secants of the graph r of f do not intersect K. Indeed, suppose that £(x) is a linear function such that

e(X1) = f(xl), £(x2) = f(x2);

x1 < x2.

(1)

The graph Q of £(x) is a secant of I'. We have

It'(x)I = ItanaI < 1 because of (c), where a is the direction angle of £. If a = 0, then Q either coincides with the x-axis or lies above it because of (b); hence it does not

meet K. If a > 0, then the zero x0 of t satisfies x0 < xl < x2 by (b) and (1). In this case, the assumption 0 < x0 together with a > 0 and (1) implies f (xi) < P X2), 0 < x1 < x2i which is in contradiction with (d). If, however, x0 < 0 and I tan aI < 1, then Q does not meet K. We can prove similarly that £ has empty intersection with K also in the case

-1 < tana < 0. We conclude that the secants of F do not meet K, or, in other words, the tangents and secants of K have at most one point in common with F. Those secants and tangents of K that are parallel to the x-axis (denote the set of them by x) obviously have empty intersection with V. Other tangents and secants with nonzero direction angle intersect r since F divides the plane into two connected parts, one of which contains

the half-plane y < 0 while the other contains the half-plane y > 1. A straight line with nonzero direction angle has points in both half-planes, hence it must cross the graph F of f. Thus, the curve r of the Euclidean plane has the required properties. We augment the Euclidean plane to a projective plane by adding an ideal point to each family of parallel lines. Attaching the ideal point P,,. of the x-axis to IT, we get a closed curve F on the projective plane, which

3. SOLUTIONS TO THE PROBLEMS

290

is continuous by (e) and is intersected in a single point, namely in Pte, by the straight lines of X. The points of r are simple since the graph r of the continuous function f consists of simple points and we have closed r by adding one single boundary point. Finally, we have to present a nonlinear function that has properties (a)(e). For example, y = e_x2/2. This function gives a solution according to the previous considerations. But one can easily construct other suitable functions. The simple, though not differentiable, function whose graph is obtained from the x-axis by replacing the segment between the points A(-2, 0), C(2, 0) by the broken line ABC, where B has coordinates (0, 1), also does the job.

Problem G.26. Let T be a surjective mapping of the hyperbolic plane onto itself which maps collinear points into collinear points. Prove that T must be an isometry.

Solution. (A) First we show that T is injective. The proof (based on the idea of Nandor Simanyi) consists of three steps.

1. We claim that the inverse image of a point is convex. For this

purpose, we need to show that if T(A) = T(B), then every point C of the segment AB is mapped to T(A). Suppose it is not so: T(C) # T(A) for some C E AB. Let e and f be the straight lines perpendicular to T(A)T(C) at T (A) and T (C), re-

Let us choose the points P and Q in such a way that their images are points of e and f, respectively, different from T(A) and T(C) (see Figure G.19). There are no three collinear points among A, B, P, Q since there are obviously no three collinear points among their images T(A), T(B), T(P), spectively.

T(Q). According to this, the straight line CQ intersects side AB of DABP in an inner point and thus must cross one of the two other sides. The image of this second intersection point must lie

on both e and f, which is a contradiction since e and f do not intersect each other.

T(Q)

Figure G.19.

T(C)

3.4 GEOMETRY

291

2. Now we show that the inverse image of a point X contains no interior point. Suppose to the contrary that the interior of T-1(X)

is not empty. Draw a straight line f through X. We claim that int (T-1(f \ {X})) # 0. Indeed, consider a point Z of f different from X and denote by P a point of its pre-image (see Figure G.20). Then the images of straight lines that go through P and intersect T-1(X) lie in f . In other words, T-1(f) contains a pair of opposite angular domains with common vertex P. Since T-1(X) is completely contained in one of these angles, it is disjoint from the other; therefore, int(T-1(f \ {X})) # 0, as we claimed. Now rotate f about X and consider the sets T-1(f \ {X }). They form an uncountable family of disjoint subsets of the hyperbolic plane such that each set has an interior point, which is a contradiction.

f X

Z=T(P)

Figure G.20.

3. Now we are in the position to prove that the preimage T-1(X) of every point X consists of one single point. Again we use the indirect method. If T-1 (X) is not a point, then it is a segment, half-line, or a straight line according to the previous considerations. In any case, T-1(X) is contained in a straight

line, which will be denoted by e in the following. Let A and B be two points of e that are mapped to X (see Figure G.21). It is easy to see that T(e) is a part of a straight line g. Draw a straight line f through X other than g. One can show as above that the interior of the set T-1(f \ {X}) is not empty. Rotating f about X, we can again get a contradiction.

g DT(e)

Z=T(P) T(A)=T(B) f

Figure G.21.

292

3. SOLUTIONS TO THE PROBLEMS

(B) The major difficulties of the proof are over. As a next step, we show that the images of noncollinear points are noncollinear. Indeed, if the noncollinear points P, Q, and R were mapped to a straight line e, then the images of the straight lines PQ, QR, and RP would also be contained in e. Consequently, every straight line that twice crosses the boundary of LPQR twice would be mapped into e. However, such a straight line passes through any point of the plane; therefore, the image of the whole plane should be contained in e, which contradicts the fact that T is surjective. (C) From this it follows that T-1 also takes collinear points to collinear points; the image of a straight line (under T) is a straight line, and the images of intersecting straight lines are intersecting, the images of nonintersecting straight lines are nonintersecting. (D) T preserves the ordering on straight lines. Indeed, if A, B, and C are three different points of a straight line e, then, as is known, C does not separate A and B if and only if one can find straight lines a, b, and c through A, B, and C, respectively, such that a and b intersect each other but c intersects neither a nor b (see Figure G.22). T preserves the latter property by (C), hence preserves ordering.

Figure G.22.

(E) Therefore, T preserves half-planes, half-lines, and segments. Furthermore, it preserves the ordering of pencils. Thus, the images of asymptotic half-lines are asymptotic. (F) Now we show the existence of a length d such that T maps segments of length d onto segments of length d. Let us perform the following construction (see Figure G.23.a). We denoted the two half-lines of e determined by P by f and g. Let us take a point X not lying on e and draw the half-lines h and k starting from X and asymptotic to f and g, respectively. Take a point Y on the elongation of k beyond X; let the segment YP intersect h at U. Draw an asymptotic half-line to g from U, and suppose that it intersects the segment XP in V. Denote by Q the intersection point of YV and g.

3.4 GEOMETRY

293

Figure G.23.a

We claim that Q does not depend on the choice of X and Y.

It is very easy to check this statement in the Cayley-Klein model of hyperbolic geometry (and it is also sufficient, since we know that

if a proposition is true in every Cayley-Klein model, then it is a true theorem of hyperbolic geometry). The above construction is a construction of a complete quadrangle in the model (see Figure G.23.b); consequently, denoting by I and J the horizontal points of f and g, respectively, the pairs JP and I, Q are conjugate, and the points P, J, and I determine the fourth harmonic Q uniquely.

Figure G.23.b

It is clear that the length d of the segment PQ does not depend on the choice of P either. Since T transforms Figure G.23.a into a similar one, it moves segments of length d into segments of length d, as we stated.

(G) Of course, the segments of length nd are also invariant under T, as

3. SOLUTIONS TO THE PROBLEMS

294

are the angles of regular triangles of side nd. Because the angle of these triangles tends to zero as n tends to infinity, there exist arbitrarily small invariant angles. As the sum of invariant angles is also invariant, a dense set of angles will be invariant under T. Taking into consideration that T preserves the ordering on pencils, it follows that T preserves angles. However, in hyperbolic geometry, a triangle is determined up to isometries by its angles, so the image of every triangle

under T is a triangle congruent to the original one, hence T is an isometry.

Problem G.27. Let X1,.. , X,, be n points in the unit square (n > 1). Let ri be the distance of Xi from the nearest point (other than Xi). Prove the inequality .

r2+...+r2 <4 Solution. We shall prove by general induction. The base clause is trivial

for n = 2. Let n > 2, and suppose that the proposition is true for any system of points consisting of less than n points. This assumption means also that if the points Y1i ... , Yk are enclosed in a square of side a and qj denotes the distance of Yj from the other points, then for 1 < k < n we have

q < 4a2.

Let us divide the square into four congruent squares by the medians. Denote the small squares by N1, . . . , N4 (see Figure G.24). We shall separate some cases according to the possible distributions of the points Xi in the small squares and then study each case individually. If a point Xi lies on the common boundary of two squares, then decide which square it belongs to; it does not matter how.

Figure G.24.

Case A. None of the small squares contains exactly one point. Here we may distinguish two further subcases: Either all points are in one small

3.4 GEOMETRY

295

square or the points are distributed in more than one square. In the first case, applying the induction hypothesis f o r X 1 ,.. . , X,,,_1i we get

Since r2 < 2, the proposition of the problem is true in this case. In the second subcase, however, each small square contains less than n points (and if it contains a point at all, then it certainly contains more than one). Therefore, by the induction hypothesis, 2

<4(1) =1

(1)

X;EN

for 1 < v < 4, from which the proposition of the problem follows. Case B. There is exactly one small square, say N1, that contains one point. For example, let X1 E Ni. Inequality (1) holds also in this case for

v = 2,3,4. Therefore, if ri < 1, then we are ready. We may suppose for this reason that r1 > 1. Of course, r2 < 2, and this estimation would be enough if one of the small squares were empty. Thus, we may suppose in the following that each small square contains a point; from this it follows that r1 < 5/4 because of the point in N4.

. X, N2

N3 N43

N44

Figure G.25.

Let us divide N4 into four smaller squares with the help of its medians, and we will denote these squares by N41, N42, N43, N44 (see Figure G.25).

As r1 > 1, the shaded domain contains no point. We shall show that 5 r2

XiEN4j

5 16

(j _ 3,4).

(2)

This follows from the induction hypothesis applied for N4a if the number of points contained in N4j is not exactly one. If, however, N4a

3. SOLUTIONS TO THE PROBLEMS

296

contains exactly one point, then its distance from another point in N4 is

at most '/4, that is, (2) holds in any case. Then n

r?=r2+ E r?+ > r?+

i=1

XiEN2

XiEN3

XiEN4

< 4+1+1+216 <4. Case C. The number of points is equal to one in exactly two small squares. In this case, the squares can be coupled in such a way that squares in a couple are neighbors and one of the squares in a couple contains exactly one point while the number of the points in the other is not one (see Figure G.26). Assume, for example, that N1 and N4 are in one couple and N1 is the one that contains exactly one point. Obviously, it suffices to show

that

E ri < 2. XiEN1UN4

This can be done in the same way as in case B.

1 point

1 point i

1 point Figure G.26.

Case D. Exactly three squares, say N1, N2, and N3, contain one point. Call these points X1, X2, and X3, respectively. Let us divide N4 into four small squares, as shown in Figure G.27. First of all, we remark that r2 < 5/4. We distinguish three subcases. Assume first that the rectangles N41 UN42 and N43 U N42 both contain a point. Then r3<C312+(1)213

Consequently,

1 i=1

2 3

+ E r? < Xi EN4

13 16

5 4

13 16

+1<4

3.4 GEOMETRY

297

. x1

.x2

N41

N42

N43

Figure G.27. Second, suppose that, for example, N41 U N42 contains no point but N43

and N44 do contain a point. In that case, r3 < 13/16, ri < 1 + 1/16, and furthermore, (2) is now valid. Hence, n

r?=ri+r2+r3+ i=1

is+2i <4.

rZ < XiEN4

Third, it remains to study the case when neither N41 nor N42 contains a point and one of N43 and N44 is also empty. Applying the induction hypothesis to the other (nonempty) square we get n

i=1

ri=ri+r2+r3+ E ri- -

=4.

XiEN4

Case E. The last case that remains is when each of the small squares contains one point. Let Xi E Ni. We prove that d(X1, X2)2 + d(X2, X3) 2 + d(X3, X4) 2 + d(X4, X1)2 < 4.

(3)

((d(P, Q) denotes the distance between the points P and Q.) This already implies the inequality of the problem.

Let us think of (3) as a strictly convex function defined on the set

xv=1N C R8. Such a function can attain its maximum only at the extremal points of the boundary of its domain. An extremal point corresponds to such a configuration of the points Xi in which every Xi is positioned at a vertex of the corresponding square Ni. Since for these cases (3) can be easily verified, (3) holds in general.

Remark. Contestants B. Brindza, V. Komornik, P.P. Palfy, V. Totik, and Zs. Tuza described those configurations of the points for which the sum of squares in question is equal to 4. They found that there are only

298

3. SOLUTIONS TO THE PROBLEMS

two possibilities: n = 2 and the points are positioned at opposite vertices of the square, or n = 4 and the points are put to the vertices of the square.

Problem G.28. Give an example of ten different noncoplanar points

Pi, ... , P5, Q1, ... , Q5 in 3-space such that connecting each Pi to each Qj by a rigid rod results in a rigid system.

Solution. First, we describe two constructions. 1. Let g be a rigid structure. Take a further point P, and connect it by a rigid rod to three points Q1, Q2, Q3 of g, for which P, Q1, Q2, Q3 are not coplanar. Then a rigid system is obtained. II. Let g be a rigid system and PQ a rod in it. Take a point T on this rod and connect it to two further points R, S of g, for which P, Q, R, S are not coplanar. The system obtained will also be rigid. Putting it the other way around, these two statements mean that to get a rigid realization of a graph g, it is enough to have (a) a rigid realization of a graph obtained from g by deleting a point of degree 3, or (b) a rigid realization of a graph obtained from g by deleting a point of degree four and connecting two of its neighbors by an edge. In our case, the graph to be realized is as shown in Figure G.28.

We omit the edge P1Q1. Without drawing edges between Pi's and Qt's it is sufficient to find a realization of the following graphs: by (b),

again by (b),

3.4 GEOMETRY

299

by (a),

by (a)

The last graph is a tetrahedron, which is obviously rigid.

Problem G.29. Let us define a pseudo-Riemannian metric on the set of points of the Euclidean space 1E3 not lying on the z-axis by the metric tensor 1

- x2 + y2

where (x, y, z) is a Cartesian coordinate system in V. Show that the orthogonal projections of the geodesic curves of this Riemannian space onto the (x, y) -plane are straight lines or conic sections with focus at the origin.

Solution 1. Let a, 3, and y run through the indices 1,2, and denote by (x1, x2, x3) the coordinates (x, y, z). We compute the differential equation of the geodesics. We can write the metric tensor and its inverse (gik) in the following form ga(3 = bap,

g = -

gao = bao, (x1)2

ga3 = 0,

ga3 = 0, 1

+ (x2)2,

g33 = -

(x1)2 + (x2)2

Using the formula

E 3

h=1

gih

09ih + aghk axk axi

_ 89jk aXh

300

3. SOLUTIONS TO THE PROBLEMS

for the Christoffel symbols, we obtain

rap =Pap = r30 = r.3 = 0; x-"

F33 = 2 Nl_(Xl)2 + (x2)2 ' r3 73 =

r3 3ry _-

2 (x1)2 + (x2)2

By these, the equation of the geodesics,

i,j

can be written in the form

+ 2 2-zj2 =0; x +y

x//

(1)

x2+ y2

z'20

z"+x2+ Y2 x'z'+x2+ 2y'z'=0.

(2)

(3)

This system is satisfied by the curves

x=as+b, y=as+b, z=c, the projections of which onto the (x, y)-plane are straight lines. Furthermore, looking at the equations, we see that if z' = 0 at the initial point of an integral curve, then z' - 0 along the curve. Assume z' # 0. Then the third equation can be written in the form

2 ds

z'/

ln(x2 + y2)

Therefore,

lnz = In

x2 + y2

c x2 -+y2

(where c is an arbitrary positive constant). Substituting this into equations (1) and (2), we get 2

x' + 2 (x2 + y2)3/2 = 0 ; c2

y =0. Y + 2 (x2 + y2)3/2

3.4 GEOMETRY

301

This system is non other than the well-known system of differential equations 8

x" _

aTX

c2 2

x2 _+V 2

e (C2

x2 + y2

of the Kepler problem, the solutions of which are conic sections with focus at the origin.

Solution 2. The problem of finding geodesics of the given space is equivalent to the problem of finding stationary curves of the variational problem corresponding to the integral VX72

+ y2 -

x2 + y2z2 dt

Since the variational function is positively homogeneous of degree one in the variables x, y, z, it is well known that those stationary curves of it that satisfy the condition x2

+ y2 - v/x2 + y2z2 = 1

(4)

coincide with the stationary curves of the problem corresponding to the integral x2

+ y2 -

x2 + y2z2 dt.

Changing over to the coordinate system (r, cp, z) with the coordinate transformation x = r cos cp, y = r, z = z, we obtain the variational problem

(r2 + r2cp2 + rz2)dt.

The integrand function is independent of cp and z. Therefore, the EulerLagrange equations OF

d aF

dt acp'

dt az

imply that OF = A,

aF = B az

(A and B are constant),

that is, we obtain the first integrals 2r2cp = A,

-2rz = B.

3. SOLUTIONS TO THE PROBLEMS

302

We introduce the new unknown function u(cp) = 1/r(cp). For this, we have

1 du . 2 dcp

1 du

A du

A 2r2

u2 dcp

2 dcp

Therefore, by the relation (4), 1 = T + r2cp2 - rz2 22

\ u/2 +

(du 12

z u4I

B2 u2

B2 4

Let us differentiate with respect to cp: z

{ 2 L2+uJ-2 }dcp=0. The equation du/dcp = 0 characterizes circles centered at the origin. Dividing by du/dcp, we get the equality d2u d

+u = 2A2

the general solution of which can be written in the form B2 u(cp) = 2A2 (1 + e cos(cp + w))

that is, r(cP)

p 1 + e cos(cp + w)

2A2)

and this is the focal equation of a conic section with focus at the origin.

Remark. Analogously, the trajectory of a point moving in a conservative force field can be obtained as the spacelike projection of a geodesic curve of a suitable Riemannian metric on the four-dimensional space-time plane.

Problem G.30. Let us divide by straight lines a quadrangle of unit area into n subpolygons and draw a circle into each subpolygon. Show that the sum of the perimeters of the circles is at most 7r ,,fn- (the lines are not allowed to cut the interior of a subpolygon).

Solution. It is known that the area of an n-gon circumscribed about a circle of radius R is at least R2n tan(ir/n) . Cutting the quadrangle by one straight line after the other, we determine the number of the domains

produced and the sum of the angles of the domains. When we draw a

3.4 GEOMETRY

303

new straight line, the number of polygons increases by k, and the sum of the angles increases by at most 2k7r, depending on how many nodes of the previous decomposition the new straight line goes through. This way, if the number of sides of the polygons in the resulting decomposition: k1, k2, ... , kn, then n

>(ki-4) <0. i=1

Denote by Ti the area of the ith domain and by Ki the perimeter of the inscribed circle. We know that 2

K? <

ki tan k i

Then n

KK <7r i=1

i=1

ki tank i

Ti < 1r N

i=1

ki tan : i

by the Cauchy-Schwarz inequality. Therefore, it suffices to show that n i=1

ki tan'.i

> n.

If ki = 4, then the corresponding term in the sum is just equal to 1. If ki > 5, then by >,(ki - 4) < 0, this term comes together with (ki - 4) triangles, and for this reason, it is enough to see that for k > 5,

+(k-4)3tan3 <k-3.

ktan2 Since

3 tan 3

it suffices to show that for k > 5, 4

k tan

< 0.23k + 0.079.

The left-hand side is less than 4/-7r, and this is enough for k > 6. When k = 5, the left-hand side is less than 1.11 and the right-hand side is greater than 1.2.

3. SOLUTIONS TO THE PROBLEMS

304

Problem G.31. Let K be a convex cone in the n-dimensional real vector space Rt, and consider the sets A = KU(-K) and B = (R'" \A)U{0} (0 is the origin). Show that one can find two subspaces in Rtm such that together they span RT, and one of them lies in A and the other lies in B.

Solution. First, we show that if x c int K then x E K. Take a simplex with vertices in K that has x in its interior. The vertices can obviously be moved to K by a small perturbation in such a way that x remains in the simplex; thus x E K. Now we show that if x K, then there exists u E RT such that (u, x) < 0, but (u, y) > 0 for every y E K. Let k be the point of K nearest to x. The perpendicular bisector hyperplane of the segment connecting x to k separates x from K, that is,

there exist u E Rn and b E R such that (u, y) > b for every y E K, but (u, x) <b. Since 0 E K, b < 0. If we had (u, y) = e < 0 for some y E K, (u, Ay) _ Ae 0. Thus, (u, y) > 0 for every y E K. We prove the proposition of the problem using induction. The statement

is obvious for n = 0. Assume that n > 1 and the statement is true for W provided that 0 < s < n. If the maximal number of linearly independent elements of K is s and s < n, then K is contained in an (n-1)-dimensional subspace H. In this case, let E0 C K U (-K) and Fo C (H \ A) U {O} be subspaces with which the proposition holds in H. Taking E0 for E and the span of Fo and a vector f E R' \ H for F, we find that the proposition is true. We may thus assume that K contains n linearly independent vectors and, as a consequence, it has an interior point. Set

N = {u: (u, x) > 0 for every x E K).

If N = {O}, then K = Rn, and thus K = R. Therefore, we may assume

that N

{O}. Choose a maximal linearly independent system u1,. .. , uk consisting of vectors from N. Let H be the hyperplane

H={x: If H n K is a cone,

(HnK)U(-(HnK))=HnA, (H\(HnA))U{O}=HnB; therefore, the proposition can be applied to H n A and H n B. We obtain subspaces E0 C H n A and F0 C H n B, which span H. Let us observe

that E0 C {x: (ul, x) = (u2, x) = ... = (uk, x) = 0}. Let l1i ... , l,, be a basis of E0, and let fl, ... , f, be a basis of F0. We may E K; furthermore, m + 1 > n - 1 holds. obviously assume that 1 1 ,. .. (K \ H) has at least one interior point, call any of them e, and let E be

3.4 GEOMETRY

305

the subspace generated by E0 and e. If we show that E C A, then we are done with the choice F = F0.

First, we show that Eo C K. If x E E0 and u E N, then u = )lul + + Akuk, and thus (u, x) = Al (ul, x) + ... + Ak (uk, x)

from which x E K. Now, if to E E0, then to E K and hence to + e c K; moreover, to + e E int K C K. From this, )(1o + e) E A for every A E that is, E C A.

Problem G.32. Let V be a bounded, closed, convex set in R', and denote by r the radius of its circumscribed sphere (that is, the radius of the smallest sphere that contains V). Show that r is the only real number with the following property: for any finite number of points in V, there exists a point in V such that the arithmetic mean of its distances from the other points is equal to r. Solution. (a) Since V is bounded, it is easy to see that there exists a minimal sphere

of radius r that contains V. Call this sphere G, and denote its boundary by S, its center by 0 (the origin). We show that 0 E co (S fl v). Suppose that this is not true. Then there exists a hyperplane H that separates 0 from s n v, since s fl v is compact. We may assume that

H = {(xl, ... , x",) : X. = c}, where 0 < c < r.

Since r is minimal, the spheres of radius less then r centered at Ok = (0, . . . , 0,1/k) do not cover V. Therefore, we can find points ak E V such that Iak - OI > r. The sequence {ak} possesses a convergent subsequence

ak, - a*. Obviously, a* E V and Ia*I = r; therefore, a* E S fl V and thus a* = (a*, ... , a* ), where a* > c. Then, for i > io, aki = (at, ... , an), r2 since ak E V. From this, we get where an > c and E n-

Iak, - Ok;I =

(:(a2 + (a- 1/k)2

< (r2-2c/ki+1/ki2)1/2 <r, but this is impossible. This shows that 0 E co (S fl V), and therefore 0 E co (V) = V as well. If al, . . . , am E V, then Iai I < r, and thus the average of the distance between

0 and the points ai is < r. On the other hand, s = (1/m) Em l 1 ai E V, and by the minimality of r we can find a point y in V such that I s - yI > r. Thus,

Iai - yI

(ai - y)I = Is-yI>r.

3. SOLUTIONS TO THE PROBLEMS

306

By the convexity of V, the segment [0, y] lies in V, and one can obviously

find a point z on this segment for which (1/m) E jai - zl = r. (b) If u > r, then for a1 = 0, there is no such point x E V for which Ia1 - xI = u. (c) Now let u < r. Since 0 E co (S fl v), there exist points a1, ... , ak E

S fl V and numbers t1i ... , tk such that for 0 < ti, E

ti = 1 and

Ek 1 tiai = 0. Set 6 = (r - u)/kr > 0, and choose the natural numbers q 1

and pi in such a way that k

(i=1,...,k).

pi =q, and Ipilq-til <E i=1

Let the system of points b1, . . . , b9 consist of the points ai, so that ai is taken pi times for each i. Then Ib; I = r (since bj E S) and k

E(pi/q - ti)ai < CE lai I = krE. i=1

i=1

We show that for any x E V, (1/q) >9=1 Ibi - xI > u. Indeed, denoting by (a, b) the dot product, for x E V, we have gEIbi-xI> i=1

=1 r =1

E iE1(bi-x,bi)=Q (q T

r2_ ( x,bi q

i=1

=r- 1qr x,bi >r - it qr 9

i=1

>r-krE>u.

i-1

Problem G.33. Show that for any natural number n and any real number d > 31/(3n - 1), one can find a covering of the unit square with n homothetic triangles with area of the union less than d.

Solution. Let us circumscribe congruent homothetic triangles about the faces of a regular hexagonal tessellation such that the area of the triangles is 3/2 times larger than the area of the hexagonal face. These triangles cover the plane with density 3/2. Let the radius of the circle inscribed in one of the triangles be equal to 1. Let us translate the sides of every triangle toward its center by e. The density of the new triangles is equal to (3/2) . (1 - E)2. There are holes produced in the covering: upside-down triangular holes circumscribed

about a circle of radius e. The relative number of the holes is twice the number of the triangles, so the density of the holes is (3/2) 262 = 362. Let us cover each hole with some density d > 1. By this we mean that we cover each hole with some plates, whose total area is equal to d times the area of the hole. The joint density of the triangles and the plates is

3.4 GEOMETRY

307

D = (3/2) (1- e)2 +3de2. D is minimal ate = 1/(2d - 1) . The minimum is D = 3d/(2d+ 1) . Let D,,, denote the infimum of the densities of the coverings with homothetic triangles of n different sizes. As we have seen, D1 < 3/2 . Since the holes in the above construction can be covered by homothetic triangles of (n - 1) different sizes with density arbitrarily close to Dn_1i we have

D2 <

3D1

2D1+1'

D3 < 2D2+1'...' 3D2

that is, Dn<33n1.

Problem G.34. Let R be a bounded domain of area t in the plane, and let C be its center of gravity. Denoting by TAB the circle drawn with the diameter AB, let K be a circle that contains each of the circles TAB (A, B E R). Is it true in general that K contains the circle of area 2t centered at C?

Solution. We shall show that the answer to the question is negative. Let R be the semicircle in the plane with Cartesian coordinates (x, y) defined

by the inequalities x2 + y2 < 1, x > 0. The area of R is t = it/2, and the radius of the circle of area 2t is equal to 1. Assume that P E TAB for some pair of points A, B E R. Denote the position vectors of the points P, A, B by p, a, b, respectively. The condition P E TAB is equivalent to the inequality Ip - (a - b)/2I < Ia - bI /2. Since

Ip- (a+ b)/21 ? IpI - I(a+b)/2I, we have

IpI < Ia + bI /2 + Ia - bI /2 < ((Ia + b12 + la= ((Ia12 + Ib12 + 2(a, b) + Ia12 + Ib12 - 2(a, b))/2) 1/2 _ (Ia12 + Jb12)1/2 <

Therefore, if we choose for K the circle of radius' centered at the origin, then K will contain all the circles TAB (A, B E R). Let us compute the coordinates (xC,yC) of the center of gravity C of the figure R. By the symmetry of the figure, yC = 0. To compute xc, we use the well-known method

xC = t

1x2

1 - x2 dx =

t [_(1_X2)3/21' 1= 10

- 34

3. SOLUTIONS TO THE PROBLEMS

308

We want to show that the circle of radius 1 centered at (4/3ir, 0) does not contain the circle K. For this purpose, we have only to show that 4/3ir+ 1 > v/-2-. Using it < 3.2, 2.4.0.416 < 1, and (1.416)2 > 2, we obtain 4

4 +1>

37r

3.2+I

.4+1>1.416>V2.

Therefore, R is indeed a counterexample to the proposition in question.

Remark. Competitors disproved the proposition with many different counterexamples. Most constructions resembled the set-theoretic union of the circle x2 + y2 < 1 and the ellipse x2/(1 + e)2 + y2/E2 < 1 for some suitably small positive E.

Problem G.35. Let M' C R"+1 be a complete, connected hypersurface embedded into the Euclidean space. Show that M" as a Riemannian manifold decomposes to a nontrivial global metric direct product if and only if it is a real cylinder, that is, M'n can be decomposed to a direct

product of the form M' = Mk x R"-k (k < n) as well, where Mk is a hypersurface in some (k + 1)-dimensional subspace Ek+l C Rn+1 the orthogonal complement of Ek+1

Rn-k is

Solution. The solution rests upon the two fundamental equations of classical surface theory, namely, the Gauss equation

R(X, Y)Z = g(Z, A(X))A(Y) - g(Z, A(Y))A(X) and the Codazzi-Mainardi equation

(VxA)(Y) = (VyA)(X) In these formulas, X, Y, Z are vector fields tangential to the hypersurface,

g(X, Y) is the first fundamental form, A(X) = dxm is the Weingarten map (m is the unit normal vector field of the hypersurface, dx is the directional derivative in R"+1), VxY is the covariant derivative, and, finally, R(X, Y)Z is the Riemannian curvature tensor. A(X) is a self-adjoint linear mapping on each tangent space. It is also called the extrinsic geometrical fundamental form, since with its help one can describe the shape of the surface in the space. For example, the following statements are known and can simply be derived from the CodazziMainardi equations. Let WP be the kernel of Ar at the point p E M", and assume dim WrÂ°

= constant = k on some open subset U C M'. Then the distribution of the subspaces WrÂ° is integrable, and the integral manifolds are open subsets of a subspace Rk C R'n+1 a The Riemannian curvature of the hypersurface M" is 0 if and only if the rank of A is at most 1 at any point. For such a surface let U denote the open set of those points at which A 0 0, and let n be a unit vector field on

3.4 GEOMETRY

309

U consisting of eigenvectors of A such that A(n) = An holds with A Y1- 0. Then, by the previous proposition, integral manifolds perpendicular to n are open subsets of some subspace Jn-1 in JRn+1 Let c(s) be a curve on such an integral manifold parameterized by arc length, and consider the

function A(s) = A(c(s)). Codazzi-Mainardi equations yield by a simple computation that A' = V(s)A, where V(s) := -g(c(s), Vnn), and thus A(s) = A(0)efo (t) dl Consequently, if A 0 at one of the points of the integral manifold perpendicular to n, then A 54 0 along the whole integral manifold and also at

its boundary points. For this reason, if M' is complete, then the maximal integral manifolds perpendicular to n must fill the whole subspace Rn-1. Consider a bending of such a complete hypersurface having Riemannian curvature 0 onto Rn. (If Mn is not simply connected, then we bend its universal covering space.) The above integral manifolds will be bent onto parallel hyperplanes of Rn; Thus the distance between the points of an integral manifold and another integral manifold is constant. This means, that the integral manifolds themselves are parallel, (n-1)-dimensional subspaces JRn-1 in Rn+1 since otherwise the above distance would not be constant. The orthogonal complement JR2 of the subspaces Rn-1 cuts Mn in a curve M1, and we obviously have the cylinder decomposition M' = M1 X Rn-1 To sum up, we have the following propositions. Proposition 1. A complete hypersurface of Riemannian curvature 0 is always a cylinder of the form Mn = M1 x Rn-1, and M1 is a curve in the orthogonal complement 1[82 of

Rn-1

Proposition 2. If a complete hypersurface Mn decomposes into a metric

direct product in the form Mn = Mk X Mn-k, then either Mn has Riemannian curvature 0, or the Weingarten map vanishes at each point on the tangent spaces of one of the manifolds.

Proof. Denote by TT the tangent space of Mn at p E Mn and by Tp = T1 x T the decomposition of the tangent space corresponding to the direct product decomposition of Mn. Since for a metric direct product the Riemannian curvatures are also multiplied in a proper way, R(T I , T,2)X = 0 holds. Combining this with the Gauss equation we get

g(X, A(Tr1))A(TP) = g(X, A(Tr2))A(T')

This identity can hold for a self-adjoint mapping A only if A has rank 1 or A vanishes along one of the subspaces T. We have to show that if the Riemannian curvature of the surface Mn is not identically equal to zero, then A vanishes along Ti at each point of the manifold.

Let p E Mn be a point at which R(X, Y)Z 54 0. Then, as we have seen above, A vanishes on one of the subspaces T. Suppose that A(T2) = 0 holds. By Gauss's equation, the Riemannian curvature R(2) (X, Y) Z of the

3. SOLUTIONS TO THE PROBLEMS

310

manifold Mn-k vanishes, while the curvature RM (X, Y) Z of the manifold Mk is different from 0 at p. Since the tensor R(1)(X,Y)Z is constant along

each copy of M,-k in the direct product, A(T2) = 0 along the copy of M,-k that goes through p. This means that the Riemannian curvature of Mn-k is equal to 0 identically. It follows also that A(T9) = 0 in every point q = (ql, q2) E Mk X M,-k such that the tensor RM is different from

oatg1. It remains to show that A(Tg2) = 0 also at points q = (ql, q2) for which R(1}

1q, $ 0.

Let Vk C Mk be a maximal connected open subset such that RM = 0 at each point of Vk. The Riemannian curvature vanishes on the part U = Vk X M,-k C Mn of the hypersurface, and for this reason, the rank of A is at most 1 here. We have to show that the only eigenvector n of A corresponding to the nonzero eigenvalue lies in the tangent space V. Suppose to the contrary that n VT1 at some point p. Let N be the maximal integral manifold perpendicular to n and going through p. Since the submanifolds N, Vk, and M,-k are subspaces (totally geodesic submanifolds with 0 curvature) of the locally Euclidean space U, the angle between n and the subspace T1 is constant and nonzero along N. Since the eigenvalue A of n does not vanish on the boundary of N either, A $ 0 on the boundary of N and n VT1 holds at these points as well. In any neighborhood of such a boundary point, one can find a point q = (ql, q2) such that Rill Iql $ 0, and thus the image A(Tq) lies in the subspace TQ while at the boundary points A(TP) is not contained in T 1. For this reason, the existence of such a boundary point contradicts the continuity of A. However, such a boundary point does exist obviously, otherwise N would be complete and its orthogonal projection onto the manifold Vk would give a complete open subset of Vk. This is impossible because of the definition of Vk. This proves Proposition 2. The theorem follows immediately from the two propositions. Indeed, it is clear that cylinders decompose into a metric direct product, and conversely,

if Mn decomposes into a metric direct product, then either R(X,Y)Z - 0, and then by Proposition 1 Mn is a cylinder, or A(T2) = 0, and then the copies of M"-k are parallel (n - k)-dimensional subspaces. Mn is obviously a cylinder in the latter case as well.

Problem G.36. Among all point lattices on the plane intersecting every closed convex region of unit width, which one's fundamental parallelogram has the largest area?

Solution. Let r* be the lattice generated by the vertices of a regular triangle of altitude 2. The area of the fundamental parallelogram is equal

to 1/(2f). Let r be another lattice such that r is not congruent to r*, but the area of the fundamental parallelogram of r is also 1/(2"). We shall show that

3.4 GEOMETRY

311

in this case, there exists a regular triangle of altitude 1 which shares no point in common with r. Let P*, Q* and P, Q be the two nearest points in the lattices r* and r, respectively. Obviously,

a=PQ<a=P*Q*= i3

Let the straight line PQ be horizontal. Place a regular triangle with altitude 1 onto the plane in such a way that one of its sides lies horizontally below PQ, while the other two sides go through P and Q, respectively. The distance of the straight line PQ from the upper vertex of the triangle is (av/J/2), and its distance from the base side is (1 - av/3-/2). The distance between two horizontal ranges of points in r is equal to 1/(2va). Since

1-a2

0<a<

the triangle contains no lattice points other than P and Q. Translating the triangle a little downward, we obtain a triangle of unit width that contains no lattice points. It remains to show that r* has a point on every closed convex figure k of width 1. For this purpose, let us remark that the radius of the maximal inscribed circle of a convex figure of width 1 is at least 1/3. This can be proved from the facts that every convex figure is contained in a triangle or a band circumscribed about the inscribed circle of the figure and that among the triangles circumscribed about a given circle, the width of the regular triangle is maximal. As a consequence, the width of k is at most three times the radius of its inscribed circle. Now we need only remark that every circle of radius 1/3 contains a lattice point from r*. Therefore, r* has a point in common not only with every convex figure of unit width, but also with the inscribed circle of such a figure. 0 Problem G.37. Let S be a given finite set of hyperplanes in RT, and let O be a point. Show that there exists a compact set K C_ W containing 0 such that the orthogonal projection of any point of K onto any hyperplane in S is also in K.

Solution. We may assume without loss of generality that 0 is the origin of the space R' and that the normal vectors of the hyperplanes in S generate R' (if this latter condition is not fulfilled, then we may add further hyperplanes to S until we reach the desired property). Denote b y P1 the set of orthogonal projections onto the (n - 1)-dimensional linear subspaces of R" parallel to a hyperplane in S. F o r i = 2, ... , n,

let Pi be the collection of all projection operators, the image of which is (n - i)-dimensional and can be obtained as the intersection of the images of some operators from P. Then, by the finiteness of S and the assumption we made for the normal vectors of the hyperplanes, Pi is finite and nonempty

312

3. SOLUTIONS TO THE PROBLEMS

for every 1 < i < n. Thus, P,, has one element, P,1 = {Pn}, where P,, is the zero operator. For the sake of unified notation, set Po = {P0}, where Po is the identical transformation. If Pi E Pi and P3 E Ps, and the image of Pi contains the image of Pj, or equivalently, if the kernel of Pi is contained in the kernel of Pj, then we write Pi > P3. The orthogonal projection onto a plane s from S can be given in the form x F-, Px + p3, where P8 E P1 is the operator of the orthogonal projection

onto the (n - 1)-dimensional linear subspace parallel to s, and p8 is the vector drawn from 0 perpendicular to s. Obviously, P8p8 = 0 for every s E S.

Lemma. One can find a sequence of numbers 0 = Ro < R1 < such that for any 1 < j < n, Pj E P;, and R > R3, the set

. < R,,,

G(Pj, R) := {x E W'': Pox = 0 and < R2 - Ri for every 0 < i < j - 1, P3 <P2EP2} IIPixII2

is mapped into itself by any projection x H P8x + p8 onto a plane s E S for which P8 > Pj . Since P,,, = 0, applying the lemma for j = n and R = R,,, we obtain that the closed set K = G(0, R,,,) is mapped into itself by any projection onto a plane in S. On the other hand, this set is bounded, for it is contained in the sphere of radius Rn centered at the origin. Thus, it satisfies the requirements formulated in the problem.

Proof. We prove the lemma by induction on j. For j = 1, let P1 E P1 be an arbitrary projection. Then

G(P1iR)={xE1Rn: Plx=0,IIxII <R}. If, for some s E S, P8 > P1, then, since the images of the operators P1 and P, are (n - 1)-dimensional, P8 = P1. Therefore, P8x0 + p., = p8 for any x0 E G(P1, R). Thus, if we take supSES Ilp,, II for R1, the claim of the lemma holds.

Now assume that the lemma has been proved for the values 1,2,. .. , j - 1 and that the numbers Ro < Rl < . . < Rj_1 have al.

ready been constructed. Let P3 E P; be an arbitrary projection and s E S such that P8 > Pj. If Pjxo = 0, then Pj(P8xo + pe) = Pjxo + Pjp8 = 0. Thus, projection onto s preserves the first defining property of the elements of G(Pj, R). Second, we show that if R > Rj_1 and x0 E G(Pj, R), then IIP (P8xo +ps)112 < R2 - Rz

(1)

holds for0<i< j-2 andPj <PiEPi. If P. > Pi, then Pi(P8xo + p8) = Pixo, and there is nothing to prove. If P8 > Pi, then let us denote by Pi+1 the projection operator whose image is the intersection of the images of P8 and Pi. Consider the set G(Pi+1, R*) = G(Pi+1,

R2 - IIPi_1x0112)

3.4 GEOMETRY

313

Then

R*2=R2-IIPi+lxol12>R2-(R2-R+1)=R+1 and therefore, R* > Ri+1. Thus, i + 1 < j - 1, and by the induction hypothesis, the projection x H P,x+p, maps G(Pi+1, R*) into itself (since P, >- Pi+1). However, (Po - Pi+1)xo E G(Pi+1,R*), since for Pi+1(Po Pi+1)xo = 0 and Pi+1 < Pk E Pk (k = 0, ... , i), we have IIPk(Po - Pi+1)xo112 =11(Pk - Pi+1)xo112 = IIPkxo 112 -IIPi+1xo112

<R2-R2 k-IIPi+lxoll2=R*2-Rk. Therefore, P3(Po - Pi+1)xo + p, = (P, - Pi+1)xo + p, is also contained in G(Pi+1, R*). Thus, 11p

[(p.'

- Pi+1)xo +po]112 < R*2 - R?,

that is, IIPi(P9xo +ps) - Pi+1xo112 < R2 - IIPi+1xoII2 - R2

from which we get (1). To prove the Lemma we have only to show that if R > Rj

is sufficiently

large, then (1) holds for i = j - 1 and any x0 E G(Pj, R) and any Pj_1 E P;_1 such that Pj_1 > Pj. If P, > Pj _1i then we are done because Pj_1(P,xo +p,) = P3_1xo.

If P, > P3, then the intersection of the images of P, and Pj_ 1

(n - j)-dimensional and consequently coincides with the image of Pj.

We show that Pj_1P8 is a contraction on the kernel of the operator Pj. For this purpose, it suffices to show that if x # 0, Pox = 0, then IIPi _ 1 Psx I I < I I x I I The assumption that the two sides of the latter inequality are equal yields IIP,xlI = IIxII, from which we obtain P,x = x. Repeating

the same argumentat yields Pj_1x = x. Therefore, x is in the image of P, and Pj_ 1i that is, in the image of Pj. However, this contradicts x # 0 and Pox = 0. This means that Pj_1P, is indeed a contraction, that is, there exists a number 0 < q < 1 such that Pox = 0 implies I I Pj_ 1 P, x l l< g l l x l l Now let us choose the numbers rj = rj (P Pj_1i Pj) in such a way that

R > rj.

qR + R1 < JR2 -

(2)

(This is possible since if both sides are divided by R, the left-hand side tends to q and the right-hand side tends to 1 as R -+ oo.) Now, if R > rj and x0 E G(P;, R), then IIPj_1(Psxo+p,)11

gllxo11+11p9II <qR+R1;

therefore, by (2), (1) holds for i = j - 1. Now let Rj be the maximum of the numbers rj _1iPj) for all P8,Pj_1iPj. Then we obtain that (1) holds for R > R, with i = j - 1, if Pj E P xo c G(Pj, R), Pj < Pj_1 E P,_1,

sES

and P,>P3. This completes the proof of the lemma.

314

3. SOLUTIONS TO THE PROBLEMS

Problem G.38. Let k and K be concentric circles on the plane, and let k be contained inside K. Assume that k is covered by a finite system of convex angular domains with vertices on K. Prove that the sum of the angles of the domains is not less than the angle under which k can be seen from a point of K.

Solution. Let 0 be the common center of the circles, and r and R the radius of k and K, respectively. Let us define a function F on the interior of k as follows. If the distance from 0 to P is p, then set

F(P) = f(P) =

R2 - r2

7r(R21

P2)

r2 - p2

If T is a measurable set in the interior of k, then let

m(T) = fF(P)dP. The function m is obviously a measure on the interior of k. We show that m has the following property:

If g and h are half-lines starting from a point A of K such that g is tangent to k, h intersects k, and the angle between h and g is a, then the angular domain bounded by g and h cuts the interior of k in a domain T, the measure m of which is a.

Figure G.29.

Let us compute m(T) by successive integration. Let e be the half-line starting from 0 perpendicular to h. Let us introduce a polar coordinate system on the plane, with polar axis e. Then a point P(p, cp) belongs to T

3.4 GEOMETRY

315

if and only if d < p < r, cp < 0, where d = p cos ?9 = R sin(E - a) is the distance of 0 from h, E is the angle between the half-lines AO, and g (see Figure G.29). Based on these, we get

m(T) = JT F(P) dP = =

parccos(d/p)

f (p)pdcpdp

JJ d

arccos(d/p)

2f (p) p arccos(d) dp

r R sin(e-a)

2 (P)P arccos

Rsin(E - a) P

d Pp

Therefore, we have to show that, for 0 < a < 2e, we have IT

2f (P)P arccos

R- sin (E

Rsin(e-a)

-a))

dP= a.

Substituting R sinn(e - a) = t, this means that fr

2f (p) p arccos()

dp = E- aresin(R )

if -r < t < r. This relationship is obviously true for t = r. Thus, it is enough to show that the derivatives of the two sides with respect to t are the same: 2f(p)p dP =

Jt Jr

p2 - t2

2f (P)P

Jr it

_t2dp

1 t2 ' R2_

-r < t < r

R2 - r2

r 7r R2-p2

(r2 - p2)(p2 - t2)

p=r 2 Tr

R2-t2

tan

R2 - r2 2 - t2

R2- t2 r2-p2

vfR2 -

p=t

The proposition proved, combined with the additivity of the measure m, yields immediately that if both half-lines bounding a convex angular domain of angular measure a intersect k, then the m measure of the intersection of the angle and k is equal to a (since the intersection can be obtained as the difference of two segments of k). Now consider a covering of k with a finite number of convex angular domains. Let us denote by a1, ... , an the angular measure of the angles, and by T1i . . . , Tn the intersections of the domains with k. Then, by the proposition just proved, ai > m(Ti), i = 1, . . . , n. (Equality holds here if and only if the both boundary lines of the ith angle have a point in common with k.) Therefore, by the subadditivity of measures,

al+

+On >m(TI)+

+m(Tn)>U -

TO =2E

(since U' l Ti fills the interior of k and the m measure of k is exactly 2E). This completes the proof.

3. SOLUTIONS TO THE PROBLEMS

316

Problem G.39. Let (E, 7r, B) (7r : E -> B) be a real vector bundle of finite rank, and let TrE =VC®HC

(*)

be the tangent bundle of E, where V = Ker d7r is the vertical subbundle ofTE. Let us denote the projection operators corresponding to the splitting (*) by v and h. Construct a linear connection V on VC such that

Vx V Y - Vy V X =v[X,Y] -v[hX,hY]. (X and Y are vector fields on E, [.,.] is the Lie bracket, and all data are of class C°°).

Solution. Let CO°(E) be the ring of differentiable (of class C°°) functions E ---* R, and denote by X (E) and Xv(E) the C°° (E)-module of all vector fields on E and the submodule of vertical vector fields, respectively. A mapping

D: Xv(E) x Xv(E) -.-4 Xv(E),

(Z,W) F-> DZW

will be called a pseudoconnection on V if it has the formal properties of linear connections, that is, it is C°° (E)-linear in Z and an R-linear derivation in W (DZ f W = (Z f )W + f DZW, f E C°° (E)). Assume that D is a pseudoconnection such that DZW

- DWZ = [Z, W] for every Z, W E Xv(E),

and define the mapping

V: X(E) x Xv(E) --> Xv(E),

(X,Y) --*VxY

by the formula

VxY :=

v[hX, Y].

Then V is a linear connection. It can be seen immediately that V is linear in X and Y and that for f E C°°(E) we have (vX) fY + fv[hX,Y] + v(hX)fY VxfY = = fVxY + (vX + hX)fY = fVxY+(Xf)Y (applying linearity at each point and vY = Y). Finally, a similarly simple

computation shows that V fxY = fVxY. We show that V satisfies the required relationship. Indeed, v[hX, vY] - v[hY, vX] VxVY - VYVX = = [vX, vY] + v[hX, vY] - v[hY, vX] = v([vX, vY] + [hX, vY] - [hY, vX]) = v([X,Y] - [hX, hY]) = v[X,Y] - v[hX, hY] .

3.4 GEOMETRY

317

It remains to show that an "auxiliary" pseudoconnection D does exist. We shall prove this by a simple local construction. Assume dim B = n, rank = r. By the local triviality of the vector bundle, each point of B has a neighborhood U such that 7r-1(U) -+ U is diffeomorphic to the trivial bundle U x R' --+ R'. If V): x-1(U) - U x RT is a vector bundle isomorphism, (u') 1 is a local coordinate system on U, and (la)a=1 is the dual basis of the canonical basis of R'', then

xi:=uio7r

(1<i<n),

y:=laopr2o0 (1<a<r) is a local coordinate system on 7r-1(U), and a direct computation shows that the vector fields a/aya form a local basis of Xv(E). Let us define the pseudoconnection D by the requirement Die ayR = 0

(1 < a,,3 < r)

(It follows from standard methods of the theory of linear connections that 17 can be defined this way.) Then for any two vector fields Z

a Za ay a,

(9 W W,3 ayo

over it-1(U), we have o

L5-,w = Za aya ayo

and consequently, DzW - DwZ = [Z, W], which completes the proof.

Problem G.40. Consider a latticelike packing of translates of a convex region K. Let t be the area of the fundamental parallelogram of the lattice defining the packing, and let denote the minimal value oft taken for all latticelike packings. Is there a natural number N such that for any n > N and for any K different from a parallelogram, ntmli,(K) is smaller

than the area of any convex domain in which n translates of K can be placed without overlapping ? (By a latticelike packing of K we mean a set of nonoverlapping translates of K obtained from K by translations with all vectors of a lattice.)

Solution. We show that such an N does not exist. Let h be the branch of the hyperbola defined by the equation xy = 1 that lies in the first quadrant of the plane. Let A be the intersection point of the x-axis and the tangent of h at (a, 1/a), and let B be the intersection point of the y-axis and the tangent of h at (1/a, a). Let us replace the arcs of the hyperbola "beyond" (a, 1/a) and (1/a, a) with the segments of the tangents between A, (a, 1/a)

318

3. SOLUTIONS TO THE PROBLEMS

and B, (1/a, a), respectively. Denote by g the curve consisting of the two segments and the arc of the hyperbola between them. Then the area of any triangle bounded by the coordinate axes and a tangent of h equals 2. On the other hand, if x -* oo, then the area of the domain between the coordinate axes and g tends to infinity as well. Therefore, we can choose a in such a way that the ratio of the two areas is equal to an arbitrarily small e. Let us round two opposite corners of the unit square with the help of two suitably scaled copies of the arc g cutting off the square two domains of area 6. The area of the rounded square q equals 1 - 26. Since q is central symmetric, tmin(q) is the area of the circumscribed hexagon of q having minimal area, that is, tmjn(q) = 1 - 2eb. On the other hand, n translates of q can be packed into a domain of area n - 26. Thus, for any n, we can choose an E > 0 such that en < 1, and then ntmin(q) = n-2e6 > n-26.

Problem G.41. Show that there exists a constant ck such that for any finite subset V of the k-dimensional unit sphere there is a connected graph G such that the set of vertices of G coincides with V, the edges of G are straight line segments, and the sum of the kth powers of the lengths of the edges is less than ck. Solution 1. Let G be a connected graph such that the set of vertices of G is V, the edges of G are straight line segments, and the sum of the length of the edges is minimal. By the characterization of G, G is a tree. Take an open ball about the midpoint of every edge of G with radius r(V/3- - 1)/4,

where r is the length of the edge. We prove that these spheres do not intersect one another. Assume, to the contrary, that the segments AB and CD are edges in G, p(A, B) = R, p(C, D) = r, and

S((A+ B)/2, R(V' - 1)/4) n S((C + D)/2,

r(V - 1)/4)

where p is the distance and S(a, r) is the open ball of radius r centered at a. By symmetry, we may assume R > r. Then

p(A, (C + D)/2) < p(A, (A + B)/2) + p((A + B)/2, (C + D)/2)

< R/2 + R(vf3- - 1)/2 = Rv /2 . Since at least one of the angles (A, (C + D)/2, C) and (A, (C + D)/2, D) is not obtuse if A :A (C + D)/2, we get min{p(A, C), p(A, D)} < (p2(A, (C + D)/2) + r2/4)112

< (3R2/4+R2/4)1/2 = R. If the edge AB is removed from the graph G, then the subgraph is the union of two trees. By symmetry, we may assume that B is contained in the same component as C (and D as well). But then removing from G the edge AB and adding to it the shorter of AC and AD, the length of which is less then R, we obtain a connected graph G' satisfying all the requirements for G, but the sum of the length of its edges is less then the sum for G, and this is a contradiction.

3.4 GEOMETRY

319

Since the balls we constructed are disjoint and are contained in a sphere of radius 2, their total volume is less than the volume of the ball of radius 2. Thus, k

2kwk ?

Pk(A, B)

AB is an edge in G

where Wk is the volume of the k-dimensional unit ball. Therefore, Pk(A, B) < AB is an edge in G

(

V3- -1

) 2k = ek

as was to be proved.

Solution 2. It suffices to show the following claim:

Claim. If T is a box such that the ratio of any two edges of T is less than or equal to 3, and V C T is a finite set of points, then there exists a connected graph G = G(V) set of vertices V such that the sum of the kth power of the edges of G satisfies S(G) < Ck Vol (T),

where ck is a constant depending only on k, Vol (T) is the volume of T.

Using induction on the number of points in V, we shall show that one may choose ck = 3k+lkk/2. Let IVl = n. The statement is true for n = 1, 2, and assume it has already been proved for 1, 2, ... , n - 1. Let us shrink T,

that is, move the faces of T inside as long as the assumptions on T are not violated and T contains V. This way, we make the inequality to prove sharper. If T cannot be shrunk further and e is one of the longest edges of T, and furthermore, Q1 and Q2 are the faces perpendicular to e, then V has a point on both Q1 and Q2 (otherwise, we could shrink T further). Let T1, T2, and T3 be the consecutive boxes obtained by cutting T into three equal parts by hyperplanes perpendicular to e. We distinguish two cases depending on whether T2 contains a point of V or not. Case 1. V has no point in T2. Let V1 = V fl T2, V3 = V fl T3. Since Qi fl v 0 for i = 1, 2, we have l Vi l < n. Applying the induction hypothesis for T1, V1 and T3, V3, we obtain connected graphs G1 = G(V1) and G3 = G(V3), for which S(Gi) < ckVol(Ti) = (1/3)Vol(T), i = 1, 3. Connecting a point of Gl to a point of G3i we obtain a connected graph G such that S(G) < S(G1) + S(G3) + (V/k-)k jell

< 2ckVol(T) + (j)k3kVol(T) < ckVol(T) by the choice of ck.

Case 2. V has a point in T2. Choose x E V fl T2 and cut T into two closed boxes T1 and T2 by a hyperplane passing through x perpendicular

320

3. SOLUTIONS TO THE PROBLEMS

to e. Then T* also has the property that the ratio of any two of its edges is not more than 3. Thus, the induction hypothesis can be applied again to Tl , Tl fl V and T2*, T2 fl V since the sets Tl fl V and TT fl V have less than n points. Therefore, one obtains the connected graphs Gi = G(V fl Ti ) and G2 = G(V fl T2*) such that S(G2) < ckVol(T*)

(i = 1, 2).

Setting G* = Gi U GZ yields a connected graph on the set of vertices V and

S(G*) = S(Gf) + S(G*) < ckVol(T1) + ckVol(T2) = ckVol(T).

Remarks. 1. We may choose for G a Hamiltonian circuit as well, but the proof of this fact is more difficult.

2. One can prove that c2 = 4 is the smallest suitable constant and that it is good for the Hamiltonian circuit problem also. It is an interesting problem to find the exact constants for k > 2.

Problem G.42. Let us draw a circular disc of radius r around every integer point in the plane different from the origin. Let Er be the union of these discs, and denote by d, the length of the longest segment starting from the origin and not intersecting Er. Show that 1

lim (d,. - 1) = 0.

r--.o

Solution. Let the second endpoint (x, y) of one of the longest segments be on the circumference of the disc centered at (k, n). Reflecting the segment in the point (k/2, n/2), we obtain that the segment connecting the points (k - x, n - y) to (k, n) has a point in common only with the circle centered

at (k, n). Since the points (0, 0), (k - x, n - y), (x, y), and (k, n) are vertices of a parallelogram, the two other sides of which have length r, this parallelogram contains no lattice points in its interior. The segment connecting the origin to (k, n) has a point in common only with the disc about its endpoint. The length of this segment differs from dr by at most r.

Let Dr be the length of the longest segment connecting the origin to a lattice point and having a point in common only with the disc about its endpoint. As we observed, I Dr - dr 1< r. According to this, it suffices to show that

lim(Dr-1)=0. r

r 0+

For this, we shall apply the well-known theorem of lattice geometry saying that a triangle that has lattice points for its vertices but contains no further

3.4 GEOMETRY

321

lattice points in its interior and on the boundary (except for the vertices) has area 1/2.

Let V = (n, k) be the endpoint of a segment of length Dr, and let H be the domain containing those points of the plane whose orthogonal projection onto the straight line OV lies in the closed segment OV. Since translates of H with integer multiples of the vector (n, k) cover the whole plane, the distance of any lattice point P E H from the straight line OV is at least r if P is different from 0 and V. Let Q be a lattice point in H different from 0 and V such that the distance of Q from the straight line OV is minimal. Then there are no further lattice points on the segments OQ and OV, nor in the interior of triangle OVQ, because these points are closer to OV than Q. Furthermore, there are no lattice points in the interior of the segment OQ since every disc of radius r centered at a lattice point different from 0 and V is disjoint from the segment OV. Therefore, the area of triangle OVQ is equal to 1/2. Consequently, the distance of

Q from the straight line OV, that is, the altitude of triangle OVQ, is 1/ 1 OV 1. We find that the distance of Q from OV is at least r if and only if I OV 1< 1/r.

This yields another characterization of Dr: Dr is the length of the longest segment connecting 0 to a lattice point without going through further lattice points (that is, the coordinates of the endpoint are relative primes) and having length at most 1/r. By this, Dr < 1/r, of course. It is well known that for any e > 0, if n is large enough, then there is a prime number between n and n + en. Let 0 < e < - 1, and choose a prime p satisfying 1

(1+e)r<p< Such a prime exists by the theorem mentioned above, provided that r is small enough.

Let q be the largest nonnegative integer such that p2 + q2 < r2 Then q p, then by p2

2p2

+ q2 > 2r < (1 + E) r < would hold. On the other hand, q > 1, as for r < 1, p2+12 < (1/r-1)2+1 < 1/r2. Thus, 1 < q < p, and since p is a prime, p and q are relative primes.

This implies that the segment connecting the origin to (p, q) is suitable. Therefore 2

D2 > p2 + q2 > p2 +

(

r2 - p2 - 1

)

- 2

p2 + 1

r2 2

1 - (1+6)2 +

1 > (r - 1 -

(1 + E)2

)

322

3. SOLUTIONS TO THE PROBLEMS

that is,

Dr >

r - 1'1 -

(1 +

E)2

- 1, and for any sufficiently small r,

We conclude that for any 0 < E < we have 1

(1 + E)2

<D,.<1. r

Since

(1+E)2

-'0

e -+ 0,

the proposition follows immediately.

Problem G.43. We say that the point (al, a2, a3) is above (below) the point (b1, b2, b3) if a1 = b1, a2 = b2 and a3 > b3 (a3 < b3). Let e1i e2, ... , elk (k > 2) be pairwise skew lines not parallel with the z-axis, and assume that among their orthogonal projections to the (x, y)-plane no two are parallel and no three are concurrent. Is it possible that going along any of the lines the points that are below or above a point of some other line ei alternately follow one another? Solution. Assume that it is possible. From this, we derive a contradiction.

Let Pi,j be the point of ei, which is above or below the line ej (1 < i, j < 2k, i # j). In the first case, we call Pi,3 an upper point, and in the latter case we call it a lower point. In the text below, by a point of ei we mean one of the points Pi,j. Those two points of ei that have all points of ei on one side will be called the "ends" of ei. Two points of a line will be called neighbors if there is no point of the line between them.

Lemma. Every straight line contains at least two points that are above or below an end of another line. Proof. Obviously, it is enough to consider the straight line el. We may also assume that el is horizontal (that is, parallel with the (x, y)-plane), since this can always be managed applying a transformation of the form (x, y, z) --> (x, y, z + ay +,3z), which leaves the location of upper and lower

points unchanged. Let S be a plane perpendicular to e1, and denote by f2,f3,...,f2k the orthogonal projections of the lines e2ie3,...,e2k onto S (see Figure G.30). Let us introduce a coordinate system on S having horizontal and vertical axes and take from among f2, f3, ... , f2k the ones, say f2 and f3i whose slope is minimal and maximal, respectively. We claim that if e2 is below (above) e1, then each point of it lies on the negative (positive) side of el with respect to the positive direction of the horizontal coordinate axis. Assume that this is not so and, for example, e2 is below el and has a point on the positive side of el (the other cases can be treated analogously). Let Pi,,,2 be the nearest to P2,1 among these points. This

3.4 GEOMETRY

323

is an upper point since it is a neighbor of a lower point. The point P2,i0 below it lies on the straight line ei,,. It is easy to see that one can find a sequence of indices i1 , . , i,, such that 1. in = 1;

2. the points and are neighbors on ei, (j = 1, 2, ... , n - 1); is in the negative direction from 3. (j = 1, 2, ... , n - 1).

Figure G.30.

It is also easy to see, however, that in this case, the slope of one of the straight lines fi...... is less than the slope of f2, and this is a contradiction. This way, we proved that one of the ends of e2 and e3 is above or below el. Therefore, the lemma holds.

We remark that there cannot be more than two ends above or below a straight line, because it would mean that the 2k straight lines would have more than 4k ends. Therefore, there are exactly two points on every straight line that are above or below an end of another line, and these latter lines are the ones giving the minimal and maximal slopes after the above transformation and projection. Let the ends of el be above or below e2 and e3, respectively. Since the straight line contains 2k - 1, that is, an odd number of points, either both

e2 and e3 are above el or both of them are below el. We may assume that they are above el. We also suppose that e2 and e3 are horizontal since this can be reached by applying a suitable transformation of the form (x, y, z) H (x, y, z + ax +,3y). Let S be a plane perpendicular to e2, and introduce a coordinate system on S as shown in Figure G.31. By the previous remark, among the orthogonal projections of the straight lines ej(j # 2) onto S, the projection of el has maximal slope, el has maximal slope, since el goes below e2 and has a point in the positive direction from P1,2, and since the projection of e3 is horizontal, that of el has positive slope. This implies that P1,3 is higher than P1,2 (its z-coordinate is

3. SOLUTIONS TO THE PROBLEMS

324

P3 Figure G.31.

greater). Repeating the same argument but changing the role of e2 and e3, we obtain that P1,2 is higher than P1,3. However, these two implications contradict one another. We conclude that the indirect assumption is false: upper and lower points cannot follow one another alternating on each straight line.

Problem G.44. Suppose that HTM is a direct complement to a vertical bundle VTM over the total space of the tangent bundle TM of the manifold M. Let v and h denote the projections corresponding to the decomposition TTM = VTM ÂŽ HTM. Construct a bundle involution P : TTM -> TTM such that P o h = v o P and prove that, for any pseudoRiemannian metric given on the bundle VTM, there exists a unique metric connection V such that

VXPY-VYPX=Poh[X,Y] if X and Y are sections of the bundle HTM, and

VXY-VyX = [X, Y] if X and Y are sections of the bundle VTM.

Solution. As usual, call the sections of the bundles VTM --> TM and HTM -> TM vertical and horizontal vector fields on TM, and denote their modules by XvTM and . HTM, respectively. If X(TM) denotes the module of vector fields in TTM -+ TM, then

X(TM) = XvTM ÂŽ XHTM, and, for simplicity, the projections corresponding to this decomposition are also denoted by v and h.

3.4 GEOMETRY

325

Let (U, (ul, ... , un )) be a chart for M, and consider the induced chart

for TM, where 7r : TM --+ M is the projection of the tangent bundle, and

xi=uio7r, y'(v)=v(u') (vETM, 1<i<n). When this chart is fixed, there exist unique smooth functions

Nk:ir-1(U)--+R (1<i,k<n) such that the vector fields

(1 < i < n)

Nk aa

Sri

form a local basis in XHTM. Then a

ay i' axi

(1<i<n)

is a local basis in .A(TM), and on the coordinate patch 7r-1(U) every vector field X E .A(TM) can be written in the form

X Xi 6xi +Yi

where the first term is horizontal and the second is vertical. Define the map P by the formula Z

S2i

+ YZ ayi H YZ

Ski

+ Xx yi

An easy calculation using the above formulas shows that this map is independent of the choice of the chart. From the formula, it is immediate that p2 = 1 and P o h = v o P. It is also clear that P is a bundle involution.

Since .A(TM) _ XvTM ÂŽ . HTM, and a connection V :.A(TM) x . vTM ---+ XvTM is linear over the function ring in its first variable, it suffices to define V on the summands .XvTM x . vTM and XHTM x

.vTM.

Case 1. Consider the map F : (3vTM)3 - XvTM defined by F(X, Y, Z) = X (Y, Z) + Y(Z, X) - Z(X, Y)

- (X,[1',Z])+(Y,[Z,X])+(Z,[X,Y]), where (., .) denotes the pseudo-Riemannian metric given on the vertical bundle. For fixed X and Y, the map Z -4 F(X, Y, Z) is linear over the

326

3. SOLUTIONS TO THE PROBLEMS

function ring, and, since the form (. , .) is nondegenerate, there exists a unique VxY E XvTM such that (2VxY, Z) = F(X, Y, Z)

for all Z E .vTM. An argument analogous to the proof of the theorem on Levi-Civita connections shows that the map (X, Y) ---> VxY satisfies all requirements. Case 2. The restriction of the connection V to XHTM x XvTM is defined by the requirement that

(2VuY, Z) = U(Y, Z) + PY(Z, PU) - PZ(PU, Y) - (PU,Po h[PY,PZ]) + (Y,Po h[PZ, U])

+(Z,Poh[U,PY]) hold for all U E tHTM, Y, Z E XvTM. A simple but lengthy calculation shows that this restriction of the connection also has all required properties. In order to prove its uniqueness, suppose that the map

V' :.CHTM x XvTM -4 XvTM also fulfills the requirements of the problem. Then, for all U E .CHTM, Y, Z E XvTM, we have

U(Y, Z) + PY(Z, PU) - PZ(PU, Y) (1) (VUY, Z) + (Y, VUZ) + (V' Z, PU) + (Z, V PYPU) - (VZZPU, Y) - (PU, VPZY) =2(V' Y, Z) + (V' PU - VUY, Z) + (VUZ - VPZPU, Y) + (VPYZ - VPZY, PU) (?)

2(VUY, Z) - (P o h[U, PY], Z) - (P o h[PZ, U], Y) + (P o h[PY, PZ], PU),

where at (1) compatibility of V with the metric is used, and at (2) the assumption V'XPY - V',PX = P o h[X, Y] is used. Therefore, it follows that (V UY, Z) = (V UY,

Z),

that is, (VUY - VuY, Z) = 0

for all

Z E . vTM.

Since the form (., .) is nondegenerate, this implies that V = V, which completes the proof.

3.4 GEOMETRY

327

Problem G.45. Let A be a finite set of points in the Euclidean space of dimension d > 2. For j = 1, 2, ... , d, let Bj denote the orthogonal projection of A onto the (d-1)-dimensional subspace given by the equation

xj = 0. Prove that

d d

I Bj I> I AI d-1.

j=1

Solution. We use induction on the cardinality of the set A. If JAI = 0 or 1, then the statement is obvious. If JAI > 1, then the projection of A onto a suitable coordinate axis contains at least two distinct points. Without loss of generality, we can assume that this coordinate is the dth. This means that a suitable hyperplane with equation Xd = a divides A into two disjoint nonempty subsets: A = X U Y.

Let Cj and Dj denote the projections of X and Y, respectively, onto the hyperplane xj = 0; then ICj I + JDj I = IBj I for each j E {1, ... , d - 1}, and ICdI < IBdI and IDdI < JBdI. By the induction hypothesis, we have d-1

Ix Id-1<IBdI.fl ICjI

and

j=1

d-1

IYId-1 < IBdI

f IDj I.

j=1

Using the inequality 1/k

(al ..... ak) k + (bl ..... bk) k <

(H(ai+bi)) %=1

which is an equivalent form of the usual inequality between arithmetic and geometric means, we have

f d

d-1 IBjI11(d-1)

IBdI11(d-1)

II (ICj I + IDj

I)1/(d-1)

j=1

BdII/(d-1)

d-1

j=1

. r ICj I1/(d-1) + H IDj I d111

>- IXI + IYI = JAI,

that is, d

IB,I > IAId-1, j=1

which is what we wanted to prove.

3. SOLUTIONS TO THE PROBLEMS

328

Problem G.46. Let Al , ... , An be a sequence of n > 3 points in the Euclidean plane R2. Define the sequence A('),.. . An(') (i = 1, 2.... ) by ,

induction as follows: let A(z) be the midpoint of the segment A(i-)

A12-1).

where n+l Show that, with the exception of a set of zero Lebesgue measure, for every initial sequence (A(O), ... , A(Â°)) E (R 2)n, there

exists a natural number N such that the points A(lN), ... , An) are consecutive vertices of a convex n-gon.

Solution. If R' is naturally identified with the complex plane C, then our sequences can be considered as vectors in the space Cn, and the induction step corresponds to the linear transformation L(z1,...,zn) =

zl + z2 zn + zl 2 ,..., 2

In Cn there exists a basis consisting of eigenvectors of L: we have

L(ej)_Ajej for the vectors ej =

(j=1,2,...,n) (1,ej,e2j,...,e(n-,)j) and scalars Aj

= (1 + ej)/2

1,.. n) where e = e2"i/n is a primitive nth root of unity. All vectors v in Cn can be written in the form v = E; 1 ajej. Since en = (1, 1, ... ,1), adding the vector an en means a translation of our system of points in the original plane R2 by a fixed vector. Therefore, it suffices to consider the problem restricted to the subspace generated by the vectors e1,. .. , en-l- Suppose that n-1

(A10),

... , 4 0 ) ) = v = ajej. j=1

Applying the induction step N times, we obtain n-1 A1N),...,A,NI)

= LN(v) _ a3A ej. j=l

An elementary calculation shows that I AjI < IA' for 1 < j < n - 1, and IAn_1I = IA, I. So, for 1 < j < n - 1, we have IAjI N/IA,IN -> 0 when N -> 0. Therefore, it suffices to show that for almost all v the vector

ale, +an_len_, corresponds to successive vertices of a convex n-gon. This is indeed the case if Iai l : Jan-11, because e1 is a vector corresponding to successive vertices of a regular n-gon, and the vector ale1 + an_1en_1 corresponds to the sequence of points obtained from successive vertices of this regular n-gon by applying the linear map z - a1 z + an _ 1 z in R2. This transformation is nondegenerate, since a,z+an_12 = 0 can only hold when z = 0 or jai I = Jan-11, and thus it takes convex n-gons into convex n-gons. To summarize the property required in the problem is true for all vectors Enj=1 ajej of Cn except when la1I = lan-,I, and this exceptional set does indeed have zero Lebesgue measure.

3.4 GEOMETRY

329

Problem G.47. Suppose that n points are given on the unit circle so that the product of the distances of any point of the circle from these points is not greater than two. Prove that the points are the vertices of a regular n-gon.

Solution. Let the circle be the unit circle around the origin in the complex plane. Let z1, z2, ... , zn be the complex numbers that represent the points. By a suitable rotation, we may assume that z1 Z2 Zn = 1. Consider the following polynomial: P(W) _

(W-z1)(W-z2)...(W-Zn)

=wn+Q(w)+1. Then IP(z)I is the product of the distances of the point represented by the complex number z from the given points. So, if z is a complex number of absolute value 1, then IP(z) I < 2. Let w1i w2, ... , wn denote the nth roots of unity. It is well known that w1 + w2 + + Wk = 0 for all k = 1, 2, ... , n - 1. This implies that Q(wl) + + Q(Wn) = 0. If Q(w) is not identically zero, then, for some j, the complex number Q(wj) is different from zero and has nonnegative real part. Then P(wj) = 12 + Q(wj) l > 2, which contradicts the assumption. Therefore, the polynomial Q is identically zero, that is, P(Z) = zn+1. Then the roots z1i z2, ... , zn of the polynomial P(z) form a regular n-gon.

3. SOLUTIONS TO THE PROBLEMS

330

3.5 MEASURE THEORY Problem M.1. Let f be a finite real function of one variable. Let D f and D f be its upper and lower derivatives, respectively, that is, D f (x) = limsup h,k-.0

f (x+h) - f (x-k) f (x+h) - f (x-k) D f (x) = liminf h+k h+k h,k-.0 h, k> 0

h, k> 0

h+k>O

Show that D f and D f are Bore]-measurable functions. Solution. It is obviously sufficient to prove the assertion for D f . It is also clear that 00

{x: Df(x)>c}=.U n Aik Z=1 k_1 for all real values c, where Aik denotes the union of intervals [a, b] shorter

than 1/k satisfying (f (b) - f (a))/(b - a) > c + 1/i. We show that Aik is a set of type FQ which implies the assertion. More generally let {I.y} (ry c F) be a system of open intervals, and let A = U.yErL (where 7, stands for the closure of Ly). Obviously,

A=BuCuD, where B = U yErly, C is the set of all points that are left-hand endpoints of at least one Ly but are not interior points of any of them, and D is the set of those points that are right-hand endpoints of at least one I. but do not lie in the interior of any Ly. Clearly, B is an open set. If x E C, let rx be a rational point in the interior of the interval Ly that has the left-hand endpoint x. Points rx corresponding to distinct points x are distinct since

in the case x < y, rX = ry, the point y would lie in the interior of the interval I.y starting at x. Consequently, C, and similarly D, is countable. So each of the sets B, C, D, and therefore A as well is of type Fo.

Remark. It is interesting that for the left-hand and right-hand derivatives a similar theorem does not hold. For instance, if E is a nonmeasurable

set that along with its complementary set is dense, and f (x) is the char-

acteristic function of E, then D+ f (x) = 0 or +oo if x E E or x V E, respectively, and so D+ f (x) is nonmeasurable in this case. It should be noted that in the book by S. Saks (Theory of the Integral, Hafner, New York, 1937, Vitali's covering theorem, pp. 112-113), the author proves only Lebesgue measurability of D f and D f with much more powerful tools, although generalized to arbitrary dimension.

3.5 MEASURE THEORY

331

Problem M.2. Let E be a bounded subset of the real line, and let Sl be a system of (nondegenerate) closed intervals such that for each x E E there exists an I E 1 with left endpoint x. Show that for every e > 0 there exist a finite number of pairwise nonoverlapping intervals belonging to Sl that cover E with the exception of a subset of outer measure less than e. Solution. Let A be a bounded open interval containing E. Denote by Sn the set of all points of A having neighborhoods not containing points of E that are starting points of intervals belonging to Sl and longer than 1/n. Obviously Sn is an open set, Sn+1 C Sn and, furthermore, (n',Sn)fE = 0 since at each point of E there begins at least one nondegenerate interval belonging to I and if 1 is already less than the length of the latter, then n the point cannot belong to S,,. It is a well-known property of the outer measure that for any set A the outer measure A* (X fl A) as a function of X is a measure on the Lebesgue-measurable sets. Thus, taking A = E and using the relation Sn+1 C Sn as well as the finiteness of the outer measure of E, we obtain n

limo A* (Sn fl E) = A'(( _ 1 Sn) n E) = A* (0) = 0.

Fix no so that A* (Sno fl E) < e/2. If the set El = E - (Sno fl E) is nonempty, which we may assume, then in each neighborhoood of any of its points there is a point of E that is also a point of El and at which an interval longer than 1/no begins. There is a point a1 E El such that A*((-oo, al) n El) < 2(noA(A) + 1)' and by the former arguments we may assume at once that some interval [a1, b1] starting from al and belonging to I is longer than 1/no. Then we choose (if still possible) a point a2 E El such that a2 > b1, *((bl, a2) fl El) < 2(no\(0) + 1)' and

[a2, b2] E Il,

b2-a2>

and so on. The procedure ends after a finite number of steps, namely we can construct at most no.\(A) + 1 intervals [ak, bk] in this way (since bk - ak > 1/no). What is left out from E by these intervals is covered by Sno nE and the not more than no\(0)+1 sets (bk_1i ak)flEl (bo = -oo) of outer measure not exceeding e/(2(no\(0) + 1)). Their total outer measure is less than e, which completes the proof.

3. SOLUTIONS TO THE PROBLEMS

332

Remarks.

1. In addition to being nonoverlapping, the intervals we have chosen are disjoint.

2. The assertion cannot be replaced by the stronger statement that it is possible to choose an at most countable number of nonoverlapping intervals such that the part not covered by them has measure 0. This is shown by the following simple example: E = (0, 1), S2 = {[x,1] : x E (0, 1)}.

Problem M.3. Let f (t) be a continuous function on the interval 0 < t < 1, and define the two sets of points At = {(t,0): t E [0,1]},

Bt = {(f(t),1): t E [0,1]}.

Show that the union of all segments AtBt is Lebesgue-measurable, and find the minimum of its measure with respect to all functions f.

Solution. We first show that the set A = Uo<t<1AtBt is closed, hence Lebesgue-measurable. Let Pn be a convergent sequence of points in A, P,,, -> Po. By the definition of A, to every n there is a to such that Pn E According to the Bolzano-Weierstrass theorem, the sequence to contains a convergent subsequence: tnk -> to. The distance from the point Pnk to the segment AtoBto is not greater than max{lf (tnk) - f (to)l, Itnk Atn Btn .

toI}, which tends to 0 by the continuity of f (t). It follows that Po E AtoBto C A. Since A contains the limit point of any convergent sequence of its points, A is closed.

A simple calculation shows that the point of the segment AtBt that lies on the straight line y = c has abscissa (1 - c)t + cf(t), which, by the continuity of f (t), is a continuous function of t. Consequently, if two points

of the set A lie on the line y = c, then A contains the segment that joins these points. Now we can determine the minimum of the measure of A. If f (t) is

constant, then A is a triangle of unit base and unit altitude, so it has measure 1/2. If the segments AoBo and A1B1 do not intersect, then the trapezoid with vertices Ao, Bo, A1, and B1 is a subset of A, so the measure of A is not less than 1/2. If the segments AoBo and A1B1 intersect at some

point C, then the triangles AOCA1 and B0CB1 are subsets of A, so the measure of A is not smaller than 1 1 d _ 11+d2 t(d) =

2 1+d'

where d denotes the distance from Bo to B1. By a simple calculation, we - 1. Thus, obtain that the minimum of t(d) on the positive half-axis is

the measure of A cannot be less than - 1. On the other hand, for - 1. f (t) = (v/2-- 1)(1 - t), the measure of A is exactly Remark. I. N. Berstein (Doklady Acad. Nauk. SSSR 146 (1962), 1113) refers to the result of the problem but he states erroneously that the minimum is 1 .

3.5 MEASURE THEORY

333

Problem M.4. A "letter T" erected at point A of the x-axis in the xyplane is the union of a segment AB in the upper half-plane perpendicular to the x-axis and a segment CD containing B in its interior and parallel to the x-axis. Show that it is impossible to erect a letter T at every point of the x-axis so that the union of those erected at rational points is disjoint from the union of those erected at irrational points.

Solution 1. We call width of the letter T erected at A the minimum of the lengths CB and BD. We devide the irrational points into countably many classes H1, ... , H,, .... The class Hk consists of all irrational points for which the widths of the letters T erected at them are greater than 1/k. Baire's theorem implies that the set of all irrational points cannot be represented as the countable union of nowhere-dense sets, so some Hk is dense in a suitable interval I. Selecting an arbitrary rational point A of the interval I, denote by b the width of the letter T erected at it. The class Hk contains an element Al whose distance from A is less than min{b,1/k}. The letters T erected at A and Al obviously intersect each other since they

both have widths greater than the distance from A to A1. The proof is complete.

Solution 2. Project on the x-axis the segment CD parallel to the x-axis of each letter T, and denote the projection by I. The projection II is an interval containing x in its interior. We prove the existence of an irrational

a and a rational r such that I,, and Ir contain r and a, respectively, in their interior. This already implies that the letters T erected at a and r intersect. For each rational number r = p/q (we shall write all rational numbers in irreducible form), we choose a closed subinterval Jr of I, containing r in its

interior and having length not greater than 1/q2. Consider a sequence of rational numbers where the denominators are strictly monotone increasing and each number lies in the intervals J, corresponding to the previous terms: P1 rl = -, r2 =

,. .. , rn, _ Pn , ... ; qn

q1 < q2 < ... < qn, < ...; r,nEJrinJr2n...nJrn_1.

Such a sequence obviously exists and, moreover, it is a Cauchy sequence, so it has a limit limes-,* rn, = a. Since Jr, contains the points r7,+1, rn,+2, . . and is closed, it follows that a E Jr,a and, in particular,

Ia - r,I <

From the theory of numbers it is well known that in this case a can only be irrational. I, contains a in its interior, so for sufficiently large n we have

334

3. SOLUTIONS TO THE PROBLEMS

rn E I,,. On the other hand, a E Jr, C Ir,. Thus, for sufficiently large n, the letters T erected at the points a and rn intersect each other. Solution 3. Assume that the assertion is false, that is, we have succeeded

in erecting a letter T at each point of the real line so that none of the letters T erected at rational points intersects a T erected at an irrational point. We may assume that the letters T are symmetric, that is, B is the midpoint of the segment CD (let C always have the smaller abscissa among

C and D). We define a monotonically increasing sequence of numbers al, a2.... an, ... by recursion. Let al be any real number. If an is already defined as a rational number, then let an+1 be irrational, while if an is irrational,

then let an+1 be rational, and choose an+l(> an) so that an+1 - an is smaller than one-quarter of the length of BnDn. Denote the abscissa of Dn

by bn. Making use of the fact that the letters T erected at an and an+1 do not intersect, we obtain the following sequence of inequalities:

an+1 < an+1 < an + 2(an+1 - an) < an + 2

bn - an 4

an + an 2

< 6n.

It follows that {an} is bounded, and so a = liman exists, and also that a < 6n for every n. Then, however, for sufficiently large n, the letter T erected at an intersects the T erected at a, which contradicts the initial assumption. Remarks.

1. In addition to the assertion of the problem, a contestant has proved that in any interval there is a continuum of irrational numbers a such that the letter T erected at a intersects a T erected at a rational point. Really, from the reasoning of Solution 2 it turns out that if r1, r2.... is a sequence of rational numbers such that Jrl 3 Jr2 D ... and ql < Q2 < .... then {rn} is obviously convergent and its limit is a "suitable" irrational number. Let r = p/q be a rational number belonging to the

interval (a, b) such that Jr C (a, b). Let ro = Po/qo and r1 = pl/ql be rational numbers satisfying qo > q, ql > q, Jro C Jr, Jr1 C Jr, and Jro n Jr1 = 0. The existence of ro and r1 of this kind is evident. Once we have defined the rational numbers ri1,,..,ik (i1 = 0,1; ... ; ik = 0,1), let ril,,..,ik,o and ril,,,,,ik,1 be two rational numbers such that gil,...,ik,o > qil,..., ik, JT'il...... k,0 C Jril...... k ,

gi1,...,ik,1 > gi1,...,ikI Jri1,..... k,1 C Jril...... k

Jril....,ik,0 n Jri1...... k'1 = 0.

With each number 0 < x < 1, we associate an irrational number a E (a, b) such that the letter T erected at a surely intersects a T erected at some rational point. If the infinite dyadic form of x is

x = 0.iJ2 ....

3.5 MEASURE THEORY

335

we associate with x the limit of the sequence

This will be a "suitable" irrational number, and the construction guarantees that a1 and a2 associated with distinct numbers x1 and x2 are distinct.

2. The union of a segment AB in the upper half-plane that is erected perpendicularly at point A of the x-axis and a segment BC that is parallel to the x-axis will be called a [ or a I depending on whether the

abscissa of C is greater than or less than that of B. Two contestants proved that if we erect a [ at each rational point and a I at each irrational point, then the union of the [ intersects the union of the 1. This assertion

is obviously sharper than that of the problem. Two other contestants showed by examples that it is possible to erect a [ at each point so that the I are pairwise disjoint. Indeed, if the points B of the [ lie on the graph of a strictly monotone decreasing positive function and the length of BC is arbitrary, then clearly any two [ are disjoint. 3. It is easy to see that the next theorem generalizes the statement of the problem: Let X be a complete metric space of second category, and let

X = P U Q, where P and Q are disjoint dense sets. Then it is impossible to define a real function f on X such that each point p E P has a neighborhood Vp for which x E Vp fl Q implies f (x) < f (p) and such that each point q E Q has a neighborhood Vq for which x E Vq fl P implies

f(x) < f(q)

This was proved by some contestants for complete metric spaces and compact metric spaces, respectively. It was also shown that the hypothesis of completeness cannot be dropped.

Problem M.S. Let f and g be continuous positive functions defined on the interval [0, oo), and let E C [0, oo) be a set of positive measure. Prove that the range of the function defined on E x E by the relation X

F(x, y) =

f (t)dt + fo g(t)dt

h as a nonvoid interior.

Solution. Put O(x) =

f(t)dt,

fi(x) =

g(t)dt.

The function 0 and its inverse 0-1 (which obviously exists) are absolutely continuous, so they map sets of measure 0 onto sets of measure 0. Therefore, the image O(E) = A of the set E of positive measure is measurable and has positive measure. Similarly, the set O(E) = B has positive measure. Since the range of the function F(x, y) considered on E x E is exactly

336

3. SOLUTIONS TO THE PROBLEMS

the set A+ B = {a + b : a E A, b E B}, it is sufficient to prove that for sets A and B of positive measure, the sum A + B contains some interval. Let u and v be points of density 1 of A and B, respectively, and let e > 0 be chosen so that the relations 0 < 6 < 2e, 0 < 6' < 2E, 6 + 6' > 0 imply

p(An [u-6,u+6']) > 2(b+6'), p(Bn [v -6,v+6']) > 2(6+6'), where p stands for Lebesgue measure. We show that (u+v-E,u+v+E) C

A+B. Let t E (u+v-E,u+v+E) and

A*=t-A={t-a: aEA}. Then A* n [t - u, v + E] is congruent with A n [t - v - E, u], so

p(A* n [t-u,v+E]) > 2(u+v+E-t)

and

p(Bn [t-u,v+E]) > 2(u+v+E-t), from which it follows that A* n B # 0. Let x E A* n B; then t - x E A, x E B, and, consequently, t = (t - x) + x c A + B. Remarks.

1. Two contestants showed that it is sufficient to assume that f and g are positive functions integrable in every finite interval. 2. A contestant remarked that if we know of E only that it has positive outer measure, then the statement becomes false.

3. Denote by Al and B1 the set of all points of density 1 of A and B, respectively. If u c Al and v E B1, then by the density theorem of Lebesgue u and v have density 1 also in Al and B1, respectively, so by the solution above, Al + B1 contains some neighborhood of u + v. This means exactly that Al + B1 is an open set (see J. B. H. Kemperman, A general functional equation, Trans. Am. Math. Soc. 86 (1957), 28-56, Theorem 2.2).

Problem M.6. In n-dimensional Euclidean space, the union of any set of closed balls (of positive radii) is measurable in the sense of Lebesgue. Solution. Let {Ga} (a E A) be an arbitrary set of closed balls of positive radii. Put H = UQEAGO. Denote by G the set of all closed balls (of positive radii) that are contained in some Ga. Clearly, the elements of G constitute a cover of H in the sense of Vitali. Therefore, by Vitali's covering theorem (see for example S. Saks, Theory of the Integral, Hafner, New York, 1937,

p. 109), we can choose an at most countable number of pairwise disjoint balls S1, S2,... belonging to G such that H \ UiSi has Lebesgue measure 0. Since Si C H, the relation

(H\us)u(us)

implies that H is Lebesgue-measurable.

3.5 MEASURE THEORY

337

Remarks.

1. Most contestants solved the problem with the help of Vitali's covering theorem.

2. Several contestants remarked that H can be obtained as the union of at most countably many Jordan measurable sets. 3. One contestant proved the following generalization of the problem: The union of any set of convex sets with nonvoid interiors is measurable in the sense of Lebesgue.

Problem M.7. In n-dimensional Euclidean space, the square of the two-dimensional Lebesgue measure of a bounded, closed, (two-dimensional) planar set is equal to the sum of the squares of the measures of the orthogonal projections of the given set on the n-coordinate hyperplanes.

Solution 1. We solve the problem in the generalized form where a bounded, closed subset of H or an r-dimensional subspace L is projected to the r-dimensional coordinate subspaces. H is the difference of two bounded open sets N2 E N1. Each of them is the union of an at most countably infinite number of disjoint parallelepipeds t1i and t2j. Their images under a linear transformation T (the orthogonal projection) are parallelepipeds whose volume (Lebesgue measures, denoted by A) satisfies

A(Ttli) = c)1(tli)

and

A(Tt27) = cA(t2.7),

where c is a constant depending on the position of the two subspaces and the direction of projecting only. Further, since a projection takes the union and the difference of sets into the union and the difference, respectively, of the image sets, from the additivity of the measure it follows that TH is

measurable and .(TH) = c.(H). If Til...i,, is the orthogonal projection to the coordinate subspace [xi1, ... ,

xi,.] and cil...i,, denotes the corresponding constant, then the sum of the squared measures of the projections is A2(Ti1...i,.H)

ci1...i,.A2(H) = \2(H)

Ci1...i,.,

(1)

where the summation is extended to all combinations i 1, ... , i, of order r of the numbers 1, 2, ... , n. If e1,.. . , en are unit vectors of a Cartesian system of coordinates in the n-dimensional Euclidean space and H is the r-dimensional cube spanned by the orthogonal unit vectors n

uk = E ai k)ei i=1

(k = 1, ... , r)

3. SOLUTIONS TO THE PROBLEMS

338

belonging to L, then ).(H) = 1, and the value cil...i,. _ A(Ti1...irH) is equal to the determinant IAi1,..ir I formed of the columns i1 i . . . , it of the matrix A = {ack)}. Then, however, by the Binet-Cauchy formula, c21... it

IAil... it I2 = IAA* I

= 1,

which, owing to (1), verifies our assertion.

Solution 2. We prove the assertion in the previous generalized form where a (not necessarily compact) Lebesgue-measurable set H lying in the r-dimensional subspace L is projected to the r-dimensional coordinate subspaces.

Let e1, ... , en be the orthonormal basis vectors of a Cartesian system of coordinates in the n-dimensional Euclidean space En, and u1,. .. , ur be the orthonormal basis vectors of a system of coordinates in the subspace L. We may assume that L contains the origin since this can be achieved

by translation, and translation means translation of the projections, so the measure of H and the measures of its projections remain unchanged. Consequently, we may choose n

uk =

(k = 1,...,r),

i=1

satisfies IAA* I = 1. where the matrix A = Denote by Ail...ir the operator of orthogonal projection from L to the coordinate subspace [xil, ... , xir] of En. The operator Ail...i,, sends the basis vectors Uk to the vectors of the latter subspace with components Let (As,...i,,) be the matrix of this projection operator (that

is, the square matrix formed of the columns i1 i ... , i, of A), and let IAi1...ir I

be the determinant of this matrix. The Lebesgue measure of H is given by

IL = fXHdA, L

where XH denotes the characteristic function of H. By the formula for transformation of integrals, the measure of Ail ...i,.H is

XAi1...i,.H dA = IAil...ir I I XH dA = IAil...ir 11L-

(This is valid also when IAil...ir1 = 0, since in this case Ai,...i,.H lies in an at most (r -1)-dimensional subspace, that is, a subspace of measure 0 with

respect to A.) Summing for all combinations i1, ... , i,. of order r, by the Binet-Cauchy formula we obtain {

fXA1...I,.HdA}2 = /L2

[x,l.... ,xirl

which proves the assertion.

IAil...ir

= Âľ2 IAA.I = /U2,

3.5 MEASURE THEORY

339

Problem M.B. Let us use the word N-measure for nonnegative, finitely additive set functions defined on all subsets of the positive integers, equal to 0 on finite sets, and equal to 1 on the whole set. We say that the system 2l of sets determines the N-measure p if any N-measure coinciding with p on all elements of '21 is necessarily identical with p. Prove the existence of an N-measure p that cannot be determined by a system of cardinality less than continuum.

Solution. We shall need a definition and a lemma.

Definition. We say that the system A consisting of certain subsets of a set is independent, if for any distinct members X1i ... , Xj, Y1, ... , Y,, of A, the set is infinite (here U denotes the complement of U).

Lemma. Any countable set has an independent system of subsets that is of power continuum.

Proof. Indeed, take the following set for the countable basic set. Let

M be the set of all sets of real numbers that can be obtained as finite unions of intervals belonging to any type of closedness but having rational endpoints. Clearly, there is only a countable number of such sets. Denote by Ma the set of those sets of real numbers described above that contain the real number a. Then the system {Ma} of subsets of M, for a running

through the set of all real numbers, has power continuum. It is easy to see that {Ma} is independent. Really, for given real numbers al,... , aj, bl,... , bk of A, the set Ma,

nMb, n...nMbk

consists of those elements of M, that is, those sets of real numbers described above, that contain a1, . . . , aj but do not contain b1, ... , bk. The set of these sets, however, is countably infinite.

Next we construct an N-measure that takes the values 0 and 1 only. To this end, we shall need Zorn's lemma, so our proof is not purely constructive.

A system of subsets of the set N of all real numbers is called a filter if it is closed for taking finite intersections and if together with any set it contains all larger sets as well. To exclude the filter consisting of all subsets,

we also require that a filter should not contain the empty set. Since the union of an increasing chain of filters is again a filter, to any filter with the help of Zorn's lemma we can find a maximal filter containing it. This is called an ultrafilter. If a filter contains neither the set A nor q, then it can be extended by A. If it contained both, then it would also contain their intersection, that is, the empty set. Therefore, an ultrafilter contains exactly one of the sets A and A. If we now associate with an ultrafilter

340

3. SOLUTIONS TO THE PROBLEMS

M the N-measure p that takes the value 1 on elements of M and 0 on their complements, then we obtain a set function with values 0 and 1 that is defined on all subsets of N and is additive. Additiveness could fail only if neither of the sets has measure 0, but this is impossible: as the empty set does not belong to our ultrafilter, two disjoint sets of measure 1 do

not exist. We shall prove for a measure of this kind that it cannot be determined by less than a continuum number of sets. From the subsets of N, in the manner described in the lemma, we construct an independent system of power continuum, and denote its elements by Na,. If c is an infinite sequence consisting of distinct real numbers, we denote by S, the union of the complements of the corresponding sets Na,. The filter generated by the Na,, the S,, and the complements of finite sets consists of all sets containing finite intersections of these. To see that the latter is really a filter, we have to show that it does not contain the empty set. In the opposite case, we could find al, . . . , am and cl, . . . , c,,, so that the set

x =Nal n...nNam nscl n...nSc is finite. Here ci is an infinite sequence of distinct real numbers, and consequently we can choose real numbers bl, . . . , b,. from the elements of cl, ... , c,, so that al, ... , am, and bl, . . . , b,, are distinct. Independence implies that the set

Nal n...nNam nNbl n...nNb =Y is infinite. On the other hand, the definition of the Sc yields Y C X. This contradicts our assumption. It follows that the system of sets considered does not contain the empty set; hence it is a filter. We extend this filter to an ultrafilter M and define the N-measure p with the help of the latter. We have to prove that /.c cannot be determined by a system 2i of power less than continuum. If % is a determining system containing, among others, sets of measure 0, then these can be replaced by their complements: it is sufficient to consider determining systems 2i all of whose elements belong to M. Taking a % of this kind, we assume that it contains less than a continuum of sets. If we consider the finite intersections of the sets in 2i, their power is still less than continuum. Therefore, we may assume from the beginning that 2i is closed for taking finite intersections. If there were a Z E M that contained no element of 2i, then Z would not be disjoint from any set in % and, consequently, the sets containing sets of the form ZnA (A E 2() would constitute a filter. This filter could be extended to an ultrafilter, and the latter would define an N-measure v. Then Âľ and v would coincide on the elements of %, whereas Âľ(Z) = 0 and v(Z) = 1. This is impossible since 2( is a determining system. So each element of M contains a set from 2i. Then at least one element W of 2t is contained in infinitely many elements of {Na}; let c be a countable sequence of the corresponding subscripts. The set Na is disjoint from W for each element of c, from which it follows that Sc = UNa is also disjoint from W. But this means that M contains

3.5 MEASURE THEORY

341

two disjoint sets, which contradicts our assumption. The contradiction has

been caused by the assumption that p can be determined by less than a continuum of sets. For the completion of the proof, it remains to show that p vanishes on every finite set. This is evident since p(D) = 1 whenever D is the complement of a finite set. Remarks.

1. If the continuum hypothesis is assumed, the problem becomes trivial. In fact, it is easy to prove that a countable set cannot determine an N-measure. 2. It would be a sharpening of the problem to prove that no N-measure can be determined by less than a continuum number of sets. This, however, is undecidable within the usual Zermelo-Fraenkel system of axioms for set theory.

3. The problem can be generalized to other cardinal numbers. For any cardinal m, there exists a finitely additive 0-1 measure that vanishes on all sets of power less than m and cannot be generated by less than 2' subsets. To prove this, it is sufficient to give an adequate generalization of the lemma. One contestant proved this generalization.

4. One may raise the question of whether there are relatively many or relatively few N-measures for22m the determination of which 2' subsets are N-measures, and there are equally so needed. There are altogether

many N-measures that cannot be determined by less than 2' sets. In the original case where the basic set is countable, one contestant proved this.

Problem M.9. Let {q5 (x)} be a sequence of functions belonging to L2(0,1) and having norm less than 1 such that for any subsequence {0nk (x)} the measure of the set {x E (0,1) :

> 0nk (x)

tends to 0 as y and N tend to infinity. Prove that Â˘n tends to 0 weakly i the function space L2(0,1).

Solution. Suppose there is a function f E L2(0,1) such that II

0.(x)f(x)dx

does not tend to 0 as n -> oo. Then in view of the relations

10.(x)f(x) dxl J1 0

Il0nll llf II

<_ 11111

(n =1, 2, ... )

3. SOLUTIONS TO THE PROBLEMS

342

(replacing, if necessary, f (x) by e"" f (x) with a suitable 0), there is an a > 0 and a subsequence {Â˘flk(x)}k 1 such that r1

ReJ 0lk(x)f(x)dx>a 0

for every k (I If II stands for the norm of the function f (x) in L2(0, 1)). Obviously, to every e > 0 there is 6 > 0 such that the relations m(A) < 6, A C (0, 1) imply IIXAf II2 < E (here m(A) denotes the measure of the set A, and XA(x) denotes the characteristic function of A). Let E be a positive

number less than a, let 6 be a positive value corresponding to e in the sense just indicated, and choose y(E) and N(E) so that for N > N(E) the measure of the set

E = xE(0,1):

N E 0-k (x) k=1

is smaller than 6. Then, for N > N(E), it follows that N

Ref

Na <

f (x)0.,, (x) dx

k=1 r

k=1

=Ref f(x)E0..k(x)dx+Re>J f(x)0.k(x)dx J If (x) I

k=1 E

0.k

xdx +

< vy(E)IIf II2 + NIIXEf II2 <

1: fE I f(x)Onk (x) I dx

VrN-y(e)IIf

II2 + NE,

that is, V-N-(a - e) < y(e) IIf 112, which is impossible for sufficiently large N

since a - e > 0. Consequently, 1

0.(x) f (x) dx - 0

00)

for all functions f E L2(0,1), as stated. Remarks. Essentially all acceptable solutions have followed this way. 1. Two contestants noted that it is sufficient to require the following: for any subsequence {(pf,k (x)} and any number c > 0, the measure of the set

xE(0,1):

0-k (x) k=1

tends to0asN--roc. 2. One contestant showed by a counterexample that without boundedness

of the sequence {0.(x)} in the norm of L2(0,1) the assertion is not

3.5 MEASURE THEORY

343

always true. He also mentioned that the hypotheses of the problem do not imply strong convergence. 3. Another contestant pointed out that it is sufficient to make the following assumption: to any subsequence {nk} of the natural numbers, there is a function f (N) 54 0 for which f (N)/N -> 0 as N -> oo and the measure of the set N

X E (0,1) :

1 k-i E 0nk (x) > y

)

tends to 0 as y -* oo, N --> oc. He also noted that the sequence {0",(x)} does not necessarily converge either in measure or in mean. .4. A contestant observed that the hypothesis may be weakened as follows: from any subsequence of the sequence {0,,,(x)} one can choose a subsequence {cbl,.(x)} for which the measure of set (1) tends to 0 as y --> oc, N -* oo. 5. One contestant called attention to the fact that if one assumes boundedness of the sequence {0, (x)} in the norm of LP(0,1), where 1 < p < oo, then it can be shown that, under the hypothesis of the problem,

0,,.(x) f (x) dx --* 0

(n -> oc)

follows for all functions f (x) E L9(0,1), where 1/p + 1/q = 1. 6. Four contestants treated the problem in the complex case.

Problem M.10. We say that the real-valued function f (x) defined on the interval (0, 1) is approximately continuous on (0, 1) if for any xo E (0, 1)

and e > 0 the point xo is a point of interior density 1 of the set

H= {x : If (x) - f (xo) I< E}. Let F C (0, 1) be a countable closed set, and g(x) a real-valued function defined on F. Prove the existence of an approximately continuous function f (x) defined on (0, 1) such that f (x) = g(x)

for all x E F.

Solution 1. Let F = {c1, C2.... }. Denote by I(x, n) the closed interval of center x and radius 1/n. By the "central p-multiple" (0 <_ p < 1) of the interval (a, b) we shall mean the closed interval

[a+(b_a)(1_P),b_ 2(b-a)(1-p)] We shall define a system Jn of intervals by recursion so that it has the following properties. J,, consists of countably many disjoint closed intervals, none of which intersects F. Further, c, is the only accumulation point of

344

3. SOLUTIONS TO THE PROBLEMS

the set of endpoints of the intervals forming Jn and, finally, K,,,, and Kn are

disjoint if m # n, where Kn stands for the union of the intervals forming Jn. Suppose that for n' < n we have already defined J,,, so that they have the properties required. Then the closure of the set U'-i Ki = Ln is Ln U {cl,...,cn_l},

and it does not contain cn. Therefore, it does not intersect a suitable neighborhood I (cn, rn) of cn either. We may assume that rn > n. For each

j (j = 1, 2, ... ), consider the interior of the set I (cn, rn + j) \ I (cn, rn + j + 1) \ F. This is the union of a (finite or) countable number of open intervals. Obviously, the total length of these open intervals is equal to the measure of the set I (cn, rn +j) \ I (cn, rn +i + 1) = M (n, j ), so the union of a finite number of them has at least (1-1/ j) times this measure. Take the

central (1 - 1/j)-multiples of these finitely many intervals, and let these new intervals constitute the system J,3. Put Jn = U _ l Jn, j . Clearly, the systems Jn of intervals so obtained possess the required properties (in particular, the recursion is correct). Further, the intervals

forming Jn,j fill at least (1 - 1/j)2 times the measure of M(n,j) and, therefore, Kn, the union of the intervals forming Jn, has density 1 at cn. Define f in the following way. Let f (x) = g(cn) if x E Kn or x = cn. If we prove that the set U°°_1KnUF is closed, then f can be defined at points not belonging to this set so that on the countable number of disjoint open intervals forming the complement of this closed set, f is linear, and at the endpoints of the intervals it takes the values already fixed. It is obvious that the function f defined in this way is approximately continuous in points of F and continuous in all other points. To see the latter, observe that for x V F the distance from x to F is positive, say greater than 1/m, where

m is a positive integer. Thus, by the assumption rn > n, Kn does not intersect the 1/2m-neighborhood of x if n > 2m. Consequently,

I(x, 2m) fl (U Kn U F) = I(x, 2m) fl (U Kn). n=1

n=1

The right-hand side, however, is the union of a finite number of closed intervals. This proves, on the one hand, that U°_iKn U F is closed and, on the other hand, that the restriction to I(x, 2m) of the function f constructed with the help of this set is a continuous polygon function.

Solution 2. Let F = {rl, r2.... }. We start from the idea that the closer x is to ri, the more f (x) should "feel" the value taken at rti. We try to achieve this by a definition of the following type:

f(x) -_

u( )lid *:

Lti=1 u(i)Ix-r;

if x

(1)

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345

We now impose various conditions on the order of u(i) to ensure convergence of this series and approximate continuity of the function obtained, and finally we show that these conditions can be fulfilled. The series will always be convergent if, for instance, (2) u(i) > i2lg(ri)I and u(i) > i2 (i = 1, 2, ... ). Really, for x 0 F, the closedness of F gives mini I x - riI > 0. f will be continuous in all points of (0, 1) \ F (in a small neighborhood, both the numerator and the denominator are sums of uniformly convergent series of continuous functions), hence it is also approximatively continuous there.

Consider therefore an ri. Let d = minj<i Iri - rj1, 6 < d/2. We show that in the interval (ri - 6, ri + 6), for 6 sufficiently small, the numerator in (1) will generally be around g(ri)/(u(i)Ix - riI), and the denominator around 1/(u(i) I x - riI ). In any case,

9(rj)

= 0(1) (3)

j<i u(j)Ix - rjI 1

j d - 6/2 > d/2). We divide the values x into two classes. One consists of those x that satisfy the relation 1 + I9(rj) I <

u(j)Ix - rjI

(4)

for all j > i. For these x, by (3) and (4), u()Ix-r:l +O(1) g(ri)+O(Ix-riI) 1+0(Ix-riI) u(05-r+l +O(1) which lies in the e-neighborhood of g(ri) if 6 is sufficiently small. Thus, we have to achieve that the measure of such x in (ri - 6, ri + 6) be (2 + o(1))b. If, however, (4) does not hold, then

f(x) =

ix - rj l < (j) (1 + I9(rJ)I) = Ej. The measure of such "bad" x in (ri - 6, ri + 6) is <2

Ej. Ir i -r9 I <b+ej

If now Ej < Iri - rjl2/j2, then this measure is j2 I2 Iri -rj < 2(46)2

<2 z

I r; -rj l <6+

j2 1

= 0(62) = o(6)

as 6 - 0 (since Iri - rj l2/j2 < Iri - rj I/4 and so (3/4). Iri - rj l < 6). Consequently, together with (2), we have only a finite number of conditions for each j, so they can be realized.

3. SOLUTIONS TO THE PROBLEMS

346

Problem M.11. Let {fn}°D 0 be a uniformly bounded sequence of realvalued measurable functions defined on [0, 1] satisfying

ff2 l

Further, let {cn} be a sequence of real numbers with 00

cn = +00. n=0

Prove that some re-arrangement of the series E°° 0 cn fn is divergent on a set of positive measure.

Solution. We first show that 00

2 f2

(1)

Cn n

n=0

is divergent on a set of positive measure. Suppose that (1) yields a function g that is finite a.e. By Egorov's theorem there is a set A C_ [0, 1] on which

the series (1) is uniformly convergent and for which .\(A) > 1 - (/2K), where K is a common upper bound of the functions fn. Then, by the uniform convergence, the limiting function g(t) is integrable on A, and Co

c cn n=0

fn2(t) dt =

g(t) dt < +oo.

(2)

On the other hand,

JAfn(t) dt = Jo 1 f,2, (t) dt - 1,0,11\A fi(t) dt > 1 - K2K = 2 which, in view of (2), yields 1

°O

21: cn<oo, n=0

a contradiction.

As a second step we show that the series E° 0 cn fn has a subseries divergent on a set of positive measure. Let e1, ... , En.... be independent random variables on some field (1, A, P) taking the values 0, 1 with probability 1/2, and let 0 < t < 1 be a number for which (1) is divergent. Then for the random variables Encn fn(t) we have Var (Encnfn(t)) = n=0

E n=0

+oo

3.5 MEASURE THEORY

347

and therefore, by Kolmogorov's three-series theorem, E°° o encn fn(t) is divergent with positive probability. Thus, the set of all elements (w, t) in the probability field (S2, A, P) x [0, 1] for which 00

1: (en(w)Cnfn(t)

(3)

n=0

is divergent has positive measure. Consequently, there exists w E f for which the series (3) is divergent on a set of positive measure. It follows that there is a set D of positive measure and an infinite sequence n1 < n2 < ... such that cc

1: Cnk fnk (t)

(4)

n=0

is divergent for all t E D; so there exists b(t) > 0 such that for arbitrarily large indices the partial sums of (4) have oscillation not less than b(t). Define next the numbers N1 = 1 < N2 < ... as follows. Let the number Nk+1 be such that the measure of the set FL

f ED:

max

Nk <v<µ<Nk+l

Cnifni(t)

8=v

is greater than (1 - (1/2k+1)) , A(D). Such an Nk+1 exists since for sufficiently large Nk+l, all values t E D belong to Dk. The set

D* = fl Dk k=1

has a positive measure, and >

max

Nk<v<µ<Nk+l

Cnifni(t)

b(t)

i=v

if t E D* and k > 1. Therefore, any rearrangement of the original series >°°=o cn fn, for which the sums r_NNk cni f,,, consist of consecutive terms, is divergent at all points t E D*.

Problem M.12. Let { fn} be a sequence of Lebesgue-integrable functions on [0, 1] such that for any Lebesgue-measurable subset E of [0, 1] the sequence fE fn is convergent. Assume also that limn fn = f exists almost

everywhere. Prove that f is integrable and fE f = limn fE fn. Is the assertion also true if E runs only over intervals but we also assume fn > 0? What happens if [0, 1] is replaced by [0, oo)?

Solution. Let E _ [0, 1] or E = [0, oo), and E = U' 1Ek, where Ek is a measurable set of finite measure and on Ek the convergence of the sequence

348

3. SOLUTIONS TO THE PROBLEMS

{fn} is uniform. By Egorov's theorem, such a sequence {Ek} obviously exists. It is clear that fEk f exists for every k, and fF f = limn fF fn if F is a measurable subset of Ek. The formula p(G) = limn fG fn defines a finitely additive signed measure

on measurable subsets G of E. The theorem of Beppo-Levi shows that f together with the relation Âľ(G) = IG f for each the existence of JE measurable subset G of E are equivalent to the a-additivity of P. But the latter follows from the well-known fact that the limit of a pointwise convergent sequence of finite signed measures is a finite signed measure (in

particular, it is a-additive; see P. R. Halmos, Measure Theory, Springer, New York, 1974, p. 170, relation (14)). It can also be seen that fn -

--+ 0 (n - oo). Thus, we have answered the first and third questions of the problem. The answer to the second question is negative, as shown by the example fn(x) = n if 0 < x < 1, and fn(x) = 0 if 1 <X<1. fI

Problem M.13. Let 0 < c < 1, and let q denote the order type of the set of rational numbers. Assume that with every rational number r we associate a Lebesgue-measurable subset H,. of measure c of the interval [0, 1]. Prove the existence of a Lebesgue-measurable set H C [0, 1] of measure c such that for every x E H the set

{r:xEHr} contains a subset of type q.

Solution. We give two solutions. Both make use of the following simple lemma:

Lemma. Let M be a system of sets that consists of certain subsets of the interval (0, 1), and suppose that for each A E M there is an x E A such that A n (0, x) E M and A n (x, 1) E M. Then every element of M has a subset of type q. Proof. Arrange the rational points of (0, 1) in a sequence Ti, r2, .... Let A E M be arbitrary. We shall define by recursion a sequence of points xn E A having the following properties: 1. It is ordered in the same way as the sequence {rn}; 2. A n (0, xn) E M, A n (xn, 1) E M for every n, and A n (xi, xj) E M for every pair xi < xj.

By assumption, there is an x1 E A such that A n (0,x1) E M and An(xi,1) E M. Next suppose that X1, X 2 ,-- . , xn have already been chosen, and let the immediate neighbors of rn+1 from among r1i r2, ... , rn be ri and

rj, ri < ri. By the induction hypothesis, we have xi < x; and An(xi, xj) E M , so there is an xn+1 E A n (xi, xj) such that A n (xi, xn+1) E M and A n (xn+1, xi) E M. A similar choice of xn+1 is possible in the cases where rn+1 > ri (i < n) or rn+1 < ri (i < n). The proof of the lemma is complete.

3.5 MEASURE THEORY

349

Solution 1. We need the following auxiliary theorem:

Auxiliary theorem. Let the sets Hn E [0, 1] be measurable and have measure not less than c (n = 1, 2.... ). Then the set H of all points x for which the sequence In : x E Hn} has positive upper density is also measurable and has measure not less than c.

Proof. Denote by fn the characteristic function of Hn, and put 1

gn = 1 (f1 + f 2 + ... + fn).

Obviously, x E H if and only if g, ,(x) does not tend to 0 as n -> oo, which yields the measurability of H. Since 0 < gn(x) < 1 for every n, we have

A(H) >

f,1]\H

g. (x) dx = fo g. (x) dx -

=c-

[0,1]\H

J[0

gn(x) dx

gn(x)dx.

Now x E [0, 1] \ H implies limn-oo gn(x) = 0, so by Lebesgue's theorem on the passage to the limit under the integral sign, lim

g (x) dx = 0.

o,1]\H

Thus, from the previous inequality, we obtain .(H) > c, which proves the auxiliary theorem. Let {rn} be an arrangement of the rational points of (0, 1) in a sequence

of uniform distribution, that is, for any subinterval (a, b) E (0, 1), the density of the sequence In : rn E (a, b)} is b - a. (A simple example of an arrangement of this kind is the following. We first fix an arbitrary arrangement in a sequence {sn}, then from {sn} we consecutively choose the elements of minimal index lying respectively in the intervals [0,1], 0,

2,1

[0,1

n, 1n

,...,

n-1 ,1 ,..., n

meanwhile, we should make sure that each element is selected only once.) Let A E (0, 1), and assume that the sequence In : rn E Al has positive upper measure. We show that A contains a subset of type g. By the lemma we have presented in advance, it is sufficient to prove the existence of an

x E A such that both In : rn E A fl (0, x)} and In : rn E A n (x,1)} have positive upper density.

Denote by d the upper density of the sequence in : rn E Al, and let < xm = 1 be a subdivision of [0, 1] finer than d/3. Since {rn} is uniformly distributed, the upper densities of the sequences in : rn E A n (xi_1i xi)} are smaller than d/3 for every i. Consequently,

0 = xo < x1 <

350

3. SOLUTIONS TO THE PROBLEMS

denoting by di the upper density of the sequence in : rn E A n (0, xi)}, we

have 0 < di - di_1 < d/3 for i = 1, 2, ... , m. Since d,n = d, there is an i with 0 < di_1 < di < d. Then di_1 < di implies that the set A n (xi_1, xi) is infinite, and it is easy to see that for any element x E An(xi_1, xi)}, the sequences in : rn E A n (0, x)} and In : rn E A n (x, 1)} also have positive upper measures.

Now consider the sets Hr appearing in the statement, and let H1 be the set of those x for which the sequence in : x E Hr,, } has positive upper measure. By the previous observation, if x E H1, then {r x E Hr} contains a subset of type q. Applying our auxiliary theorem to the :

sequence of the sets Hr,,, it follows that H1 is measurable and ) (H1) > c, as required. Solution 2. If the closure of a set A has positive measure, then A contains a subset of type i since the system M = {A C (0, 1) : \(A) > 0} obviously satisfies the condition of the lemma. It is therefore sufficient to prove that the set

H2={x: A({r:xEHr})>0} is measurable and \(H2) > c. Consider the function f (x) = A({r : x E Hr}). Clearly, 0 < f (x) < 1; if we show that f (x) is Lebesgue-measurable and fo f (x) dx > c, then we shall be done. For any A E [0,1], put

An =

U 2

2-1

, 2n

Obviously, Al D A2 D ... and n= nlAn = A; hence limn . a(An) = A(A)So let fn(x) = \({r : x E H,.}n); then fn f, and therefore we need only to show that fn (x) is Lebesgue-integrable with fo fn (x) dx > c. We have the relation 1

,\(An) =

9n,i (A), 2n E i=1

where gn,i(A) = 1 or 0 depending on whether A has or does not have a point in [(i - 1)/2n, i/2n]. Clearly,

fn,i(x):=gn,i({r: xEHr})= sup{k,.: rE I

i-1

i 11

2n J 1

where k, is the characteristic function of Hr. The function k, is integrable and fo kr(x) dx = c, so fn,i is integrable and ff fn,i(x) dx > c. Since fn = (1/2n) Ei= fn i, it follows that fn is integrable and ff fn(x) dx > c. Remark. Denote by Ho the set of those x for which the set {r : x E Hr} contains a subset of type 77. Let {rn} be a fixed arrangement of the rational

3.5 MEASURE THEORY

351

points of (0, 1) in a uniformly distributed sequence, and denote by H1 the set of those x for which the sequence {n : x E Hr,, } has positive upper density. Let H2 be the set of those x for which the closure of the set

{r : x E H,. } has positive measure. Finally, let H3 be the set of those x for which {r : x E Hr} is dense in a subinterval of (0, 1). It is easy to see that for any set A of rational numbers, the upper density of the sequence in : r,,, E Al is at most .(A). Therefore, H3 C H1 C H2 C Ho holds for any system Hr (thus Solution 1 proves a deeper assertion than Solution 2, since it shows from a narrower set that its measure is at least c). On the other hand, the statement .(H3) > c is false: it can be shown that, for any 0 < c < 1, there exists a system Hr satisfying the conditions of the problem and such that the set {r : x E H,.} is nowhere dense for every x E [0, 1], in particular, H3 = 0. We also note that if the sets Hr are measurable, then so is Ho. Actually, it can be proved that if the Hr are Borel (or, more generally, analytic) sets, then Ho is analytic, hence measurable. From this it follows easily that Ho is also measurable in this general case.

Problem M.14. Find a perfect set H C [0, 1] of positive measure and a continuous function f defined on [0, 1] such that for any twice differentiable function g defined on [0, 1], the set {x E H : f (x) = g(x)} is finite. Solution. Delete an interval of length 1/6 from the center of [0, 1], intervals of length 1/18 from the centers of the remaining two intervals, and, following the procedure, intervals of length 1/2 3n+1 from the centers of the 2' intervals remaining after the nth step. This is possible because the length of the remaining intervals is -1

2-n 1- 1: 2k

2 3n+1

>2'

Â°Â°

=2 n(1

2)=2

1>2.3n+1

Denote by In,i (n = 0, 1.... ; i = 1, ... , 2") the open intervals of length 1/2.3n+1 deleted in the nth step. Put H = [0, 1] \ UIn,i. Then H is perfect and has measure 1/2. Denote by h the function that vanishes on H and whose graph on In,i is an isosceles triangle of altitude 1/n for n = 0, 1, .. . and i = 1, . . . , 2n. Then h is continuous on [0, 1], and so the function

AX) = f h(t) dt 0

is continuously differentiable on [0, 1] and satisfies the relation f'(x) = 0 for all x E H. Let g be twice differentiable on [0, 1]; we establish that the

set A = {x E H : f (x) = g(x)} is finite. Suppose A is infinite. Then

3. SOLUTIONS TO THE PROBLEMS

352

there is a convergent sequence xk -> x0 in A. Since f'(xo) = 0, we have g'(xo) = 0; hence by l'Hopital's rule, lim 9(x) - 9(xo) = lim

x-,xo

(x - x0)2

g'(x)

x--.xo 2(x - x0)

g" (X0)

9'(x) - g' (X0) - 1'm x-,xo

2(x - xo)

It follows that lim f (xk) - f (xo) k-oo (xk - xo)2

- 9"(xo) 2

The contradiction will arise from the fact that f(x) - f(xo)

lim

x-.xo,xEH

=oo.

(x - xo)2

Indeed, let x c H, x # xo, be arbitrary. Then there is an interval I,',i that separates x and x0. Let n be the smallest subscript for which there exists an In,i of this kind. It is easy to see that Ix - xoI < 2 On the other hand, if, say, x > x0, then f (X) - f (xo) =

JxQ 1

h(t) dt > 1

Jr 1

2'2.3'+' *n

h(t) dt

12.3n .n.

Hence 1 f(x) - f(xo) > (4) n (x-xo)2 - 12.3" .n.4-'t - (3) I

12n

If x -> x0, then n -4 oo and (4/3)n (1/n) -+ oo, which proves the assertion.

Problem M.15. Prove that if E C R is a bounded set of positive Lebesgue measure, then for every u < 1/2, a point x = x(u) can be found so that

I(x-h,x+h)nEI>u,h

and

I(x-h,x+h)n(R\E)l>uh

for all sufficiently small positive values of h.

Solution. We shall prove a stronger statement, namely, we verify the conclusion for u = 1/2. Let

F(x) = m ((-oo, x) n E), where m denotes the Lebesgue measure. Then F is continuous, increasing,

and satisfies IF(x) - F(y)l < Ix - yI for all x, y C R. By Lebesgue's

3.5 MEASURE THEORY

353

density theorem, there exists x E R, which is a point of density 1 of E, and hence there is a positive k E R with F(x + k) - F(x) > k/2. Since for sufficiently small x we have F(x + k) - F(x) = 0, continuity ensures the existence of an x0 satisfying F(xo + k) - F(xo) = k/2. Consider the function G(x) = F(x) - F(xo) - (x - xo)/2. Then G(xo) = G(xo + k) = 0. As almost every point of R is a point of density 1 of either E or ][8 \ E, we have I G'(x) = 1/2 a.e. Thus, G is not identically 0 on [xo,xo + k]. Consequently,

either

max

tE[xo,xo+k]

G(t) > 0

min

tE[xo,xo+k]

G(t) < 0.

It is sufficient to consider the first possibility since the second is similar to it. Let x1 E (xo, x0 + k) be a point at which G assumes its maximum just mentioned. Then for every

0 < h < min{xl - xo, xo + k - xl},

we have G(xl -h) < G(xl) > G(xl+h), that is, F(xl -h)+h/2 < F(xl) >

F(xl+h)-h/2. Thus, F(xl+h)-F(xl-h) < F(xl)+h/2-(F(xl)-h) _

3h/2 and F(xi + h) - F(xl - h) > F(xi) - (F(xl) - h/2) = h/2. From these two inequalities, we obtain

m ((x1 - h, x1 + h) fl E) > 2 and

m((xi-h,xl+h)fl (]I8\E))> 2 The proof is complete.

Remark. One contestant did not make use of the boundedness of E and

for u = 1/2 proved the assertion in the case where E is measurable and

neither E nor R \ E is a null set, and also in the case where E C m* (E) 0, and m* (E) # 0 (here m* and m* denote the exterior and interior Lebesgue measures, respectively).

Problem M.16. Show that there exist a compact set K C ][8 and a set A C R of type FF such that the set

{xER:K+xCA} is not Bore]-measurable (here K + x = {y + x : y c K}). Show that there exist a compact set K C ][8 and a set A C ][8 of type FQ such that the set

{xEJ:K+xCA} is not Bore]-measurable (here K + x = {y + x : y E K}).

Solution. Let P and K be bounded perfect sets such that the sums x + y (x E P, y E K) are pairwise distinct, that is, the relations x1i x2 E P,

3. SOLUTIONS TO THE PROBLEMS

354

y1, y2 E K, xl + yi = x2 + y2 imply x1 = x2 and yl =Y2- It is easy to see that the sets

a2=O,1;

i = 1, 2, .. .

lO;

a,=O,2;

i = 1, 2, .. .

l i=1

and

K i=1

are of this kind. It is well known that there exists a set U C P x K of type G5 such that the set

B = {x E P : there exists y E K with (x, y) E U} (which is the projection of U onto P) is not Borel-measurable.

Put V = (P x K) \ U and A = {x + y : (x, y) E V}. The mapping O(x, y) = x + y (x E P, y E K) is continuous and, by the choice of P and K, one-to-one. Therefore, 0 is a homeomorphism of the compact set P x K onto {x + y : x E P, y E K}. Since V is of type FQ, it follows that A = O(V) is an Ftype subset of O(P x K) = P + K. Since the latter set is closed in R, the set A is a subset of type FQ of R as well.

It is easy to see that K + x c A if and only if 0-1(K + x) C(A), that is, if x E P and

({x}xK)nU=0, (that is, if x E P \ B). So

{xER: K+xcA}=P\B, which is not Borel-measurable.

Problem M.17. For which Lebesgue-measurable subsets E of the real line does a positive constant c exist for which sup

< c

-00<t<00

sup

eanx f(x)dx

for all integrable functions f on E ?

Solution. We show that (*) holds for those and only those Lebesguemeasurable sets that - apart from a subset of measure 0 - can be covered by a set consisting of the finite union of closed intervals and containing no pair of congruent points modulo 2ir. Denote by E the system of all sets satisfying the requirements of the

problem and by F the system of all sets having the properties just described.

3.5 MEASURE THEORY

355

If E E E, then those points of E for which there is a congruent point in E (mod 27r) form a set of measure 0. Indeed, if, for example, the set E' = E n (E + 2j7r) n (2k7r, 2(k + 1)7r)

had positive measure (j # 0 integer, E + u denotes the translate by u of E), then with

ifxEE',

f(X) -1 ifxEE'-2j, 0

otherwise,

it would follow that

L eZnx f (x) dx = 0 while, as f

for every n,

0 a.e., sup

eitx f(x) dx >0.

The same property trivially holds also for every E E F. Denote by * the reduction modulo 2 with range space (-7r, 7r]. That is, u* is the (signed) deviation of u from the nearest multiple of 2-7r (if there are two such multiples, then u* = 7r). Furthermore, put

E*={u*: nEE}, h*(x) = h(u), where x E E*, x = u*, u E E. By the property above, the latter definition is correct for a.e. x E E* since there is only one such u. Therefore, we may write

(eitx) f* (x) dx,

0(t) =

eitx f(x) dx =

0(n) =

einx f(x) dx = E (einx)*f* (x) dx.

Here f *(x) E £ l (E* ), and every element of Gl (E*) occurs among the f *(x),

where f(x) E £1(E). We show that the relation JE a (x)f*(x)dx

< cmax 1¢(n)I = cmaxJ n

einx f*(x)

is valid for every function f * E G1(E*) if and only if there exists a function b(x) = E aneinx, EIan I < +oo, such that b(x) = a(x) for a.e. point

x E E*. Then the choice c = E _. Ian1 is possible.

3. SOLUTIONS TO THE PROBLEMS

356

Suppose first that there exists a function b(x) of this kind. Then a(x)f*(x)dx

JE' 00

n=-oo

einx f* (x) dx

an I *

E ano(n)

n=-oo IanI mnax I0(n)I.

n=-oo

Conversely, consider the sequences {jn}n+_° 0o that tend to 0 as Inl -> oo; they form a Banach space with respect to the norm max l in I . The set of all sequences of the form {0(n)} is a subspace of this space, and the

functional fE* a(x) f * (x) dx is linear and, by our assumption, bounded on it. According to the Hahn-Banach theorem, this functional can be boundedly extended to all sequences { jn }. But the bounded linear functionals on the latter have the form E an jn, where E I an I < +00. Consequently, restricted to the sequences {0(n)},

J a(x)f* (x) dx = > anq5(n) _ *

n=-oo

f einx f* (x) dx E*

b(x)f*(x) dx fE*

(where b(x) _ aneinx), for every f* E £1(E*). Hence, a(x) = b(x) for a.e. x E E*. Apply the proposition just proved to the function a(x) = eitx with fixed t. Let E E F. If the finite number of intervals covering E is somewhat augmented, then the set obtained will not contain any pair of points congruent modulo 27r either. Let k(x) be a "sufficiently smooth" function that equals 1 on the original intervals, and 0 outside the augmented intervals. Finally, put 00

b(x) _

k(x +

2j)eit(x+2jn).

j=-00 Since at each point at most one term of the series is different from 0, the definition of b(x) makes sense. Furthermore, b(x) = a(x)* = (eitx)*

if x E E*.

The function b(x) is periodic with period 27r and Fourier coefficients bn = 27r

foo

°°

27r

e-"x 1` k(x + 2j)eit(x+2i7r ) dx j=-00

°°

27r

e-inxb(x) dx

-oo

eitx-inxk(x) dx.

3.5 MEASURE THEORY

Then

357

'00

Ibnl <

Ik(x)I dx < cl,

27r

or, after integrating by parts twice, Ibnl C

27r(n - t)2

Ik"(x)I dx < 00

(n - t) 2

(for a suitable choice of k(x)). Hence, 00

Y:Ibnl <c1+c2 E

n=-oo (n --t)2

= c3.

Applying the statement we have proved previously, it follows that E E E. Conversely, suppose that E E ÂŁ. We establish the existence of 6 > 0 such that for any two Lebesgue density points u, v of E, the relation I u - v I > 7r implies I(u - v)*I > S. Otherwise, indeed, there are sequences {un}, {vn} of density points with Iun - vnI > 7r but (un - vn)* -+ 0. It can be shown

that there is a number t such that [t(un - vn)]* 74 0. To see this, first let un - vn = wn be a bounded sequence; then t = 1/ sup I un - vnI is an appropriate value. If wn is unbounded, then we may assume that wn -p 00 and, what is more, wn+l/wn - oo. We form a sequence to as follows:

to=0,

to=tn_1-

(tn_lwn)* - 7r/2 ,

n=1,2,...

Here to - tn_1 = O(1/wn), and by the fast increase of wn the limit of the sequence to exists. Put 00

t = limtn = E(tn - tn_1). n=1

Then

t-tn=0 Consequently, wn

twn = (t - tn)wn + tnwn = 0

wn+1

+ tn_lwn - (tn_lwn)* + ', 2

and, since tn_lwn - (tn_lwn)* is a multiple of 27r,

(twn)*=01 wn 1+W=7r +0(1)740. wn+l /

So, this t is suitable.

358

3. SOLUTIONS TO THE PROBLEMS

Using this t, we form the function a(x) = (ettx)*,

x E E*,

and apply the statement proved earlier. Omitting a set of measure 0, a(x) can be extended to a 27r-periodic continuous function (even a function with absolutely convergent Fourier series), to be denoted by b(x). Since 1

un --,un+ J flE, n n

as well as its *, has positive measure, and similar statements hold for vn, after the omission of the set of measure 0 mentioned above, there remain some points in these sets. Let u'' , v' be such points. Then u'n - un -> 0, 0, hence 0, and a (u'*n) = b (u';,), a b (v'n). Therefore, by the continuity of b(x), it follows that a (u' ,) - a (v'*) , 0. But a (u'*) = eitun, a (v'*) = eztv;, n

whence

0 = lim[t(un -

lim[t(un - vn)]*,

and this contradicts the construction of t. Thus 6 exists. For each density point of E, consider its neighborhood of radius 6/4. Their union S is open and almost covers E. For any two points ul and vl of S, the relation I (ul - vl)*I < 6/4 implies Jul - v1I < 6/4; otherwise (assuming that 6 < 7r), Jul -vi > (3/2)7r would follow, and by the construction of S there would exist density points u and v in E with I u - ul I < 6/4, 1v-viI < 6/4, hence iu - vJ > 7r, while I(u-v)*I < I(ul -vl)*I+2.6/4 < b, which contradicts the property of S. Thus, the closure of S cannot have two congruent points modulo 27r. Hence, it also follows that the measure of S does not exceed 27r, and since each component of the open set S has length greater than or equal to 6/2, the number of the components must be finite, that is, the closure of S is a finite union of closed intervals. Consequently, E E F.

Problem M.18. Show that any two intervals A, B C R of positive lengths can be countably disected into each other, that is, they can be written as countable unions A = Al U A2 U ... and B = Bl U B2 U ... Of pairwise disjoint sets, where Az and Bz are congruent for every i c N.

Solution 1. In R, consider the equivalence relation

{(x,y)ER2: x - yEQ}, which induces the disjoint classification R = U.yErQ.y. Obviously, each equivalence class Q.y is dense in R, so A fl Q7 and B fl Q.y are countably

3.5 MEASURE THEORY

359

infinite sets. Consequently, assuming the axiom of choice, there exists a bijection

O.y: AnQ., between them for each -y. Then O.,(x) - x for every -y E P, x E A n Q. Let Q = {q1, q2, ... } be a numbering of the rationals, and consider the sets

YEP{xEAn Q.,: q.y(x)-x=qi},

Bi= U {(a-'(x): xEAnQ.y,0ry(x)-x=qi}, ryEP

for i E N. From the definitions, we see that

and B =

A = U Ai i=1

i=1

are partitions, and Bi = Ai + qi for every i E N.

Solution 2. Denote by a the relation that holds between the subsets of if and only if they can be countably dissected into each other. It is easy to verify that a is a translation-invariant equivalence relation satisfying

1'S"/

n1T")

if Sn and Tn, respectively, are pairwise disjoint sets (n E N), and SnaTn for every n.

We now observe that any interval is the disjoint union of a countably infinite number of open intervals and a countably infinite set. So, by the above remarks, we may assume that A =]0, a[, B =]0, b[, 0 < a < b. In R, we now consider the equivalence relation

- { (x, y) E R2 : x - y E Q1. Its equivalence classes are dense in R, so - assuming the axiom of choice - there exists a set X C ] 0, a/3[ containing exactly one element of each class. Put a 2a

QA=Qnlo,

3 [,

QB=Qn]O,b-

[

and

A* = U (X + q), gEQA

B* = U (X + q), qEQB

where X + q stands for the set X translated by the number q E R. Obviously, A* and B* are countably infinite, disjoint unions of sets congruent with X, so A*aB*. One also easily verifies the inclusions

1a 2a 3' 3

c A C A,

[3,b

C B

360

3. SOLUTIONS TO THE PROBLEMS

Put

A-=]0,a[\A*, A+_] 3,a[\A*, B- =]0, a [\B*,

B+ =]b- 3,b[\B*.

A simple calculation shows that B- = A- and B+ = A+ + (b - a). Consequently, the partitions

A=A-UA*UA+ and B=B-UB*UB+ prove the statement. Remarks.

1. A direct consequence of the statement is the significant fact that there is no o-additive, translation-invariant measure defined on all subsets of R for which the unit interval has measure 1. 2. The first proof can immediately be applied to subsets with nonvoid interior of nondiscrete, separable, Hausdorff topological Abelian groups. 3. By the usual tools of measure theory, the statement can be extended to Lebesgue-measurable sets as follows. Any sets A, B C_ R of positive Lebesgue measures can almost be countably dissected into each other, that is, they have partitions 00 A= i=o U Ai

and B = U Bi i-o

such that A0 and Bo have measure 0, while Ai is congruent with Bi, for every i E N. The assertion of the previous remark cannot be sharpened by deleting the sets of measure 0, since - as noticed by a contestant - a set of first Baire category cannot be countably dissected into a set of second Baire category; however, among sets of positive Lebesgue measure, there are examples of this kind.

Problem M.19. Let H C R be a bounded, measurable set of positive Lebesgue measure. Prove that

limofa((H

>0,

where H + t = {x + t : x E H} and A is the Lebesgue measure.

Solution. We show that

limofA((H

)\H)>1.

3.5 MEASURE THEORY

361

Lemma. For any 0 < E < 1, there is an interval [a, b) such that A (H n [a,b)) > (1 - E) (b - a). Proof. Assume that the conclusion of the lemma is false. Let us cover H by a countably infinite number of intervals (closed from the left), so that the sum of the lengths of the intervals is less than A(H)/(l - E). By the definition of the Lebesgue measure, this is possible. Let the ith interval be [ai, bi). Then by the indirect assumption, A(Hn [ai,bi)) < (1 - e)(bi - ai) and therefore M

A(H) <

A(H n [at, bi)) < (1 - E) E(bi - ai) < A(H), i=1

i=1

which is a contradiction. The proof of the lemma is complete.

Let 0 < e < 1, and let the interval [a, b) be such that

A(Hn [a, b)) > (1 -E)(b-a). Let 0 < t < b - a and n = [(b - a)/t] + 1. Then n

(1- E)(b - a) < A(H n [a, b)) < E A(H n [a + (k - 1)t, a + kt)). k=1

Hence, there exists an integer k, 1 < k < n, such that (1 - E)(b - a) A(H n [a + (k - 1)t,a + kt)) > n (1 - E)(b - a)

bta+1

(1 - E)t

+bta

Put

Ai=Hn[a+(k+i-1)t,a+(k+i)t)

(i=0,1,2,...).

Since, H being bounded, Ai is empty if i is sufficiently large, we have 00

A ((H + t) \ H) > E a ((A. + t) \ Ai+1) > E (.A(Ai) i=o

i=o

- \(Ai+1))

- E)t =A(A)=A(Hn[a+(k-1)t a+kt))> (1 0 1+bta

Quite similarly,

\((H-t)\H)>-

(1 - E)t

1+bta.

It follows that o

liminf), ((H tt)\H)>l

o l + bta

But this is true for every 0 < E < 1. Thus, limionf

\ ((H Itlt) \ H) > 1.

=1-E.

362

3. SOLUTIONS TO THE PROBLEMS

3.6 NUMBER THEORY Problem N.I. Let f and g be polynomials with rational coefficients, and let F and G denote the sets of values of f and g at rational numbers. Prove that F = G holds if and only if f (x) = g(ax + b) for some suitable rational numbers a # 0 and b. Solution. By a "polynomial" we shall always mean a polynomial with rational coefficients, and by the "range" of a polynomial we shall mean the set of its values assumed at rational numbers. We use two classifications of polynomials. We write f " g if the ranges of these polynomials coincide, and f : g if f (x) = g(ax + b) for all x with suitable rational numbers a # 0, b. These are clearly equivalence relations, and they are compatible with multiplications by a constant, that is, if f - g or f g, then c f " cg or c f cg, respectively (c # 0 rational). Our aim is to show that these relations are identical. One implication is clear: if f g, then f - g holds obviously. The converse will be proved in several steps. 1. It is sufficient to prove the statement for polynomials with integral co-

efficients. Indeed, assume f - g, and let c be an integer such that the coefficients of c f and cg are all integers. We have c f - cg, thus c f assuming the integer case, which yields f g as wanted.

cg,

2. If f - g are polynomials with integral coefficients, we show that they must be of equal degree. Indeed, assume that

f(x)=ax' +...,

g(x) = bx k +

. . ,

where a # 0, b # 0, and n < k. Choose a prime p/ab. We have f (1/p) = c/p'', where p/c. By assumption, g also takes on this value at some number u/v with (u, v) = 1. Now if p%v, then p cannot divide the denominator of g(u/v); consequently, p must divide v. Now, by (p, b) = 1, the exponent of p in the denominator of b(u/v)k is higher than in any other term of g(u/v), thus the denominator (in the reduced form) of g(u/v) contains p with an exponent > k. This yields n > k, hence n = k as asserted. 3. It is sufficient to prove the statement for polynomials with integral cofficients where at least one of the leading cefficients is 1. Indeed, let f - g, f (x) = aXn + .... Then we\ have

an-l f (x)

an-1f

(x) '" a"'-lg(x)'

and here an-l if (x/a) still has integral coefficients while its leading coefficient became 1. By assumption, we conclude an-l f (x/a) an-lg(x), which yields f .: g.

4. Let f - g be polynomials with integral coefficients, f (x) = xn + ... , g(x) = bxn+.... We claim that every prime divisor of b has an exponent

3.6 NUMBER THEORY

363

> n. To show this, take a p1b and consider the polynomials

fl (x) = Pn(n-1) f

( Pn-1 )J ^'

Pn(n-l) f

( X ) ^' gl (x) pn

Pn(n-1)g ().pn

We know that all coefficients of f1 as well as all but the leading coefficient of g1 are integral, and the leading coefficient of f1 is 1. Hence every value of f1 has the following property: if (in the reduced form) its denominator is divisible by p, it is divisible by pn. By f1 - g1 the same must hold for

g1. Now g1(1) = (b/pn) + c with an integer c, and by pub the exponent of p in the denominator is strictly less than n, thus it must be 0, that is, pn I b.

5. It is sufficient to prove the statement for polynomials with integral cofficients where both leading cefficients are 1. Let f - g be polynomials with integral coefficients, f (x) = xn + ... , g(x) = bxn + .... Let cn be the largest nth power dividing b. Consider the polynomials

fl (x) =

Cn(n-1) f

C-1

= xn + ... ,

x 1 = b J Cx+ n ... C-1 We have f1 - g1. By step 4, the exponent of any prime in the leading coefficient of g1 must be either 0 or > n, and by the definition of c this coefficient must be 1 or -1. We can exclude the case -1 for even n, since 91(x) = C n(n-1) g

then one domain would be bounded from below and the other from above. Finally, if n is odd, then replacing c by -c we can change a -1 into 1. So if the statement is true under this restriction, then we obtain f1 .^s g1, which yields f g. 6. We can also assume that the coefficient of xn-1 is 0 in both polynomials. Indeed, if f(x) = xn + axn-1 + ...,v g(x) = xn +

bxn-1 +

...,

then the polynomials fi (x) = nnf ( x-7.

- a) - g1(x) = nng (n - al

have this property, and f1 ~ g1 again implies f g. 7. Combining the previous arguments, we find that to solve the problem it is sufficient to prove the following statement:

Statement. If f - g are polynomials of the form f (X)

= xn + axn-2 + ... ,

g(x) = xn +

bxn-2

+ .. .

with integral coefficients, then f Pz g.

Proof. Observe that, since the leading coefficient is 1, the only rational numbers where f and g assume integral values are the integers.

3. SOLUTIONS TO THE PROBLEMS

364

Since f (x + 1) - g (x) = nxn-1 + ... , we have g (x) < f (x + 1) for large x, and similarly we obtain g(x) > f (x - 1). Thus, for a large positive integer

x, the only positive rational y that can satisfy f (y) = g(x) is y = x. If n is even, then there can also be a negative y, and similarly we find that its only possible value is y = -x. This means that one of the equations f (x) = g(x) and f (-x) = g(x) has infinitely many solutions; hence it must be an identity. This concludes the proof.

Problem N.2. Show that

(x2+y2) = (-1)[8]

(mod p)

1<x<y<pzl

for every prime p =_ 3 (mod 4). ([.] is integer part.)

Solution 1. Write p = 4k + 3. Consider all the sums x2 + y2, 1 < x, y < 2k + 1. Let r denote the number of those pairs x, y for which this sum = 1.

First, we show that the number of those pairs for which x2 + y2 - -1 (mod p) is r + 1. Since every quadratic residue has a unique representation

in the form x2, 1 < x < 2k + 1, while every nonresidue has a unique representation in the form -y2, the solutions of x2 + y2 = x2 - (-y2) = 1 count the consecutive numbers in the form (nonresidue, residue) within the sequence 1,2,... , p - 2,p - 1. Similarly, the pairs with x2 + y2 = -1 correspond to pairs of type (residue, nonresidue). Since this sequence starts

with a residue and (recall that p - -1 (mod 4)) ends with a nonresidue, the second case must happen r + 1 times. Next, we prove that the number of solutions of x2 + y2 = a2 (mod p),

1 < x, y < 2k + 1, is r for every a. Indeed, the mapping x - fax1, y - Âąayl, where the signs are chosen so that x, x1i y, y1 are all in [1, 2k+ 1], provides a one-to-one correspondence between solutions of x2 + y2 =- a2

and x2 + yl = 1. Similarly, we obtain that the number of solutions of

x2+y2--1 is r + 1. The above observations mean that these sums x2 + y2 represent every quadratic residue r times and every nonresidue r + 1 times. Since the total number of these pairs is ((p - 1)/2)2 while the number of residues and the number of nonresidues are both equal to (p-1)/2, we find r = (p-3)/4 = k. Now, the product we want to compute is not over all these pairs but only over those with x < y. Consider a quadratic residue a. We know that the total number of solutions of x2 + y2 = a is k. If among them there are

u with x < y, v with x = y, and w with x > y, then u = w by symmetry and v is 0 or 1, hence u = [k/2]. Analogously, for a nonresidue the number of solutions with x < y is [(k + 1)/2]. The product of all quadratic residues is 2)2

12.22

...

((P_ )!)2 2 1

(mod p),

3.6 NUMBER THEORY

365

and the product of nonresidues is

-f

= (-1)

1 /I2

- -1

(()

!12

(mod p).

Hence, the original product is 1[k/2](-1)[

]

= (-1)[ 8 1

Solution 2. Denote this product by P. In the product, all sums of pairs of quadratic residues are multiplied. Let g be a primitive root modp, and put h = g2. The numbers 1, h, h2, ...,h 2k represent every quadratic residue once; thus we have

P-

(hi + hi)

(mod p).

O<i<j<2k

This implies

P fi

(hi-hj)=

O<i<j<2k

(h2i-h2j) fi O<i<j<2k

(mod p).

Here each product is the value of a nonzero Vandermonde determinant. Thus, this equation can be rewritten as

2,...,h 2k)=V(1,h 2,...,h 4k)

(mod p).

Since h2k+1 = 94k+2 = 1 (mod p), the generators of the second Vandermonde determinant are congruent to 1 h2 ,..., h2k h h3 ,... ,

h2k-1.

Thus, the second determinant can be obtained from the first by k+(k-1)+ + 2 + 1 = k(k + 1)/2 transpositions of rows. By the familiar properties of determinants, this means

P = (-1)

(mod p).

2}1

Problem N.3. Let p be a prime and let lk(X, Y) = akx+bky

(k = 1,...,p2),

be homogeneous linear polynomials with integral coefficients. Suppose that for every pair i7) of integers, not both divisible by p, the values lk( , ri), 1 < k < p2, represent every residue class mod p exactly p times.

3. SOLUTIONS TO THE PROBLEMS

366

Prove that the set of pairs {(ak, bk) : 1 < k < p2} is identical mod p with

the set{(m,n):0<m,n<p-1}.

Solution 1. Assume that the statement does not hold. Then there are numbers i j such that ai - aj and bi - bj (every congruence is meant mod p). Consider the number of those triplets (k, x, y), (x, y) # (0, 0) that satisfy lk(x, y) Â° li(x, Y) -

By the assumption, for every fixed pair (x, y) (0, 0), the number of solutions in k is p; thus the total number of these triplets is p(p2 - 1). Now consider the solutions in x, y for a fixed k. If k = i or j, this is an identity, which means 2(p2 - 1) solutions. For any other k it is easy to see that the number of solutions is at least p - 1. This means that the total number of solutions is at least

(p2-2)(p-1)+2(p2-1)=p(p2-1)+p(p-1)>p(p2-1), a contradiction.

Solution 2. It is sufficient to prove that for every u and v there is exactly one k with ak - u, bk - V. The uniformity of representations implies that for every (, rl) 0 (0, 0), we have P

k=1

to get

We multiply both sides by P

2 C

e2pt((ak-u)S+(bk-v),7) = 0, k=1

for

rl) 0 (0, 0). For C - rl - 0, the same sum obviously gives p2. Now

summing these sums for all possible values of l; and 77, including (0, 0), we obtain p2 P-1 p-1

S-

U e'

-v)n) = p2.

r;=0 77=0 k=1

Changing the order of the summations, we get P2 P-1 p-1

S =k=1Er;=077=0 E P2

(:e_

(e_71)) 77=0

=J C

3.6 NUMBER THEORY

367

Now, a typical factor in this product vanishes, unless ak - u (for the first) or bk =_ v (for the second). Thus, the whole product is 0, unless ak - U and bk _= v, in which case it is equal to p2. Since the sum of these products is p2, this case must happen for exactly one value of k. Remarks. 1. The method of the second solution can be applied to prove the following

generalization of the problem: Let p be a prime, r a positive integer, and the lk's homogeneous linear polynomials in n variables of the form

lk (x1, ... , xn) = akx1 + ... + akxn

(k = 1, ... , rpn).

Assume that for every k-tuple (1 i ... , W of integers, not all divisible by p, the values of ( 6 ,- ... , n) represent every residue class mod p exactly

rpn-1 times. Then every n-tuple (ml, ... , mn) is represented mod p among the n-tuples (ai, ... , ak) of coefficients for exactly r values of k. 2. If we relax the requirement of uniform representation to those pairs 77) where neither nor q is divisible by p, then the following can be asserted: Assertion. Let f (m, n) denote the number of those pairs (ak, bk) that satisfy ak = m, bk =_ n. Then the values lk (C 77) will be uniformly disif and only if tributed for all of the p2 - 2p + 1 admissible pairs f (m, n) can be represented as f (m, n) = g(m) + h(n).

Problem N.4. Let p be a prime, n a natural number, and S a set of cardinality pn. Let P be a family of partitions of S into nonempty parts of sizes divisible by p such that the intersection of any two parts that occur in any of the partitions has at most one element. How large can I P I be ?

Solution 1. This maximum is (pn - 1)/(p - 1). Let H be the set to be partitioned, and let C1i . . . , Ck be its partitions into sets whose cardinalities are multiples of p, such that any two classes may have at most one common element. Consider an h E H, and let Ci(h) be the class of Ci that contains h. By the assumptions, the sets C1(h)\{h},..., Ck(h)\{h}

are pairwise disjoint, and each has at least p - 1 elements. Consequently,

wehavek(p-1)<p'-l, that is,k<(pn-1)(p-1).

Now we prove the corresponding lower estimate. Consider the n-dimensional projective space Pn over a finite field K of p elements, a hyperplane 0' (pn+1-1)/(p-1) in it, and the affine space P, = P,, \o,. Observe that Pn has

points, o, has (pn - 1)/(p - 1) and P,P has pn. For an arbitrary point P E a, consider those lines of Pn that contain P but do not lie in or. If we omit the point P from each such line, the remaining affine lines (as sets of points) form a partition Cp of the affine space P' into sets of cardinality p. The number of these partitions is the same as the number of points of or, that is, (pn - 1)/(p - 1).

3. SOLUTIONS TO THE PROBLEMS

368

Solution 2. We keep the upper estimation from the first proof, and present a different construction to show the lower bound. For H we take the set of all n digit numbers in base p. This set has pn elements. The partitions will be given in forms of integer-valued functions;

f (x) will mean the number of the class containing a given x E H. The functions will be indexed by pairs (j, a), where j = 0, . , n - 1, and for a given value of j, the possible values of a are a = 0, . . , pj - 1. The number . .

of possibilities for a is pj; thus the total number of these functions is 1+p+p2+...+pn-1

= pn

p-1

as wanted.

We define the function fja(x) as follows. Let the representations of x and a to base p be

+... + bn-lpn-l ' a = ao +alp+ ... + aj_1Pj -1, Let [m] denote the (smallest nonnegative) residue of an integer m mod p. x = So + Slp

Now we put C [S1 + a1Sj]p + ... +

fja(X) = C[So +

+ Sj+1P' + ... +

[Sj_l +

aj-1Sj]Lj-1

tn-lpn-2.

In the case j = 0, a = 0, we interpret this as foo(x) = S1 + S2P + ... +

Sn-1pn-2.

We show that every class contains p elements. Indeed, given the value of

fja(x)

=Y=770+...+17n-2p

n-2

we have G = 71k-1 for k = j + 1,...,n, we can choose the value of j that will give p possibilities, and after fixing j we can uniquely determine 0i ... , j from the congruences z + c

j = q:

(mod p)

(i = 0,...,j - 1).

Next, we have to show that the intersection of two classes has at most one element, that is, the system of equations

fj'a'(X) = y

fja(X) = Y,

has at most one solution. Assume first that j # j', say j' < j. Then we obtain 77'. from the second equation, and we already know that fja (x) and j determine x uniquely.

Finally, consider the case j = j'. We must have a # a', say ak # ak for some digit k < j - 1. Then the congruences Sk + Qkej = 77k

(mod p),

(mod p)

77k

determine j, which together with faj (x) determines x uniquely.

3.6 NUMBER THEORY

369

Problem N.S. Let f be a complex-valued, completely multiplicative, arithmetical function. Assume that there exists an infinite increasing sequence Nk of natural numbers such that f (n) = Ak

provided Nk < n < Nk + 4 Nk.

Prove that f is identically 1. Solution. We show a slightly stronger statement, where the 4 is replaced by 2 + e. Assume that f is (nonzero and) constant on the intervals Ik = [Nk, Nk + Mk], where Mk = (2 + e) Nk.

First, we prove that f (n) $ 0 for any n. Indeed, if Mk > n, then nx E Ik for some x; hence f (n) f (x) = f (nx) $ 0. Next, we use an interval

of constancy to create another. Let f be constant on I = [N, N + M]. If for an n we can find an integer x such that nx, (n + 1)x E I, then f(n)f(x) = f(nx) = f((n + 1)x) = f(n + 1)f(x), hence f(n) = f(n + 1). Now this condition can be reformulated as

N<x<N+M n - - n+1 Such an integer x can be found if

N+M-N>1 n+1

or, after rearranging, n2+n(1-M)+N < 0. This means that n lies between the roots of the corresponding quadratic equation, that is, (M - 1)2 > 4N (which shows that our method does not work with M = (2 - e)vrN-) and

M-1- (M-1)2-4N <n< M-1+ (M-1)2-4N 2 - 2 If M = (2+E)v/N-, then the above interval includes an interval of the form I' = [c1M, c2M], where 0 < cl < c2 are constants depending on E. Now we repeat this argument for the interval P. The result is an interval

I" = [c3i e4M] of constancy. As I runs over all the intervals Ik, these intervals I" cover the whole half-line [c3i oo]; thus f (n) is constant for n > e3. Now take an arbitrary positive integer m, and select an n > c3. We have f (n) = f (mn), hence f (m) = 1 as asserted.

Remark. The statement will not hold if the 4 Nk is reduced to exp (c log Nk flog log log Nk) with a suitably small positive c.

To see this, consider a sequence Ik = [Nk, Nk + Mk] of intervals, Nk >

Mk > Nk_1 + Mk_1i with the property that every number n E Ik has a prime factor p > Mk. Choose the numbers Ak $ 0 arbitrarily. We claim that there is a completely multiplicative function f that is identically Ak on Ik; if not, every Ak is 1, and then our function will not be identically 1.

3. SOLUTIONS TO THE PROBLEMS

370

We construct our function recursively. Assume that f (p) is fixed for every prime p < Nk_1 + Mk_1i we define f (p) up to Nk + Mk. For every n E Ik, let pn > Mk be the largest prime divisor of n. These primes are distinct, since for n n' we have (n, m) < In' - nj < Mk. Choose f (p) arbitrarily for every prime that does not occur among the pn's, and then set f (pn) to achieve f (n) = Ak. Now we have to find such a sequence of intervals. Assume that I1, .. . Ik_I are given. Take a large N, say N > expNk_1i we try to find an Ik in [N/2, N].

Let R(x, y) denote the number of those integers n < x whose prime factors are all less than y. If M is such that M(R(N, M) + 1) < N/2, then the interval [N/2, N] contains a subinterval of length M that is free of these numbers. This interval can serve as our Ik. The feasibility of taking M = exp(c log Nk log log log Nk) follows from Rankin's inequality:

R(x, y) < x exp

(

- logloglogy y log x + O(log log y) J

Problem N.6. If c is a positive integer and p is an odd prime, what is the smallest residue (in absolute value) of P-1 2.

C2n1

n=O

(mod p)?

Solution. Let q = (p-1)/2. With this notation, we have 2j+1 - -2(q-j) for every j. Consequently, we have 1 . 3 ... n!2

2n n!

(2n - 1)

(2n)!

n&2 _ 2n.1.3.....(2n-1) n!

_ n!

= (-4)n l

Hence,

(q) (-4c)n = (1 -

(2:) cn = 0

This is congruent to 0 if 1 - 4c is divisible by p. Otherwise, it is 1 if 1 - 4c is a quadratic residue mod p and -1 if it is a nonresidue.

3.6 NUMBER THEORY

371

Problem N.7. Find a constant c > 1 with the property that, for arbitrary positive integers n and k such that n > ck, the number of distinct prime factors of (k) is at least k. Solution. Write t = [1, 2, ... , n]. We shall show that the number of prime factors of (k) is at least k for n > t + k. Take a prime p < k, and let p9 < k < pk+1 The exponent of p in the decomposition of k! is E8=1 [k/pi], and in t it is s. Among the numbers n, ... , n - k+ 1, there are at least [k/pi] multiples of pi. For i < s, these also enter the decomposition of the corresponding (n - j, t); thus the exponent of p in the product (n, t) (n - 1 , t) ... (n - k + 1, t) is at least as high as in k!. Since this holds for every p, we conclude that k!I (n, t)

(n - 1, t) ... (n - k + 1, t).

This implies that k-1

n-i

ri (n - i, t)

(n).

The factors on the left side are pairwise coprime, since (n - i, n - j) I (i - j) It for i # j, and if n > t + k, they are all larger than 1. Taking a prime factor of each, we find k different prime factors of (k).

Since t = exp(k + o(k)) by the prime number theorem, every c > e satisfies the requirement of the problem for large k. If we want to exhibit a concrete value of c that works for every k, we can argue as follows. Applying Chebyshev's estimate,

Hp<4x, p<x

we obtain

fl p3< fl k fl pa<k<p-+'

p<kvk-14k

p<v"k- f<p<k

Therefore, k

t + k < k-k4 k = (4k1/-k ) < 9k, since the function x1lvf'- assumes its maximum at e2, and its value is e2/e < 2.1.

Remark. The value c = 9 we gave can be further reduced with a little extra effort; one of the contestants gave c = 4.3. It seems to be difficult to decide the improvability of c = e + E for large k. If the "prime k-tuple conjecture" is true, then there are arbitrary large values of n for which (k) has exactly k prime factors.

3. SOLUTIONS TO THE PROBLEMS

372

Problem N.8. Let f (n) be the largest integer k such that nk divides n!, and let F(n) = max2<m,<n f (n). Show that log n 1oo F(n) nloglogn - 1

Solution. First, we find an upper estimate for f (n). Let n = fk 1 p'i be the prime factorization of n. Since of (n) In!, we have n

aif(n) < E

7- 1

that is, ai log pi <

n log pi f(n)pi-1'

for all i. Summing these inequalities, we obtain log n =

n E log pi

ai log pi <-

f(n)

pi-1

Let q, < q2 < ... be the sequence of all primes. Since (logp)/(p - 1) is decreasing, we have

E log Pi < r. loggi = logk+O(1).

i=1pi-1

Combining the last two inequalities, we get

f(n) <

(1+o(1))nlogk

log n

On the other hand,

n=p"...pk'` >g1...gk>2k Consequently, k << log n, log k < log log n+O(1), and the previous estimate

of f(n) yields f (n) < (1

+o(1))nloglogn

log n

as wanted. Now we construct a number m < n with a large value of f (m). Define m1 by (m1+1)! < n < (m1+2)!. We have m1 - (log n)/(log log n)

as n --+ oc. Let p be the largest prime up to (n/m1!)1/3and put m = p3m1!. We have m < n obviously, and, since n/m1! > m1

p - (n/m1!)1/3, that is, m - n.

oc, we have

3.6 NUMBER THEORY

373

Since (mi!)m/ml gym!, the exponent of any prime other than p is at least m/ml times as high in m! as in m. We have to compare the exponents of p. The exponent of pin ml! is F_Â°__1 [ml/pi] < ml/(p - 1), hence in m

it is less than ml/(p - 1) + 3. On the other hand, the exponent of p in m! is [m/pi] > [m/p] > (m/p) - 1. Therefore, the quotient of these exponents is at least 1

p +3m1' because p -> oc, but

p:5 (n/mi!)1/3 < ((m1 + 1)(ml + 2)) 1/3 = o(m). Consequently, we have (1+o(1))nllogloogn

F(n) > f(m) > (1+o(1))ml = (1+o(1))m1 =

as asserted.

Remark. It can be proved that

F(n) =

n(log log n - log log log n) log n

(n log n

Problem N.9. Prove that a necessary and sufficient condition for the

existence of a set S C {1,. .. , n} with the property that the integers 0, 1, ... , n - 1 all have an odd number of representations in the form x y, x,y ES, is that (2n -1) has a multiple of the form 2 -1.

Solution. Let S be a set with this property, and write xa-1

1(x) = aES

g(x)=E xn-a aES

which we regard as polynomials over GF(2). Then we have f (X)

= xn-1g(1),

(1)

and 2n-2

f(x)g(x) _

1 = 1 + x + x2 +

j=0

x2n-2,

(2)

a,bES a-1+n-b=j

by our assumptions on S. Conversely, if f and g are polynomials over GF(2) that satisfy (1) and (2), then the set S = {a : the coefficient of xa-1 in f is 1} has the desired property.

3. SOLUTIONS TO THE PROBLEMS

374

Now put

(3) 1 + x + x2 + + x2n-2 = p1(x)...p1 (x), where the factors pi, ... , pj are irreducible over GF(2). With a suitable choice of the suffices, we have f =P1 ... Pr,

9 = pr+1...pt

by (2).

Observe that the roots of 1 + x + x2 + + roots of unity except 1. Furthermore, we have

x2n-2

are the (2n - 1)th

(1+x+x2+...-x2n-2)'=1+x2+...+x2n-4

and

and hence this polynomial has no multiple roots. Since the roots of f are the reciprocals of the roots of g by (1), no pi can have two roots that are reciprocals of each other. Conversely, if no pi has reciprocal roots, then the polynomials p1i ... , pi can be coupled so that the roots of any pi are the reciprocals of the roots of its mate. We now multiply one element from each pair to get the polynomial f and the others to get g; these obviously satisfy (1) and (2). Thus, such a set S exists if and only if for every root a a and 1/a belong to different irreducible factors. of l+x+x2+ We construct this decomposition (3). +x2n-2,

Lemma. Consider the permutation of the roots of the equation x2n-1 1 = 0 given by the operation of squaring. Let C1, ... , C1 be the cycles of this permutation. The irreducible factors of x2n-1 - 1 are the polynomials

i=1 ...,1.

11 SEC;

We prove this lemma at the end of the solution; see also E. R. Berlekamp, Algebraic Coding Theory, McGraw-Hill, New York, 1968, Chapter 6. Hence, our condition for the existence of the set S is satisfied if and only

if the roots a and 1/a cannot be tranformed into each other by repeated squarings for any a # 1, that is, a2k+1 54 1

for all k > 0 and (2n - 1)th roots of unity a as for all k. Since

(x2n-1 - 1, x2k+1

1. This can be reformulated

- 1) = x - 1

(xu-1,x"-1)=x(u,")-1,

the set S exists if and only if

(2n-1,2k+1)=1

(4)

3.6 NUMBER THEORY

375

for all k. Now let d be the smallest exponent for which 2n - 112d - 1. If d is even, then 2n - 11(2d/2 -1)(2

d/2

+ 1),

but 2n - 1X(2d/2 - 1), and thus (2n - 1, 2d/2 + 1) > 1, which contradicts (4). Conversely, if d is odd, then we have

(2n-1,2k+1)=(2d-1,2k+1)=1. Consequently, S exists if and only if d is odd, which is clearly equivalent to the condition given in the problem. Proof of the lemma. Let N akxk

O(x) _

k=0

be a polynomial over any field of characteristic 2. We have N

(O(x))2 =

akx2k k=0

Thus, the equation (4'(x))2 = O(x2)

holds identically if and only if ak = ak for all k, that is, every ak is 0 or 1. On the other hand, the above equality is clearly equivalent to the assumption that for every root a of 0, a2 is a root as well.

Problem N.10. Prove that the set of rational-valued, multiplicative arithmetical functions and the set of complex rational-valued, multiplicative arithmetical functions form isomorphic groups with the convolution operation f o g defined by

f(d)9(d)

(f o9)(n) _ din

(We call a complex number complex rational, if its real and imaginary parts

are both rational.)

Solution. Let Q be either the field of rational numbers or the field of complex rational numbers, and let GQ denote the group of Q-valued multiplicative functions. It is well known that GQ is an Abelian group. Now we prove that for every natural number n and every f E GQ,the equation

gogo...o9=9

n times

is solvable and the solution is unique.

3. SOLUTIONS TO THE PROBLEMS

376

Indeed, we have g(1) = 1 by multiplicativity. For every prime power pk, we have

f (PI) _ E

g (pk 1) ... g

(pk )

g(pk1) ... g(pkn) + ng(pk)

k; <n

This yields a recursion for f (pk), which shows the unicity. To show the existence, consider the function g whose values are defined by the above recursion at prime powers and that is extended multiplicatively to the other o g is a multiplicative function that coincides with f at numbers. go prime powers; thus they must be identical. Thus, both groups are divisible, torsion-free Abelian groups. By the fundamental theorem of divisible Abelian groups, both are isomorphic to some discrete direct power of the additive group of rational numbers. Since we can prescribe the values of a multiplicative function at prime powers arbitrarily, we conclude that the cardinality of GQ is 2xÂ° for both choices of Q. This implies that the direct power must have 2xÂ° factors in both cases. This proves the isomorphy of the two groups.

Problem N.11. Let H denote the set of those natural numbers for which -r(n) divides n, where r(n) is the number of divisors of n. Show that (a) n! E H for all sufficiently, large n, (b) H has density 0.

Solution. (a) We show that r(n!)In! for all n # 3, 5. We have 00

n! = 11 pap,

ap = E[n/p'],

p<h

i=1

and consequently,

-r(n!) = 11(ap + 1). p<n

To prove rr(n!) In!, it is sufficient to find, for every prime p

n, a

natural number h(p) < n such that ap + 11h(p) and h(p)

h(q)

whenever p # q. If p < V/n, put h(p) = ap + 1. We have 00

h(p) = 1 + ap < 1 +

[n/21] < 1 +

n/2' = 1 + n,

hence h(p) < n. Also, if p < q < \, then

n_n=(q-p)n> n >1; p

3.6 NUMBER THEORY

377

consequently, [n/p] > [n/q]. Since also [p/p'] > [n/q'], for all j, we can infer that h(p) > h(q). For the primes < p < n, we define h(p) by recursion. Assume that h(q) is already defined for every prime

q < p. We need to find a multiple of ap + 1 that is not among the already distributed numbers h(q), q < p. Observe that [n/pi] = 0, for j > 2; thus ap = [n/p]. Hence, the number of multiples of ap + 1 up to n is

> n - [n/p] > n - (n/p)

1 + [n/p] - 1 + (n/p)

1 + [n/p] n

Lap + 1

n+p>p-1

=(p-1)

with strict inequality unless p = n. The number of already occupied

values is 7r(p) - 1 < (p - 1)/2, with strict inequality unless p = 3, 5, or 7. Thus there is a free number for h(p) except possibly in the cases n = p = 3, 5, 7. In these cases, the divisibility T(n!)In! holds for n = 7 and does not hold for n = 3 or 5. (b) We split H into two parts. Choose a parameter K > 1. We put those numbers in which the exponent of every prime is < K into H1 and the rest into H2. Consider first an n E H1. If n = f,-1 a < K, then T(n) = fl(aj + 1) consists exclusively of primes < K, and by T(n)In, the exponent of every prime is less than K. Consequently,

T(n) <

< KK

On the other hand, we have T(n) > 28, and these inequalities together imply s < [K21og2 K] = r.

It is well known that the sequence of numbers that have at most r prime factors has density 0, hence so does H1. Every element of H2 is divisible by the Kth power of a prime. Hence, the number of elements of H2 up to x is at most 00

>[- -]< x> i-K < x Jf i t-k dt = x E[ K-1 1

i=2

Hence, the density of H is at most 1/(K - 1). Since K was arbitrary, H must have density 0.

Remark. Let H(x) denote the number of elements of H up to x. It can be shown that x(logx)-(1/2+E) << H(x) << x(log x)-1/2,

for x > xo(E). We outline a proof of these inequalities.

378

3. SOLUTIONS TO THE PROBLEMS

We start with the upper estimate. Write n in the form n = 21p1p2 ... pkm2,

where pi, ... , pk are odd primes and m is an odd integer. The number of possibilities for m is < Vx-, for lit is O(log x). Hence, the

number of those n for which pl

pk <

x1/3

is O(x5/6log x) _

o(x(logx)-1/2). In what follows, we assume that P1 ... pk > x113 The exponent of each pj in n is odd, hence 2k Jr(n), so k < 1 if n E H. For fixed 1 and m we have x P1 ... pk

y = 21m2 ,

and y > x1/3 by the previous assumption. Now by a classical theorem of Hardy and Ramanujan, the number of integers < y that are products of k distinct primes is <<

(c+loglogy)k-1

(k - 1)!

logy

with an absolute constant c. Since log x > logy > (1/3) log x in our case, this is <<

y (c + log log x)k log x (k - 1)!

We have to sum this for all possible values of l and m. Write 1 = k + j; we know that j > 0. With this substitution, our sum is x

E 2-k-9m-2 log x

(c + log log X) k-1

(k - 1)!

Here the sum over m and j contributes only a constant factor, while the sum over k is just the power series expansion of log log x = c (log x)1/2. exp c + 2

This concludes the proof of the upper estimate. The lower estimate can be obtained by considering numbers of the form n = 2q-1gP1...pk,

where q, pi, ... , Pk are distinct odd primes. This number has r(n) = 2k+lq divisors, thus n E H if k < q - 2. For a fixed q, q < (1 - e) log log x, the number of such n < x is approximately

q-12-q x log x

(log log x)g-3

(q - 3)!

The lower estimate follows by taking q -

log log X. 2

3.6 NUMBER THEORY

379

Problem N.12. Prove that if a, x, y are p-adic integers different from 0 and pox, paixy, then

1 (1+x)Y-1

log(1+x)

(mod a).

Solution. Every p-adic number a can be uniquely represented in the form

a = P'(')E, where a is a unit and v(a) is an integer; p-adic integers are characterized by v(a) > 0. Hence, a divisibility al,l is equivalent to v(a) < v(0). In particular, the condition palxy means 1 + v(a) < v(x) + v(y), and the statement to be proved means v(A) > v(a), where

A- 1y (1+x)y-1 x

log(1+x) x

By the previous observation, it is sufficient to show that v(A) > v(x) + V(Y) - 1. The series

(1+x)y=1+Il)x+...( )xn+... and

Xare convergent since pi x, Q) is a p-adic integer, and n - v(n) -+ oc. Substituting these expansions into A, we obtain

A=I:00Bn, n=2

where

Bn = (1 (y) y

{(y

1)(y

(-1)n-1

n) xn-1 /J

xn -1 2)...(y

1))

(- 1)n-1(n

1)!}

Here the first factor is a polynomial in y whose constant term vanishes;

thus it is a multiple of y. Consequently, Bn = cyxn-1/n! with a p-adic integer c, thus

v(Bn) > v(y) + (n - 1)v(x) - v(n!). To prove that v(A) > v(x) + v(y) - 1, it is sufficient to show that v(Bn) > v(x) + v(y) - 1 for every n; by the previous inequality, this would follow from v(n!) < (n - 2)v(x) + 1. Since v(x) > 1 by assumption, we need only to show v(n!) < n - 1. But the exponent of p in n! is v(n!)

째째

nl P3J

<j=1i'= 째째 n

n P-1

< n.

Consequently, v(n!) < n - 1 since it must be an integer. This concludes the proof.

3. SOLUTIONS TO THE PROBLEMS

380

Problem N.13. Let 1 < a1 < a2 < < an < x be positive integers such that En 1/ai < 1. Let y denote the number of positive integers smaller than ix not divisible by any of the ai. Prove that cx

Y > logx

with a suitable positive constant c (independent of x and the numbers ai).

Solution. The number of multiples of aj up to x is [x/aj]. Hence, the total number of multiples is

[X1 < a<x. aj We have to improve on this trivial bound. Choose an x0 such that the number of primes between x/2 and x is at

least x/(3logx) for all x > x0; such an x0 exists by the prime number theorem. For x < x0, we use the trivial estimate y > 1. For x > x0, we distinguish two cases.

If the number of aj > x/2 is less than x/(6logx), then there are at least x/(6logx) primes x/2 x/(6logx). If the number of aj > x/2 is at least x/(6logx), then we have

aj <x/2

a <1- aj>x/2 a <1-x6logx

6logx.

Hence, the number w of integers up to x/2 that are divisible by some aj satisfies w

<x/2 Lx/2J :5ai E

This implies

x 2

-w>

121ogx

12logx

Problem N.14. Let T E SL(n, Z), let G be a nonsingular n x n matrix with integer elements, and put S = G-'TG. Prove that there is a natural number k such that Sk E SL (n, Z).

Solution. Let d be the absolute value of the determinant of G, a positive integer by the assumption. Every element of G-1 is a fraction with denominator d. Since we have Sk = G-1T kG = G-1(T k - I)G + I,

3.6 NUMBER THEORY

381

it is sufficient to prove that there is a positive integer k for which every element of T'k - I is a multiple of d (the determinant of Sk is automatically 1).

By the box principle, there are two matrices among To, T', ... , TC2 say T' and T', 1 > m, such that every element of TI is congruent to the corresponding element of T. This means that every element of Tl -T' = Tm(Tl-m - I) is a multiple of d. This property is preserved if we multiply it by the matrix T-m, which has integer elements by T E SL(n, Z).

Problem N.15. Let

f(n) = E pa p1n

pÂ°<n <p"+1

Prove that

lim sup

n-oo

)

log log n = 1. n log n

Solution. It is well known that the number w(n) of different prime divisors of n is at most (1 +o(1)) (log n)/(log log n). Since obviously f (n) < nw(n), we conclude immediately that f (n) log log n < 1. lim sup n-oo nlogn To prove the other inequality, we select a parameter k and try to compose

an n from primes below k. The number of these primes is - k/ log k by the prime number theorem. Now take another number m, which we shall later specify in terms of k. Every prime p < k has a power between m and km. We cover the interval [m, km] with intervals of the type [m(1 + c)'- 1, m(1 + E)t], where i = 1, ... , [c log k], c depends only on e. For a

suitable i, there are at least l = [clk/(logk)2] primes p < k that have a power in this interval. We select 1 of them; let P be the product of these 1 primes. If P < Em, then there is a multiple of P in the interval [m(1 + E)t, m(1 + e)i+1]; let n be such a multiple. We have f(n) > lm(1 + E)i-1 > ln(1 - E)2.

We have to estimate n in terms of 1. We have P < k' anyway; thus the choice m = k'+1 guarantees P 1/E. Further, we have n < m(1 + E)t+i < km(1 + E)2 < k1+2,

hence l > (log n) /(log k) - 2. Also n < kk from the above inequality. Thus k >> (log n) /log log n, and we obtain log n 3 n log n l > (1 - E) f (n) > (1 - e) log log n'

log log n'

for large k.

Remark. It can be shown that max f (n) n<x

n log n log log n

382

3. SOLUTIONS TO THE PROBLEMS

Problem N.16. Let a and b be positive integers such that when dividing them by any prime p, the remainder of a is always less than or equal to the remainder of b. Prove that a = b.

Solution. By taking a p > max(a, b), we immediately see that a < b. Assume that a < b, and write

k = min(a, b - a). By a theorem of Sylvester and Schur, there is a prime p > k that divides (b). We show that this prime contradicts the assumption. k We know plb(b - 1)...(b - k + 1),

and thus the residue of b is a number r < k - 1. Now if a < b/2, then k = a < p, and thus the residue of a is a = k > r, as claimed. If a > b/2, then we have k = b - a. Write b = qp + r. We have

a - (q - 1)p = (b - k) - (q - 1)p = p + r - k > r, while p + r - k < p - 1, and thus this is the residue of a and it was larger than r.

Problem N.17. Call a subset S of the set {1, ... , n} exceptional if any pair of distinct elements of S are coprime. Consider an exceptional set with a maximal sum of elements (among all exceptional sets for a fixed n). Prove that if n is sufficiently large, then each element of S has at most two distinct prime divisors.

Solution. We prove the following slightly more general statement.

Statement. We call a subset S of the set {1, ... , n} k-exceptional if any k distinct elements of S are coprime. Consider a k-exceptional set with a maximal sum of elements (among all exceptional sets for a fixed n). If n is sufficiently large, then each element of S has at most two distinct prime divisors.

By maximality, every prime p < n must have a multiple in S. We divide the primes p < n into two classes: into P we put those for which there is an m > 1 such that pm E S, and into Q those for which the only multiple of p in S is p itself. We can estimate the number of p E P, p > x, as follows. For every

such p, take an m > 1, pm E S, and then a prime q1m. These primes satisfy q < n/x and any prime is represented at most k - 1 times among them; thus the number of our p's is at most (k - 1)7r(n/x). Consequently, if 1 < x < y < n, then the number of primes q E Q, x < q < y, is at least 7r(y) - -7r(x) - (k - 1)7rn. x

(1)

3.6 NUMBER THEORY

383

Now assume that there is a t E S that has at least three distinct prime divisors, say t = u"vQw7r, where u < v < w are primes, and a, /3, ry, and r are positive integers. Our plan is to find two suitable primes p, q E Q and replace t, p, q by up and vq. We must have p < n/u, q < n/v; thus the loss is

n n n n 11 t+p+q<n+u+v <n+2+3 = sn.

If we can achieve up > cn and vq > cn, where c = 11/12, then the inclusion of up and vq offsets this loss. If there are at least two elements of Q in both (cn/u, n/u] and (cn/v, n/v], then this is possible. (We need two elements to guarantee that p q.) By (1), for this it is sufficient that

7ru-7r-> (k-1)7ru+2 u c U

and

n r--7r->(k-1)-7rv+2. V V c

By the prime number theorem, these inequalities hold if v b f , so that the previous procedure fails, we argue as follows. We have

n n U vw<-<-<-. 1

We try to find a prime p E Q and replace t and p by u3 p and vw with a suitable exponent j. The new number is admissible if p < n/ui. The loss is at most p + n, while the sum of the new terms is pug + vw > pug + 62n. Thus the process yields a profit if p(ui - 1) > (1 - 62 )n, that is, we need to find a p E Q satisfying

1-b2 n<p<-. n

u3 - 1

By (1), such a p exists if 7r ( n U3

)

(

bl n) > (k - 1)7r

U3 -

(1,- b2)

(2)

Recall that u < 1/b2, hence there is a power of u in the interval (b-4, b-'] With this as uj, the left side of (2) is of order n/ log n, while the right side stays bounded, so (2) will hold for every sufficiently large n.

Problem N.18. (a) Prove that for every natural number k, there are positive integers al < a2 < < ak such that ai - aj divides ai for all 1 < i, j < k,

i# j.

3. SOLUTIONS TO THE PROBLEMS

384

(b) Show that there is an absolute constant C > 0 such that a1 > kck for every sequence al, ... , ak of numbers that satisfy the above divisibility condition.

Solution. (a) For k = 1, we can put a1 = 1. Now we show the inductive step from

k to k + 1. Assume that 0 < a1 < < ak is such a set. Write b = nk1 ai. Then the numbers b < b + al < ... < b + ak form a suitable collection of k + 1 numbers. Indeed, we have

(b + ai) - (b + aj) = ai - aj jai lb + ai, for 1 < i, j < k, i # j, and also (b + ai) - b = ai I b.

(b) Consider any prime p < k. If ai - aj (mod p), then ai - 0 (mod p) by the divisibility condition. Consequently, there are at most p - 1 numbers ai that are not divisible by p. Consider now the divisible ones. Dividing them by p, we again get an admissible system. Thus there can be at most p - 1 numbers not divisible by p among them.

Repeating this argument, we find that the exponent of p in A =

]lk1 ai is at least

k (k - (p - 1)) + (k - 2(p - 1)) + ... + (k - [

p-1

](p - 1)1 1

> - p21(1 + Lp-l]I)3p

if p < -vfk-. Consequently, we have (with c = 1/3)

A > [ pck2/P, p<v and hence

ak = max aj > Al/k > 11 pck/p p<Vk-

!2g P > exp c'k log k = kc'k

= exp ck

p<f

for k > 4. From the relation ak - a1 jai, we infer a1 > ak/2 > 2, hence

a1 =

ai >

2a1 >

ak > kck

with C = c'/2. We have proved the inequality a1 > kck for k > 4. For k = 3 it is easy to see that the minimal possible value of a1 is 2, hence it holds if C < (log 2)/(3log 3). For k = 1 and 2, it fails by the examples {1} and {1, 2}.

3.6 NUMBER THEORY

385

Problem N.19. Let n1 < n2 < ... be an infinite sequence of natural tends to infinity monotone increasingly. Prove numbers such that that Ek_1 1/nk is irrational. Show that this statement is best possible in a sense by giving, for every c > 0, an example of a sequence n1 < n2 < ... k c for all k but >k 1 1/nk is rational. such that

Solution. Assume indirectly that E 1/nk = p/q, where p and q are positive integers. Then, for arbitrary k, we have Â°Â°

z-1 ni

Multiplying both sides of this equality by qnl ... nk, we see that 00

1ni

qnl ... nk i=k+1

is a positive integer for all k. Now we show that cc

n1 ... nk

1 --> 0

(k -' oo),

i=k+1 ni

and this contradiction will disprove the indirect assumption. By the monotonicity condition, we have 2-(k+1) 2-1 2-k 2-(k-1) > ... > n1 > nk > nk-1 nk+1

which yields nk+1+...+2

n1 ... nk <

On the other hand, by applying the inequalities 2-(k+2) 2-(k+3) 2-(k+1) < ... < nk+3 nk+1 < nk+2

and obvious estimations, we obtain 1

nk+1

nk+2

nk+3

+...<

nk+1

+ n3 nk+1 + n2k+1 k+1 1

nk+1-1 These two estimations together imply k

= nk+1 _2-(k+1) n < nk+i nl...nk ) (nk+l nk+1 - 1 nk+1 - 1 i=k+1 i

2 .

3. SOLUTIONS TO THE PROBLEMS

386

Here the first term tends to 1 and the second to 0 as k -4 oo; thus the whole expression tends to 0. To solve the second part of the problem, we are going to construct a sequence such that

>c (k=1,2.... )and E 1

is rational.

k=1 nk

Take an arbitrary integer n1 > c2 + 1, and let nk+1 = nk - nk + 1 for k = 1, 2, .... The definition implies

nk+1 - 1 > (nk - 1)2 > ... > (n1 - 1)2k > c2k}1. Consequently,

nk-k

> c for all k. On the other hand, an induction shows

that k

n1 - 1

- nk(nk - 1).

Thus, the sum of the series is the rational number 1/(n1 - 1).

Problem N.20. Prove that for every positive number K, there are infinitely many positive integers m and N such that there are at least KN/ log N primes among the integers m + 1, m + 4, ... , m + N2. Solution. The basic idea of the proof is to average for several values of m while excluding divisibility by small primes. Let Q = 3 5 ... pl, the product of the first 1 odd primes, and let c

be a natural number < Q such that -c is a quadratic nonresidue for the moduli 3, 5, ... , pl; such a c exists by the Chinese remainder theorem. We try to find m in the form m = c + kQ, where 1 < k < N2.

Consider an 1 < i < N; by the choice of c, we have (i2 + c, Q) = 1. By the prime number theorem for arithmetical progressions, the number of primes among the first N2 elements of this progression is asymptotically equal to N2Q 1 O(Q) log N2Q' thus for sufficiently large N (for a fixed value of Q), it exceeds

30(Q) log N'

Thus, the total number of primes (counted with multiplicity) among the integers

i2+c+kQ, is at least

1<i<N, 1<k<N2 Q

3Â˘(Q) log N

3.6 NUMBER THEORY

387

Consequently, for a suitable value of m = c + kQ, there are at least

30(Q) log N

primes among the integes m + 1, m + 4, ... , m + N2. Since i

O(Q)

j['1Pi

> 1

and the sum E 1/p is divergent, we can select Q so that Q/30(Q) > K. Then the previous arguments yield that for every sufficently large N, the integer m can be chosen so that there are more than K K. (N/ log N) primes in the sequence m + 1, m + 4, ... , m + N2.

3. SOLUTIONS TO THE PROBLEMS

388

3.7 OPERATORS Problem 0.1. Let a, bo, b1, ... , b,i_1 be complex numbers, A a complex square matrix of order p, and E the unit matrix of order p. Assuming that the eigenvalues of A are given, determine the eigenvalues of the matrix

boE

b1A

b2A2

...

abn_1An-1

boE

biA

... bn-2An-2

abn_2An-2

abn-iAn-1 b0E

\ab1A

ab3A3

ab2A2

bn_1An-1

...

bn_3An-3

...

boE

Solution. We show that if the eigenvalues of A are a1 i ... , AP then the eigenvalues of B will be the numbers

0(akAi)

(k = 1,...,n; j = 1,...,p),

where a1 i ... , an denote the roots of the equation Zn - a = 0, and

O(x) = bo + b1x + ... +

bn_1xn-1.

In fact, if AO, A,,-, An_1 are quadratic matrices of order p having complex entries, a is an arbitrary complex number and

... An-1

aAn_1

...

An-2

aA1

aA2

aA3

...

f Ao

then

det C = det M(a1) ... det M(an),

(1)

where

M(x) = Ao + A1x + ... + For a = 0 the assertion of this theorem is trivial. For a # 0, the validity of the theorem follows from the simple fact that M(a1) ... (0) An-lxn-1.

C=W (0)

... M(an)

3.7 OPERATORS

389

where W is the Kronecker product with E of the Vandermonde matrix built from the roots of the equation z' = a, that is,

f E

a1E

... anE

W =V®E= n-lE ... an-1E/ n 1

Apply (1) to the matrix B - AE (where £ is the unit matrix of order np) :

det(B - A£) = 11 det (q(akA) - AE). k=1

On the other hand it is well known that the eigenvalues of the matrix

q(akA) = b0E + blakA + ... +

bn_1ak-1An-1

are

0(akAl), ... W(akAp)' This proves the assertion.

Remark. Another group of solutions is based on the theorem stated, but not proved, by one participant in the following general form:

Theorem. If the eigenvalues of the quadratic matrix A of order p are (i, j = 1, ... , n) are regular functions on the disc Izi < A, then the eigenvalues of the matrix

1 i ... , gyp, further maxi JA2 I = A, and fib (z) f11(A)

... fin (A)

fnl(f1)

fnn(A)

are given by the eigenvalues of the matrices f11(Ai)

fln(Ai)

(i = 1, ... , p) fn1(Ai)

...

fnn(Ai)

One participant proved this theorem in the less general case where the functions fib (z) are polynomials; this proof can be applied in the more general case as well.

3. SOLUTIONS TO THE PROBLEMS

390

Problem 0.2. Let U be an n x n orthogonal matrix. Prove that for any n x n matrix A, the matrices Am

U-jAU'

j=o

converge entrywise as m -> oo.

Solution 1. For an arbitrary n x n matrix A, let IIAII denote the norm of A, that is, IIAII = sup IIAxII, IIxII=1

where x varies in the n-dimensional Euclidean space. If U is orthogonal, then U-1 is also orthogonal, and IIUAII = IIAUII = IIAII Entrywise convergence and convergence in norm of a sequence of matrices are equivalent. Thus, by the Bolzano-Weierstrass theorem, from a norm-convergent sequence of matrices it is possible to select a normconvergent subsequence.

For a given orthogonal matrix U, natural number m, and arbitrary ma-

trix B, put

1: U-'BU'.

qp + 1

i=O

We shall need the following lemma:

Lemma. For any natural number m,

II(Am)p-ApII=0. Proof. Let p > m. Then

+1m+1

(A"")p

E i=o j=o

U-i-j AUi+j

p+m

s(k)U-kAUk, p +lm+1 E k=0 where

1<m+l (k=0,...,p+m);

s(k) = o<i<p,O< j <m

if m < k < p, then s(k) = m + 1. Hence

pE II (Am),, - ApII =

p+lm+1 1

s(k)U-kAUk

k=O

M-1 1

p+lm+1

- p+lk_o 1

U-kAUk

p+m

(s(k) - m - 1)U-kAUk + k=0

<P+IIIAII-0 as p -p oc. This proves the lemma.

s(k)U-kAUk k=p+1

3.7 OPERATORS

391

Now the assertion of the problem can be proved as follows. Since I I A,,,. II < All for all natural number m, the sequence {A,,,,} contains a subsequence {Am, } that is convergent in norm to a matrix H. Then for k oo we have

11H - U-1HUII < IIH - Amk II + IIAmk - U-1AmkUII

+ IIU-1(Ank - H)Ull <

2IIH-AmkII+mk+IIIAII -'0,

whence H = U-1HU. Thus, Hp = H for every natural number p. We prove that limp-. Ap-HII = 0. Actually, for every natural number k we have IIAp - HII = IIAp - HHII

IIAp - (A-k)pll + II(Arnk)p-HplI

IIAp - (A+m)pll + IIAmk - HII,

whence by the lemma we obtain limsup IIAp - HII < IIAmk - All.

p-oo

For k -* oo, the assertion follows.

Solution 2. Consider the natural embedding of the n-dimensional real space in the n-dimensional complex space. In the latter, the n by n real orthogonal matrices are unitary matrices. Thus, we prove more than required if in the statement of the problem we replace the orthogonal matrix by a unitary matrix, and real matrices A by matrices with complex entries.

So let U = (Uij) be a unitary matrix in the n-dimensional complex space. Then, as we know, in a suitable basis the matrix U has the form (Uij) = (Eibij), where bi.7 is the Kronecker symbol and Ie1I = Obviously,

(U')ik = Eibik,

= IEnI = 1.

(U-7)ik = Ei'bik

Now, if A = (aik) is an arbitrary matrix, then (U 7AU7)ik =

Ei jbirarsEBbsk = Ei 3aikEk = Et 3E3aikr,s

Therefore,

(Am)ik = m+1

1.` E-3Eaik =

If Ei = Ek, then Ei jEk = 1, so limn .

aik

m+1

m E.-aei

(Am)ik = aik.

3. SOLUTIONS TO THE PROBLEMS

392

On the other hand, if Ei # Ek, then (EiEk)m+1

aik j aik (Am)ik = m + 1 La (EiEk) = m + 1 j=0

- 1 -+0

EiEk - 1

as m -+ oo, since (EiEk)+1

-1

EiEk - 1

- (EiEk -1I .

We see that in this case limm_oo (Am)ik = 0.

It remains to note that if for a sequence of matrices (ai, i)

M=1

know that for each pair of indices i, k the numbers tend to some aik as m -+ oo then the sequence of matrices, evidently, is entrywise convergent to the matrix (aik).

Solution 3. As a generalization of the problem, we prove the following theorem.

Theorem. Let H be a complex Hilbert space, and U a unitary operator in H. Then, for any compact operator A of H, the sequence of operators (Dm (A)

(m = 0,1, ... )

E U-jAU'

m+1

j=0

is weakly convergent, that is, for every pair of elements f, g in H (Dn(A))f,g) = 0,

lim

m,n-oo

where (f, g) denotes the inner product of the elements f, g E H.

(This statement obviously contains the statement of the problem. Indeed, the n by n real orthogonal matrix appearing in the problem induces a unitary operator of the n-dimensional complex Hilbert space; in this ndimensional space all linear operators, in particular those induced by n by n real matrices, are compact, and in finite-dimensional spaces weak and entrywise - so-called strong - convergences of operators coincide.) Proof. For proving the theorem, we first observe that the mappings A 4Dm(A) (m = 1, 2, ... ), defined for all bounded operators A of H, have the following properties:

a. For any pair of bounded linear operators A, B of H and any pair of complex numbers a and ,Q,

(Dm(aA+,3B) = 0- m(A) +,3(m,(B)

(m = 0,1, ...) .

b. For any bounded linear operator A of H, I I,Dm (A) I <_ IIAII I

where IIAII stands for the norm of A.

(m = 0,1, ...

3.7 OPERATORS

393

c. If the sequence {Ak}k 1 of bounded linear operators is uniformly convergent to a bounded linear operator A, that is, I Ak - A l -> 0 as I

k -+ oo, and the sequence is weakly convergent for each k, then the sequence {4'm(A)}m=o is also weakly convergent. From these three properties only c. needs to be verified. So let {Ak} 1

be a sequence of operators with the properties above, and let f and g be two elements of the Hilbert space H. Without loss of generality, we may assume that Ilf II = llgll = 1. Then l

(A)

- m.(Ak) + 4'm(Ak) - ,Dn(Ak) + Ak)f, 9) 1 +

< 211A - Akll +

,Dn(A)] f,9)l f, 9) 1 +

A) f, 9)1

P.(Ak)) f,9)1

(we have made use of properties a and b and applied the Schwarz inequality).

Now let e > 0 be arbitrary. Then, by one of the assumptions on the 1, there is an index k = k(e) such that

sequence {Ak}

IlAk(e) - All < 4.

Next, let N = N(k(E), E) be a positive integer satisfying Pn(Ak(e))) f,9) I <

for m, n > N. By the assumptions, such N exists. Then, in view of the foregoing, we obtain -P. (A)) f,9)I < E

for m, n > N. This proves property c. It is well known that every compact operator is the uniform limit of finite rank operators, further every finite rank operator is a finite linear combination of rank 1 operators. Consequently, by properties a. and c., it is sufficient to prove our assertion for operators of rank 1. So let A be an operator of rank 1. Then there are two elements 0 and ii in the space H such that Af = (f, O) for all f in H. Now, if h and g are any two elements of H, then (,Pm(A)h, 9) =

1 E (Ujh, 0) (U-j0, g) . 1 E (AU' h, U39) = m+1 m+1 j=0 j=0

Denote by H 0 H the tensor product of H with itself. (The definition of the Hilbert space H 0 H is the following. Denote by Ho the algebraic

394

3. SOLUTIONS TO THE PROBLEMS

tensor product of H with itself. As is well known, this is the free module generated by the symbols x ® y (x, y E H) and having the complex field for operator domain. By the inner product of two elements m

Xi ®yi,

iV = > j=1

i=1

®y7

of H0, we mean the number n

(Xi,Xj)(yi,yj,)

(1)

i=1 j=1

Two elements', zb' E Ho are considered to be identical if (i - ', - ') = 0. On the factor space Ho that arises after this identification, (1) defines a norm. Completing Ho in the metric induced by this norm, we obtain the Hilbert space H = H ® H.) Let U0 be the operator of H satisfying the relation

Uo(x ®y) = (Ux ®U-1y)

(x, y E H).

U0 is uniquely determined in this way and is unitary in H. On the other hand it is clear that

m+1 j

(Uo(h®g)

By a classic theorem of ergodic theory (see for example F. Riesz and B. Sz. Nagy, Functional Analysis, Blackie, London, 1956, §144), the righthand side is convergent for every pair of elements ,, z 1' E H and so, in particular, for the pair = h ® , i' = 0 ® g. The proof is complete.

Problem 0.3. Prove that if a sequence of Mikusiriski operators of the form pe-as (A and µ nonnegative real numbers, s the differentiation operator) is convergent in the sense of Mikusiriski, then its limit is also of this form. Solution. Let µne-A^s be the convergent sequence of operators considered.

We may assume thatµn -ppandA-* A (0<p,A<+oo)as n-oo;in fact, passing to a suitable subsequence, this can always be achieved without any influence on convergence and limiting operator. By the definition of convergence, there exists an operator {p}/{q} ({p}, {0}) such that {q} E C[0, oo), {p}, {q} µne-Ans

{P} {q}

(1)

3.7 OPERATORS

395

is an almost uniformly (that is, uniformly on each finite interval) convergent sequence of functions in C[0, oo). We may assume that p(0) = 0, since otherwise {p}/{q} in (1) can be replaced by the equal expression ({p}{1})/({q}{1}). Define p to be zero on the negative half-line; then p will be a continuous function on the whole real line and not identically zero on the positive half-line. From (1) it follows that the sequence of functions µne-lns{p}

= {µnp(t - An)}

(2)

is almost uniformly convergent. If A = +oo, then we see that (2) is almost uniformly convergent to the zero function; thus µne-An9 --+ 0 = 0

e_8

(n -ti oo).

If A < +oo, then we show that µ is also finite. Let to be a point with p(to - A) # 0. Then p(to - An) - p(to - A) (n -' oo), so µnp(to - An) can only be convergent if p < +oo. Then, however, µnp(t - An) tends almost uniformly to the function {µp(t - A)} = pe-a9{p}. Consequently, µne-ans

pe-as

(n -+ oo)

Problem 0.4. Prove that an idempotent linear operator of a Hilbert space is self-adjoint if and only if it has norm 0 or 1.

Solution 1. If T is self-adjoint, then it is an orthogonal projection operator; it is well known that such operators have norm 1 or 0. Conversely, let IITII < 1. For any element x of the space and any number p, we have

T (µTx - (x - Tx)) = pTx and therefore I,zI2IITxII2 < II uTx - (x -Tx)II2 = -2Re [,a(Tx, x - Tx)] + IPI2IITx1I2 + II x - Tx 112. Consequently,

Re[p(Tx,x-Tx)] <

2IIx-TxII2.

But this is possible for every p only if

(Tx,x-Tx)=0. Hence, it follows in a simple and well-known manner that T* = T. The proof applies to real as well as complex spaces.

3. SOLUTIONS TO THE PROBLEMS

396

Solution 2. The kernel (null space) of T, in view of the property T2 = T, coincides with the range of I - T, that is, with the set of all elements of the form (I - T)x where x runs through all elements of the space. It is a wellknown fact, and easy to verify, that the orthogonal complement of the range of I -T is equal to the kernel of I -T*. By a well-known result of Nagy (see, for example, B. Sz. Nagy and C. Foiaq, Harmonic Analysis of Operators on Hilbert Space, Akademiai Kiado and North-Holland Publ. Co., 1970), from the inequality ITII < 1 it follows that the invariant elements of T and T* are the same. So, the kernel of I - T* coincides with the kernel of I - T, which, in turn, coincides with the range of T (the latter fact follows from

the property (I - T) 2 = I - T). Thus (Tx, x - Tx) = 0 for every x, which implies the statement.

Problem 0.5. Let T be a bounded linear operator on a Hilbert space H, and assume that I ITn II < 1 for some natural number n. Prove the existence of an invertible linear operator A on H such that II ATA-1 II < 1.

Solution. Let (.,.) denote scalar multiplication in H. It is easy to verify that the formula n-1

[x,y] = 1: (T`x,Tiy) i=o

defines a new scalar product on H. Let H denote the space equipped with the scalar product [., .]. Obviously, n-1

(x, x) < [x, x] <

1 II"` 112

(x, x)

i=O

for all x c H, that is, the norm in H is equivalent to that in H. Consequently, H is also a Hilbert space. Since the dimension of a Hilbert space is the minimal power of complete sets, that is, sets whose closed linear span is the whole space, the dimen-

sions of H and H are equal. But Hilbert spaces of equal dimension are isomorphic. So, consider a Hilbert space isomorphism A : H -> H. Since the underlying vector space for H and H is the same, A is an invertible linear operator on H. Let x E H be arbitrary, and put y = A-1x. Then (x, x) - (ATA-1x, ATA-1x) = (Ay, Ay) - (ATy, ATy) = [y, y] - [Ty, Ty] n-1 n-1 _ [ (Tiy Tiy) - E(Ti+ly,Ti+ly) _ (y, y) - (Tny,Tny), i=0

i=ol

which is nonnegative since IITnII < 1. Thus, we have obtained

(x, x) > (ATA-1x, ATA-1x)

3.7 OPERATORS

for all x E H, that is,

397

IIATA-111 < 1. Therefore, the

operator A satisfies

the conditions of the problem.

Remark. One contestant showed that the positive square root of the positive, bounded, self-adjoint, linear operator E o T*''Ti can be chosen for A.

Problem 0.6. Is it true that if A and B are unitarily equivalent, self-adjoint operators in the complex Hilbert space f, and A < B, then A+ < B+? (Here A+ stands for the positive part of A.) Solution. We first show that there are self-adjoint operators A and B in C2 with eigenvalues al, A2 and .3i.\4, respectively, such that A < B but A+ B+. For instance, let A and b have matrices 0

-1

( 1 -1)

and

(0 0)

respectively. Then for (x, y) E C2, we have (B(x, y), (x, y)) - (A(x, y), (x, y)) = Ix12 + xy + ty + Iy12 = Ix + y12 > 0,

that is, A < B. From the matrices, it is clear that A and b are selfadjoint, B positive semidefinite, A indefinite. Thus b+ is equal to b, and A+ is positive semidefinite with A+(0,1) # (0, 0), since (0, 1) is not an eigenvector of A. Therefore,

(A(0, 1), (0,1)) > 0 whereas (B+(0,1), (0, 1)) = 0. B+. Now let f = 12 (complex). For simplicity, we write each element of N as a series EÂ°Â° 1 anen convergent in 12, where So A+

en = (0,...,0,1,0,...).

Let A

(anen) = A(al, a2) + n=1

Ananen, n=3

where An = .k if n = 4k + s, 1 < s < 4, and A1, A2, )3, A4 are described above. Here el and e2 are identified with (1, 0) and (0, 1). Further, let

= B(ai, a2) + E Ananen. B(anen) E 00

n=1

n=3

3. SOLUTIONS TO THE PROBLEMS

398

Both A and B are sums of two self-adjoint operators, so they are selfadjoint. Since B - A = B - A, we have o0

K (B - A)

anen,

anen n=1

n=1

= \(B -

A)(al, a2), (al, a2) J > 0,

that is, A < B. Similarly, B+ - A+ = B+ - A+ and, consequently, A+Â˘ B+ We now prove that A and B are unitarily equivalent. Let 00

U 1(anen) E

= U1(a,,a2) +

anen, n=3

n=1

where U1 denotes the unitary transformation of coordinates that reduces the matrix of h to diagonal form: ((JiEU1 1)(a, b) = (A3a, A4b).

Ul is obviously unitary, and 00

(U1BU1 1) E anen) = A3alel + A4a2e2 +

Ananen

n=3

n=1

Let

(fanen)

= > af(n)en7

n=1

where f denotes the following permutation of N: f (n) = n-4 for n = 4k+3

or n = 4k + 4 with k > 1. Further, f(3) = 1, f(4) = 2, and f (n) =n+4 for n = 4k + l or n = 4k + 2 with k > 0. Then 00

n=1

bnen) _ E af(n)bn n=1

(U2 E anen n=1

_ E anbf-i(n) = (> anen, U2 1 E bnen n=1

n=1

which implies that U2 is also unitary. Further, we have (U2U1BUj 1U2 1)

(anen) n=1

Ananen. n=1

Finally, let 00

U3 Eanen =U3(a,,a2)+Eanen, n=3

n=1

where (U3 1AU3)(a, b) = (Ala,.2b), and U3 is unitary. With this third unitary transformation, we have

(U3U2U1BUUU) (anen) =A (anen) 1

and so A and B are unitarily equivalent. Consequently, the assertion is false.

n=1

3.7 OPERATORS

399

Problem 0.7. Let K be a compact subset of the infinite-dimensional, real, normed linear space (X, II . II). Prove that K can be obtained as the set of all left limit points at 1 of a continuous function g : [0,1 [---+ X, that is, x belongs to K if and only if there exists a sequence tn, E [0, 1[ (n = 1, 2.... ) satisfying limn_,,. tn, = 1 and limn. II9(tn) - xII = 0.

Solution. The compact set K C X has, for every E = 1/n (n E N), a finite E-net zl (n), z2 (n), ... , zin (n) E K, that is, if x E K, then I I x - zi (n) 11 < E = 1/n for some 1 < i < in. Denote by xo, x1, X2.... the sequence obtained by writing the finite (1/n)-nets z1(1), z2(1), ... , zjl (1), zl (2), Z2(2).... , ziz (2), zl (3).... one after the other. It would already be possible to define a piecewise linear function g: [0,1[-+ X such that g(1 - 1/k) = xk (k E N), and therefore each point of K is a left-hand limit point at 1 of the function g. The problem is that g may have further limit points that do not belong to K. So, we have to improve the procedure by interpolating a new sequence of function values yo, y1, y2, .... To define the latter, we shall need the following proposition (which is essentially the same as the well-known Riesz lemma on almost orthogonal vectors):

If L C X is a linear subspace of finite dimension, then there exists a vector y E X \ L such that I I y I I = 1 and dist(y, L) = 1, where dist(A, B) = inf { II a - bI I : a E A, b c B}

denotes the distance between the subsets A, B C X, and the singleton {a} is replaced by a. Indeed, since X is infinite dimensional, it contains a vector v E X \ L, and the subset

D={u: uE L,IIv - uII < IIvII} of the finite-dimensional subspace L is bounded and closed. Thus D is compact, so it contains a vector uo E D, which lies closest to v. Then, setting vo = v - uo, IIvoII = min{IIv - u I : u E D} = dist(v, L) = dist(vo, L). I

Consequently, the vector y = vo/IIvoII has the required properties. Next, we define a sequence in X by induction. Let Lo = lin {x0, xl }, and using the proposition just proved, choose a vector yo E X \ Lo such

that IIyo II = 1 = dist(yo, Lo).

(k E N) are already defined, then set

Lk = lin{xo,yo,xl,yl,...,xk-1,yk-1,xk,xk+1}, and using the preceding proposition again, choose the vector so that II yk 11 = 1 = dist(yk, Lk).

ykEX\Lk

3. SOLUTIONS TO THE PROBLEMS

400

For simplicity, we first define a continuous function 0: [0, oo[ -> X that has K for the set of limit points at oo. Fork E No = NU{0} and 0 < r < 1, put

0(3k + r) = (1 - r)xk + r(xk + yk), 0(3k + 1 + r) _ (1 - r)(xk + yk) + r(xk+l + yk),

0(3k + 2 + r) = (1 - r) (xk+l + yk) + rxk+l. Let us verify the desired properties. a. By the construction of the sequence xO, x1i x2, .... for any x E K there is a subsequence xk -* x; setting T,,, = 3k,,, we have

llm 0(7-,,,) = llm xk = x. b. On the other hand, we show that no x Â˘ K can be a limit point of at oo. In fact, the closedness of K yields 6 = min{dist (x, K),1} > 0, and if for some To E [3ko, 3ko + 3] (ko E No) we have dist (x, 0(To)) <

6/3, then for every r > 3ko+3 the relation dist (x, 0(r)) > 6/3 is valid. Now there exists a natural number ko < k E N such that bl. either 3k+(26/3) < T < 3k+3- (26/3), in which case the relation P(TO) E Lk and the definition of 0 imply dist (0(To), 0(T)) >

3 26

and therefore

dist (x, (p(T)) > dist (0(ro), (b(T)) - dist (x, 0(TO)) >

b2. or 13k - T1 < 26/3, in which case for xk E K, by the definition of Â˘, we have dist (xk, (b(T)) < 3

and so dist (x, (a(T)) > dist (x, K) - dist (K, (b(T)) > 3.

Thus, there is indeed no sequence r -* oo with 0(T,y) --+ x. Finally, the required function g : [0,1[--+ X is provided by the transformation

tE[0,1[.

3.7 OPERATORS

401

Problem 0.8. Denote by B[0, 1] and C[0, 1] the Banach space of all bounded functions and all continuous functions, respectively, on the interval [0, 1] with the supremum norm. Is there a bounded linear operator T:B[O,1]->C[O,1]

such that T f= f for all f c C[0,1]? Solution. Suppose the existence of such an operator T. For any 0 < a < 1, consider the function

fa(x)

if x < a,

ifx>a.

The function ha = fa - T fa is, apart from the jump of size 1 at point a, continuous and satisfies T(ha) = T(fa - T fa) = 0. Replacing ha by -ha is necessary. We may assume that ha > 1/3 in a small, punctured, right-hand or left-hand neighborhood of a. Let a1 = 1/2, and define the sequence a1, a2i ... so that ai E (0, 1) and ai belongs to a corresponding small neighborhood of ai_1 for all i > 2. Then for the function gn = Ii 1 hay, we have gn E B[0, 1], gn is continuous apart from the jumps of size 1 at the points a1, . . . , an, and gn > n/3 in a small, one-sided, punctured neighborhood of an. It is easy to construct a continuous function fn satisfying the relation 1

I1fn-9nII sup Ifn(x)-9n(x)I < _. 2 xE [0,1]

(To speak exactly, the sign of equality is valid.) Then, for the norm of the operator T, we obtain the estimate

IIT(fn -9n)II > 2IITfn -T9nII 1, = 2IIfn!I > 2nIlfn-9nII 3

IITII >

valid for any natural number n. Consequently, T cannot be a bounded operator. It follows that there is no operator T satisfying the requirements of the problem.

Problem 0.9. Does there exist a bounded linear operator T on a

Hilbert space H such that 00

n Tn(H) _ {0} but n Tn(H)_ # {0}, n=1

where - denotes closure?

n=1

3. SOLUTIONS TO THE PROBLEMS

402

Solution. There exists an operator T of this kind. Let H be a Hilbert space of countably infinite dimension, and let {en}n 1 be an orthonormal basis in H. We define T on the basis vectors first. Let

ifn=1,

10 Ten =

if n > 2 is not a square, ifn = j2 for an integer j > 2.

en+1

ajel +

en+1

2, we assume that 0 < Iaj I < 1 for each j, and

For the sequence {o

aIaj12 <oo. j=2

We next extend the definition of T to all of H. For an arbitrary vector x E H, put 00

Tx :_

CnTen,

where Sn = (x)en).

n=1

First, we have to prove that the series that defines Tx is convergent, that is, the sequence of the partial sums is a Cauchy sequence. Let k < 1 be positive integers. Then 2

2 l

SnT en

ISn12 +

n=k

k<j2<l

n=k

since, substituting the defining expressions of the vectors Ten in the sum Etn=k SCnTen, in the linear combination of basis vectors so obtained the coefficients of the vectors en with k + 1 < n < 1 + 1 have absolute value less than or equal to 1, the coefficient of e1 is Ek<j2<! and the remaining basis vectors do not appear in the linear combination. Using the Cauchy inequality, it follows that 2

nTen n=k

< n=k

1e3212

IajI2

IenI2 + k<j2<1

k<j2<1

<(1+a)1: ICn12. n=k

Since En 1 Ieni2 = IIxI12 < oo, the series EÂ°Â° 1bnTen is convergent. The linearity of T is now obvious, and if in the preceding inequality we write

k = 1 and let 1 tend to infinity, then we obtain the boundedness of T, namely, IITII < 1+a. We show that e1 E T1(H)

for every positive integer 1. In fact,

ej2 = T2j-2e(j_1)2+1

for

j>2

3.7 OPERATORS

403

and therefore e1 + 1ej2+1

= a Tej2 =

T2j-1

e(j-1)2+1)

Thus,

el + ej2+1 E T2 1(H) C T1(H) if 2j - 1 > 1, and letting j tend to infinity, we obtain ET1(H)

It remains to show that n,1T1(H) = {O}. Suppose x E ni_Â°1Tt(H). From the definition of T it is clear that, for any 1, the vectors e,,, with 2 < n < 1 + 1 are orthogonal to Tl (H). Thus (x, en) = 0 for all integers n > 2. On the other hand, if Ty 0 for some y E H, then there necessarily exists an integer n > 2 with (y, en) # 0. Then, however, (Ty, en+1) :34 0 and, by the foregoing, x = TO = 0.

3. SOLUTIONS TO THE PROBLEMS

404

3.8 PROBABILITY THEORY Problem P.1. From a given triangle of unit area, we choose two points independently with uniform distribution. The straight line connecting these points divides the triangle, with probability one, into a triangle and a quadrilateral. Calculate the expected values of the areas of these two regions.

Solution. First, observe that an affine transformation does not change the ratio of the areas, therefore the expectation of the ratio does not change. We will use this fact by choosing the type of the triangle in the most convenient way for our purposes. Denote the vertices of the triangle by A1, A2, A3, and the chosen points

by X and Y. The probability that the line e connecting the points X and Y crosses one of the points A2 (i = 1, 2, 3) is equal to zero because of the independence of X and Y. So the probability that we will get a triangle and a quadrilateral by dividing the triangle by the line e is equal to one. Denote by ..4,t the event that the point Ai is a vertex of the small triangle. The events .At (i = 1, 2,3) form a complete system of events, so by the theorem of "complete expectations," 3

E(t I A-)P(A.),

E(7,) t_1

where t denotes the area of the small triangle, and T denotes the area of the original triangle. If the original triangle is equilateral, then E(t/TIA,t) is independent of i (so does P(.A1) for i = 1, 2, 3), hence 3

E(T) = E(TIAiP(.A..) = E(1IA1) i=1

Suppose now that we have a right-angled isosceles triangle, Ai is the right-angled vertex, and the equal sides have unit length. Denote by F(a, b) the (conditional) probability, that for the line segments C and q of the sides cut by e, we have < a and q < b, assuming that < 1 and q < 1. That is,

F(a,b)=P(t; <a,77<blr; <1,q<1). Since a < 1 and b < 1, F(a, b)

P(t; < 1, 77 < 1)

If X = (x1i x2) and Y = (y1i y2), thenr

P( < a, r7 < b) = c

J(a,b)

dxldx2dyldy2i

3.8 PROBABILITY THEORY

405

where c is a constant, independent of a and b, and 0(a, b) is the domain of the four-dimensional space such that, for the corresponding line, < a and y < b. By substitutions xi = axi and yi = by, the domain L(a, b) is transformed to A(1,1), and therefore

P( < a, y < b) = ca2b2

dxtdx2dyldyI

0(1,1)

= a2b2P( < 1, y < 1), that is, F(a, b) = a2b2. So the density function is f (a, b) =

a2F(a, Saab

= 4ab,

and thus the ratio of the areas of the triangles is given by g(a, b) = 1 2 ib 1 = ab. 2

The expectation of the ratio is

jj 1

E(T) = E g(a, b) = 1

g(a, b) f (a, b) da db 0

4a2b2 da db = 4

So the expectations of the areas of the triangle and the quadrilateral are 4/9 and 5/9, respectively.

Problem P.2. Select n points on a circle independently with uniform distribution. Let Pn be the probability that the center of the circle is in the interior of the convex hull of these n points. Calculate the probabilities P3 and P4.

Solution 1. Assume that the circumference of the circle is one. Fix a point on the circle, and introduce the arclength parameter. Let Tl...... n be the parameter values associated with the chosen points. Denote by An the event that the center of the circle is not inside the convex hull of the points. Furthermore, denote by Bi the event that there is no point on the arc (Ti, Ti + 1/2). By the independence and uniform distribution of the points, it is obvious, that, except for an event of probability zero, the events + Bn, P(Bi) = 1/2n-1, and Bi are mutually disjoint and An = B1 +

therefore Pn = 1 - P(An) = 1 - n/2n-1. Hence P3 = 1/4, P4 = 1/2. Solution 2. Denote by TiTj the length of the shorter arc with endpoints Ti and r3 and by Bi3 (1 < i < j < n) the event that all of the points are

3. SOLUTIONS TO THE PROBLEMS

406

on the shorter arc with endpoints ri and rj. (If ri and rj are the endpoints of a half-circle, then Bi, denotes the event that all of the points are on one of the half-circles defined by them.) It is obvious that An = Li,7 Bid If the points are distinct, the events Bi3 are mutually disjoint. Since the points are independent and uniformly distributed, they do not coincide with

probability one, and so P(An) = >i i P(Bij). Furthermore, P(T x) = 2x, so (P(TiTj < x))' = 2 (0 < x < 1/2) and P(BijJ'r < x) = xn-2. Because of the theorem of total probability,

P(Bij) = 2

1/2

x'L-2dx

= (n

1)2n-2

Hence,

Pn = 1 - P(An) = 1 - (2)

(n -

1)2n-2

=1-2n-1 .

Problem P.3. Let El, E2,- - -, E2n be independent random variables such that P(ei = 1) = P(Ei = -1) = 1/2 for all i, and define Sk = 'i 1 Ei, 1 < k < 2n. Let N2n denote the number of integers k E [2, 2n] such that either Sk > 0, or Sk = 0 and Sk-1 > 0. Compute the variance of N2n.

Solution. It is known (see, for example, W., Feller, An Introduction to Probability Theory and Its Applications, Wiley, New York, 1957, Vol. I, p. 77) that the distribution of N2n is given as 12k2(n-k))

(k)(

P(N2n = 2k) =

(k = 0, 1, ... , n),

22n-k n

(k = 0,1, ... , n - 1).

P(N2n = 2k + 1) = 0

(1) (2)

The symmetry of the distribution implies that E(N2n) = n, and from (1) and (2) we conclude that 12k) (2(n-k)\

E(N22n) _ E(2k)2 k

n-k 1

22n

k=0

Assume that JxJ < 1, and consider the functions 2k)

F(x) _

- (2k)2 2k

x2k and G(x) = (1 -

x2)-1/2.

k=1

It is easy to see that CIO

(2k)

G(x)-= E 92k kx2k k=0

(3)

3.8 PROBABILITY THEORY

407

The generating function of E(N2n) is F(x)G(x), that is, F(x)G(x) = 1: E(N2

)x2k

k=0

Using (3), it is easy to verify that

F(x) = x(xG'(x))', which implies that F(x) = x(x2(1 - x2)-3/2)t = 2x2(1 - x2)-3/2 + 3x4(1 - x2)-5/2.

Therefore, F(x)G(x) = 2x2(1 - x2)-2 + 3x4(1 - x2)-3.

(4)

Simple calculations show that x2)-2 =

(1 -

0" kx2k-2 k=1

and

x2)-3 = [ k(k2 1)x2k-4

(1 -

k+1

So from (4), we get

F(x)G(x) =

2kx2k + k=1

00 3 k=1

k(k - 1)x2k =

oc'

k=1

k2 + 2 k I x2k

Therefore,

E(N2n) = 3n2 + 2n . Hence,

Var (N2n) = E(N2n)

- E2(N2n) = 12n2 + 2n J - n2 = n(n2+ 1)

Problem P.4. A gambler plays the following coin-tossing game. He can bet an arbitrary positive amount of money. Then a fair coin is tossed, and the gambler wins or loses the amount he bet depending on the outcome. Our gambler, who starts playing with x forints, where 0 < x < 2C, uses the following strategy: if at a given time his capital is y < C, he risks all of it; and if he has y > C, he only bets 2C - y. If he has exactly 2C forints,

3. SOLUTIONS TO THE PROBLEMS

408

he stops playing. Let f (x) be the probability that he reaches 2C (before going bankrupt). Determine the value of f (x).

Solution 1. We are going to prove that f (x) = x/2C (0 < x < 2C). First, we show that f (x) is nondecreasing. Let 0 < x1 < x2 < 2C, and suppose there is a sequence of tossings such that he can reach 2C from x1. We show that by the same sequence of tossings he can reach 2C from x2 (maybe earlier), so f (X2) > f (xl) . Let us see how the amount of the player's money changes after one toss: (a) if x1 < x2 < C, then 2x1 < 2x2 for a win, and 0 < 0 for a loss; (b) if x1 < C < x2i then 2x1 < 2C for a win, and 0 < 2(x2 - C) for a loss;

(c) if C < x1 < x2, then 2C < 2C for a win, and 2(x1 - C) < 2(x2 - C) for a loss. Thus if x1 < x2, then, for the new amounts xi and x2, we have xi < x2. Define x = (a/2n) 2C (1 < a < 2' -1 , a is odd, n = 1, 2,... ). We will

prove by induction that for such an x, f (x) = a/2n = x/2C. For n = 1, f((1/2) 2C) = f (C) = 1/2, since starting from C the player will have 2C with probability 1/2. Suppose the formula holds for n = k (k > 1), that is, f (x) = f ((a/2 k) 2C) = a/2k. If x = (a/2 k+1) 2C and a < 2k (that is, x < C), then the player will win 2x forints or lose everything with probabilities 1/2, 1/2, respectively. So f (x) = 2 f (2k 2C) + 2 f(O) = 2k+1

If x = (a/2 k+1) 2C and 2k < a < 2k+1 (a $ 2k since it is odd), then the inequality x > C implies

if (a-2 2k

A( xX )) = 2 f (2C) + 2 f 1

- 2C I =

1a-2k

a_ x 2k+1

Suppose now xo 0 (a/2n) 2C (1 < a < 2' - 1, a is odd), and suppose f (xo) # xo/2C, but, for example, f (xo) > xo/2C Then there should exist a number having the form a/2n such that 2C < 2n < f (xo),

(1)

and it contradicts the fact that f (x) is nondecreasing. Namely, because of (1), xo < (a/2n) 2C, and, on the other hand, 2n

= f (2n 2C) < f (xo)

We can obtain a similar contradiction for f (xo) < xo/2C, therefore f (x) = x/2C (0 < x < 2C).

3.8 PROBABILITY THEORY

409

Solution 2. Introduce the notation g(x) = f (2Cx) (0 < x < 1). Since f (0) = 0 and f (2C) = 1, we get g(0) = 0 and g(1) = 1. It is easy to verify

that g(x)

2g(2x - 1)

if x E (2',1].

(2)

To see this, observe that if x E (0, 1/2] , the gambler wins or loses 2Cx forints with probability 1/2, so g(x) = 29(2x) + 2g(0) = 2g(2x).

If x E (1/2, 1], the gambler will have 2C or (2x - 1)2C forints with equal probabilities 1/2 , so

g(x) = 29(1) + 2g(2x - 1) = 2 + 2g(2x - 1). We will next show that the only bounded solution of the functional equation (2) is the identity function g(x) = x (from which we get the same solution as before).

Put h(x) = g(x) - x. Then h(x) is bounded and 2h(2x)

h(x)

if x E [0, 2],

h(2x - 1) if x E (2,1].

(3)

We are going to show that h(x) = 0. Since h(x) is bounded, both M = supXE[0,1] h(x) and m = infxE[o,1] h(x) are finite, and

h(2x) < M

h(x)

if x E [0, 1], 2

2h(2x - 1) < M if x E c1,1].

The definition of M implies that M < M/2, that is, M < 0. A similar argument shows that m > 0, that is, h(x) - 0. Problem P.S. For a real number x in the interval (0, 1) with decimal representation

0.a1(x)a2(x)...an(x).... denote by n(x) the smallest nonnegative integer such that an(x)+lan(x)+2an(x)+3an(x)+4 = 1966.

Determine fo n(x)dx. (abcd denotes the decimal number with digits a, b, c, d.)

Solution 1. The integral is understood as a Lebesgue integral. Introduce the following notation:

Ln = {x : x E (0, 1), n(x) = n}, An = {x: x E (0, 1),

an+1(x)an+2(x)an+3(x)an+4(x) = 1966}.

3. SOLUTIONS TO THE PROBLEMS

410 )\

is the Lebesgue measure, An = \(L,,), an = \(An), where (n = 0,

1, 2, ... ). Observe that Ln and An are the unions of finitely many intervals, so they are measurable sets. First, we show that Y-Â°O o )1n = 1, that is, the integrand is defined almost everywhere. If

H={x:xE(0,1), ak+1(x)ak+2(x)ak+3(x)ak+4(x) # 1966, k = 1,2,...}

and

Cn={x:xE(0,1), a4k+1(x)a4k+2(x)a4k+3(x)a4k+4(x) Â˘ 1966, k = 1,2,...},

then H C Cn (n=0,1,2.... ) and A(Cn)

- (104 104-1)n+1

Therefore, limn . A(Cn) = 0, \(H) = 0. The lengths of the intervals in An are equal to 10-(n+4), the number of these intervals is 10" (we can choose the first n decimal digits in ion different ways), and therefore an = 10-4.

Obviously, An f1 Ln = Ln (n = 0,1.... ), and if n - 3 < k < n - 1, then An fl Lk = 0 since two sequences 1966 cannot overlap. If 0 < k < n - 4, then ,\(An n Lk) = \(An-k-4),\(Lk) = 10-4\k, since the assumption that x is in Lk restricts only the first (k + 4) decimal digits of x. So n-4

10-4 = \(An) = E A(An n Lk) = : 10-4\k + \n, k=0

k=0

that is, n-4 104I\n = 1

ak k=0

By the equality Ek o Ak = 1, we get

= \k = 104I\n+4

(1)

k=n+1

For the integral of n(x), 00

n'\n =

n(x)dx = n=0

1 00

n=Ok=n+1

\k.0

3.8 PROBABILITY THEORY

411

Hence, by (1), o0

f n(x)dx = E0"104An+4 = 1O4 0

n=0

An n=4

=104(1-A0-A1-A2-A3), and since )o = Al = A2 = A3 = 10-4, we clearly get 1

n(x)dx = 104 - 4 = 9996.

Solution 2. We solve the problem in a more general form. Let a = s1iS2,...,Sk be an integer given in the decimal system, and for x E (0, 1) define the function n(x) as follows: let n(x) be the smallest nonnegative integer such that an(x)+lan(x)+2 ... an(x)+k = S1S2 ... Sk.

We are going to calculate the value of ff n(x)dx. Let (Il, A, P) be the probability space of the Lebesgue-measurable subsets of the interval (0, 1) endowed with the Lebesgue measure. Then on the set S1 = (0, 1), the digits al, a2i ... , ak, ... are independent and

P(ak-a)

(k=1,2,...; i=0,1,...,9).

On this space, n(x) is a random variable, and its expectation is

n(x) dx.

Put Pn = P(n(x) = n) and Qn = P(an+lan+2 ... an+k = a) = 1110k

(n=0,1,...).

Let l be the smallest natural number such that shifting a by 1 positions,

the digits on the same positions coincide. (That is si = si+1 for all i = 1, 2, ... , k - 1). Obviously, 1 < 1 < k . Rewrite kin the form k = pl + r, where 0 < p, 1 < r < 1. The a consists of p blocks r = T1-s2 .... sl of length 1, and the first r digits of this block r. Notice that Qn = (POQn-k + PiQn-k-1 + ... + Pn-kQO)

(Pn-k+r10-(k-r)

Pn-k+r+110-(k-r-1)

(n = k,k + 1,...) since the event an+lan+2 ... an+k = a

can occur in two different cases, as follows:

+... + Pn_110 ' + Pn)

(2)

3. SOLUTIONS TO THE PROBLEMS

412

(a) Block a appears only from the (i + 1)th index (0 < i < n - k), and after n - k - i digits it occurs again. (The probability of this case is (b) Block a appears from the (n - k + r + il)th index (i = 0,1,. .. , p), thus an-k+r+il+1an-k+r+il+2 ...an+r+il = a,

and it appears again starting from the (n + 1)th index. Note that besides the previous assumption, it is also necessary that the values of the undefined (k - r - il) digits are given. The probability of this case is Pn_k+r+il10k-l-il Assume that Pn = Qn = 0 if n = -1, -2, .... Then (2) holds also for 0 < n < k. Multiply (2) by zn, add up the resulting equations for all values of n, and introduce the notation 00

Q(Z) = E

P(z) _ E Pnzn,

Qnzn

n=0

n=0 to get P

Q(Z) = zkp(z)Q(z) +

zk-r-i1P(z)10-(k-r-ti).

i=o

Using the obvious formula 00

- = lok zn = 10k(1 - z)'

Q(z) _

we have

P(z) _

zk + iok (1 - Z)

zk-r-il10-(k-r-il)

From P(1) = 1 we can see that n(x) is defined almost everywhere, and P

n(x)dx = P'(1) _ -k + E 10r+il 10

i=0

This last equality shows that the integral is always an integer. If a = 1966,

then k=l=r=4,p=0, and thus 1

n(x)dx = 104 - 4 = 9996. Jo

Ifa=1961,then k=4,l=3,p=r= 1, and thus n(x)dx = 10 + 104 - 4 = 10,006. Ji

3.8 PROBABILITY THEORY

413

Ifa=1919,thenk=4,l=r=2,p=1,thus

= 102 + 104 - 4 = 10,096,

and, finally, if a = 1111, then k = 4, 1=r= 1, p = 3, thus 1

n(x)dx = 10 + 102 + 103 + 104 - 4 = 11, 106.

Remarks.

1. The problem can obviously be generalized for any number system not just the decimal one. 2. One can reduce the problem to compute the expectation of the first return time of a recurrent sequence of events. In this case, the value of the integral can be determined using the theorem by Erdos, Feller, and Pollard. Problem P.6. Let f be a continuous function on the unit interval [0, 1]. Show that 1

moo

...

I dx1 ... dxn = f 2 I (1

and li

JJ0' 0

...J

f (e

Solution 1. Let k be an arbitrary positive integer and n > k. Consider the multinomial expansion of (E 1 xi)'. The number of terms that contain all variables with powers not exceeding one is

n(n- 1)... (n-k+1)

=nk+O(nk-1)

The integrals of these terms on the unit cube

Rn={(xl,...,xn): 0<xi<1} (i=1,...,n) are equal to 1/2k. The number of terms containing at least one xi with a power higher than 1 is not greater than n nk-2 = nk-1. The integrals of these terms on Rn are not greater than 1. From this observation, we get

x1+...+xn n

asn -+oo.

dxl...dxn -*

\2/

3. SOLUTIONS TO THE PROBLEMS

414

An easy calculation shows that 1

...

(n xl ... xn)kdxl ... dx,, _

i=1

k/indxi

= (1 + n)

( 1\k e

asn ->oo. Hence, we have shown that the statements are valid for f (x) = xk. It is also obvious that the statements hold for constant functions, and if they are valid for two continuous functions, they are valid for their linear combinations as well. So we have proved the theorem for polynomials.

If the function f is continuous on [0, 1], then by the Weierstrass approximation theorem, for any positive e there is a polynomial p(x) such that I f (x) - p(x) I < E/3 (0 < x < 1). Since the statements are valid for polynomials, there is a K such that f 1...

Cxl+'n +xnl

dx1...dxn,-7 (1) <

p(xl...x)dxl...dx-pJ1...

for any n > K. Therefore, for any n > K,

f1 f(x1+ n+xn)dxl...dxn-.1 x

f1 f1

...I p\ 1

dx1...dxn

2(1)

f (21)

(1 dxl...dxn-p\2)

< E,

and similarly,

j...jf(xixn)dxidxn_f() 1

< E.

Now the proof is completed.

Solution 2. Let Sl, ... , fin, ... be mutually independent, uniformly distributed random variables on the interval (0, 1). Then the variables rin = log en (n = 1, 2, ...) are also mutually independent, and they have the same distribution. Since E(en) = 1/2 and E(rjn) _ -1 (n = 1, 2, ... ), a theorem by Kolmogorov implies that

51+...+en

3.8 PROBABILITY THEORY and

771+...+17n

415

-1

with probability 1. By the assumptions the function f is bounded on the interval [0, 1] and continuous at x = 1/2 , and the function g(x) = f (ex) is also bounded on (- 00, 0] and continuous at x = -1. Using the Lebesgue theorem, we get

E(f(cn)) -* E(f(1)) = .f(2),

E(g(V')) -> E(g(-1)) = E(f(e)) _ f(e), and, finally, the definitions of cn, cp;,, and g imply that \ f(1)=E(f(Vn))=J ... 2 f1

f(X1+...+xn\dxl...dxn,

and

f (e J =

11. ..

1 f( n xl ... xn)dx1 ... 0

Remark. The statements also hold in the more general case when f is bounded, integrable, and continuous at x = 1/2 and x = 1/e, respectively. In the first statement we can also drop the condition of boundedness (as was pointed out by Laszlo Lovasz).

Problem P.7. Let A1, . . . , An be arbitrary events in a probability field. Denote by Ck the event that at least k of A1i . . . , An occur. Prove that n

11 P(Ck)

fj P(Ak)-

k=1

Solution. We start with three remarks: Comment A. The statement is true for n = 2, that is,

P(A + B)P(AB) < P(A)P(B). To see this, use the notation AB = x, AB = y, and AB = z. Then our inequality becomes

(P(x) + P(y) + P(z))P(x) < (P(x) + P(y))(P(x) + P(z)), which obviously holds.

Comment B. If Al D ... D An, then Ck = Ak.

3. SOLUTIONS TO THE PROBLEMS

416

Comment C. Ck is identical to the event that at least k occur from the events A, + A2, A, A2, A3, ... , An, that is, from the events A, + A2, A, A2, A3, ... , An, exactly as many occur as from the events A1, A2, ... , An. This statement is trivial. Our proof is based on the above comments. Put A° = Ai (i = 1, . . . , n), and assume that the events Ai , ... AA are already defined. Next, define the events ,A`+1 as follows: choose a pair Aq , A of events, i < j, Ai+',

for which Al' +1

A' , A AZ (that is, Aµ and A' are incomparable) and = Aµ + A , A +1 = A A and Ak+1 = Ak for k # i, j.

define Ai Continue this procedure until there exists at least one pair of incomparable events. This procedure will terminate in finitely many steps, since in each step the number of incomparable pairs decreases. This observation follows from the facts that Ai +1 and A +1 become comparable, and any event is comparable to at least as many of the events Aµ + A, Aa AJ , as of Ai and By comment A, n

H P(Ak+1)

11 P(Ak).

k=1

(1)

Observe that comment C implies that the events Ck remain unchanged, and because of comment B they are identical to the final events Aq . Therefore, (1) implies the original statement.

From the proof of comment A, we see that if 11k=1 P(Ak) # 0 and P(Aq A7) # 0, then strict inequality holds in (1). Hence, equality holds if either 11k=1 P(Ak) = 0 or the original system of events {Ak} are already "ordered." Remarks.

1. If multiplication is replaced by addition, then in the statement we always have equality. That is, n

P(Ai).

P(Ci) i=1

i=1

2. Let Sk(x1, ... , xn) be the kth elementary symmetric polynomial of the variables x1, ... , xn. Then Sk(P(Cl), ... , P(Cn))

Sk(P(A,),... , P(An)).

Slight modifications of the above proof can be applied to show the validity of this more general statement.

Problem P.8. Let A and B be nonsingular matrices of order p, and and ri be independent random vectors of dimension p. Show that if , ri and A + 913 have the same distribution, if their first and second let

3.8 PROBABILITY THEORY

417

moments exist, and if their covariance matrix is the identity matrix, then these random vectors are normally distributed.

Solution. Put C = A + r7B. Without loss of generality, we can suppose that E(i7) = E(() = 0. Since and 77 are independent, Var

77B)] = A* Var

B* Var (77)B,

where Var (c), Var (77), and Var (() denote the covariance matrices of , 77, and C, respectively. They are equal to the identity matrix, so

A*A+B*B = I. Thus, for arbitrary vector t, (tA*, tA5) + (tB*, tB*) = (t, t),

and by the regularity of A and B, ItA*I< Its

ItB*I < Its

(t # 0),

(1)

where 1.1 denotes the Euclidean length of vectors. Since and 77 are independent, E(ei(t,tA+vB))

= E(ei(tA'O)E(ei(tB',l1)).

If Â˘(t) denotes the common characteristic function of , 77, and c, then W(t) = cv(tA*)cv(tB*)

(2)

We are going to show that W(t) # 0. Assume that cp(t) has real roots. Then its continuity implies that there is a root to with smallest absolute value. Since cp(0) = 1, to $ 0. Thus, (1) implies that co(toA*)

co(toB*)

and this contradicts equation (2). Consider the function log cp(t) + 2 (t, t),

and choose the branch of the logarithmic function for which 0(0) = 0. The assumptions imply the existence of vectors dd

VI(t) =

`P(t),

(0) =

3. SOLUTIONS TO THE PROBLEMS

418

and matrices fz

cP (t)

dtz

co"(0) = -Var (

co(t),

Therefore,

) _ -E.

[(t)l dz

J t=o

t-o

= 0,

and hence li(t) = o((t,t))

Itl , 0.

(3)

We are going to show that O(t) - 0. Suppose O(t) 0 0. Then there exists a positive E and a t # 0 such that IV,(t) I > E(t, t). Because of the continuity of co(t), there exists such a t1 with minimal absolute value , and (3) implies that t1 # 0. From inequality (1), we conclude that I,0(t1A*)I < E(t1A*)t1A*)

and

I',(t1B*)l < E(t1B*,t1B*). From (2),

,O(t) =,O(tA*) + ,(tB*), so

I'(tl)I <- ,(t1A*)I+I ,(t1B*)I <E[(t1A*,t1A*)+(t1B*,t1B*)] =E(tl,t1). I

This inequality contradicts the definition of t1. Consequently, fi(t) - 0, and so co(t) = e- 2L (t,t).

Problem P.9.. Let 1, z, ... be independent random variables such that Eln = m > 0 and vz < oo (n = 1, 2, ... ). Let {an} be a sequence of positive numbers such that an -+ 0 and >Â°O_1 an = oo. Prove

that

Jim

1: akSk = 00 n- oo

= 1.

k=1

Solution. We can solve the problem using the Kolmogorov inequality, but in the solution given below we only use the Chebyshev inequality. Instead of an -+ 0 and Var (ÂŁn) = v.2, it is sufficient to assume that {an}

is bounded and Var ( n) = a2. For instance, suppose 0 < an < 1. Put Sn = Ek=1 ak, and define the sequence NJ by S,.,, < n2 < Sr..+1. Then

nz-1<Sr,.<nz.

(1)

3.8 PROBABILITY THEORY

419

If limnk -1 akG = oo does not hold, then there is a K such that n CC

akSk < K

(2)

k=1

occurs infinitely many times. It is sufficient to show that for any fixed K, this event has probability 0 since the union of these events for n = 1, 2.... includes all event sequences for which limn E akG = oo does not hold.

Let Ak be the event that there is an n between rk and rk+1 such that (2) holds. Note that (2) holds infinitely many times if infinitely many Ak occur. Denote the probability of Ak by Pk. We have to show that with probability one only finitely many Ak occur. We prove this by verifying that E' 1 Pk is convergent. In this case, >k>m Pk --> 0, and >k>m Pk is greater than the probability that infinitely many Ak occur. Therefore, this latest probability is equal to 0. (Actually, here we have proved and applied the Borel-Cantelli lemma.) Put 7)k = Eik aiSi and Ok = Ei k rk+1 ai j. If (2) holds for some rk < z < rk+1 with z replacing n, then either rk

aie

kzm

i=1

E aiSiC-k23, i=rk+1

supposing k2m > 6K. In the last case, O k = Ei k rk+1 ai I fi I > k2m/3, so at least one of the inequalities Ok > k2m/3 and 11k < k2m/2 is valid. Thus, Pk <

P(qk <

l k2mI+P

k2m

(3)

We have to show that on the right side both terms are small. It is obvious that E(77k) = mSrk and Var (7)k) < cr2Srk since ai < 1 implies that Var (7Ik)

Q2 E ai

i=1

= o2Srk.

Furthermore,

E(Ieil) <E(1+E1) =

1+m2+Q2 =A

and

Var(Ie

<E(ei) <A.

3. SOLUTIONS TO THE PROBLEMS

420

Since the absolute values of independent random variables are independent, E(V)k) <-

(Srk+l - Srk)

max

rk <i<rk+l E(IeiI) <

(Srk+l -

Srk)A = 0(k)

and rk+1 q/'

Var (00 <

i=rk+l

+l Var(Iei1) C

(Srk+l

-Srk)A=0(k).

Now we can use the Chebyshev inequality and (1) to see that E(rlk) k2z )

P(77k

Var (ilk) - k2Q,

k2m,

k3 2

P(177k

- E(ilk)I ?

P(ibk > 3) < P(I'k - E('k)I ?

) = 0(

z k3 ) = O(T2

if k is large enough. From these relations, the convergence of E Pk follows.

Remarks.

1. One can sharpen the proof to see that >k<n ak4 tends to infinity as fast as >k<n ak with probability one. 2. The condition an = O(Sn) is weaker than boundedness, and it does not guarantee the convergence to infinity with probability one, but the slightly stronger condition

an = 0

i-(1+E1'-4l og (S.

(which is still weaker than boundedness) does.

P.10. Let

be independent, uniformly distributed, random

variables in the unit interval [0, 11. Define

h(x) = n #{k: t9k < x}. Prove that the probability that there is an x0 E (0, 1) such that h(xo) = xo,

is equal to 1 - n.

Solution 1. Let us denote by Ik the interval ((k - 1)/n, k/n). Consider the elementary events that for every i, i9i belongs to Ik,, where ki is given for every i. So we decomposed the original sample space to the union of

n' disjoint events of probability 1/nn (events with probability zero are omitted). These events will be called atoms.

3.8 PROBABILITY THEORY

421

1 = h(x) is an integer multiple of 1/n, Since the value (1/n) in order to check whether h(xo) = xo holds for some x0, it is sufficient to know which intervals Ik, contain t9i, that is, which atom will be the result of the experiment series. They have the same probability; therefore, it is sufficient to determine the number of atoms for which there is no k such that exactly k i9ti occur until k (1 < k < n). Denote by An (1 < k < n) the number of atoms for which there is no k such that the first k intervals contain exactly k t9i values. Let Bk denote the number of atoms for which there are exactly k i9t values in U3<kI. but for any t < k, Uj<tI9 does not contain exactly t t9i values. So every atom that has exactly k i9ti values in the first k intervals belongs to exactly one Bk. Next, we determine the value of Bk. We can choose the t9is belonging to U3<kIi in (n) different ways. We can choose the first k intervals Ik; containing these t9i values in Ak different ways, and we can arrange the remaining n - k Vi values in the remaining n - k intervals in (n - k)n-k different ways. So we have the recursion n-1

nn - An = y k=1

n k

Ak(n - k)n-k,

Al=1.

(1)

We are going to verify that An = nn-1 by mathematical induction. It is easy to check that Al = 10 and A2 = 21 or A3 = 32. Suppose that we have already verified this relation up to n - 1. In order to show this equation for n, we have to show that n-1

nn -

nn-1

k=1

nkk-1(n - k)n-k (k)

(2)

It is easy to see by interchanging k and n - k that this equation is equivalent to n-1

()k2(n

nn-1(n - 1) _

k)kk-2(n

- k)n-k-2

(3)

k= 1

We will prove this equality by a well-known theorem of Cayley. Consider the trees having the numbers 1, 2, ... , n as vertices. (A tree is a connected

graph not containing a circle.) Two graphs are different if they do not have the same (i, j) pairs as edges. By the Cayley theorem, there are nn-2 different such trees. Now choose in every tree in all possible different ways one vertex and one edge. We will call these graphs vertex-edge-signed trees, and consider these graphs different if they are different before our vertexedge selection, or the selected vertex-edge pair is different. Since a tree has n - 1 edges, we have nn-1(n - 1) different vertex-edge-signed trees, and this number is exactly the left-hand side of (3). Let us count the number of these vertex-edge-signed trees in another way. By omitting the signed edge, we get a tree with k vertices with one signed vertex and a normal tree with (n - k) vertices (k = 1, 2, ... , n - 1). On the other hand, if we choose

3. SOLUTIONS TO THE PROBLEMS

422

k vertices out of n, define a tree on them, select a vertex and define a tree from the remaining n - k vertices, and connect the two trees by an edge, which will be called the signed edge, then we get a vertex-edge-signed tree. Thus, the two procedures are the reverse of each other. In this way, we get n-1

11 1 k=1 (k)

k kk-2 (n - k)n-k-2 k(n - k)

different graphs, which is the right-hand side of (3).

Solution 2. The first part of the solution is the same, but here we give an algebraic proof of (2).

Lemma 1. If 0 < j < m, then

E ()-k 1)kk' = 0. k=1

Proof. Denote by D the operation of differentiation followed by a mul-

tiplication by x. Thus Dm(xk) = kmxk. It is easy to see by induction that if x0 is a root of the polynomial p(x) with multiplicity mo, then x0 will also be a root of Dm(p(x)) with multiplicity at least mo - in. Apply this observation to the polynomial p(x) = (1 - x)m and the operator Dj, j < m. Then we get that x = 1 is a zero of the polynomial D1.((1 - x)m) _ 1: k=1

(k)(-1)kkjxk. /J

This completes the proof of Lemma 1. Lemma 2. n-1

(n)

kk-1(z

- k)n-k =

nzn-1 - nn-1

Proof.

n-1

1: I k kk-1(z - k) k=1\ /

n-k

n-1

n-k

( n - k 1 zj(-k)n-k-.7

j1=0

{\n

(n)kk-1

k=1

n-ln-k

E k!j!(n k=1 j=o

- k - j/J

z".(-1n-k-jkn-l-j

k - j)!

3.8 PROBABILITY THEORY

n-i n-k

1: 1: k=1 j=O

(n) (m;3)Zi(_1)n-k-ikn-1-i

n n-k (n)

= 1:1: k=1 i=0

n-1 n-j

423

k')Z,(-1)n-k-jkn-1-j _ nn-1

(n3)(n-j)zj(_,)n-k-jkn-1-i _ nn-1

= j=0 k=1

n-2n-j

(;)

n - 3) i (+ nzn-1

n -2

if (n) -1)n j=0

nzn-1

k=1

(n -71)kkn-j-1) -

nn-1 +

nzn-1

- nn-1

since, by Lemma 1, the inner sum is equal to zero.

Substitute z = n to get (2).

Problem P.11. We throw N balls into n urns, one by one, independently and uniformly. Let Xi = Xi(N,n) be the total number of balls in the ith urn. Consider the random variable

y(N,n)= min Xi - NI. 1<i<n

Verify the following three statements:

(a) If n -+ oo and N/n3 -> oo, then

P y(N' n) < 2 n

1 - e-x

2/'r

for all x > 0.

(b) If n -> oo and N/n3 < K (K constant), then for any e > 0 there is an A > 0 such that P(y(N, n) < A) > 1 - E.

Solution. The basic idea of the proof is the following: Xi is a B(p, N) (binomial distribution with parameters p, N), where p = 1/n. If the Xi's were independent, the statements could be proved easily. But of course

3. SOLUTIONS TO THE PROBLEMS

424

+ X,,, = N. However, we will show that k of they are not, since X1 + them can be considered independent for sufficiently small values of k. Suppose that the Xis are independent and their distribution is B(p, N). Let al, ... ak be natural numbers in the interval [0, NJ,

a=P(Xi=ai,i=1,...k), a'=P(X'=ai,i=1,...,k). Introduce some additional notations: k

a=Eai, ai=pN+bi, b=>bi i=1

i=1

We will show that a - a' under the conditions k = 0(1), bi = o(-,I-N), and

N > n. Obviously,

pa (1

(fl=1 ai!)(N - a)!

- kp) N-a

and k

a'_

a4))

i=1

ftINpa(1_p)kN-a

a' _ N!k-1(N - a)!(1 7k a (l 1i-1(N - ai)!)(1 - kp)N-a p)kN-a

The value of ,3 will be estimated by the Stirling formula:

t! =

27rt tte-t+0(1/t)

The error of log t, that is, the difference of the logarithms of the true and estimated ,Q values can be determined from the O(1/t) terms as O

N - ai

+k-1 N

N-a

0( 1 N

()'

since

ai = pN + o(v), a = kpN + o(VN--), p = 1 -> 0 n

and N > n implies N --+ oc . The terms a-t and the value of the f terms can be given as

Nk-1(N-a) _ r1k 1(N-ai)

V1- r,

1(1- f)

27r are cancelled, and

3.8 PROBABILITY THEORY

425

So, finally,

N N(k-1) (N - a) N-a (1 f 1(N - ai)"`-ai (1 - kp)N-a k N-ai . lli=1(l - Ni - Ni (1 p)kN-a (1 b N-a k b N-ai =(1-N(1-kp)) .11(1-N(li p)) p)kN-a

(la N-a 7 kp)N-a

Since each term converges to 1, we can apply the formula log(1 + e)

ce + O(e2). The value of the second-order term is k

b + Ei=1

= o(1).

The value of the first-order term is

-b(N-kpN-b) N(1-kp)

bi(N - pN - bi)

N(1-p)

+ b2

N(1-kp)

N(1-p) -o(1),

which implies that log/3 = o(1).

Solution to part (a). Put

ak=P(IXi - pNI < x

c= P

k),

(Nn) < x VnNs

Recall that if A1, ... , An are arbitrary events and

Si =

P(Ai1,...,AZJ),

1 <i 1 <... <i j <n then

2s+1

(-1)S3 < P(A1 ... An) < 1:(-1)iSj j=0

j=0

(see, for example, K. Bogndrne, J. Mogyorodi, A. Prekopa, A. Renyi, D. Szdsz, Exercises in Probability Theory (in Hungarian), Tankonyvkiado, Budapest, 1971, Problem 1.2.11.c). So 2s+1

(-W (n),j

<1-c<

()aj

3. SOLUTIONS TO THE PROBLEMS

426

Note that

ak-a'k=P lXi'-pNl<x 3;i=l,...,k

since we can apply the previous asymptotic relation for any system Xi pN = bi, when IbiI < x N/n3. Since a' = a'k, with

a' =P IX' - pNj < x

((k

)k+1)

+(nj) 7

The first sum is the first k terms of (1 - a')', and substituting them by

(1 - a')'" itself, the error will be only O((k+l)a'k+1) From the local De Moivre-Laplace theorem we conclude that

(here we use the fact that N/n3 -+ oo). Select an arbitrary, small e > 0. If k is sufficiently large, then the first error term is less than e, the second term is o((l+a')") = o(1), and finally,

log(1 - a') = -a' + O(a'2) , n log(1 - a') = -1/ x + o(1), which proves the assertion.

Solution to part (c). We are going to prove for an arbitrary, small e > 0

that there exist b and no such that for n > no and N/n3 < b, we have P(y(N, n) < 1) > 1 - E. If we wish to apply the previous reasoning, we face the difficulty that is not small enough, and (1 + a')" is not bounded if a' is too large. We can, however, overcome this difficulty since these quantities can be replaced by (k+i)a'k+l

k+1

and

k+1 ]) (c N

(i+C)

[A/]

VVVVVV N

respectively. Consider the probability that for any of the first [A.fN/n] urns, I Xi - N/nI < 1. (If b is small enough, then A N/n < n). Here

P(,Xi -

<1) =c N,

3.8 PROBABILITY THEORY

427

where c remains between positive bounds for N > n. (We can assume this since otherwise the assertion is obvious.) Then the proof can be completed as it was demonstrated in the proof of (a).) Solution to part (b). We can assume that with the 6 given in the previous part N/n3 > 6, otherwise, the previous result implies the desired inequality. Now a' = cA/n, where c is between positive bounds (they depend on 6 and K). The proof then can be completed with this a' in the same way as was shown at the end of the solution to part (a).

Problem P.12. Determine the value of

sup [log E - E log

1<ÂŁ<2

where l= is a random variable and E denotes expectation.

Solution. Since log a is concave in

and 1 < < 2,

logy = log[(2 - l;) + ( - 1) 2] > ( - 1) log2, which implies that E(logl;) > [E(l;) - 1] log 2. Note that 1 <

and therefore log

E(log ) < log E(l;) -

1] log 2

< max [log t - (t - 1) log 2] 1<t<2

=logto-(to-1)log2=K where to = 1/ log 2 E [1, 2]. We will next show that K is the least upper bound. Consider the random variable l=o defined by P(eo = 1) = 2 - to and P(lo = 2) = to - 1. Then log E(lo) = log to and E(log o) = (to - 1) log 2, therefore, K = - log log 2 - 1 + log 2. Remark. A similar proof shows that if W(t) is a concave and continuous function on the closed interval [a, b], then

sup {co(E(e)) - E(co(e))} = max

-b-

.a co(a)

A further generalization is also possible for continuous but not necessarily concave functions cp.

Problem P.13. Find the limit distribution of the sequence q,, of random variables with distribution P (?ln = arccos(cos2

(2j - 1)7r)) = 2n

(arccos(.) denotes the main value.)

(i = 1...... n).

3. SOLUTIONS TO THE PROBLEMS

428

Solution. Let

be defined as 272n1 =

Then 7 1 n

(j=1,2,...,n).

Since n E (0, 1) (n = 1, 2, ... ), and the

arccos(cos2

function arccos(cos2 7rx) maps the interval (0,1) into (0, 7r/2), P(7?n < x) = 0 for x < 0 and P(7?n < x) = 1 for x > 7r/2. Suppose x E [0, 7r/2]. Since cos x is strictly decreasing in the interval [0,7r], arccos x is strictly decreasing in the interval [-1, 1]. The distribution of Sn is symmetric with respect to 1/2, therefore, P(7jn < x) = P(arccos(cos2 7rG) < x) = P(COS2(7rG) > COS x)

= P(cos(7ren) >

cOS x) + P(cos(7rG) < - COS X)

= P(7rG > arccos(- cos x)) + P(7rG < arccos( COs x)) arcco cos x )

= 2P(7rn < arccos cos x) = 2P(n <

For xe(0, 7r/2), arccos cos x E (0, 7r/2); thus only the value of P(n < y)

for ye(0,1/2) is to be determined. Let k be the largest integer such that k/n < y. Then n

2j-1 < k

P(n<k+1k+1

Therefore, limn-o. P(en < y) = y, and consequently,

ifx<0,

lim P(qn < x) _ n-oo

arccos cos x

1 1

if x E (0, 2 ),

ifx> z.

Remarks.

1. The random variable qn may take the same value twice.

2. The numbers \j = cos((2j - 1)/2n)7r (j = 1, 2, ... , n) are the zeros of the Chebyshev polynomial Tn = cos(n arccos x), and the corresponding Cotes numbers are An,j = 7r/n. Using the characteristic function of 7jn, one can easily prove the following.

Theorem. Let f be a continuous function in [-1, 1], and let n be a random variable defined by n

(j=1,...,n), (n=1,2,...).

Assume that is a random variable with density function 11(7r VI - x2) in the interval (-1, 1) and zero otherwise. Then the sequence converges weakly to the random variable f ().

3.8 PROBABILITY THEORY

429

In the problem, select f (x) = arccos x2 so that the limit distribution of the sequence {rij is the distribution of ri = arccosl;2. Simple calculations show that the density function of 77 is 1

1-t.22

xe(0, 2) and 0 otherwise.

Problem P.14. Let p and v be two probability measures on the Borel sets of the plane. Prove that there are random variables bl, S2, 711, 772 such

that (a) the distribution of ( i, 2) is p and the distribution of (771, r/2) is v, (b) i < 111, C2 < n2 almost everywhere, if and only if µ(G) > v(G) for all sets of the form G = U%` 1(-oo, xi) x (-oo, yi).

Solution. The necessity part of the assertion is obvious, since if _ (Sl S2) and ,q = (Mi rj2), then qiG implies eG, so

v(G) = P(rifG) <

µ(G).

First, we prove the sufficiency part of the assertion for the special case and yi, ... ,yn, where it and v are concentrated to the finite set x1i ... yl,... , yn and respectively. Define the graph G with vertices x1,.. . , where the vertices xi and yj are connected by an edge if and only if xi _< yj ((a, b) < (c, d) means a < c and b < d). Define a random variable ( , ri) that

takes the values (xi, yj) with probability ai,j. These numbers aij satisfy the following relations:

(a) aij > 0; (b) Ean--1 aij = v(y,); (c) Ej1 aij = p(xi); (d) If aij > 0, then there is an edge between xi and yj. By the KSnig-Egervary theorem, such numbers aij exist if and only if f o r any set Y C {y1, ... , y, } the p measure of the set X consisting of all points xi connected with the elements of Y is at least v(Y). Assume that Y = {yi, ... , yk}

with yi = (yi, yi )

and

G = Uk 1(-0o, y) x (-00,y ) Then a point xi is connected to Y if and only if xi E G. Therefore,

E p(xi) = µ(X n G) = p(G) > v(G) > v(Y). xiEX

Consequently, there are numbers aij having properties (a)-(d), and any vector variable (, that has the values (xi, y3) with probabilities aij satisfies the requirements of the theorem.

3. SOLUTIONS TO THE PROBLEMS

430

Now consider the general case. Denote by F(x, y) the distribution function of the underlying variable and put F1,

(X/' x") = !-W-oc, x') x (-oo, x"))

F,.(y', y") =

v((-oo, y') x (-oo, y")).

The existence of a distribution function F satisfying

(ii)

F(x, oc) = F. (x) F(oc, y) = F, (y)

(iii)

F(y, y) = F, (y)

(i)

solves the problem, since for a random variable (C, r7) having distribution F, ÂŁ < q holds with probability one because F(y, y) = F(oo, y).

Divide the square [-n, n] x [-n, n] into n4 small squares of sides 1/n. Define An as follows: concentrate the /t measure of every point into the closest node. (More precisely, consider the set of all points of the plane that are closest to a given vertex, and concentrate the Fc measure of this set to the given vertex. If a point is in the same distance from two or more vertices, then select the one that has the smallest abscissa among the vertices with smallest ordinate.) Let vn be defined analogously. Then (Âľn, vn) satisfies the requirements of the theorem, since they are measures concentrated in finitely many points. i]n) such that n < qn , the distribution Therefore, there is a variable

of G is An, and the distribution of 77n is vn. Let Fn be the distribution Then function of Fn (Y, Y) = Fn (00, Y),

furthermore, Fn(x', x", 00, oc) = An

00, X') X (-00, x")),

Fn (00, 00, Y', y") = vn((-oo, y') x (- 00, y"))

Let S be a countable and everywhere-dense set of the continuity points of

F and F. Select a subsequence nti such that F(a, b) = limi .

Fn; (a, b)

exists for all a, b E S. Extend the definition of F by the relation

F(x, y) = sup{ lim Fn; (a, b) : a < x, b < y, a, b e S}. 8-00

It is easy to verify that F is a distribution function and that it satisfies (i), (ii), (iii). Thus, the proof is completed.

3.8 PROBABILITY THEORY

431

Problem P.15. Let X1, X2, ... , Xn be (not necessarily independent) discrete random variables. Prove that there exist at least n2/2 pairs (i, j) such that H(Xi+X1)> 1 min {H(Xk)}, 3 1<k<n

where H(X) denotes the Shannon entropy of X.

Solution. Consider the graph G with vertices {1, 2, ... , n}, and assume that i and j are connected with an edge if and only if the inequality H(Xi + X3) > 1 min H(Xk) 3 1<k<n

(1)

does not hold. There is no loop in G, since for all i

H(Xi + Xi) = H(Xi) > 1 min H(Xk). - 3 1<k<n

We are going to show that there is no triangle in G. Let i, j, and 1 be different elements of {1, 2, ... , n}. Obviously, Xi = 2 ([Xi + Xj] + A + X1 - [Xi + X1]), that is,

H(Xi) = H([X2 + Xj] + [Xi + X1] - [Xj + Xi])

H(Xi+X3)+H(Xi+Xi)+H(XX+X1) < 3 max{H(XX + X3), H(XZ + X1), H(XJ + Xi)}. So

1 min H(Xk) < 1 H(Xi) 3 31<k<n

< max{H(Xi + Xj), H(X2 + X1), H(XJ + X1)}.

Therefore, there are two vertices among i, j, and l that are not connected. By a well-known theorem of Pal Turin (see, Mat. Fiz. Lapok 48, (1941), pp. 436-452), the number of edges of such a graph is not larger than n2/4.

Consequently, the number of pairs (i, j) such that (1) holds is at least n2 - 2. (n2/4) = n2/2. Remark. The bound n2/2 is the best possible. Let X be a random variable such that H(X) 0, and define X1=X2=...=X[2]

= X,

X[1+1 = ... = Xn = -X2 If either 1 < i < n/2 < j< nor 1 < j < n/2 < i < n, then H(Xi+X3 The number of such pairs (i, j) is equal to [n2/2].

= 0.

3. SOLUTIONS TO THE PROBLEMS

432

Problem P.16. Let SI, b.... be independent, identically distributed random variables with distribution P(61 = -1) = P(61 = 1) = 2

WriteSn =6+6 +

+ G n (n= 1,2,...), So =0, and max Sk .

Tn = V

O<k<n

Prove that lim inf (log n) Tn = 0 with probability one. n-oo

Solution. Denote maxa<k<b Sk by (Sb - Sa)*, and maxo<k<n Sk by S. First, suppose that we have two sequences g(n) and f (n) such that 00 (a)

E P{(f(n))-1,2'Sf(n) > g(f(n))} < oo n=1

and 00

P (Sf(n+1) - Sf(n))* +

f (n)g(f(n)) <

f(n+1)

n=1

logf(n+1)

_ 00

(b)

for all E > 0. If such sequences g(n), f (n) are found, then by the BorelCantelli lemma with probability one, only finitely many of the events Sf(n)

> g(f(n))

f(n)

and infinitely many of the events (Sf(n+1) - Sf(n))* +

f(n)g(f(n))

< log f(n + 1)

f(n + 1)

occur. Consequently, infinitely many of the events Sf(n+1)

< (Sf(n+1) - Sf(n))* + Sf(n)

f(n+1)

logf(n+1)

will occur with probability one. Hence, it is sufficient to find sequences g(n), f (n) satisfying (a) and (b). We are going to apply the approximation

P I Sn <

x\ 1

edu

3.8 PROBABILITY THEORY

433

(see, for example, A. Renyi, Foundation of Probability, Holden Day, San Francisco, 1970). This approximation holds if x is replaced by a sequence xn = o(n 3 ). So

(Sf(n+i) - Sf(n))* +

f(n)9(f(n))

f(n + 1)

> P (Sf(n+1) - Sf(n))* <

f(n+1)- f(n)

< log f (n + 1)

f (n)9(f (n))

logf(n+1)

f(n+1)

e_ 22/2du > 2 A,

where A

.f (n)9(f (n))

logf(n+1)

f(n+1)

is small enough. Therefore, we have to find sequences f (n) and g(n) such

that Â°O

1: logf(n+1)

(i )

if (n)92(f (n)) < o0

(ii)

f(n + 1)

n=1

f (f (n))

(iii)

e-"2/2du <

oo.

It is easy to see that, for example, g(f (n)) = n3 and f (n + 1) = (n!) 14 satisfy these relations.

Problem P.17. Let the sequence of random variables {Xm, m > 01, X0 = 0, be an infinite random walk on the set of nonnegative integers with transition probabilities

i>0' 4i=P(Xm+1=i-1IXm=i)>0, i>0.

Tai = P(Xm+1 = i + 1 I Xm. = i) > 0,

Prove that for arbitrary k > 0 there is an ak > 1 such that

Pn(k)=PI\o<j max Xf=k <n satisfies the limit relation L

lim

Pn(k)ak < oo

L-+oo L

n=1

3. SOLUTIONS TO THE PROBLEMS

434

Solution. Introduce the notation

Pn (k) = P( max X. < k), 0<j<n

P*n(k)=P(maxXj<k and Xn=i) (0<i<k). o<j<n Obviously, Pn(k) < PP (k) _

oP*n(k). It is easy to see that the

following inequality holds: k

P*n(k)pipi+l ... Pi+k+1 + PP+k+l (k) < Pn (k). i=O

Define Ek = omin jPiPi+1 ... Pi+k+l }.

Then 0 < Ek < 1 and P`+k+l(k) < (1 - Ek)Pn(k)

Using mathematical induction, one can easily show that for all positive integers n,

,n(k) < (1 -

Ek)11(k+1)-2

If 1

< crk < [

]1/(k+1)-2

1-Ek

then

P.(k)ak < (1 -

Ek)2n-2

and this inequality implies the statement.

Problem P.18. Let Yn be a binomial random variable with parameters n and p. Assume that a certain set H of positive integers has a density and that this density is equal to d. Prove the following statements: (a) limn-,,, P(Yn E H) = d if H is an arithmetic progression. (b) The previous limit relation is not valid for arbitrary H. (c) If H is such that P(Yn E H) is convergent, then the limit must be equal to d.

Solution. (a) By the assumptions of the problem,

P(Yn=k)= (fl)pkqn_k

(q = 1 -p).

For any given k, limn- 0 P(Yn = k) = 0; consequently, limn-.". P(Yn E H) does not change if finitely many elements of H are changed. We

3.8 PROBABILITY THEORY

435

can therefore assume that H is a complete residue class with respect to a module D:

H={k: k=-amodD}. Then

(fl)pkqn_k.

P(Yn E H) _

(1)

k=-a( D)

Denote by e a primitive Dth unit root. It is known that D-1 (k-a)v

V=0

D if k { 0

if k

a(D), a(D).

We can write (1) in the following form: 00 D-1

P(Y E H) _ E

k=0 v=0

(fl)pkqn_kE(k_a)v k

D-1

L_00

kk=O

_ D E -av 1:

(n) (W)kgn-k

D-1

E-av(q

'=0

+ EVp)n -' D = d,

since in the last sum, for v > 0, the absolute values of the corresponding terms are less than one, so they converge to zero as n -* oo. b) We construct a sequence H with zero density such that

I'm supn-00 P(Yn E H) = 1. The expectation and variance of Yn are np and npq, respectively, so the Chebyshev inequality implies that

P(I Yn - np > A npq) 5 a2 If An is a sequence such that limn-0 An = oo, then

lim P(I Yn - npJ<A/)=1.

n-oo

Let nk be an arbitrary increasing sequence of natural numbers, and let

c/ H = U (nk - [nk/4], nk + [nk/4]) k=1

By the previous observation, k

E H) = 1,

and so lim sup P(Yn E H) = 1. n--+oo

3. SOLUTIONS TO THE PROBLEMS

436

Finally, if nk increases sufficiently fast (for example, nk = 10k), then H has zero density. (c) The majority of the contestants verified that if H has density, then lim P(Yn E H) = d in the sense of Cesaro, that is,

En P(Y H) v=1

(n--goo).

From this property, the original statement follows, since for arbitrary convergent sequence, the Cesaro limit is the same as the usual limit. One can also prove that we do not need to assume the existence of the density of H, since if P(Yn E H) is convergent, then H automatically has a density that equals limn-,,. P(Yn E H). A more general result can also be verified: for an arbitrary set H of natural numbers, define h(x) = I In E H : n < x}1. Then Sn

lim

n-oo n

- h(np) = 0. np

Thus H has a density if and only if P(Yn E H) is convergent in the sense of Cesaro.

After these remarks, we can start to prove part (c). We are going to prove that if the density of H is d, then pn = P(Yn E H) --+ d in the sense of Abel, that is, E°° o(pn -Pn-1) = d in the sense of Abel, = d. If we prove this result, then we are ready, since limn-00 P(Yn E H) = c implies P(Yn E H) --+ c in the sense of Abel, that is, c = d. For JxJ <1, )xn

namely, lima-.l-o EnO°_o(Pn -Pn-1

G(x) = E(Pn - Pn-1)xn = (1 - x) E pnxn n=0

n=0 00

x)n =0 ExEEH x)

(fl)pkqnkxn k

kEH

n=0

00 n(n - 1)...(n - k + 1)(gx)n-k

(Px)

x)kE

k! 1

(1-qx)k+1

1-x

1-qxk

-qx) .

Substitute z = px/(1 - qx). If x --+ 1 - 0, then z ---f 1 - 0, and further-

more (1 - x)/(1 - qx) - 1 - z. So X

lim G(x) = z--.1-0 lim (1 - z) 1-0

(2)

kEH

if this last limit exists. Let an be the indicator sequence of H, that is, an equals 1 if n E H,

3.8 PROBABILITY THEORY

437

and equals 0 if n H. The limit of an in the sense of Cesaro is the density of H, which is d. Then, by a theorem of Frobenius, the limit of the sequence an in the sense of Abel also exists and equals d. This Abel

limit is the right-hand side of (2). Consequently, lim.,_l_o G(x) = d, which was to be proved.

Problem P.19. Let {G1}Â°1=1 be a double sequence of random variables such that (i, j, k, l = 1, 2, ... ).

Eeijek1 = O((log(21i - ki + 2) log(2Ij - lI + 2)) -2)

Prove that with probability one,

m n kl ---> 0

as max(m, n) -+ oo.

k=1 1=1

Solution. Let b+m c+n S(b, c; m, n) =

where

E G1

and m, n > 1.

b, c > 0,

k=b+11=c+1

Using assumption (1) and the identity thickmuskip=1.4mu b+m c+n

(

E(S2(b,c,m,n)) _ E E

b+m c+n-lc+n-1

E E E E(G14,1+j)

k=b+ll=c+1 j=1

k=b+ll=c+1

c+n b+m-lb+m-k

EEE

E(Glek+i,1) I

i=1

1=c+1k=b+1

b+m-1c+n-1b+m-kc+n-1

+2k=b+ll=c+1 EEE i=1

E(GiG+i,l+j) j=1

+E(G,l+jG+i,l)) , it is easy to see that

m-1 n-1 1 E(S2 (b, c, m, n)) = O{mn F F i=0 j-0 (log(1 + 2i) log(1 + 2j))2 }

= O{mn (log

mn 2m)2 (log 2n)2

}

(m, n = 1, 2, ...) .

Put M(b, c; m, n) = max max IS(b, c, k, l)1, 1<k<m 1<1<n

(2)

3. SOLUTIONS TO THE PROBLEMS

438

and use the following result (Theorem 4 in F. Mdricz, Momemt inequalities

for the maximum of partial sums of random fields, Acta. Sci. Math. 39 (1977), pp. 353-366). Assume that the nonnegative function f (b, c; m, n) (b, c, m, n E N) satisfies the following inequalities:

f(b,c,m,n), f(b,c;m,n),

f(b,c;h,n)+ f (b + h, c; m - h, n) f(b,c,m,n)+ f (b, c + i; m, n - i)

forb>0,c>_O,and1<h<m,1<i<n. Let X(m)andA(n)betwo nondecreasing sequences, and define K and A as follows:

K(1) = X(1),

A(1) = A(1).

Form> 2 andn>2, K(m) = X(h) + K(h - 1), A(n) = A(i) + A(i - 1),

i = [2 (n + 2)]

([. ] denotes the integer part of real numbers). Assume that for some y > 1, and for all b, c > 0 and m, n > 1, E(IS(b,c;m,n)I") <_ K^1 (m) XT (n) f(b,c;m,n).

Then E(M"(b, c; m, n)) < K7(m)A"(n) f (b, c; in, n)

for all b, c > 0 and in, n > 1. In our case, choose f (b, c; in, n) = mn and X(m) = V/m--/ log 2m, .fi(n) _ V,'n-/ log 2n, ry = 2. Note that f (b, c; m, n) is an additive set function on the rectangle [b + 1, b + m] x [c + 1, c + n]. Furthermore, for 2P < m < 2P+i P

K( m ) < K(2P+' - 1)

= E X (2P) _ k=0

2P" '° = O ( log2m =(p+1) O

k=0

2k/2

Consequently, E(M2 (b, c; m, n ) = O{ mn

(log 2m)2 (log 2n)2

}

In the case b = c = 0, define S(m, n) = S(0, 0; in, n) = Em i E

(3) Sk1

Since (2) holds, °°

°°

22k221

k=01=0

E(S2 (2 k 21)) = O(1)

°)

°°

22k221

E 22k221 (k + 1)2(1 + 1)2 k=01=0

< 00

3.8 PROBABILITY THEORY

439

A theorem of B. Levi implies that with probability one, 1

2k21

S(2k21) --+ 0

max(k, l) --+ oo.

(4)

To complete the proof, we have to show that 1

max

2k21 2k <m<2k+1 21<n<21+1

IS(m, n)

- S(2k 21)1 -- 0 ,

(5)

with probability one, as max(k, 1) -+ oo. Consider the following decomposition: S(m, n) = S(2k, 21) + S(2k, 0; m -2 k , 2') + S(0, 21; 2k, n - 21) + S(2k, 21; m - 2k, n - 21),

from which we see that the left-hand side of (5) is not greater than 221 1 {M( 2k 0.2 k21)+M(021.2 k 21)+M(2 k 21.2 k 21)}.

By (3), 00

E(M2 (2k,

21)) < 21; 2k,

22k221

oo,

k=01=0

and the theorem of B. Levi again implies that 2k21

M(2k21, 2k21) -4 0

(6)

with probability one, as max(k, 1) -+ oc. Notice that 1-1

M(2 k' 0; 2k, 21)

M(2k, 2q,

2k,

2q),

4=-1

where, by definition, M(2k, 2-1; 2k, 2-1) = M(2k, 0; 2k,1). A Toeplitz lemma and (6) imply that with probability one l-1

2k21

M(2k, 0; 2k, 21) <

q=-1

211q 2k2q M(2k, 24; 2k 2q) -, 0

as max(k, 1) --+ oc. A similar argument shows that 2k21M(0,21;2k21) , 0, as max(k, 1) -+ oo, with probability one. Combining (4) and (5), we complete the proof.

3. SOLUTIONS TO THE PROBLEMS

440

Problem P.20. Let P be a probability distribution defined on the Borel sets of the real line. Suppose that P is symmetric with respect to the origin, absolutely continous with respect to the Lebesgue measure, and its density function p is zero outside the interval [-1, 1] and inside this interval it is between the positive numbers c and d (c < d). Prove that there is no distribution whose convolution square equals P. Solution. Assume indirectly that there exists such a distribution Q. Then Q([-1/2,1/2]) = 1, and its moments satisfy the relations l/2

Al =

10 xkdQ(x)

xkdQ(x) <1. ÂŁ112

Thus,

limsup'` I kkl M1-

Therefore, the characteristic function coQ of Q is analytic on the real line. Since P is symmetric, cpp is real. The equality cpp(0) = 1 and the continuity of cop imply that, for any x with sufficiently small absolute value, cpp(x) > 0. For such x values cpp(x) = (cpQ(x))2, so cpQ(x) is also real. Since coQ is analytic, it is real everywhere, which implies that c 2 p (x) > 0 for all x. By a well-known theorem (see for example E. Hewitt, and K. Stromberg, Real and Abstract Analysis, Springer, 1965, p. 409), we know that if a density function is bounded and its characteristic function is nonnegative, then its characteristic function is integrable. Therefore, cpp is integrable, so the density function p should be equal to a continuous function almost everywhere, but because of its "jumps" at -1 and 1, this is impossible.

Remark. For many related negative and positive decomposition results see I. Z. Ruzsa and G. J. Szekely, Algebraic Probability Theory, Wiley, New York, 1988 and G. J. Szekely, Paradoxes in Probability Theory and Mathematical Statistics, Reidel (Kluwer), Dordrecht, 1986.

Problem P.21. Let po, p,.... be a probability distribution on the set of nonnegative integers. Select a number according to this distribution

and repeat the selection independently until either a zero or an already selected number is obtained. Write the selected numbers in a row in order of selection without the last one. Below this line, write the numbers again in increasing order. Let Ai denote the event that the number i has been selected and that it is in the same place in both lines. Prove that the events

Ai (i = 1, 2,...) are mutually independent, and P(Ai) = pi. Solution. We will show that for any k (k = 1, 2, ... ), and any sequence P(Ai1Ai2

... Aik) = pilpi2 ...pik.

3.8 PROBABILITY THEORY

441

Modify the experiment in such a w a y that in addition to zero and a repetition of a number, we also stop a t selecting a n y of the numbers il, i2 ,- .. , ik. Consider now a (finite) outcome of the original experiment in which Ail Ail ... Aik occurs. (The probability of infinite outcome is equal to zero.) From the original sequence, omit the values i 1, i2, ... , ik. In this way, an arbitrary outcome of the modified experiment can be uniquely obtained. On the other hand, it is easy to see that there is only one w a y to insert the numbers il, i 2 ,- .. , ik in such way that event Ail Ail ... Aik occurs, that is, the numbers i1, i 2 , ... , ik are placed in their increasing locations. This one-toone correspondence implies that the probability of the event AilAi2 ... Aik is equal to pilpi2 ... pik 1.

Problem P.22. Let X1,. .. , Xn be independent, identically distributed, nonnegative random variables with a common continuous distribution func-

tion F. Suppose in addition that the inverse of F, the quantile function < Xn:n be Q, is also continuous and Q(O) = 0. Let 0 = Xo:n < Xl:n < the ordered sample from the above random variables. Prove that if EX1 is finite, then the random variable y

1 [ny]+1

O = sup o<y51

(n + 1 - i)(Xi:n -- Xi-1:n) -

n i=1

fo(1 - u)dQ(u)

tends to zero with probability one as n -+ oo.

Solution. Introduce the notation [ny]+1

HH(y) =

(n + 1 - i)(Xi:n - Xi-l:n) 7E

i=1

and

HF(y) =

y - u)dQ(u). f (1 0

Thus, HF(1) = E(X1) and Hn(1) = limHn(y) yti

= - EXi,

(1)

n i=1

that is, the problem generalizes the strong law of large numbers for nonnegative random variables. Because of the continuity of F, the independent random variables Yi = F(Xi), 1 < i < n, are uniformly distributed on the interval (0, 1). Denote by En(y) = n-l x {k : 1 < k < n , Yk < y} the empirical distribution function of Y1, Y2,. .., Yn, and let Un(y) = { l

Yk:n

if kn l < y!5 n k = 1, ... , n,

Yn:n

ify=1

3. SOLUTIONS TO THE PROBLEMS

442

be their empirical quantile function. Further, let

1<k<n, Xk<x} (0<x<00)

Fn(x)=n-1x{k and

Qn(TJ) _

{

if k-' <y< k- k=1,...,n,

Xk:n Xn:n

if y = 1

be the empirical distribution and quantile function of the original sample, respectively.

Consider the random function U.. (y)

Gn(x) = f

(1- En(u))dQ(u)

By the continuity of F and Q, JQ(un(y))

Q,.(y) (1 - Fn(x)dx

(1 - En(F'(x)))dx =

Gn(y) =

with probability one, where the exceptional set (where the original sample

elements coincide) is independent of y. If (k - 1)/n < y < k/n for some integer 1 < k < n, then this last integral is the following: X1-

(1 - Fn(x))dx =

Xs n

(1 - Fn(x))dx,

where k

(1 i=1

1)(Xi:n - Xi-l:n) = Hn(y)

Since Gn(1) = Hn(1) with the common value given in (1), we conclude

that for n=1,2,..., P{ sup IHn(y) - Gn(y)I = 01 = 1. 0<y<1

It is sufficient to show that for n -+ oo,

An = 0<y<1 sup IGn(y) - HF(y)I - 0 with probability one. Obviously,

An* < sup IGn(y) - HF(U(y))I + sup IHF(Un(y)) - HF(y)I 0<y<1

= Q(1) +,N(2) ,

O<y<1

(2)

3.8 PROBABILITY THEORY

443

Notice that for 0 < y < 1, Un(y) has only n different values from the interval [0, 1]; therefore, Only < On 3)

= sup f y(1- En(u))dQ(u) - f y(1- u)dQ(u)I. 0<y<1

Let 0 < e < 1 be an arbitrary value. Then

i$)

< J (1 - En(u))dQ(u) + J 1

(1 - u)dQ(u)

1 1

+ Q(1 - e) sup Iy - E. (y)

Since Q(1 - e) < oo, the third term tends to zero with probability one (Glivenko-Cantelli theorem, see, for example, A. Renyi, Probability Theory, Akademiai Kiadd, Budapest, 1970, VII. ยง8). The first term (I(A) denotes the indicator function of the event A) can be written as

(1 - En(u))dQ(u) =

1 E

1ni=1 Jl

(1 - I({Ui < u}))dQ(u),

1 E

and because of the strong law of large numbers, this quantity tends to 1 fi E(1 - u)dQ(u) with probability one.

In summary,

P{limsupLn3) <2 (1 - u)dQ(u)} = 1, 1_E n-oo where the upper bound can be made arbitrarily small by choosing a sufficiently small E since E(X1) < oo. Thus, we have proved that the first term of (2) tends to zero (.41) -> 0) with probability one as n - oo. It is known (see, for example, M. Csorgo and P. Revesz, Strong Approximations in Probability and Statistics, Akademiai Kiado, Budapest, 1981, p. 162) that the Glivenko-Cantelli theorem is also valid for the quantile function of a uniformly distributed sample. Therefore, sup lUU(y) - yI --+ 0

0<y<1

with probability one, as n -> oo. Using the fact that the integral fo (1 u)dQ(u) is a continuous function of y, for the second term in (2) we get P{ lim

n-.oo

One)

= 0} = 1,

which completes the proof.

Remark. Suppose n machines start working at time t = 0, and let Xl:n < X2:n < ... be the failure times for these machines. Then nHn(y) is the total time until the ([ny] + 1)th failure. The best published result (N. A. Langberg, R. V. Leon, and F. Proschan, Characterization of nonparametric classes of life distribution, Annals of Probability 8(1980), pp. 1163-1170, Theorem 3.2) shows only pointwise convergence, that is, for all fixed 0 < y:5 1, Hn(y) -+ HF(y) with probability one, as n -+ oo.

3. SOLUTIONS TO THE PROBLEMS

444

Problem P.23. Let X0, X1, ... be independent, identically distributed,

nondegenerate random variables, and let 0 < a < 1 be a real number. Assume that the series k=0

is convergent with probability one. Prove that the distribution function of the sum is continuous.

Solution. Define 00

Z = > ak-1Xk

and

Y = E akXk. k=0

k=1

Then Y = Xo + aZ, where Y and Z have the same distribution, and X0 and Z are independent. Assume that the distribution function of Yis not continuous. Define

p=maxP(Y=a), a and denote by a1, ... , an the points for which

P(Y=aj)=p. Then

p=P(Y=aj)=P(Xo+aZ=aj)

P(Xo = x)P(Z = a

P(Xo=x)>O

P(Xo = x) max P(Z = at - x) < p.

P(Xo=x)>O

Therefore, we have equality everywhere. This implies >xER P(X0 = x) 1, that is, X0 has a discrete distribution, and

P(Z= a,-x)=p

P(Xo=x)>0.

Put P(Xo = x) > 0},

K = {x :

Pa={x: P(Z=aax)=p}. Then

C n

nP,

j=1

3.8 PROBABILITY THEORY

445

Since IPal=nand P,,,=P0+a, K C Po +a.,,

that is, K - aj C Po

j = 1,...,n.

This relation can hold only if IKI = 1, that is, if X0 is degenerate.

Remark. For another proof and some related results see I. Z. Ruzsa and G. J. Szekely, Algebraic Probability Theory, Wiley, New York, 1988, Section 5.5.

Problem P.24. Let X1, X2, ... be independent random variables with the same distribution:

P(Xi = 1) = P(Xti = -1) = 2

(i = 1, 2, ... ).

Define

So =0,

e(x,n)_ Ilk: 0<k<n, Sk=x}j

(n=1,2 .... ), (x=0,Âą1,Âą2,...),

and

a(n) _ I {x : l; (x, n) =1} 1

(n = 0,1, ... ).

Prove that P(lim inf a(n) = 0) = 1

and that there is a number 0 < c < oo such that P(limsup a(n)/logn =

c)=1. Solution. Represent the values of So, S1i ... in a two-dimensional coordinate system as follows: starting from the origin, make a step to the right, and move up or down if Xn = +1 or Xn = -1. Note that a(n) counts the points whose second coordinates are visited exactly once by this random walk.

First, we show that

P(lim inf a(n) = 0) = 1, which means that the events An = {a(n) = 0} occur infinitely many times with probability one. In the moments when the random walk returns to zero, there are infinitely many such points with probability one, the value

of a(n) is at most one, since only the extreme values can occur once. However, the probability that between two returns the random walk visits the extreme values at least twice is positive. Therefore, among the infinitely many instants, there will be one, and thus there will be infinitely many ones, when the value of a(n) is zero.

3. SOLUTIONS TO THE PROBLEMS

446

Let us consider the second statement. We will proceed via several lemmas. Denote by An the set of second coordinates visited only once, that

is, An = {x : there is a k, 1 < k < n, such that Sk = x, and for j $ k, Sj # x}. Then a(n) = IA 11.

Lemma 1. limn-. P(Sj > 0, 0 0, 0 < j < n) _ c*>0. Proof. Define the following conditional distribution: µn (dx, dy) = P

Sn = dx,

sup Sk = dy

V/n o<k<n

jSj >0

1 <_ j< n

If n = 2m + 1, then

P(Sj>0,0<j<n,Sn-Sj>0,0<j<n) =P(Sj>0,0<j<m,Sm> SUP (Sn-Sk)-(Sn-Sm), m<k<n Sn-Sj>0,m<j<n,Sn-Sm> sup (Sk-Sm)) o<k<m

= P(Sj > 0, 0 < j < m) P(Sn - Sj > 0, m < j < n),um * ,um(A), where

A = {(x1, yi, x2, y2) : xi > y2 - x2, x2 > yi - xl}. We know that the sequence µn of measures tends weakly to a measure p* without atoms and has a positive value on the open subsets of the set {(x, y) : x > 0, y > 0}. On the other hand, 9 P(Sj>0,0<j<m)P(Sn-Sj>0,m<j<n)- 47r n' 1

so the statement of the lemma holds for integers of the form 2m + 1. For integers of the form 2m, the proof is similar.

Lemma 2. Suppose k - alogn, ak(n) = (°kn)). Then for arbitrary y > 0, there is an no = no(k,77) such that for any n > no, [(c* - 77) logn]k < Eak(n) < [(c* + ii) logn]k.

Proof. Put

C(r,t)={S,.<S3<St,r<j<t}, D1(t) = {Sj < St,

0 < j < t},

D2(t)={Sj>St,

t<j<n}.

Then

P(C(r, t)) = P(C(0, t - r)) = t c* r (1 + o(1)), P(Di(t)) =

-(1 +o(t))

P(D2(t)) =

VIn - -t ('+o(')).

(K = 2/3

447

3.8 PROBABILITY THEORY

Put

A+ =Anf1{z:z>0}, a (n) _ (a+k(n))

a+(n) = I A+ I ,

< 3k < n, use the notation

and f o r 0 < jl < j2 <

Bjl,...,jk = Dl(jl)C(jl,j2) ... C(jk-l,jk)D2(jk) Then

Eat (n) =

P(Bj1,...,jk),

O<jl<...<jk<n

since the event Bj1 ,..,jk means that

(il, ... ,jk) C A,+, and the event a(n) = l contains exactly (k) such events. Put

P(C(j1,j2)) ... P(C(jk-l,jk)) = U(0,l - j).

U(j,l) _ j=j1 <...<3k=t So

n-r

U(0, r) E P(D1(j))P(D2(r + j)) j=0

r=0

n-r 1 const E U(0, r) E n

j=1

r=0

7n=---3 ---r

< const E U(0, r) r=0

k-1 n

const

< [(c* + 17) log n] k.

P(C(01 A) j=0

On the other hand,

Ea+ (n) >

P(Dl(jl)C(jl, j2) ... C(jk-1,jk)D2(jk)) 0<j1 <n/3

n/3k>j=-j=-1>0

conSt

n/3

E P(DI (j)) j=1

> const [(c* - 2) log

k-1

P(C(O1 i)) j=1

nk-1 3k 1

> [(c* - rl) log n] k

since k - a log n, so [(c* - 77/2)/(c* - q)] a 10g n > log n if n is large enough.

3. SOLUTIONS TO THE PROBLEMS

448

We can get the same result for An in a similar way. So

Al plus maybe one more point, if Sn > 0,

I So the statement of the lemma is valid for a(n).

Lemma 3. For all K > 0 and e > 0, there is an no = no(K,e) such that for all n> no, n-(K/c')-e < P(a(n)

< K logz n) < n-(K/c*)+e.

Proof. Put k = (K log n) /c*. Then

P (c(n) > K logz n) = P(ak(n) > (Klo2n))

< Eal kg(n))

If n is large enough, then we can use Lemma 2 to get tog((c'e)/(c'+n)) < n-(K/c*)+e

P(a(n) > K logz n) <

For the proof of the lower bound, put q,,,, = P(a(n) = m). Then m

Eak (n) _ E q. m

(k)'

Put k' = (K/c*) (1 + E2) log n. Then for any sufficiently small e, 1

(M)

Eak, (n)

m>(k+e) loga n and

qm k, < 3 Eak' (n) m<K log a n

because if k" = ((K + e) /c*) log n, then k" > k' and 4m

m>(K

loga n

qm (kmil(mkl)

Ck'/

(k")

((K+e) log 2 n)

((K+) toga n)

((K+ek) 'logs n)

4m m

< l((K+e) logs n) Eak (n) k"

Ck J

3.8 PROBABILITY THEORY

449

Again using Lemma 2 with 77 = e3, we get qm

2 25Kc*

(M) < (c` log n)k' exp{(-

+ 0(e:2)) log n}

m>(K+c) log2 n 3 Eakw (n)

if n is sufficiently large. On the other hand, K loge n

(m qm1 k/) 1: m< K log2 n \

)

qn+1mk

((K logs n) ( k / m<K log2 n

Using Lemma 2 again, we get

qm(M) ki < 3 Eak, (n). m<K log2 n

These inequalities imply

q , . k J > 3Eak,(n). K log2 n<m<(K+E) loge n So

. T,

P(a(n) > K log 2 n) >

qm (M)

log n k'

K loge n<m<(K+E) loge n

(K+E)

)

E(ak'(n)) > in i(1+E3)log[er T(1+E3)-2 IKC}}ee

- 3 ((K+E) loge n) - 3 k'

This last expression tends to 1 for E > 0 and 77 > 0, so we have checked the lower bound. Using the above lemmas, we are able to prove that

P Clim sup a(n) = c = 1. loge n

Put

Mn=inf{j:Sj >n}. Then lim Mn = 0

and

M(n+l)2 - Mn2 > 2n + 1.

Consider the instant when the random walk reaches the height k2 for the first time. Start a k-step walk from this point, and denote it by Uk , that is,

Uk = {0, SMk2+1 - SMk2 , ... , SMk2+k - SMk2 }.

3. SOLUTIONS TO THE PROBLEMS

450

These walks are independent from each other. Put Bk

the walk Uk visits exactly once at least (c* - E) log2 k points with positive coordinates

Then P(Bk) > k-1+E/2 if k is large enough. So > P(Bk) = oo, hence P(lim sup Ak) = 1. But on the event Ak a(Mk2 + k) > (c* - E) log2 k, and since Mk2 + k < ks + k for sufficiently large values of k,

a(n)

c* - e

loge n >

for infinitely many n with probability one. The upper bound of Lemma 3 with K = c* + e shows

P(a(n) > (c* + e) loge n) < 00,

that is, lim sup a(2) < c* + e,

log n

for large values of n with probability one. Recall that the Kolmogorov 0-1 law implies that lim sup a (n) /log 2 n is constant with probability one; therefore,

P (limsup

a(2) =cl=1 log n

with a suitable c*/64 < c < c*.

Remark. Unfortunately, the original formulation of the second statement of the problem was incorrect. It contained a log divisor instead of loge. The proof of the correct statement appeared in P. Major, On the set visited once by a random walk, Probab. Th. Rel. Fields 77, (1988), pp. 117-128. The present proof follows this work. Problem P.25. Let (S2, A, P) be a probability space, and let (Xn, .'n) be an adapted sequence in (Q, A, P) (that is, for the a-algebras.Fn, we have ,r1 C_ .F2 c ... C_ A, and for all n, Xn is an Fn-measurable and integrable random variable). Assume that E(Xn+1I.Fn) =

2Xn +

2Xn-1

(n = 2, 3 ... ).

Prove that supnEIXnI < oc implies that Xn converges with probability one as n -4 oc.

Solution. Put Yn = Xn + (1/2) Xn_1 (n = 2, 3, ... ). Then Yn is an Fn-measurable and integrable random variable. Furthermore, E(Yn+1I.Fn) = E

(x+1 + 2XnIFn) = 2Xn + 2Xn-1 + 2Xn = Yn

for n > 2. That is, (Yn,.Fn,

n = 2, 3 ...) is a martingale.

3.8 PROBABILITY THEORY

451

Since supra E(IYnI) < (3/2) sup, E(I XnI) < oo, the martingale convergence theorem implies that Yn is convergent with probability one, and lim Yn(w) = Y(w) n-oo if w E S2', where P(S2') = 1. Here Y is a random variable that is finite with probability one, and therefore we may assume that it is finite for w E W. We will show that for w E S2', the sequence Xn(w) is also convergent. Let w E 1' be given, and an = X, (w). We will prove that if the sequence bra = an+(1/2) an_1 converges to c, then an converges to (2/3) - c. Assume

first that c = 0. Then for any E > 0, there is an N such that Ibnl < e/2 as

n > N. So 1

Ian+1I = Ibn+i - 1anl < Ibn+1I + 2IanI <

E+2anl

for all n > N. Using this inequality for n = N,. .. , N + k - 1, we get IaN+k I

+ - + ... + 2k + 12 I < E + 12 I 4

< 2E

if k is large enough. However, for all n > N + k, Ianl < 2E. That is, the sequence converges to zero.

If c # 0, then the same proof applies for the sequence an = an - (2/3) C.

Problem P.26. Let X1, X2,... be independent, identically distributed random variables such that Xi > 0 for all i. Let EXi = m, Var(Xi) = a2 < oo. Show that, for all 0 < a < 1, a2a2 / X1 + ... + Xn li nVarQ n mz( 1-a )

Solution. Put Xn = (X1 +

+ Xn)/n and Sn = X1 +

+ Xn. Notice

first that

[(Xi +..+Xn/l) n

=nE[(X-m+ma-E(X))2] J

= nE(Xn - ma)2 - n(E(Xn - ma))2. First we show that the second term tends to zero as n - oo. The quadratic Taylor polynomial of (1+z)a with remainder term implies that for z > -1, 1- (1 +z)a < (1- a)z2 - az. From the Jensen inequality and this relation, we get

0 < ma - E(Xn) = maE (1

- (;y) M 2

3. SOLUTIONS TO THE PROBLEMS

452

which implies the assertion. Consider next the first term. Introduce the

notation Zz = (Xi/m)-1 and Zn = (Z1+Z2+ +Zn)/n. Then Zi > -1, E(Zj) = 0, Var (Zi) = Q2/m2, furthermore for arbitrary 0 < e < 1, nE(X" ma)2 = m2anE((1 + Zn)a - 1)2

= m2a[nE(((1 + Zn)a - 1)2llznl<e)

+ nE(((1 + Zn)a In the next step, we will use the following inequalities. In the first term,

alzl/(1 + zj) < 1(1 + z)a - 11 < ajzj/(1 - jzj) for jzj < 1. In the second term, I (1 + z) a -11 < z a for z > -1. Observe, furthermore, that if Iz j < e,

then jzj/(1 + Izi) > zI/(1 + e) and Izj/(1 - jzj) < zj/(1 - e), therefore, m 2a [n(1

+ e)2 E(Z n 1

'<E )]

< n E(Xnm a)2

< M 2a [n(1

E)2E(7'nhIZ,.I<e)

Therefore, it is sufficient to show that z

nE(ZnIlZnl<e) -+

and 2

nE(ZnallZnl>e) -' 0.

Let 0 be the standard normal distribution function. Then for large n,

r ( nZn 1

az E

- ) = m2 n > nE(Z2I n I-9nj!5e m2 = nE(Z2) a2

> mz E since

(

MZn

y2dO(y)

11\ Q/M J

nZn/(Q/m) -+ 0 in probability. In addition, e2anE

(

< ne2aE

Zn Zn

III>1 2

= ne2a-2E(Z2I n Since

mz =nE(Zn) we get 2

nE(ZnII Z..I <_e) asn -goo.

-'m 2;

) -+ 0.

... )<_emn

3.8 PROBABILITY THEORY

453

Problem P.27. Let F be a probability distribution function symmetric with respect to the origin such that F(x) = 1 - x-1K(x) for x > 5, where

K(x) -

if x E [5, oo) \ UO0_5(n!, 4n!),

if x E (n!, 2n!], n > 5, if x E (2n!, 4n!), n > 5.

3- Zn,

Construct a subsequence ink} of natural numbers such that if X1, X2, ... are independent, identically distributed random variables with distribution function F, then for all real numbers x lim P cc

X j < 7rx

l nk j=1

+ 1 arctan x.

Solution. Observe first that the independence of X1, X2, ... implies that for any nk > 1 and t # 0 the characteristic function of the random variable nk Ynk

_ ink EXj j=1

can be given as

E(ett ^k) = (1 -

(1/7) ' hnk (tllr)ItIJJ\nk. nk

The assumed symmetry implies that, for any s # 0,

(1-E(etnkx')=

link(s)= ISI

I3I J-5

nk I

1-cosnkx dF(x)

1-cosnkx I dF(x)-2

(1-cosy)d(yKl Iylnk" )5e

The above form of the characteristic function is useful since by a suitable choice of subsequence ink} we can guarantee that these functions converge to a-ItI. Note that this is the characteristic function of the Cauchy distribution, which is the limit distribution given in the problem. (See A. Renyi, Probability Theory, Akademiai Kiado, Budapest, 1970, IV. §10, VI. §2.) The first term of the above representation of hnk tends to zero if for k --+ oo, nk ` oc. It is easy to see that if K - 1, then the second term (with the choice nk = k) tends to the limit

°° 1 - cosy

2 o

dy = 7r

454

3. SOLUTIONS TO THE PROBLEMS

Thus, in this case nk = k would be a suitable choice. Therefore, it is sufficient to choose the sequence {nk} so that for any y > 0 and s 0,

K((y/lsl)-nk)- 1ask-goo. Let {ak} be an arbitrary sequence of positive integers such that ak -f 00 and ak/k --+ 0 as k -+ oo. Then for arbitrary x > 0 and sufficiently large

k, 4k! < akk!x < (k + 1)! and for any such k, K(akk!x) = 1. That is, if we choose nk = akk!, k = 1, 2, ... , then the continuity theorem of P. Levy, implies that hflk (s) -a 7r for k -+ oo. Therefore, E(eit" k) -+ a-Itl for

arbitrary t # 0 as k -+ oo, and the repeated application of the continuity theorem completes the proof.

Problem P.28. Let a E C, IaI < 1. Find all values of b E C for which there exist probability measures with characteristic function 0 satisfying 0(2) = a and 0(1) = b.

Solution. If 0 is a characteristic function with the required properties, then 0(0) = 1, 0(-1) = b, and 0(-2) = a, and the self-adjoint matrix 0(1)

o(O)

qS(2)

0-1)

0(1) 0(0)

00) 0(-2) 0(-1)

is positive semidefinite. So its determinant is real and nonnegative:

1+ab2+ab2-2bb-aa> 0.

(1)

If a = u + vi and b = x + yi, then (1) can be rewritten as (2 - 2u)x2 - 4vxy + (2 + 2u)y2 < 1 - u2 - v2.

(2)

Let a' be a complex number such that its square equals a. It is easy to see that if IaI = 1, then (1) holds for the points of the interval between a' and -a' (since IbI < 1, it does not hold for the other points of the connecting line). If IaI < 1, then (1) holds for the points in the interior and on the curve of the ellipse with major axis 2 + 2Ia1 and foci a' and -a'. These statements are shown next. Assume first that lal = 1. Let the random variable X be defined as arg a'. Then Ox (1) = a' and OX (2) = a'2 = a, so b = a' is suitable. A similar proof shows that b = -a' also satisfies the required properties. Next assume that lal < 1, and let b be a boundary point of the ellipse. In this case, we have equality in (1). If b2 - a = re2ai, then (1) implies

that r2 = (b2 - a)(b2 - a) = lb14 - ab2

= Ib14 - 21b

that is,r=1-IbI2.

+ = (Ib12 -1)2,

ab2 + Ia12

3.8 PROBABILITY THEORY

455

Put c =Re a-aib, w1 =a+ axecosc, and w2 =a -axccosc. Furthermore, let

pi =

P2 =

Im a-"b

2+21

-c2'

Im a-"b

2 1-c2

Since lbl < 1, p1 and P2 are positive. Define X as follows: P(X = w1) = p1

and

P(= w2) = p2

Then the characteristic function 0 of X satisfies

(1) = pie"' +p2ew2i = eai(piearccosc +p2earccosc) = e' ((pi + P2) cos(arccos c) + i(pi - P2) sin(arccos c)) = eai((p1 + p2)c + 2i (p, - p2)V'1 - ) = e«i((Re a-aib) + i(Im e-ccib)) = b and

0(2) = ple2wli +p2e2w2i = e2ai(Pie 2arccosc

+p2e-2arccosc)

= e2aa ((pl + p2) cos(2 arccos c) + i (p1 - p2) sin(2 arccos c) ) = e2ai((p1 + p2)(2c2 - 1) + i(pi - p2)(2c 1 - c2)) = e2«i(2(Re a-"b)2 - 1 + 2i(Re e-aib)(Im e-aib))

= e2ai((e-aib)2 + le-aibl2 - 1) = b - e2ai(1 - lbl2) =b2-re2ai=b2-(b2-a)=a. Hence 0(1) = b and 0(2) = a. Finally, we show that the set of the suitable b values is convex, and consequently, the interior points are also good. Let b1 and b2 be two suitable points, and let Q1 and Q2 be two probability distributions such that OQl (2) = OQ2 (2) = a, OQl (1) = b1i and r¢Q2 (1) = b2.

Define Q3 = )tQ1 + (1 - \)Q2 with 0 < A < 1; then Q3 is also a

probability measure, OQ,(2) = a, and 0Q,(1) = )b1 + (1 - ))b2.

Problem P.29. Let Y(k), k = 1, 2,... be an m-dimensional stationary Gauss-Markov process with zero expectation, that is, suppose that Y(k + 1) = A Y(k) + e(k + 1),

k = 1,2,...

Let Hi denote the hypothesis A = Ai, and let Pi(0) be the a priori probability of Hi, i = 0, 1, 2. The a posteriori probability P1(k) = P(H1IY(1), ... , Y(k)) of hypothesis H1 is calculated using the assumptions P1(0) > 0, P2(0) > 0, P1(0) + P2(0) = 1 .

3. SOLUTIONS TO THE PROBLEMS

456

Characterize all matrices A0 such that P{limk-,,. P1(k) = 1} = 1 if Ho holds.

Solution. Let Y(1) be an m-dimensional, normally distributed random vector with zero expectation and covariance matrix D. The so-called "white noise" e(k), k = 1, 2, ... , is independent from the past of the process Y(k), that is the s(k)'s are m-dimensional, normally distributed random vectors independent from Y(1) and from each other, and they have zero expectation and covariance matrix Q. The background of the problem is as follows: from hypothesis H1 and H2, we accept the one for which the posterior probability converges to 1. We examine the robustness of our procedure, that is, if Ho holds, then our decision must result in the hypothesis that is "closer" to H0.

Suppose that Q is regular. Since the process is stationary, each Y(k) has the same covariance matrix D, k = 1, 2, .... From (1) we obtain the following matrix equation: D = E(Y(k + 1)Y* (k + 1)) = E([A Y(k) + s(k + 1)] [A Y(k) + e(k + 1)]*) = E(A Y(k)Y*(k)A*) + E(E(k + 1)e*(k + 1)) = ADA* + Q.

Hence Q = D - ADA*; therefore, D is also regular. Since Q is positive definite and D is at least semidefinite, for any x E R" such that Dx = 0, 0 < x*Qx = x*Dx - x*ADA*x = -(A*x)*D(A*x) < 0,

that is, x = 0. Since P1(0) + P2 (0) = 1, P1(k) + P2 (k) = 1 for all k = 1, 2, ... , and therefore limk..,, P1(k) = 1 if and only if limk.,,. P1(k)/P2(k) = oo. The Bayes theorem implies that, for i = 1, 2,

P(HijY(1) =yi,...,Y(k) = Yk) =

PY(1),...,Y(k)IH;(y1,...,Yk)Pi(0), PY(1),...,Y(k) (Y1, ... , yk)

where py(1),...,Y(k) (or py(1)....,y(k)IH;) denotes the probability density func-

tion of variables Y(1), . . . , Y(k) (or the corresponding conditional density function under Hi). This relation implies that

P(HIIY(1) =y1,...,Y(k) =yk) _ PY(1),...,Y(k)IH,(Y1,...,yk)P1(0) P(H2I Y(1) = y1, ... , Y(k) = yk)

PY(1),...,Y(k)jH2 (y1, ... , Yk)P2(0) .

/ (2)

Under the hypothesis Hi, the random variables

Y(1) : N(0, D),

Y(k + 1) - A2Y(k)

N(0, D)

are independent. (Here N(M, S) denotes the normal distribution with expectation M and covariance matrix S. In the following, we will denote the density function of N(M, S) by pN(M,s))

3.8 PROBABILITY THEORY

457

In this case, PY(i),...,Y(k)I Hi (y1, ... , Yk)

= PN(o,D) (Y1)PN(o,Q) (y2 - Aiy1) ... PN(o,Q) (Yk - A:yk-1) 1

e-zyiD-1y1

(27r)mIQIe

-2(Y2-Aiyl)*Q-1(Y2-Aiy1)

(2-7r --I DI

(yk-Aiyk-1)*Q-1(yk-Aiyk-1)

(21r)jQI e 1

(21r)kmIDIIQIk-1

C-2 [yiD-1y1+E' 2(yi-Aiyi-1)*Q-1(yi-Aiyi-1)]

Substituting this relation into (2) and using the variables Y(1), . . we get the following ratio of the posterior probabilities: P1(k) =PP1 (O) exp( 1 P2 (k) 2 2 (0)

Y(k),

E[Y(7) - A2Y(j - 1)]*Qi-1[Y(j) - A2Y(j - 1)] j=2

E[Y(7) - A1Y(7 - 1)]*Q-1[Y(j) - A1Y(7 - 1)]). j=2

That is, if the hypothesis Ho holds, that is, the process is based on the matrix A0. Then with the random variable L(Y(1), ... , Y(k)) k-1

_ E[(Ao - A2)Y(7) + IF(i + 1)]*Q-1[(Ao - A2)Y(7) + IF(i + 1)]f-1 k-1

- E[(Ao - A1)Y(7) +,F(7 + 1)]*Q-1[(Ao - A1)Y(7) + E(7 + 1)] j=1

we get

=exp (L(Y(1) ...,Y(k)) \I

The question to be answered is as follows: when does the limit relation

m L(Y(1),... , Y(k)) = oo hold with probability one?

The process (Y(k), e(k+1) k = 1, 2, ...) is an ergodic Gauss-Markov process. By the ergod theorem, lim 1 L(Y(1), ... , Y(k))

k-+oo k

= E([(Ao - A2)Y(1) + e(2)]*Q-1 [(Ao - A2)Y(1) + E(2)])- E([(Ao - A1)Y(1) +s(2)]*Q-1[(Ao - A1)Y(1) +E(2)]) = tr [(A0 - A2)*Q-1(Ao - A2)D] - tr [(A0 - A1)*Q-1(Ao - A1)D] = L

3. SOLUTIONS TO THE PROBLEMS

458

with probability one. If L > 0, then limk_,,,. P1(k)/P2(k) = oo with probability 1, and this is the same as limk_ o. P1(k) = 1.

Remark. For L < 0, we get limk ,,,. P2(k) = 1 with probability 1. One

can also prove that for L = 0, neither Pl(k) nor P2(k) tends to 1 with probability one, that is, no appropriate decision between Hl and H2 can be made.

Problem P.30. Let X and Y be independent identically distributed, real-valued random variables with finite expectation. Prove that

EJX +YJ > EJX -Y1. Solution 1. For any real numbers a and ,Q, ja + /jI - la-01 = 2 sign(a/3) min{jcaj, I/3 }. Therefore,

E(JX +Yj) - E(JX -Yj) =E(IX +YJ - IX -Yj) =2E(min{IXj, Yj}sign(XY))

=2J0 'IP(IXI > t, JYJ > t,XY > 0) - P(JXI > t, JYJ > t,XY < 0)}dt.

Since X and Y are independent, identically distributed variables, and P(Z > t) = P(Z > t) almost everywhere,

E(JX +YJ - IX -Yj) =2

=21

j{P(X > t)]2 + [P(X < -t)]2 - 2 P(X > t)P(X < -t)}dt

> t) - P(X

-t)2dt > 0.

E(JX +Yj) > E(JX -Yj).

Solution 2. This problem has a simple solution, if we use the Fourier transform of generalized functions. We may assume that X has a nice density function f (x), say an infinitely many times differentiable function with finite support. Then the statement of Problem P.30 can be rewritten as

f IxI(f *f)(x)dx> J IxI(f *f )(x)dx,

3.8 PROBABILITY THEORY

459

where * denotes convolution, and f - (x) = f(-x). This formula can be rewritten by means of the Plancherel formula if JxJ is considered as a generalized function. The Fourier transform of Jxi, when it is considered as a generalized function, equals -2Q-2. (See, for example, I. M. Gelfand, G. E. Shilow, Verallgemeinerte Funtionen (Distributionen) I. VEB Deutscher Verlag der Wissenschaften, Berlin (1960) Vol. 1). The Plancherel formula, which is actually the definition of the Fourier transform of generalized functions, and the definition of Q-z as a generalized function (see formula (5) on page 60 of Gelfand and Shilow's book) state that the last formula can be rewritten as

f2(-a) - 2f2(0)] da

-2

{f(a)f-() +

(-o) - 2f(O)f- (0)] da,

where-denotes the Fourier transform. A simple calculation shows that the last formula is equivalent to the relation

' [f(ar) - f(-U)]

dar < 0

This relation clearly holds, since f(Q) - f (-v) is a purely imaginary number.

Remark. The second solution was suggested by Peter Major.

Problem P.31. Let X1, X2, ... be independent, identically distributed random variables such that, for some constant 0 < a < 1,

P {Xi = 2k/-} = 2-k,

1, 2, .. .

Determine, by giving their characteristic functions or any other way, a sequence of infinitely divisible, nondegenerate distribution functions G,,, such that sup

-00<X<00

nl/a

Solution 1. Put S,,, = X1 +

<x -Gn(x)

->0

n ->oo.

+ X,, n = 1, 2. .... The characteristic

function

p(t) = E (eitxl)

Â°Â°

kk=1

ck/n>

eit2

t E R.

Introduce the sequence ryn = n/2 flog n1 , n = 1, 2, ... , where lul = min{n E N : u < n}. Obviously, 1/2 < ryn < 1 for all n. The independence

3. SOLUTIONS TO THE PROBLEMS

460

assumption implies that for all t the value of the characteristic function of the random variable Sn/nl/a is 00

1 + Deit2(1/')k/nl/' -1) 2k

) = cpn (1/n 1/a ) =

cpn(t) = E(e

k=1 00

it2a/a)(k-rlognj)/7n/°

(

2 Flog nj

1+n

(eit2' /°/ryl,/°

2k- Flog ni

- 1) n

r=- Flog nl +1

For a given 1/2 < 'y < 1, define 00

h--1 (t) _

r=-oo

tER

and put

7(T) = eh1(t)

tER

Since for all t E IR, 00

Ih7(t)I

27 E

2-r+71-1/alt

-0 <2y+ r=0

r=1

-t

1_1/a

1-21-1/'

ItI,

the above definition is correct, and .y(.) is the characteristic function of the random variable 00 2r/a

Z.y = E r=-oo

,yl/a Yr('Y),

where the Yr(y)'s are independent Poisson-distributed random variables with expectation E(Yr(ry)) = -y/2r, r = 0,±l,±2,... . For the sake of simplicity, Z.y can be defined as the limit in distribution of the partial sums

2r/a

E ,y1/aYr('Y) r=-n By the continuity theorem of P. Levy it is easy to see that this is possible. Obviously, Zr is an infinitely divisible random variable. Observe that, for any fixed t E R, .y (t) is continuous as the function of as a bivariable function is continuous on ry on the interval [1/2, 1]. the domain [1/2] x R). We are going to prove that the distribution function H.y of Z.y, which is uniquely defined by the relation

7(t) =

eitxdH (x),

t E R,

3.8 PROBABILITY THEORY

461

is continuous on the whole real line for all -y E [1/2, 1]. Simple calculations show that 7

-1 a 2

°°

dt = 00

J 2

°°

exp

r=-oo

f f

- cos(s2))

r=-oo

exp -ry E(1 - cos(s2-k/,))2" ds

f exp

(eis2 °_ 1)

k=0 oo

-ry E (1 - cos(s2-k/,))2k

exp

ds,

(1)

k=,c(s)

where for arbitrary s > 0, ac(s) denotes the smallest integer k > 1 such that log(s/7r) < k - 1. Since ac(s) < 2 + log(s/-7r) and s2 -k < 7r/2, for all k > ic(s), s2-k/a < 7r/2. Thus, using the inequality

(0<x<2),

1-cosx> 2x2 we get 00

cos(s2-k/,,)) 2k >

k=rc(s)

4 6221-,c(s) >

Hence

k=ic(s)

(2)

f00 I

2(1 2/a)k >

.y(t)Idt < 2'y1/a

e-try/7rsds

< oo.

The continuity of H., follows from the inversion formula of characteristic functions. (See, for example, A. Renyi, Probability Theory, Akademiai Kiadd, Budapest, 1970.) Consider the random variables Zry with distribution functions G,,(x) = H7 (x), x E R , and characteristic functions XFn (t) _ 7n (t), t E IR, n

1,2,... . Since for n -* oo

(eit2'/°/7,1,x°

h 7n()

- 1) ^In2r , 0

r=- [log n) +1

for arbitrary t E R, it is easy to see that

On (t)-an(t)-*0

n ->oo.

(3)

3. SOLUTIONS TO THE PROBLEMS

462

Finally, introduce

A(F., Gn) = where

sup

-oo<X<00

I F'n(x) - Gn(x)I

Fn(x)=P j Sn <x},

xER,

and consider a sequence {n'} of positive integers tending to 00. Since 1/2 < ryn, < 1, the Bolzano-Weierstrass theorem implies that there is a subsequence {n"} C {n'} such that for some ry E [1/2, 1], ryn --> ry as n" -4 00. In this case, A(Fn,,,

A(Fn,,, Hy) + A(Gn,,, H-r)

and Tn (t) = yn (t) -+ .y (t) for all t E R as n" -> 00.

From the continuity theorem of Levy, we know that Gn (x) _ H.yn (x) --> H.7(x) at any continuity point of the limit function. Since the limit function is continuous everywhere, it converges uniformly by a well-known theorem of Polya. Hence

A(Gn", Hy) -> 0

n" -> oo.

The relation (3) and a similar argument imply that

A(Fn,,,H.y)->0

n"->oo.

Since the sequence {n'} was arbitrary, A(I''n, Gn) --> 0

n -> oo.

Solution 2. Define the sequence Z.y of random variables in the same way as above: 'Yn =

flog nl '

L,, = [logn],

Z.y =

n-1/ÂŤ

E 2(L^+r)/ÂŤyr(ryn),

r=-oo

where the Yr(ryn)'s are independent Poisson-distributed random variables The "three-series theorem" implies with expectation ryn/2r = that the sum defining Z.y is convergent. We will use estimations (1) and (2), which guarantee that the absolute value of the characteristic function

Wt) of the distribution function Gn (x) of the random variable Z.y is integrable, and this integral remains bounded independently of n. The distribution functions Gn(x) are obviously infinitely divisible, and we will show that they satisfy the requirements of the problem. This assertion will follow from the following two statements.

3.8 PROBABILITY THEORY

Put Sn = n-1/a En=1 Xk

463

and let Fn(x) denote its distribution func-

tion.

(a) There exists a sequence En -> 0 of positive numbers such that for all x E (-oo, oo), Gn(x - En) - En < Fn(x) < Gn(x + En) + En.

(b) For all n, the density function gn(x) of Gn(x) exists, and there is a real M that is independent of n, and for all x E (-oc, oo), gn(x) < M. First, we show that the requirements of the problem follow from (a) and (b). Indeed,

Fn(x)-Gn(x) = Fn(x)-Gn(x+En)+Gn(x+En)-Gn(x)

En+M En, (4)

Gn(x)-Fn(x) = Gn(x-En)-Fn(x)+Gn(x)-Gn(x-En)

En+M En. (5)

Since En -> 0, from (4) and (5) we have the limit sup., I Gn (x) - Fn (x)0,I - . which was to be proved. Statement (b) follows from estimations (1) and (2).

Proof of statement (a). Let dn be a sequence of real numbers that tends to infinity slowly enough. Put Ln -dn

Znl) = n-1/a > 2(Ln+r)/ayr(_Yn), r=-oo 00

Zn2) = n-1/a Zn3) = Z-rn

2(Ln+r)/ayr(7n)

- Znl) - Zn2) n

S(1)

= nl

EXkI(Xk<2(Ln-dn)/a)yk(7'n), k=1 n

S(2)

= n-1/a E XkI(Xk>2Ln+dn)/c)yk('Yn), k=1

Sn3) = Sn

- S(1) - S(2)

where IA is the indicator function of the set A.

The relations P(5,2) # 0) -> 0, P(Zn2) # 0) -+ 0, E(Sn1)) -+ 0, E(Znl)) -+ 0, and simple calculations imply that Sn - Sn3)

and

Zyn - Z(3)

where = denotes the stochastic convergence. Observe that Sn3) can be written in the form Ln+dn

21/avin-1/a,

Sn3) _

l=Ln-dn

(6)

3. SOLUTIONS TO THE PROBLEMS

464

where vl = 01j, 1 < j < n , xj = 21/'J, 1 = 1, 2, ... , and #A denotes the cardinality of the set A. Denote the distribution function of by F(3) (x), and that of Z,3) by Gn) (x). We are going to prove that Var (F(3) (x), G'3) (x)) -' 0,

(7)

where Var ({pi}, {qi}) denotes the distance >i Ipi - qiI of the discrete distributions {pi} and {qi}. From relations (6) and (7), statement (a) follows immediately. We only have to prove the convergence in (7). Denote by distr{Yj(ryn), IiI < dn} and distr{vl+L,,,Ill < dn} the joint distributions of the random variables {YY (yn)} and {vl+L }, respectively, (Ii 1 < dn, 111 < dn). The limit relation (7) follows from the convergence Var (distr {Yj (ryn),

distr {vl+L },

Ii 1 < dn},

(8)

111 < dn).

This last statement can be proved as follows. Since dn tends to infinity slowly enough, by restricting the range {pl , Ill < dn} of the random variables appearing in (8) by log2 n, the resulting error does not affect the validity of the following estimation:

P(vl+L..

pl , I4I<dn )=

d,.

111=-dnPL! (n-El-dnpl)!

(l_2_L+l) d

t--drift

)

11 2-P1

l=-d

dII

)n_Edn

l=-d -n2-L"-1

(n2

pl!

= (1+0(log3 n

1+0

n log3n)/

do l

f lP(Y(7n)=pl)

= (1+0 The proof is complete.

(P(Yn(-Yn)=pl ,

IlI -dn)

3.9 SEQUENCES AND SERIES

465

3.9 SEQUENCES AND SERIES Problem S.I. Let the Fourier series a 2 + E(ak cos kx + bk sin kx) k>1

of a function f (x) be absolutely convergent, and let 2

ak + bk > ak+1 + bk+1

Show that

(k = 1,2,...).

f02, (f(x + h) - f(x - h))2dx

(h > 0)

is uniformly bounded in h.

Solution. By the Denjoy-Lusin theorem, >' 1(Iak! + jbki) < oo. Hence 00

10k<00

(Pk=

ak+bk; k=1,2,...).

(1)

k=1

So we obtain El 1,02k < oo. By the Riesz-Fischer theorem and the completeness of trigonometric functions, it follows that (f (x))2 and thereby (f (x + h) - f (x - h))2 is integrable. Easy calculation shows that

f(x+h) - f(x- h) - 2E(bksinkhcoskx-aksinkhsinkx). k=1

The Parseval formula implies that 2ir

(f (x + h) - f (x - h))2 dx = 41r E Pk sin2 kh.

(2)

k=1

By the condition ok > ek+1 (k = 1, 2, ... ), we obtain kPk < 1 Pa (k = 1, 2, ... ). So by (1), kok = 0(1). Using sin2 x < lxi, by (1) we obtain 00

OK)

kQk = O(h).

ok sin2 kh < IhI k=1

k=1

By this and (2), we conclude the theorem.

Remark. Instead of the absolute convergence of the Fourier series and

Pk > Ok+1 (k = 1, 2,... ), it is enough to suppose that E' 1 kPk < oo. Gabor Halasz and Tibor Nemetz remarked that by the conditions of the theorem (1/h) fo (f (x + h) - f (x - h))2 dx = o(h) holds.

3. SOLUTIONS TO THE PROBLEMS

466

Problem S.2. Let y1(x) be an arbitrary, continuous, positive function on [0, A], where A is an arbitrary positive number. Let

(n=1,2.... ).

yn(t) dt

yn+1(x) = 2 f 0

Prove that the functions yn(x) converge to the function y = x2 uniformly on [0, A].

Solution. We will not only prove the theorem but also approximate the rate of the convergence.

Let yi(x) and yi*(x) be given continuous functions on [0,A] such that 0 < yi(x) < yl*(x),for all 0 < x < A. Denote by yi(x),y2(x),...,yn(x),... and by yi*(x),y2*(x),...,y,**(x), . the functions obtained from yl(x) and y**(x) by the above iteration. Obviously, y,*, (x) < y,** (x) for all 0 < x < A. So it is enough to prove that the claim of the problem holds for a constant function of value C > 0, since if we choose C to be the maximum (minimum)

of y1(x), then the nth element of the function series received from the constant function will be larger (smaller) than yn(x) at each x. So let Y1 (x) - C, where C > 0, and let Y2 (x), Y3 (x), . . . , Yn (x) be the functions obtained by the iteration. We can show by induction that Yn(x) =

x2-(1/2"-')

where cn is defined by the recursion cn+1

Cn 1

(n = 1, 2, ... ; c1 = C).

(1)

2nn

Take the 2n+1th power of each side of (1): 2n+1 Cn+1 = cn2n

1 1

2n11

(1 -

The multiplier of Cn2n is positive and converges to a finite positive value (to e2) as n goes to infinity. So the quotient of the neighboring elements of the , . .. cnn, ... remains between two positive constants. Therefore, series ci, c22

by appropriate constants m and M,

mn < c n < Mn (n = 1, 2, ...)

Hence, by mn/2n < cn < Mn/2n

(n = 1, 2, ... ),

it follows that cn =

eÂ°("/2n).

(2)

3.9 SEQUENCES AND SERIES

467

Obviously,

Yn(x) - x2 =

X2-(1/2'-') . (en - 1) +

(x2-(1/2n-1)

- x2).

By (2), for all 0 < x < A, x2_(1/2n

')=0.(cn-1<A

2Icn-11

2n )

It is easy to show by derivation that the function x2-(1/2n-') - x2 achieves

its maximum on the interval (0,1) at xo = (1 - 2 2"-1, and 1)

') - xo2 =

2-(1/2n-1)

- x0

/Zn

1))

< I' -

211

)J = 2

ForA>l, 1<x<A, Ix2-22"T-x2l=x2 2n rI1-x 1

Thus,

I<A2I1-A2

n(X) x= 0 y2

02 1

n 2n

Hence, it follows that, starting from the function y1(x), which satisfies the conditions of the problem, we get yn(x) -

x2 = 0 (

where the constant included in the sign 0 depends only on A, and on

the maximum and minimum of y1 (x). So the claim of the problem is proved.

Remarks.

1. Gabor Halasz proved the following generalization of the problem. If the function G(y) is monotone increasing and continuous for all y > 0,

G(0)=0,G(y)>0fory>0, and (

f G(y) dy < +oo, 0

but

dy = +oo,

then obtaining the functions yn(x) by the recursion X

yn+1(x) = f G(yn(t))dt, 0

3. SOLUTIONS TO THE PROBLEMS

468

it follows that the functions yn(x) uniformly converge to the unique solution of the differential equation y'(x) = G(y), which goes through (0,0) and does not agree with the constant function (x = 0) in any neighborhood of the origin. 2. If we suppose that yl (x) is nonnegative instead of positive, we can state the following:

Let b be the upper bound of those x values for which y1 (u) = 0 for all

0 < u < x, and let f (x, 6) be the function such that f (x, 6) = 0 for 0 < x < 6 and f (x, 6) = (x - 8)2 for b < x < A; then the functions y.,, (x) uniformly converge on [0,A] to the function f (x, b).

Problem S.3. Let E be the set of all real functions on I = [0, 1]. Prove that one cannot define a topology on E in which fn ---+ f holds if and only if fn converges to f almost everywhere.

Solution. Let I2m+k =

k1 k

[-_,] 2n,

(m = 0,1,2.... ; k = 1, 2, ... 2 ),

and denote by fn the characteristic function of I. Then If"} will nowhere converge to 0; but if If,,,, } is an arbitrary subsequence and xnk E Ink, where {xnk, } converges to a number x E I, then obviously {fnkt } -* 0 everywhere but at x, so in particular, almost everywhere. So if T is a topological space

and xn 74 x, xn E T, x E T, then there exist, a neighborhood of x such that for some subsequence Xnk , xnk Â˘ V, and hence for any subsequence xn,,, xnk{ 74 x. Remarks.

1. The solution shows that the same statement is true if in the definition of almost-everywhere convergence we give the role of the zero-measurable sets to a set system R, where for any x E I, {x} E R but I Â˘ R. 2. With a slight modification of the solution, one can show that the statement holds when I is the set of real numbers or a closed interval.

Problem S.4. Let the continuous functions fn (x), n = 1,2,3,..., be defined on the interval [a, b] such that every point of [a, b] is a root of fn(x) = fm(x) for some n m. Prove that there exists a subinterval of [a, b] on which two of the functions are equal.

Solution 1. Denote by En,,.,, the zero set of fn(x) - f,,,,(x) for n # m. These sets can be enumerated. Let

M1 ,M2,M3i...,Mk,... be an enumeration of these sets. If Ml is a nowhere-dense set on [a, b], then there exists a subinterval [al, b1] of [a, b] that is disjoint from Ml. If M2 is a

3.9 SEQUENCES AND SERIES

469

nowhere-dense set on [a, b], then there exists a subinterval [a2i b2] of [a1i b1]

that is disjoint from M2. If all of the Mk are nowhere dense on [a, b], then we can continue this method. The intersection of all the intervals [a, b] ID [a1, b1] ID ... ID [ak, bk] I ...

is not empty and does not contain a root of fn (x) = fm (x) for any n # m. This is a contradiction. So there is an MV that is dense on some [c, d] part of [a, b], that is, the roots of the corresponding equality fjy(x) = f,, (x) are dense on [c, d]. By the continuity of the functions, it follows that fw(x) = f,- (x) on [c, d].

Solution 2. The union of the sets Enm of the previous solution is the interval [a, b]. If none of the sets En,,,, is dense on any subinterval of [c, d], then their union cannot be an interval, since an interval cannot be a union

of countably many nowhere dense sets. So there is a set that is dense on some [c, d] subinterval of [a, b]; that is, the set of the roots of the corresponding fw(x) - fj (x) is dense on [c, d]. Since the functions are continuous, each point of [a, b] is a root. Remark. Many participants used the Baire category theorem to generalize the problem.

Problem S.5. If >m=_m lam I < oc, then what can be said about the following expression? 1

lim

+00

E lam-n + am-n+1 + ... + am+n

2n + 1 m=-00

Solution. Since I mÂ°=-m I am I = v < oo, Lm=-m am = S exists. Let 1

Cn =

+00

2n + 1 m=-00

lam-n + am-n+1 + ... + am+n I

We will show that limn-00 Cn = A. Let e be an arbitrary positive number. Then there is a natural number M such that Iaml < e. JmJ>M

Hence, for n > M,

...I+

(2n+1)Cn= Iml>n+M

I...I+ n+M>!mJ>n-M

I...I. lml<n-M

3. SOLUTIONS TO THE PROBLEMS

470

E lam,-n + ... + am+n l < E am-n + ... + am+n Iml>n+M

Iml>n+M

<(2n+1)

a,,, l < (2n + 1)E, Iml>M

and

+... + am+n l <

lam-n

or < 4MQ.

n+M>ImI>n-M

Since m - n < -M < M < m + n if lml < n - M, it follows that if lm l < n - M, then llam-n+...+am+nl-ISll< E

laml<E,

ImI>M so

lam_n+...+am+nl

- (2n-2M+1)lSll

Iml<<n-MI

< E Ilam-n+"'+am+nl-lSll<(2n-2M+1)E. Iml<n-M

Therefore,

1(2n + 1)Cn - (2n - 2m + 1)ISI I < (2n + 1)E + 4Ma + (2n - 2M + 1)E. Dividing by (2n + 1) and taking the limit as n oo, lim sup lCn - HSl l < E + E = 2E. n-.oo

Since e is an arbitrary positive number, we conclude that

lim Cn = ISl

n--.oo

Remark. Let (M, or, µ) be a commutative measurable group. Namely, M is an additive group, a is an invariant a-algebra of the subsets of M under addition and p is an invariant measure, where p(M) > 0. Suppose that the function S : M x M - M x M defined by S(x, y) = (x, x + y) maps measurable sets of M x M into measurable sets of M x M. Let Al, A2, ... (Ai E a) be a sequence of sets symmetric about 0 (that is, Ai = -Ai) such that UloAi = M; Ai C Ai.+1; 0 < p(Ai) < oo; p(Ai+1 - Ai) = 0(µ(A2)); Ai + A; C Ai+j. Claim. If f (x) E L1(M, p), then

p(An)

f ff

(x + y) dFiy)I dµ. =

ffdp

This can be proved by similar methods. Mikl6s Simonovits gave this generalization.

3.9 SEQUENCES AND SERIES

471

Problem S.6. Let a1, a2i -,an be nonnegative real numbers. Prove that

Eai

Eai-1

i=1

Solution 1. If one of the ai is zero, then the statement follows immediately from the known inequality between the nth power mean and lower power means. It also follows that equality holds if all the other aj numbers are equal. So we can suppose ai > 0 f o r i = 1, 2, ... , n. Denote by Ek the sum of the kth power of the numbers a1, a2i ... , an, and by Sk their kth elementary symmetrical polynomial. We generalize the inequality:

n (Il1)>1Ek_1_<m5k+(k_1)()Ek, 1 < k < n .

(1)

The problem is the special case of inequality (1) for k = n.

For n = 1, 2, k = 1, equality holds. So we can suppose n > 3, k > 2. We introduce the following notation. If a1,... , an > 0 are real numbers, then denote by [al, ... , a, ] the sum (1/n!) E a l ... a ", where we sum over all permutations i1 , . , in of 1, ... , n. Lemma. If n > 3 and v, 04,

> 0, b > 0, are real numbers, then

[v+26,0,0,a4.... ]-2[v+6,b, 0,a4,...]+[v,6,6,a4,...]>0.

(2)

Equality holds if all of the ai are equal.

Proof. We cut the sum (2) into (3) (n - 3)! terms, so it is enough to prove that for arbitrary real numbers b1, b2, b3 > 0, b1+26 + b2+26 + b3+26 - (b1+6 b2 + b2+6b1 + bi+6 b3 + b3+6b6 1

+ b2+6b3 + b3+6b2) + bib2b3 + bsb2b3 + bib2b3 > 0,

where equality holds if b1 = b2 = b3. This is straightforward by the following obvious inequality:

xµ(x - y)(x - z) + yµ(y - x)(y - z) + z''(z - x)(z - y) > 0,

for x, y, z > 0, µ > 0. Equality holds if x = y = z. We conclude the lemma by changing x, y, z, and p into ai, a2, a3, and v/b, respectively. The ideas in the proof of the lemma appears in G. H. Hardy, J. E. Littlewood, Gy. Pdlya, Inequalities, Cambridge University Press, Cambridge, 1959. To prove inequality (1), write it in the following form:

k - 1)

1: aia -1 < nSk +

((k_1)() -

())

Ek .

(3)

3. SOLUTIONS TO THE PROBLEMS

472

The following equalities hold by the definition of [al, .... an]: 1

n(n -- 1)

ajar -1 = [k - 1,1, 0 L.- 0 1#j

n-2

n=[k,0,...0, 1

n-1

(1 Sk = 1,...1,0,...0

(4)

n-k

If we substitute the inequalities (4) for (3), and making use of the relation (k) _ n . (k-1)+ we can write (3) in the following form:

(n-1)n[k-1,1,0,...0 < 1,...1,0,...0 +I n-2

n-k

k-1-

[k,0,...0. (5)

n-1

By (n - 1) (k/n) = k - 1 - (k/n) + 1, we can write (5) as

[k-1,1,0,...0 - 1,...1,0,...0 n-2

n-k

(k-1-k)[k,0,...0 -[k-1,1,0,...0. n

(6)

n-2

n-1

We introduce the following notation:

of = [t + 1,1,1, ...1, 0.... o ] - [t,1,1, ...1, 0.... 01 k-t-1 n-k+t-1

k-t

n-k+t-1

n>3, k>2, t=0,1,...,k-1. By substituting v = t - 1, 6 = 1 into the lemma, it is straightforward that

0 = 00 < Di < ... < pt -1

(7)

Equality holds if all of the a2s are equal. We can write (6) in the following form: k-2

Ck-1- knll Akk-11

This inequality holds, since by (7), k-2

E Ot < (k - 2)Ok-1 and k - 2 < k - 1 - k. t-o

3.9 SEQUENCES AND SERIES

473

It is straightforward that if n > 3 and k > 2, then equality holds if k = n and all of the ais are equal.

Remark. Inequality (1) was first proved by Laszlo Lovasz. He also proved it for the case when some of the ais are zero, but he did not examine the cases of equality. The previous proof is the generalization of Peter Gacs's proof. He used this method for proving the case k = n, that is, for proving Problem S.6. Solution 2. If one of the ais is zero, then the statement follows immediately from the Chebyshev inequality for uniformly ordered sequences. It also follows that equality holds if all the other numbers aj are equal. So we can suppose ai > 0 f o r i = 1, 2, ... , n. Let n

i=1

i_1

i=1

F(a1,... an)=nfl ai+(n-1)ai-ai>aa-1. By the homogeneity of F, it is enough to prove that

=n.

F(a1i...,an) > 0 if

(1)

F(a1i . . . , an) = 0 holds for n = 1, 2. Let us suppose n > 3. It is enough to prove (1) in those cases when F has a local minimum at a1+ +an = n, that is, when 1

dF dai

= (n - 1)(a - ai -1) + flai = Aai

(i = 1, ... , n)

(2)

i_1

for some real A.

By the Descartes sign rule and H tai > 0, the polynomial (n - 1)x (n - 1)xn-1 + Ax + fz 1 ai cannot have more than two positive roots. So by (2) it is enough to prove that

F(x,... x, y, ... , y) > 0 for x, y > 0. k n-k By the homogeneity, it is enough to prove that g(x) > 0, where

g(x)=F(x,...x,1,...,1), k

0<x<1. g(1)=1.

n-k

If we show g(x) > 0 for 0 < x < 1, k # 0, k 54 1, then it also follows that in case n _> 3, ai > 0, equality holds if all of the ais are equal. Let 0 < x < 1, 0 < k < n. By rearranging, we get g(x)=k(n-k)(xn-x"i-1-x+1)-kxn+nxk-n+k.

3. SOLUTIONS TO THE PROBLEMS

474

Hence,

9(x) = k(n - k)(1 - xn-1) - n(1 +

+ xk-1) + k(1 + ....+ xn-1)

k(n - k)xn-1 > 0, since

k(n-k) > (n-k)(1+...+xk-1) and k(xk+ +xn-1) >

k(n-k)xn-1.

Equality holds in both inequalities if k = 1, k = n-1, that is, iff n = 2.

Solution 3.

n L ai i=1

i=1

ai -1

ati -

=2 1` L(ai - aj ) (ait-2 + ait-3 aj+... + at-2 ) 2

i=1 j=1

i=1

for arbitrary t, since aj)2(ai-2

(ai -

a -

+ ... +

alai-1

(1)

i, j = 1, 2, ... n.

for

and

I:ai=1, i=1

then

flai = i=1

n (an+1 + ai -

an+1) ? an+1 + an+1 E(ai - an+l )

i=1

n-1 = an+l (1 - n) + an+1 n

(2)

We shall use induction. Equality holds for n = 1, 2. Suppose that n > 2, an+1 < an < < al, and that the inequality of the problem holds for n. We are going to prove it for n + 1. By the homogeneity, we can suppose that 1 ai = 1. By the induction, n

n +nHai-Eai-1>0.

(n-

i=1

(3)

i=1

We have to prove n+1

n+1

i=1

nEaZ+1+(n+1)flai->ai>0 >0.

(4)

3.9 SEQUENCES AND SERIES

475

The left side of (4) can be written in the following form: n

n Eai

+ na; +i + nan+1

ii=1

i=1=1 ai + an+1

an+1)(E a2 + a;+1)

- (a +

(5)

i=1

By (3), it is enough to prove that (5) is at least as large as an+1 times the left side of (3). So, by rearrangement, we have to prove that n

(n > ai +1 -

i=1

- an+1(n > ai) -

an)

ai -1) i=1

+ an+1(fl ai - (1 - n)a'n+l - a; +i) ? 0.

(6)

i=1

By (1) and ai > an+1 (i = 1, 2, ... , n), the difference of the first two terms is n

n-2

(ai - aj )2 [a n-1 + ai

aj +

+ aj n-2

i=1 j=1 n-2

- an,+1(ai

n-3

+ ai

aj + ...

n-2)] > 0.

(7)

By (2), the third term is also nonnegative. By (7), if n > 2, then equality holds iff a1 = a2 = = an. If an+1 > 0 and a1 = a2 = = an, then the third term of (6) can be zero only in the case ai - an+1 = 0; otherwise, strict equality would hold in (2).

Problem S.7. Let f (x) _> 0 be a nonzero, bounded, real function on an Abelian group G, 91, ... , gk are given elements of G and Al, ... , Ak are real numbers. Prove that if Aif(gix) > 0

holds for all x E G, then k

kAi>0. i_1

Solution 1. We can suppose that f (gl) > 0. Denote by An the set of those elements that can be written in the form gi' , ... , gk k , where the maximum

3. SOLUTIONS TO THE PROBLEMS

476

absolute value of the numbers al, ... , ak is n, where n > 0 is an integer. Denote by S(H) the sum >xEH AX), where H is a finite set. inf S(An+1) - S(An-1) - 0 S(An)

n>0

holds, since if for some s > 0 and for all n > 0,

- S(An-1)

S(An+1)

S(An)

would hold, then S(An+1) > S(An-1) + ES(An) > (1 + e)S(An-1)

and so S(A2n+1) ? (1 + E)nS(A1)

would hold, which is a contradiction, since S(An) < sup f (x) (2n + 1)k. xEG

By (1), k

S (9i A n)

i=1

1: 1: .if(9ix) ? 0,

S(An) x1A i=1

y S(An)

that is, k

iL=1

i=11

S(9iAn)

hence

S(An) - S(92An) I- S(An+1) - S(An-1) S(An) S(An)

and therefore k

i=1

S(An+1) - S(An-1) S(An)

It follows that k i=1

\i > -

i=1

,\j

inf

n>0

S(An+1) S(An-1) S(An)

- 0.

Solution 2. Let us again suppose f (g1) > 0. It is enough to consider the subgroup G' generated by the elements 91,...,gk. We are going to

3.9 SEQUENCES AND SERIES

477

define a discrete translation-invariant measure on G'. Let E > 0, and let al ... ak be integers. For x = gi' . . . g"'`, denote by µ6(x) the number +e)-(kall+ '+lakl). The measure of G' is finite since (1 00

1: XEG

E (1 +

µ6(x) <_ 2k

E)-(al+...+ak)

ak=0

a,=0

= 2k E (1 + e) a' ...

(1 + ak=0

a1=0

and so 0 < S = J f(x)dµe(x) < oo. G'

Furthermore, µe(9ix) = (1 + E)tlµe(x),

and so

pe(9ix) - µe(x)1 < spe(9ax). Integrating inequality (1) on G', we get 0 <

Aif(9ix) dye(x) C

G, z=1 k

f f(x)dµe(x) G'

/I/'

I Ai I

i=1

.i + SE

Ef (9ix) dµe (9ix) = S

i=1

So bye -+ 0, k

SEAi>0 i=1

it follows that

Ai>0. i=1

Remarks.

1. Peter Gacs and Andras Simonovits remarked that the statement of the problem holds for an arbitrary group if k = 2. 2. Peter Gacs and LAszlo Lovasz showed that the statement generally does not hold for a noncommutative group. 3. L6szl6 Lovasz, Endre Makai, and Imre Ruzsa showed that f (x) has to be bounded. /. Laszl6 Lovasz discussed situations where the converse theorem holds.

3. SOLUTIONS TO THE PROBLEMS

478

Problem S.8. Show that the following inequality holds for all k > 1, real numbers a1, a 2 ,--. , ak, and positive numbers x1, x2, ... , xk.

Exi In

i=1

EaixiIn xi < i=1

E x1-a.

i=1

Solution 1. By the known inequality relating weighted arithmetic and geometric means,

F,=i XiYi >

,i=1 xi

(ftyi)

(xi > 0, yi > 0, i = 1, 2, ... k; k > 1).

Li=1 xi

i=1

If we substitute yi = xi a` for i = 1, ... , k and take -1 times the logarithm of each side, then we get the claim of the problem.

Solution 2. We are going to prove the following generalization of the problem.

Generalization. Let (X, S, p) be a measure space, where Âľ(x) > 0. Let x(8)1-a(9), x(s) > 0 and let a(s) be measureable functions on X, and x(s), a(s)x(s) ln(x(s)) are integrable functions on X. In this case,

f x(s)dp In

f a(s)x(s) In x(s)dp

(s)1_a(9)d,a < x

f d(s)dp

(1)

Equality holds if x(s)-a(') is a constant for almost all s E X. Proof. The following inequality holds for all positive d:

d-1>Ind. Put

(2)

f a(s)x(s) In x(s)dp u = exp x f x(s)dp

By substituting d = u. x(s)-a('), multiplying the inequality by x(s), and integrating each side on X, we get the following equality:

x(s)1-a(') dp

- J x(s)dp > 0.

(3)

This is equivalent to the claim of the generalization. Equality holds in (2) if d = 1. So equality holds in (3) (that is, in (1)) if U. x(s)-a(9) is a constant for almost all s E X, that is, if x(s)-a(9) is a constant for almost

allsEX.

3.9 SEQUENCES AND SERIES

479

Problem S.9. Construct a continuous function f (x), periodic with period 21r, such that the Fourier series of f (x) is divergent at x = 0, but the Fourier series off 2(x) is uniformly convergent on [0, 21r]. The background of the problem. The following statement is a special case of the Wiener-Levy theorem. If a positive function f (x) is periodic by 27r and the Fourier series of f2 (x) is absolutely convergent, then the Fourier series of f (x) is also absolutely convergent. The following question arose. Can the absolute convergence be replaced by uniform convergence? The problem above states that it cannot be replaced without assuming the positivity. (R. Salem constructed an f (x) whose Fourier series is uniformly convergent, but the Fourier series of f2(x) is not.) There are more ways to solve the problem. Laszo Lovasz slightly modified Lipdt Fejer's known construction: By adding the Fejer polynomials under a relatively strong gap condition, he received a continuous function f (x), whose Fourier se-

ries is divergent at x = 0, and the Fourier series of f2(x) is uniformly convergent. It is also nontrivial to prove the uniform convergence of the Fourier series of f2 (x). Lajos Posa and Peter Gacs solved the problem in an essentially similar way: They considered the function f (x), instead of its Fourier series and used the observation that if the following inequality holds for a continuous, 27r-periodic function g(x), then the Fourier series of g(x) is uniformly convergent: I

y (x)-9OI< (log

Ixi-vl)i+E

(See the Dini-Lipschitz condition in I. P. Natanson, Constructive Function Theory 1-3, 1964-65, VII. 3ยง.) Now we prove the claim.

Solution 1. Denote by sn Y; x) the nth partial sum of the Fourier series of f (x), where f (x) is periodic by 21r and integrable, that is,

f 7r

sn(f;x) = it

f(2d)

sin(2n + 1)?9 dV. sinV

(0)

If, on [0, 27r], (2n + 1)x

fn(x) = sin

(1)

then

sin2 (2n + 1)V

sn(fn, 0) = it

sin O

rr/2

1 f sin2(2n + 1)V 7r

sin 79

d19

/Zsin2(2n + 1)t9

n7r

d?9 >

f 0

sin2 0

d?9 > Al log n, (2)

480

3. SOLUTIONS TO THE PROBLEMS

where Al and later A2, A3, ... are positive constants. Note that if Ig(x)I 1 is integrable, then for all x,

Isn(g;x)I <A2logn.

(3)

The function we are going to construct will have the form

F(x) = E c sin

2n, + 1

(4)

V=1

where c,,, n are still undefined. Outside [0, 27r] we will get the function by periodic continuation. The restriction (v

Ic"I < 4-"

= 1,2,...)

(5)

and F(0) = F(2,7r) = 0 will guarantee the continuity of F(x) and the absolute convergence of (4). Let cl = 1, c2 = 1/4, n1 = 1, and suppose that cl, C2.... , cµ_1, cµ and n1, n2, .... nµ_1 are already defined. Since the functions fn(x) - defined in (1) - are twice differentiable and their Fourier series are uniformly convergent, the following inequality holds at each x, for m > A3

A=1,2.... , v - 1 .

I sm(fn,.; x)I :52,

(6)

Denote by n" the smallest positive integer for which cµ log nµ >

(7)

and

nµ > 2 + max{A3i 3nµ_1} .

(8)

If cµ+1 = min

1LA

4 , log

nµ

(9)

then (5) is straightforward. So F(x) is defined. We will use the fact that the following numbers are pairwise different:

ni + nk + l and nj - n,n

(1 < i < k, 1 < m < j).

(10)

By nl = 1 and by (8), the numbers above are different if 1 < i < k < 2, 1 < m < j < 2. If they are different for ni, nk, n nm, where i, k, j, m are less then µ, then by joining nµ to ni, nk, n3, nm, we get the following sums and differences:

>nµ-n2>...>nµ-nµ_1.

3.9 SEQUENCES AND SERIES

481

Even the smallest of them, nµ - nµ-1, is larger than the largest of the previous ones, nµ-1 +nµ_1 + 1. So the numbers at (10) are indeed pairwise different.

By (4) and (5), n + 1)x}

CµCv{cos(nµ -

F2 (x) = 2 µ,v ao

E cv + i=1

-2

c, c COS(nµ 1<v<µ

cv cos(2nv + 1)x - 1 cµcv cos(nµ + nv + 1)x. 1<v<µ

v=1

Since the numbers at (10) are pairwise different, this gives the Fourier series of F2(x). By (5), this is absolutely convergent. We still have to prove

that the Fourier series of F(x) is divergent at x = 0. To see this, divide the series into three parts: µ-1

cjsn(fnj; 0) +

Sn(F; 0) = Cµsn(fn; 0) +

j=1

Cjsnµ (fn,,; 0)

j=µ+1

=U1+U2+ U3. By (2) and (7), U1 > µ.

By (6) and (8),

(12)

µ-1

2>cj<3

1U21

(13)

j=1

Finally, by (3) and (5), 00

IU31 < A2 log nµ

c3,

A4cµ+l log nµ < A4.

(14)

j=µ+1

So by (11), (12), (13), and (14), the Fourier series of F(x) is divergent

atx=0.

Solution 2. We will define the function f (x) instead of its Fourier series, such that (0) will hold for f2(x), f (0) = 0, and 7r

lim up n-oo

sin nt f (t)dt > 0. t

(15)

f (x) will be an even function. Its nth subsum Sn(0) at 0 thus has the following form:

3. SOLUTIONS TO THE PROBLEMS

482

Sn (0) =

Jf(2t)sin2+1tdt. sin t

7r 0

If a continuous function is convergent, then its Fourier series and the Fejer means will converge to the same value, to the value of the function. So to see the divergence of Sn (0), it is enough to prove that Sn (0) 74 0. To prove this, we modify the kernel:

sin(2n+1)t sint

= sin2ntcott+cos2nt =

sin2nt ( 11 +sin2nt cott - J +cos2nt. t t

cott - (1/t) can be modified to be continuous at 0, so by applying the Riemann lemma twice, we get

lim sup f f (2t)

sin(2n + 1)t sin t

n--oo

dt = lim sup J f (2t) n

sin 2nt

= lim scup

f f

f (2t)

sin 2nt

0 IT

= lim sup p

f (t)

sir nt

dt > 0.

Thus Sn(0) 74 0, that is, Sn(0) is divergent if (15) holds. Since f (n) is even, it is enough to construct it on [0, 7r]. Let an =

Ire-n3,

an = n-2,

Pn =

en2

(n > 0)

and

In = [an, do-1], g(x) = 6n sin pnx . Define gn(x) = gn* (x) between the second and the last roots of g*(x) on In, and gn(x) = 0 elsewhere. We will define f (x) as a sum of a subsequence of the sequence {gn(x)}. Construct a sequence n1 < n2 < ... of natural numbers such that k

fk(x) dx < o

if fk = >gn;,

(16)

i=1

r sin pnj x

9nk+1(x) dx

< o2 -k

for all j < k,

(17)

where p is a sufficiently small constant, say o = 10-5. To see that such a sequence exists, observe that, on the one hand, by Pn --+ oc, by the continuity of gn(x)/x, and by the Riemann lemma, all terms of the integral in (16)

3.9 SEQUENCES AND SERIES

483

converge to zero; on the other hand, by the continuity of (sin 0.,,x)/x, the integral in (17) converges to 0 for any fixed j. Define f (x) _ Ek o g,,,,. (x). Obviously, f (x) is everywhere continuous, even at 0. We will show that the Fourier series of f (x) does not converge to 0 at 0, that is, it is divergent. For the reader's convenience, denote by Sn ~ Tn if Sn/Tn -+ 1. By (16) and (17), the difference of fo (sin pnkt)/t dt and fa nk-1 911k (t) (sin pnkt)/t dt is nk a maximum of 2e. ank-1 gnk (t) sinpnkt

bnk sine enkt dt

ank

3 Ire-nk 7fenk-(nk-1)3

= bnk

sine u U

e3nk -3nk -1

bnk

2 nk_2nk

sine u du

nt=1 3-7r

31r

So the Fourier series of f (x) is divergent at 0. Now we are going to show that 2

If2(x)-f2(y)I=

(log(Ix-YD/

(18)

that is, the Fourier series of f2(x) is uniformly convergent. Let x > y and x E In, y E I7n for some n > r. We move y to the sine curve pn sin p.,, t of I7n such that y lies on the same quarter period of the sine curve as x, but f2(x) remains the same. Ix - yI does not increase if n = m or n > M. So we have to prove (18) only in the case when x and y are on the same In and Ix - yj < lr/pn. By the Lagrange mean value theorem, If2(x)

- f2(y)I = bnI sinpnx - sinpn2yl <- 2b2npnl x - yJ = 2b2npn

- yI(log Ix - yi)2 )2

(19)

(log 1=lye

Since h(log h)2 monotonically increases if h < e-2, by (19) we have to prove

(18) only in the case IX - yI = 7r/pn; but this case is trivial. This finishes the solution. Remark. Solution 1 proves more, as it proves that the Fourier series of f 2 (x) is absolutely convergent.

3. SOLUTIONS TO THE PROBLEMS

484

Problem S.10. Prove that for every 19, 0 < 19 < 1, there exist a sequence An of positive integers and a series En 1 an such that (1) An+1 - An > (An)", 00

(ii)

lim r -4 1-0

anra exists, n=1

(iii) E an is divergent. n=1

The background of the problem. Andras Simonovits noted that An+1 > cAn (c > 1) and the (C,1) summability together guarantee the convergence.

Moreover, if we substitute x = e-b in G. H. Hardy, Divergent Series, Clarendon Press, Oxford, 1949, §7.13, then Theorem 114 can be stated as follows. Given c > 1 constant and a sequence of natural numbers, such that An+1 > cAn (c > 1), En_=1 anrAn convergent in the unit circle and its limit exists at r -+ 1 + 0, then E°° 1 an is convergent. So this theorem, conjectured by Littlewood and proved by Hardy and Littlewood, states that the Abel summability and the convergence are equivalent notions in the case of sequences satisfying the Hadamard gap condition An+1 > cAn (c > 1). Our problem states that the Hadamard gap condition cannot be substituted with a much weaker condition. The participants gave two kinds of generalization for the problem. Some of them substitute the Abel summability with the stronger (C,1) summability; others showed that for any sequence An satisfying lim inf (An+1 - An) /An = 0,

there exists a sequence EO° 1 an that satisfies the conditions of the problem. This latter statement is also a generalization of the theorem, since if An = e\/n-, then

An+1-An=e n+1_e`f =(e n+1 " -1)e`r =

1) = (1 + (o(1)) 2--,

and therefore,

An+1 - An

An+1

An An

-4 m fnr

-1

Solution. We will prove the following statement, which is slightly weaker than the previous two generalizations. Theorem. For any series {Ak} that satisfies lim inf

Ak+1k Ak

= 0,

there exists a divergent series En00=1 an that is (C,1) summable and an = 0 only if n is not in {Ak}.

3.9 SEQUENCES AND SERIES Proof.

485

Let 1 < m1 < m2 < m3 < ... be a subsequence of A1, ... , Ak, .. .

such that

2k-1

k1: mn<m2k

m2k - m2k_1 <

and

m2k

n=1

Choose a,,,, to be (-1)k+1, the others to be zero. Since E; 1 aj is 0 infinitely many times and 1 infinitely many times, En 1 aj is divergent. Denote En 1 aj by sn. Since

0<tn= S1+s2+"'+sn n

< 1,

to is monotone decreasing in [m2k, m2k+1) and monotone increasing in [m2k_1, m2k). Therefore, to prove t,,,. --+ 0, it is enough to show that tm2k -p 0:

tm2k =

(m2 - m1) + (m4 - m3) + ... + (m2k - m2k-1)

< 1 +1 k

m2k

- 0.

Hence, tm --+ 0, that is, E, 1 aj is (C,1) summable.

Problem S.11. Let 0 < ak < 1 for k = 1, 2,.... Give a necessary and sufficient condition for the existence, for every 0 < x < 1, of a permutation pry of the positive integers such that 00

air. (k) ,)k

k=1

Solution. Obviously, the following condition is sufficient: inf ak = 0, k

sup ak = 1. k

We are going to prove that this condition is also necessary. Let x E (0, 1) bk and be fixed. For a permutation 7r, let dk = 2k Lx -

(bi

+... + bk

(do = x).

If we choose the permutation 7r such that 0 < dk < 1 holds for all k, then obviously

x =

2k.

k=1

Let us suppose that 7r(1), 7r(2), ... , ir(k) are already defined such that 0 < do, d1i ... , dk < 1. We want to define -7r(k + 1) such that 0 < dk+1 < 1. This is equivalent to 2dk - 1 < bk+1 < 2dk,

(1)

486

3. SOLUTIONS TO THE PROBLEMS

and also equivalent to 2(1 - dk) - 1 < 1 - bk+1 < 2(1 - dk).

(2)

We have to take care that ak will be used exactly once in the construction. Let a, be the element with the smallest index such that at Â˘ {b1, b2i ... ,

bk}. If 2dk - 1 < at < 2dk, then we can choose bk+1 = a,. If at > 2dk, then let n be the smallest natural number such that 2n+ldk > a,; then 2n+ldk < 2a1 < 1 + a,, so 2n+1dk

- 1 < at < 2n+ldk.

Since infk ak = 0, bk+l, bk+2, ... , bk+n can be chosen sufficiently small, such that by d3+l = 2dk - bj+1i dj+n will be close enough to 2ndk. So we can achieve 2dk+n - 1 < at < 2dk+n.

Thus, we can choose bk+n+l = at.

If a, < 2dk - 1, or equivalently 1 - a, > 2(1 - dk), then we repeat the above process by substituting 1 - a,,1 - aj,1 - d3 in the role of a,, aj, dj, respectively, and by using (2) instead of (1); and in the meantime, we use the equality 1 - dj+l = 2(1 - d3) - (1 - b3+1). By supk ak = 1, we can choose the numbers 1 - bj to be sufficiently small, that is, b3 is sufficiently close to 1. Repeating the process above, we obtain the required permutation.

Problem S.12. Let Al < A2 < ... be a positive sequence and let K be a constant such that n-1

ak < Kay

(n = 1, 2, ... ).

(1)

k=1

Prove that there exists a constant K' such that n-1

E Ak < K'An

(n = 1,2,...).

(2)

k=1

Solution 1. We can suppose that K is an integer. We are going to prove by induction that (2) holds if K' = 8K. The theorem is straightforward if n > 8K, since K > 1. Let n > 8K. We observe that by the monotonicity of {An} and by (1), n-1 4K A2 2

Ak < KAz

4K C k=1

3.9 SEQUENCES AND SERIES

487

SO )'n-4K > )n/2. By this, by the induction, and by the monotonicity of {An},

n-1

n-4K-1

Ale

n-1 Ak +

k=1

k=n-4K

k=1

n-1

< 8KAn-4K +

Ak < 4KAn + 4KAn = 8KAn. k=n-4K

Thus, the solution is complete.

Solution 2. The monotonicity of {an} is superfluous. We are going to prove the following. If µi > 0, then the necessary and sufficient condition of

n-1

Epj<K/bn, n=1,2,...

(*)

t=1

is that there exist c > 0 real and r natural numbers such that, for each n, µn+l > cµn µn+r > 2µn.

and

(a) (b)

This implies the statement of the theorem, since if An+1 > cAn and 2 2 An+r > 2an, then An+1 > y C)tn and''n+2r > 4an, SO )n+2r > 2An. > An+r (a) is necessary: by (*) and by µi > 0, trivially µn-1 < Kµn, so (a) holds if c = 1/K.

(b) is necessary: by (*) again, 1

/in+1 >

Kµn, µn+2 >

Kµn,

... , µn+r-1 >

Adding them together, we get

/in+1 + ... + /in+r-1 >

µn+r-1) >

Thus, (b) holds if r > 2K2 + 1. (a) and (b) are sufficient: define µo = µ-1 = n-1

i=1 j=0 r

i=1

= 0.

µ2-i 7

<21:µn-i<21: ca =Kµn i=1

r-1 µn

Pi = E E lin-i-jr < 1 1 i=1

Kµn

r-1 Ani

so by (*),

/in+r > K (/Un+l + ... +

r K=21:c-t

i=1

In the last step, we used that, by (a), A,1,+, > c1µn,, holds. This finishes the solution. Moreover, a similar argument shows that

the theorem remains true if the exponent 2 is replaced by any exponent

a>0.

3. SOLUTIONS TO THE PROBLEMS

488

Problem S.13. Given a positive, monotone function F(x) on (0, oo) such that F(x)/x is monotone nondecreasing and F(x)/xl+d is monotone nonincreasing for some positive d, let An > 0 and an > 0, n > 1. Prove

that if 00

i AnF I an L J k 1< 00,

n=1 or

k=1

(1)

00AnF >ak k A < 00, n=1

(2)

k=1

then EÂ°Â° 1 an is convergent.

Solution 1. If x > 1, then F(x)lx > F(1). If x < 1, then F(x)lxl+d > F(1). We can suppose that F(1) = 1.

F(x) > x if x > 1,

(3)

F(x) > xl+d if x < 1.

(4)

and

First, we prove the statement of the problem when (1) holds. Denote by >n the sum over those numbers n that satisfy n k

>1.

k=1 An

Denote by Ell the sum over the other numbers n. Then, by (1) and (3), n

00 > > 'AnF ant Ak/An n

E'an E Ak > >'anAi.

k=1

Since Al > 0,

Ean <00.

(5)

By (1) and (4), n

00 > E"AnF an E Ak/An > n

"an+dAnd n

k=1

Divide the sum En into two parts:

Ell =

(1) +

E"(2) ,

Ak k=1

(6)

3.9 SEQUENCES AND SERIES

489

where the first sum runs over those numbers n that satisfy (1+d)/d

an < An/ E Ak k=1

and the second runs over the other numbers n. Then by one of the theorems of Abel and Dini, and by (1 + d)/d > 1, n

(1)an < n=1

t Ak

An/

< 00.

(7)

k=1

By (6), 00 >

1+d

(2)ananAnd

E Ak n

k=1

> (2) an

(I+d)/d

nd(E Ak)I+d =

An/

(2)an.

k=1

Comparing this with (5) and (7), we conclude that En=1 an is convergent.

Now, we prove the statement of the problem when (2) holds. If an = 0 for all but finitely many n, then the theorem is trivial; otherwise, we can delete those An, an pairs for which an = 0. So we can suppose that an > 0-

Let X = In: an > 1}- If n.X, then n

akAk

> anAn > 1,

k=1 An

and so by (2) and (3), 00 >

nEX

n akAk k=1

n = > >akAk >

nEX k=1

an,, An,,,

nEX

where no is the smallest element of X. So X is a finite set and the sequence

an is bounded. Let An = anan. Then (2) can be written in the following way: 00

An F n=1 an

ran

Ak) < oo. An

By the boundedness of the sequence an, (1) is satisfied with the substitution

An = An and, as we proved before, from this it follows that En=1 an is convergent.

3. SOLUTIONS TO THE PROBLEMS

490

Solution 2. We prove the statement of the problem when (2) holds, but we will not use (1). Solution 2 is the same as Solution 1 through formula (4). Denote by >' the sum over those numbers n that satisfy

akAk X

k=1

> 1.

(5)

By (2) and (3), we obtain n

00 > E 'AnF n

akAk/An

k=1

> E /anoAno = anoAno

> akAk k=1

L: '1.

n<no

By choosing no such that ano # 0, it follows that (5) holds only for finitely many n. So, by (2) and (4), 00

1+d

00 > E n=1

An = k=1

n=1

(n: d

1+d

akAk

(6)

k=1

Hence, assuming that not all of the an's are zero, it follows that 00

An d<00, n=1 and therefore

.1n->00.

(7)

Let nq be the largest natural number n such that 2q < An < 2q+1; if such an n does not exist, then nq can be chosen arbitrarily. We will see that in the latter case the multiplier of Anq is zero. Continuing (6), A-d

00' n=1

1+d

E akAk

1+d

00 A

q=1

k=1

ak,\k k:2q<ak<2g}1

f 00

> E 2-d(q+1)2q(1+d)S1+d =

2-d [ 2gS.1+d9

q=1

where

Sq =

ak.

k:2q <.\k <2q+1

So there is a positive K for which

q=1

< K.

3.9 SEQUENCES AND SERIES

491

Hence, for any natural number q, 2gS9+d

< K.

Therefore

Sq < K2-g/(1+d)

which trivially implies 00

E Sq <

00-

q=1

By the definition of Sq and by (7), there is an no such that >9 1 Sq > Therefore, E°° 1 an < 00, which finishes the solution.

E°° no an.

Remarks.

1. Originally, Laszlo Leindler placed part (1) of this problem at the competition committee's disposal. Atilla Mate noticed that the statement also follows from (2).

2. Laszlo Babai states a similar theorem for integrals: Let F(x) be a positive monotone function on (0, oo) such that F(x)/x is monotone nondecreasing and there is a positive d for which F(x)/x1+d is monotone nonincreasing. Furthermore, let A(t) and A(t) be absolutely continuous functions on (0, oo), A(t) monotone nondecreasing, A'(t) > 0 almost everywhere, and limt_,o A(t) = limt-.o A(t) = 0; then any of 00

A'(t)F(A (t)A(t))dt

< 00

and 00

A'(t)F(A t)

A'(x) A'(x) dx) dt < 00

implies limt+00 A(t) < oc. (By the monotonicity of F(x)/x, the functions after the f signs are measurable.)

Problem S.14. Let f (x) = E00 1 a, /(x + n2), (x > 0), where 1 < oc for some a > 2. Let us assume that for some,3 > 1/a, we have f (x) = 0(e-gy as x - oc. Prove that an is identically 0.

E°°

lanln-a

Solution. Obviously, the series defining f (x) is convergent at each x 54 -n' (n = 1, 2, ... ), so f (x) is meromorphic on the whole plane. Let 00

F(x) = rl (1 + x n=1

h(x) = F(x)f (x).

3. SOLUTIONS TO THE PROBLEMS

492

Since a > 2, this product is convergent for all x. Easy calculation shows that F(x) is an entire function, so for any e > 0, 1(x) = O(eIxI1

}e)

The function h(x) is also entire, and there is a sufficiently small a such that, for x > 0, I h(x) I = If (x) I

I F(x) I = O(e-x 3)O(e-xlla+E) _ o(1).

(1)

Furthermore, for any x, 00

Jh(x)

E Ianna

E InamnIIn (i+L) n=1

O(elxll/a}e).

F(IxI) =

So if M(r) = max IF(x)I, then log M(r)

r1/ÂŤ+E

Nrr

)

= O(1)

(2)

So by a Phragmen-Lendelof type of theorem (Gy. Polya and G. Szeg6,

Problems and Theorems in Analysis, Springer, Berlin, 1976, vol. I, 111.6.332), (1) and (2) can hold for an entire function h(x) only if h(x) is constant. So by (1), h(x) = 0. Hence f (x) = 0. Therefore, if for some n, an # 0, then f (z) has a pole at z = -ne. But this is a contradiction, so a1=a2=...=0.

Problem S.15. Given a positive integer m and 0 < 6 < 7r, construct a trigonometric polynomial f (x) = ao+>m 1(an cos nx+bn sin nx) of degree m such that f (0) = 1, f6<lxl <,r I f (x) I dx < c/m, and max_, <x<, If'(x)l < c/b, for some universal constant c.

Solution 1. Let 19 = max{6/2,1/m}. Extend the following function periodically to all real numbers: O(x)

={ 1-11, 0,

if IxI <19,

if 19<1xI<'Ir.

This nearly satisfies our purpose, it just is not a trigonometric polynomial. Let us examine the Fejer kernel of its Fourier series. Let

r f (x) = co

cp(x - t)Km(t) dt,

3.9 SEQUENCES AND SERIES

493

where sin2 m'2 1 t

Km (t) _

2(m + 1) sin2

and co is chosen such that f (0) = 1. Obviously,

1/2m

,v/2

1= f(0) > cot f Km(t) dt >

-co

Km(t) dt

2 -1/2m

-,9/2

trivially if tI < 1/2m. Then

IKm(t)I>lOm. Hence,

1=f(0)>

co<20. ,

Since cp is absolutely continuous,

f'(x) = co So

V(x - t)Km(t) dt.

f Km(t) < f a

x+19

If'(x)I < co

Km(t) dt <

80 19

-7r

X-19

Estimate f6<,x,<n If (x) I dx in cases 19 = 6/2 and 19 = 1/m. If 19 = 1/m, then

7r x+6

If(x)Idx2

f If(x)Idx<2 f fKm(t)dtdx 6 x-6

6<lxl<ir < 419

Km(t) dt < 419

f Km(t) dt = 4m a

If 19 = 6/2, then

f <IxI<ir

or++9

If (x) I dx < 419

Km (t) dt = 26

6-19

2 Km (t) dt.

6/2

Using elementary estimates on the interval 0 < t < it + 6/2 (< (3/2)ir), Km (t) < r30

3. SOLUTIONS TO THE PROBLEMS

494

and so

7r+s/2 26

Mt2

s/2

dt=240.

s/2

This finishes the solution.

Solution 2. Start from the function h(x)

x(02 - x2)2, if x < 0, if 0 < I

{ 0,

< 7r,

where the parameter 0 will be defined later. Let Cneinx

h(x)

n=-oo and

s(x)

Cneinx

n=-l We will show that i can be chosen in such way that f(x) = s2(x)/s2(0) satisfies the conditions of the problem. Obviously, f (x) is a trigonometric polynomial of degree not more than m, and f (0) = 1. We will estimate the difference of s(x) and h(x). The Cauchy-Schwartz inequality yields

s(x)-h(x)I

E Icnnl <

+oo

In <

InI>1

(cnn)2

EI cnnl2

n=-oo

N n1>l

nI<1

By Parseval's formula, + oo

(cnn)2

27r

I h'(x) I2 dx < 1

Is(x) - h(x) I <

Is'(x) - h'(x) I <

l0 103.

(1)

Fa2 ls(

x)l<lh(x)I+<3.

3.9 SEQUENCES AND SERIES

495

Moreover,

Vj29j-3

zi + 2K = -A,

-o

If'(x)Is2(o)Is(x)Ils'(x)I < suffices if

0 > b.

(2)

This inequality and (1) also hold if we choose i = max{b, 9/l}. The integral of If (x) I can be estimated in two ways. If 0 = b, then

f If(x)I dx= s2(o) J

Is2(x)Idx=

6<IxI<r

6<Ixl<r

6<IxI<r

f (s(x)-h(x))2dx

82(o)

s2o) f(s(x) - h(x))2 dx =

s210)

i ICnI2

-r 21r

Inl>l

°O

21r

Ls n cn<Ql2 < 7. n=-oo

If i = 9/l ± b, then

I f (x) I dx <

6<Ixl<r

-r

l S2

(o) dx = 2(

Ian l2

n=-l

<21r EIcnI2=4 f h2(x)dx<8L = l2. r -O° This finishes the solution.

Problem S.16. Prove that if M

1: an<Nam (m=1,2,...) n=1

holds for a sequence {an} of nonnegative real numbers with some positive integer N, then ai+r > pai f o r i, p = 1, 2, ... , where iN

ai =

E n=(i-1)N+1

(i = 1, 2, ... ).

3. SOLUTIONS TO THE PROBLEMS

496

Solution 1. Since ai+p = a(i+p-1)N+1 + ... + a(i+p)N 1

(i+p-1)N

n=1

> NN E an=

(i+p-1)N+1

n=1

(i+p)N

a,+...+

an n=1

an+ E an+...+ n=N+1

(i+p-1)N

an=al+a2+

+ai+p-1,

n=(i+p-2)N+1

it follows that ai+p

al + a2 +

+ ai+p-1.

Hence ai+l > ai, and therefore ai+p > ai + ai+1 + ... + ai+p-1 > pai

Solution 2. Similar to the previous solution, i-1

(1)

2p-1E ak

(2)

k=1=1

We are going to prove that ai+p

k=1

for any i and p. Indeed, (2) is the same as (1) for p = 1, and if (2) holds for p, then by i+1

2p-1 I` ak = 2p-1 (c.i + ai+p+l >_

i k=1

k=1

i ak

> 2p-12

ak k=1

it also holds for p+1. Since ai > 0, from (2) it follows that ai+p > 2p-tai, which finishes the solution.

Remark. We can similarly prove the following, more general statement. Let {an} be a sequence of nonnegative real numbers such that for some c, n-1

an>c>ai. i=1

Define ai as we did in the problem. What is the largest constant dp such that for all such sequences the following holds? ai+p ? dpai

3.9 SEQUENCES AND SERIES

497

In the problem, c = 1/(N - 1). First, we will find a cp such that n

an+p ? ep i=1

e1 = c suffices for p = 1. Since n-1

ai > ep(1 + c)

an+p+l ? ep i=1

ai, i=1

it is not a bad idea to choose ep+l = ep(1 + c), that is, ep = c(1 +

c)P-1

Then (i+p)N

ai+P =

e(P-1)N+j

an > E e(P-1)N+j > an >j=1

n=(i+p-1)N+1

j=1

n=1

ai.

So we can choose dp such that N

dp =

C)(P-1)N+j-1 = (1

C(1 +

+ C)(P-1)N[(l + C)N

j=1

It is easy to see that if we define al = 1, an = c(l+c)n-2 (n > 2), then for any N and p, ap+l = dpal holds, so the inequality cannot be sharpened. As stated, in the problem c = 1/(N - 1), so

(P-1)N

dp- (N-1)

(N-1) -1 >eP-1(e-1),

but even for c = 1/N it is true that dp >

2P-1.

Problem S.17. Let S. = E; 1 bjz,' (v = 0, Âą1, Âą2.... ), where the bj are arbitrary and the zj are nonzero complex numbers. Prove that

ISoI < n max ISv I. o<Ivl<n

Solution. Let 11j 1(z - zj) = Ek=oakzk and maxo>k>n Iakl = Iaml. Obviously, la7z.I > 1. Then n

akSk-m

= j akbj zj -m =

k=0

bjzj

akz3 = 0. k=0

k=0

Hence n

(-a )Sk-m < E I Sk-m I< n max I Sv I. 0<Ivl<n k=O;k54m

Remark. One can show that equality holds if bl = b2 = the set of numbers zk is the same as {a e(2vri)/((n+l)j) (j where a is an arbitrary constant of absolute value 1.

= bn and n)},

3. SOLUTIONS TO THE PROBLEMS

498

Problem S.18. Let p > 1 be a real number and R+ = (0, oc). For which continuous functions g : ][8+ -+ R+ are the following functions all convex?

Mn(x)

_ E 19(

)x+1

E 19( i

X = (xl, ... , xn+l) E

1[8++1,

n - 1, 2, ...

Solution. We prove that Mn (n = 2, 3, ...) are all convex if g is a constant function. If g is a constant function, then by Minkovsky's inequality, Mme,(x+y) <- Mme,(x)+M. (y)

(x,y E R'+1),

(1)

and by Mn(ax) = aMM,,(x) (a E R+,x E R++1), it follows that Mn is convex. Conversely, if all of the MM,, are convex, then we will show the following.

If M2 (x, u, 1) is a convex function of x for all u E (1- 6,1 + 6), b > 0, then the function g is a constant. Taking each side of the inequality

M2(axl + (1 - a)x2i u,1) < aM2(xl, u,1) + (1 - a)M2(x2i u,1) (X1, x2 E R+, U E (1 - 6,1 + 6), a E (0,1)) to the power p and using the inequality

at + (1 - a)s < (atP + (1 - a)sP)1IP (t,s > 0, a E (0,1)) between the arithmetic mean and the n-th power mean, we obtain

M2(axl+(1-a)x2,u,1) <aM2(xl,u,1)+(1-a)M2(x2,u,1). Deleting uP = auP + (1 - a)uP from both sides, we get 9(u)(1 - up) 9(aX1+(u a)X2) + g(u)

< a 9(u)(1 - up)

9(!) + 9(u)

+ 1 - a g(u)(1 - up) )

9(u) + 9(u)

for x1i x2 E R+, u E (1 - b, 1 + b), a E (0, 1). If u < 1, then divide by g(u) (1- uP) and take the limit u -* 1- 0. Since g is continuous, we get 1

9(ax1 + (1 - a)x2) + g(1) < a9(x1) + g(l) + (1 -

9(x2) + 9(1)

Using the same method for u < 1, we get the opposite inequality. There-

fore, equality holds. So the function f = 1/(g(x) + g(1)) satisfies the following Jensen's equality:

f (axl + (1 - a)x2) = of (xl) + (1 - a)f (x2)

(X1, X2 E R+, a e (0, 1)).

3.9 SEQUENCES AND SERIES

499

Since f is continuous and positive, f (x) = Cx + D, where C and D are nonnegative constants. So Cx + D = g(x) + g(1) < g(1)

2 f (1) = 2(C + D).

This does not hold for large values of x when C > 0. Therefore, C = 0 and g is constant.

Remark. Some of the participants remarked that we do not have to suppose the continuity of g, since it follows from the convexity of Mn. Moreover, g is continuous if M2 (x, u, 1) is a convex function of x for u E

Problem S.19. Suppose that R(z) = -.an Zn converges in a neighborhood of the unit circle {z : IzI = 1} in the complex plane, and R(z) = P(z)/Q(z) is a rational function in this neighborhood, where P and Q are polynomials of degree at most k. Prove that there is a constant c independent of k such that 00

lanl < ck2 max IR(z)j. n=-oo

Solution. Denote by D the unit disc of the complex plane D = {z

IzI < 1}, and by 8D its boundary. We can suppose that IR(z)I < 1 on the perimeter of the unit circle. So we have to prove that EO° _. IanI < ck2. If we prove

lanl < ck2, n=- inf

then we can apply it to R(1/z)/z, so that we obtain E°° o Ianl < ck2, which finishes the solution. We know that an

y r

-(Z) dz,

(1)

where we can integrate along 8D, or along such a curve t, such that R(z) does not have a pole in the region bounded by 01) and r. Now we will try to choose skillfully r c D. By the previous results,

n=-oo

-L lanI -<27r =

27r

i r

IR(z)I(1 + zI + ... + IzI2 + ...) Idzl

J IR(z)I 1 IdzlI r

3. SOLUTIONS TO THE PROBLEMS

500

We are looking for a P for which the integrand is not too big in the rightmost integral. Because the multiplier is 1/(1 - jzl), we will have to bring P as far from the unit circle as possible. The only problem is that we do not know anything about the behavior of R(z) far away from 8D. Denote the poles of R in the unit disc by q1, q2, ... , qh, where the points are included in the sequence with appropriate multiplicity. Obviously, h < k. The Blaschke product h

B(z)

z - qv

-Vil-qvz

is holomorphic in D, vanishes at the points qv, and has absolute value not more than 1 in the points of the unit circle. So S(z) = R(z)Q(z) is holomorphic in D and has absolute value not more than 1 in the boundary of the unit circle. Thus, by the maximum modulus principle, IS(z)l < 1, for z E D, and so IR(z)I <

(IzI < 1).

7h lz-g

(2)

l 1v=1

This is the approximation we were seeking. Let

t=a{zED

-_L

1<v<hz :jzj=1}.

IfzEr,then by(2), JR(z)I

(i+)k<e. (1- k+1 )k =

(3)

The curve defined above is not necessarily a Jordan curve, but it consists of finitely many closed Jordan curves. Nonetheless, R(z) will be holomorphic in the region bordered by r and 8D. So all of the consequences of (1) will hold. Obviously, t consists of some subcurves of the curves

z - qv

1 -qvz

If qv = 0, then r is a circular arc. But it is also a circular curve when qv # 0, since IF, is obtained from the circular arc jwI = k/(k + 1) by the transformation z - qv

1-qvz

(4)

and such a transformation maps a circle onto a circle. (Of course, qv is not the centerpoint of r.. Moreover, if we are considering Q as the

3.9 SEQUENCES AND SERIES

501

Figure S.1.

Poincare model of the hyperbolic geometry, then the transformation (4) is a conformal transformation of the hyperbolic space, so q will be the nonEuclidean center of I',,. The non-Euclidean radius of r,, is approximately log k, see Figure S.1.) Since F C U"r,,, - by (3), 1

lanl <n_-oc 27f

IdzI

<1 - IzI e

27r L'-lrV

IdzI

1 - Izl

The latter sum can be approximated by substituting (4) and some calculations as

r 1-Il

j *1 `2J 1

=2 k

1z 12

12 < 1-(kk

)227r<87rk.

cr4

(Here we had to do "some calculations" to prove the equality above. The equality also shows that curve IdzI/(1 - Iz12) is invariant under the conformal transformation of the hyperbolic plane.) So IanI < 4ek2,

n=-co

and as we noted above it follows that 00

IanI < 8ek2.

n=-oc

Remark. The inequality above is not sharp. Gabor Somorjai showed with a better curve IF that 00

l an I < ck log k.

n=-oc

3. SOLUTIONS TO THE PROBLEMS

502

Problem S.20. Let Kn (n = 1, 2, ...) be periodical continuous functions of period 27r, and write

fa 2

kn(f; x) =

f (t)Kn(x - t)dt.

Prove that the following statements are equivalent: (i) fo I kn (f ; x) - f (x) I dx -+ 0 (n -+ oo) for all f E L1 [0, 27r]. (ii) kn (f; 0) - f (0) for all continuous, 27r-periodic functions f .

Solution 1. (i)- (ii): Denote by C[0; 2ir] the space of the continuous functions under

the supremum norm. Let us examine the sequence of operators L1 ---4 L1:

Ilkn(';')II =

sup

Ilkn(f;')IILI

(1)

IIfIILI<<1 27r

sup

fIL,<0 11

27r

IKn(t)I dt.

On the other hand, the functions if x E [0; 6],

f 6 (x) - { 0,

if x E (S; 27r),

by the continuity of Kn(t), satisfy 27r

IIkn(f5;')IIL1 - f IK.(t)I dt

0),

and so 27r

Ilkn('i')II =

fIKntIdt. 0

By applying the Banach-Steinhaus theorem to the sequence 2ir

we obtain f IKn(t)I dt < C, where C is the same constant for all n. 0

If f is continuous, then 27r

kn(f; x) - kn(f; x + h)I = J(f(t)_f(t_h))K(x_t)dt 0

< Csup t I f(t) - f(t - h)I, and so the functions kn (f; x) are equicontinuous.

3.9 SEQUENCES AND SERIES

503

Let us suppose that (ii) does not hold. Then there is a continuous f, an e > 0, and a sequence {vn} such that k,,,, (f; 0) > f (0) + e for n = 1, 2,... . By the uniform continuity of {k n Y; x) II there is a 6 such that, for all n,

k,,, (f; x) > P X) + 2 if O <x<6 (n= 1, 2, ... ). Hence,

f b

21r

JIkvn (f; x) - f (x) I dx > 0

(k-n (f; x) - f (x)) dx > 6

(n = 1, 2.... ),

but this contradicts (i). (ii)-*(i): It is well known that the norm of the linear functional kn(.; 0) C[0; 27r] --+ R is fo " I Kn (t) I dt. So by the Banach-Steinhaus theorem, ff " I Kn (t) I dt < C (n = 1, 2, ... ), where C is a constant. Since 2

r f(t)kn(x0

kn(f, xo) = f 0 = f27r f(xo

- t) dt

+ t)Kn (-t) dt = kn (f (x0 + t); 0) 0

is continuous for all f , kn (f; x) -* f (x) holds everywhere. But kn Y; x) C sup(f ), so by Lebesgue's theorem, 2,r

I:;

Ikn(f, x) - f (x) I dx - 0 (n - > oo, f E C[0, 27r]).

Let f E Ll [0, 27r] and let e > 0 be arbitrary. Choose a continuous function g in such a way that 119 - f IILI < e holds. Then, by (1),

27r

I kn (f, x) - f (x) I dx < I2 I kn (f, x) - kn (9; x) I dx + f27r I kn (9; x) - 9(x) I dx,

27r

I9(x) - f (x)I dx <CII9 - f 41 f27r

Ikn(9;x)-9(x)Idx+II9-f1ILI.

By the previous argument, the middle term above goes to 0 if n -* oo, so

27r

I kn(f, x) - f (x) I dx

becomes arbitrarily small for sufficiently large n. This finishes the solution.

Solution 2. We are going to use the following version of the BanachSteinhaus theorem. Let X and Y be Banach spaces, and let A, An : X -* Y

3. SOLUTIONS TO THE PROBLEMS

504

(n = 1, 2, ...) be linear bounded operators. The sequence {An} pointwise converges to A if there is a constant C and a closed system S C X such

that IIAn,II<_C

(n=1,2,...),

(1)

Anx-4Ax (n--*oo)

(2)

hold for all x E S. (S C X is called a closed system if the linear hull of S is dense in X.)

Let An : L, --4L, be the operator k.,,(.; ), A' be an identity on L1, R be the functional 0), and A2 : C[0, 27r] -+ R be

An : C[0, 27r]

the mapping f -+ f (0). As we showed in the previous solution, IIA 11 = lIAnII = fo" IKn(t)I dt, so (1) holds for {An} and {An} at the same time. Let S = {em = etmx},Â°,Â°=1. Since S is closed in L, and in C[0, 27r], and 2a

(A'em)(x) _

= eimx

etmtKn(x

- t) dt = 1 eim(x+t)Kn(-t) dt 0

etmtKn(-t) dt = em(x)A2em,

it follows that f2n

`41emIILi =

em(x)I dx

= IA22em -A 2 eml j

27rIIA2em

lem(x)I dx

- A2emIIR,

so (2) holds for {An} and A'. Furthermore, (2) also holds for {An} and A2 simultaneously. So we proved the equivalence of the pointwise convergences

An , Al and An , A2. 00

Problem S.21. Let us assume that the series of holomorphic functions

E fk(z) is absolutely convergent for all z E C. Let H C C be the set of k=1

those points where the above sum function is not regular. Prove that H is nowhere dense but not necessarily countable.

Solution. Let gn(z) = En 1 fk(z) and g(z) = i' 1 fk(z). Then gn(z) is holomorphic and gn(z) - g(z) for all z E C. To prove the first part of the statement of the problem, we have to show that inside each circle K of C, there is a circle k in which g(z) is holomorphic. Let cp(z) = supra 9n(z)1This exists by the convergence and is finite for all z E C. Let Ak = {z E KI k - 1 < W(z) < k} (k = 1, 2.... ). By Baire's theorem, there exists an N such that AN is dense inside some circle k C K. If cp(z0) > N for

3.9 SEQUENCES AND SERIES

505

some z0 E k, then Ign(zo) I > N for some n, and by the continuity of gn(z) there is a neighborhood of z0 where also ep(z) > N, which contradicts the

fact that AN is dense in k. So cp(z) < N if z E k, and so Ign(z)I < N if z E k. By the Vitali-Montel theorem, there is a subsequence of {gn(z)} that is equiconvergent inside k. Since gn(z) is convergent, this subsequence converges to g(z). Hence, by Weierstrass's theorem, g(z) is holomorphic inside k. So we proved that H is nowhere dense. Remarks.

1. In the solution, we did not use the fact that the series of functions

fk(z) is absolutely convergent. To prove the second part of the 1 statement of the problem, we will need the following result. Mergelyan's theorem. Let K be a compact set of C, such that its complement is connected, and let f be is a continuous complex function

on K, such that it is holomorphic inside K. Then for any e > 0 there exists a polynomial P, for which If (z) - P(z) I < E for all z E K. (See W. Rudin, Real and Complex Analysis, McCraw-Hill, London, 1970.)

Let A = {zjIm z < 0}, B = {zIIm z > 0}, C = {z: z is real}. There are series of compact sets {An}, {Bn}, and {Cn} (n = 1, 2, ... ), such that An C An+1, Bn 9 Bn+1, Cn C Cn+1, C\ (An U Bn U Cn) is connected for n = 1, 2, .. . and cc

UAn=A, UBn=B, 1

UCn=C.

So An U Bn U Cn is compact, its complement is connected, and (1, if z c: An U Bn,

f (z) -

ifzECn.

Hence, by Mergelyan's theorem, there is a polynomial PE,n for any e > 0 and n, such that 11 - PE,n I < e, if z E An U Bn, and I PE,n (z) I < e, if z E C n (n = 1, 2, ... ). Let Qk(z)-Qk-1(z),..., f1(z) = Q1(z), f2(z) = Q2(z)-Q1(z),...,fk(z) = where Qn(z) = P(1/2n).n(z) (n = 1, 2, ...) is a series of polynomials. Then :%=1 fk(Z) = Qn(z), so 00 1 1fz is not real, f k (z) 0, if z is real. k=1

Here H is the set of real numbers, and is hence uncountable. On the other hand, for any z e C there exists an N such that z e An U Bn U Cn if n > N, so in this point z, cc

k=1

I QN+k(z) - QN+k-1(z)I C E 22N+k1 <00;

I fN+k(z)I = k=1

hence, E' 1 fk (z) is convergent at z.

k=1

3. SOLUTIONS TO THE PROBLEMS

506

2. We can prove a stronger statement: the measure of H can be positive. Let R be the real axis, I the imaginary axis. It is enough to show that if S is a nowhere-dense subset of I, then there is a sequence if (n)}n l of holomorphic functions that satisfies the conditions of the problem and R x S contained in the H set obtained from this sequence If (n)}°°_1. So let S be a nowhere-dense, open set of I. Then I\S- is a dense set of I, so it is a union of countably many disjoint intervals; let these be ik = (ak, bk) (k = 1, 2, ... ). (We can suppose that ak and bk are finite.) k)

be monotone decreasing, {rl(k) } ° 1 monotone increasing, 1 such that rikl < I(k) and rikl -+ ak, sikl -+ bk if l -+ oc. Let {an}n 1 be a sequence of natural numbers, such that it contains each natural number infinitely many times. Let n be an arbitrary number. Let an = k, denote by l the cardinality of {m : m < n, a,,,, = k}, and let Let {r( I

Kn = {zlr2,- 1 < Imz < s2,- 1, IRezJ < n}, An = {zllmz = r2l), IRezI < n}, Bn = {zlImz = 821 , IRezI < n},

Ln = {zlr2i _ 1 > Imz > min(ak - n, -n), IRezI < n},

Mn={zl821-1 <Imz<max(bk+n,n), IRezI <n}. The set Pn = Kn U An U Bn U Ln U Mn is compact, and its complement

is connected in C-. So by Runge's theorem, there is a polynomial fn such that Ifn(z)I <

ZEKnULnUMn

and

I fn(z) - nI < 2

z E An U Bn.

(i) The functions fn are polynomials, and hence holomorphic on C.

(ii) For any z E C, there exists an No such that z E Kn U Ln U Mn for any n > No. So I fn(z)I < 1/2n for any n > no, and En'=1 Ifn(z)I < 00.

(iii) Let zo E R x S arbitrary. Let e > O and U E = { z l I z - zo I < e}. There is an index ko such that UE/2 fl (R x iko) # 0; hence, UEl{z: Imz=ako}'A 0 or U5 fl{z: Imz=bko}#0. Let us suppose that UE fl {z : Imz = ako } # 0 and z1 E UE fl {z : Imz = ako}. (The other case can be discussed similarly.) Let no > max(IRezoI +

e, IIm zo I + e) such that if n > no and an = ko, then Kn fl UE # 0. Let Eno 9no = n= 1 9n Since 9no is continuous at z1, there is a 0 < S < e such that for all z E V6 = {z: Iz-z1I < S}, Igno(z)-gn1I < 1. Let ni > no such that and = ko, n1 > 5 + 2I9no (z1) I and Kn, fl V6 54 0. Let Al E and A2 E aKnl fl U6, where aKnl denotes the boundary of Kn1.

An,

fl U6

3.9 SEQUENCES AND SERIES

If n > no and n

507

n1, then Ifn(Ai)I < 1/2n and I fnl(Al) -nil < 1/anl.

If(al) - 9n(zl) -nil < I9no(A1) - 9no(zl)I + Ifni(A1) - n1I 00

fn(A1) < 1 +

n>no,n#nl

n=no+1

< 2.

On the other hand, I fn (\2) I < 1/2n if n > no, so 00 Ifn(A2)I < 2.

If (A2) - 9no(z1)I < I9no(t2) - 9n(zl)I +

n=no+1

Hence,

If(A,)I >- I9no(zl) +n1I - If(A,) - 9no(zl) - n1I >- ni - I9no(zl)I - 2

and If(A2)1 < If(A2) - 9no(zl)I + I9no(zl)I < 2 + I9no(zl)I

If(al)I - If(A2)I >- nl - 4 - 2gno(zl) > 1. Therefore, the total variation of If I is more than 1 in U. Since b > 0 is arbitrary, If I is not continuous at zo. So If I is not continuous in the points of R x S. That is, { fn}0 1 satisfies the conditions of the problem, and the set H, obtained from If (n)}Â°nÂ°__1, contains R x S. By this, the statement of the remark is proved.

Problem S.22. Let f (x) be a nonnegative, integrable function on (0, 27r) whose Fourier series is f (x) = ao + E' 1 ak cos(nkx), where none of the positive integers nk divides another. Prove that IakI ao.

Solution. Let Kn (x) = 1/2 + (1 - +1) cos x +... be the nth Fejer kernel. It is known that Kn (x)

0. By integrating, we get 0 < fo " f (x)Kn (nkx) =

7r(ao + (1 - 1/(n+ 1))an,,), so if n - oo, then ank < ao. Kn(x - 7r) =

1/2-(1-1/(n+1))cosx+..., so 2,t

f(x)K(nkx-7r)=7r

/ao-(1-_--4-1)ank)

Therefore, if n -* oo, then an,< < -ao. 0

3. SOLUTIONS TO THE PROBLEMS

508

Problem S.23. We are given an infinite sequence of 1's and 2's with the following properties: (1) The first element of the sequence is 1. (2) There are no two consecutive 2's or three consecutive 1's. (3) If we replace consecutive 1's by a single 2, leave the single 1's alone, and delete the original 2's, then we recover the original sequence. How many 2's are there among the first n elements of the sequence?

Solution. Since, by the above conditions, the first n - 1 elements of the sequence determine the nth one, there can be only one sequence. It is easy to see that the following sequence of numbers 1 and 2 satisfies the conditions of the problem: in this sequence, the index of the nth 2 is larger

by 2 or 3 than the index of the (n - 1)th 2 if the nth element of the sequence is 1 or 2, respectively. The following condition is equivalent to (3):

(3') If we change all 1's into 12 and all 2's into 112, then we obtain the original sequence.

Denote by f (n) the number of 2's among the first n elements. We are going to prove that, for all n, f(f(n) + 2n) = n,

(a)

f(f(n) + 2n - 1) = n - 1,

(b)

f (f (n) + 2n + 1) = n.

(c)

To prove this, change the first n elements of the sequence as described

in (3'). Hence, we obtain 3 f (n) + 2 (n - f (n)) = f (n) - 2n elements, such that the number of 2's is n, so we proved (a). By (3'), this f (n) - 2n ends with 12, and by (2), the (f (n) - 2n + 1)th element is 1, so we obtain (b) and (c). Obviously,

f(1)=0 and f(2)=1.

(d)

Now we are going to prove that if function g : N -> N satisfies the conditions (a)-(d), then g = f . f (n) = g(n) follows from (d) for n = 1, 2. Let us suppose that we proved f (k) = g(k) for all k < n. Since h(k) = f (k) + 2k, h(k + 1) - h(k) < 3, and so there is a k < n natural number for which n = f (k) + 2k + En, where en = 0, Âą1. Then, by (a), (b), or (c) and the induction, f (n) = f (f (k) + 2k + En)

if En = 0, 1,

k-1, if En=-1.

Moreover,

g(n) = g(.f (k) + 2k + En) = g(g(k) + 2k + En) _

Hence, f (n) = g(n).

if En = 0, 1,

k-1, if En=-1.

3.9 SEQUENCES AND SERIES

509

Finally, we are going to show that g(n) = [(/ - 1)n + 1 - 1/vF2] satisfies the conditions (a)-(d), so f (n) = [(v'2- - 1)n + 1 - 1/'] holds. Obviously, g(1) = 0 and g(2) = 1. To verify (a), write g(n) in the form

g(n) = (v-1)n+1-1/f]-e(n), whereE(n) E [0,1]. (However, e(n) = 0 cannot hold, but we do not need this fact.) Hence,

g(g(n) + 2n) = [(/ - 1) (v - 1)n + 1 - 1 - E(n) + 2n) + 1 -

]

= [n+(1-E(n))(v12_ - 1)] =n. We can similarly verify (b) and (c).

Remark. After writing down the first few elements of f (n), it becomes plausible that f (n) is the integer part of some linear function. If we substitute the function f (n) = [an + b] into (a) and (b), we obtain that (a) - 1, b = 1 - 1/v obtained in the and (b) hold only for the pair a = solution.

Problem S.24. Let ao, al, ... be nonnegative real numbers such that

a,=00. n=0 For arbitrary c > 0, let

nj (c) =min k : c j <

E ai

j = 1, 2, ...

i=0

Prove that if E 0 a? < oo, then there exists a c > 0 for which E 째1 a,, l,i < oo, and if ,째째 0 ai = oo, then there exists a c > 0 for which E 째 1 a,, W = 00.

Solution. Let

Sk =

ai,

i=0

and denote by Xk,j the characteristic function of the interval Ik,j Let [(1/j) - Sk-1, (1/j) . Sk].

f : (0, 00) ---* (0, +00] .

f (c) = E ani (C) ; j=1

Since Xk,j(c) = 1 iff nj(c) = k, the function f = >k j akXk,j is Lebesguemeasurable. So 00

]f(c)dc= a

akA(Ik,j U [a, b]) k=oo,j=p

3. SOLUTIONS TO THE PROBLEMS

510

We are going to give a lower and upper estimate for the value of the integral.

Lower estimate. We sum over only those pairs k, j for which Ik,j C [a, b],

so if a < 1 Sk-1 < 1 Sk < b, 3

or, equivalently, Sk

Sk-1

Note that

A(Ik,i) = .

So b

sk-r1/a

ak > > ak

J f (c) do > k=O j

k=0

(sk/b)+1 00

a2 In Sk-1/a = 0 a2kIn k=0

Sk/b + 1

k=0

bSk-1 ask + ab

Upper estimate. We sum over only those pairs k, j for which Ik,i n [a, b]

1 Sk-1 < b and a< 3

Sk-1 < j < Sk a

f (c) dc < 0" ak2

k=0

'ak /I 1 + 12 + In ask-1bsk+ ab

k=0

\\\

If lim sup an = a > 0, then an > a/2 for infinitely many n. So a2 = oo. Since infinitely many terms of the sum EO_ ani(,/2) are larger than an, f (a/2) = oo. So let us suppose liman = 0. Then sn/sn_1 -> 1. Therefore, lim In asks

lim In

+ ab

as k 1 + ab

In a > 0, b

In a

3.9 SEQUENCES AND SERIES

511

So there exist real numbers A > 0, B > 0, C > 0, and D such that 00

E Cln askbsk-1 ) ak > (a2A+B, + ab k=0

k=0

k=0 \

+ln asbsab) ak < k 1+

(a) C+D. (k"=00

So if a2 < oo, then fe f (c) dc < oo. Therefore, f (c) < oo almost everywhere, and if ak = oo, then fe f (c) dc = oo. So for any a < b, the set 00

{ j=1 E ani (C) > m} is a dense set for an any m. However, the function an,,(,) is a left-upper-semicontinuous function of c, so the following sum is also left-upper-semicontinuous: 00

1: and (c) j=1

Therefore, the set {any (,) > m} is dense and open. So the set {f = oo} is of second category and therefore is not empty.

Problem 5.25. Let 2/(/+ 1) < p < 1, and let the real sequence {an} have the following property: for every sequence {en} of 0's and ±1's for which En 1 enp' = 0, we also have E°° 1 enan = 0. Prove that there is a number c such that an = epn for all n.

Solution. We start with a definition. Definition. Let L denote the set of all positive and strictly decreasing sequences {ln} for which L = En'- 1 In < oo. We call such a sequence interval filling if for every x E [0, L] there is a sequence En E {0, 1} with

x = E', Enln. We need three lemmas.

Lemma 1. If a sequence {ln} E L satisfies 00

An < E Ai

(n =

2,...

i=n+1

then it is interval filling. Proof. For x E [0, L] we define the numbers En inductively as follows: 1, En

I 0,

if Ei=1 EiAi + An < x; if En 1 X. i i=1 Ei i >

3. SOLUTIONS TO THE PROBLEMS

512

For every n for which En = 0, n-1

0 <x - > eiAi <x i=1

EiAi < An, i=1

hence if there are infinitely many such n, then x = Ei00 Ei limn_.o li. If, however, there are only finitely many such n's, then for the largest one n-1

EiAi < An <

Ai =

i=n+1

i=1

from which

EiAi' i=n+1

x <

Enln,

n=1

so our claim holds even in this case. Lemma 2. For 2/(,Vr5- + 1) < p < 1, the sequence

p, p2, p3, ... is interval filling, and for any natural number N, the sequence 2

PAP,

P N-2 ,PN-1 ,P N+1 ,PN+2

is interval filling too.

Proof. By Lemma 1, it is enough to verify that 00

pn < E pz

for all

nEN

i=n+1

and

Pn-1 <

E pi

for all

ncN

i=n+1

are satisfied. The first of these is equivalent to 1 < p/(1-p) and the second

one to 1 < p2/(1 - p), hence each of these is true if 2/(s+ 1) < p < 1. Lemma 3. Let A and B be nonempty disjoint sets such that their union

is the set of the natural numbers, and let 2/(f + 1) < p < 1. Then there exist nonempty sets A' C A, B' C B with the property

[: pi = [: pi tEA'

(1)

iEB'

Proof. Let L = p/(1 - p) and x = >iEApi Let us choose an N E A for which x < L - p.NIf A is finite, then N can be the largest element in

3.9 SEQUENCES AND SERIES

513

A; otherwise, we can choose any element in A with sufficiently large index. Since Lemma 2 says that the sequence A e

p N-2 p N-1 '

p N+1 , p N+2

is interval filling, there is a set C C N \ {N} for which

Y:pZ = x = > pi. iEC

iEA

Let A' = A \ C, B' = C \ A. Then (1) clearly holds, and because N E A', the set A' is not empty. From this and (1), it follows that B' is not empty, either.

After these preparations, we turn to the proof of the statement in the problem. For every c E R, the sequence a' = an - epn also satisfies the assumptions of the theorem. Suppose, on the contrary, that the claim is not true. Then we can choose a real c for which there is an n with a' > 0 and also another one with a' < 0. Let

A={nEN:a;,>0}, B={nEN:a'n<0}. On applying Lemma 3, we get two sets A' and B' with the properties stated there, and with the aid of these we define if n E A';

11, en = -1,

if n E B';

otherwise.

1 0,

Then

Eeip' = Epl- Epi=0, i=1

but

iEB'

iEA'

Eeiai = i=1

ai iEA'

ai > 0, iEB'

and this contradicts the hypothesis of the problem. The contradiction obtained proves the claim.

Remark. The result is true for any 0 < p < 1 (see Z. Daroczi, I. Kdtai, T. Szabd, On Completely Additive Functions Related to Interval-filling Sequences, Arch. Math. 54 (1990), 173-179).

Problem S.26. Let S be the set of real numbers q such that there is exactly one 0-1 sequence {an} satisfying cc

E anq-n = 1. n=1

Prove that the cardinality of S is 2"0.

3. SOLUTIONS TO THE PROBLEMS

514

Solution. Let {bn} be a sequence such that b1 = b2 = 1, b3n = 1, b3n+1 = 0, and b3n+2 = 0 or 1 (n = 1, 2, ... ). There is exactly one number q > 1 such that

n=1

= 1.

Such a q exists, since the series EO_1 bnxn is convergent if xj < 1, continuous, it takes the value 0 at x = 0, and goes to infinity if x --+ 1-. We will prove that if there is a sequence {an} with elements 0 or 1 and 00

a"F

Fn n=1 g then an = bn for all n.

Let us suppose that this is false. Let k be the smallest number such that ak # bk. Then

an1: n 1: 00

n=k

Since 00

g5_g2 qn - q+q2+q3n = q4+q3+q2-q bn

n=1

that is, q5 - q4 - q3 - 2q2 + q + 1 > 0 holds,

1- q3-q2-q-1 q3-1 q3-1 q2

-1-

(q3-q2-q-1)(q3-1)

(q3-1)2

=1-q(q5-q4-q3-2q2+q+1)+1 <1. (q3 - 1)2

Suppose that bk = 0 and ak = 1. If n > 1, then at least one of bn, bn+1, and bn+2 is zero, so bn+1

bn+2

qn + gn+1 + qn+2 - qn

qn+1'

Hence CK)

n=k

bn_ [qn

L qn Lm=01\ n=k+1

q< qk m=1

and this is a contradiction.

(q3r2+q31)

qk+3+n+2 J 00

n=k

3.9 SEQUENCES AND SERIES

515

Otherwise, suppose that bk = 1 and ak = 0. Since at least one of bn, bn+1i and bn+2 is 1, bn+1 bn+2 qn + qn+1 + qn+2 bn

qn+2'

n=k

m=1

O°

> qk

> qk +

qn+3m

°°

(i+

m=1

q3m °°

Y- g3-2 + q3-1 + M=1

n=k+1

M=1 cc

n=k

q3m

°°

- qk

n=1

an an

and this is also a contradiction. So only the sequence {bn} satisfies the conditions. The cardinality of different sequences {bn} is 2x0. If two {bn} sequences

are different, then the corresponding numbers q are also different. So the cardinality of S is maximum 2k°. But it cannot be more than this, since the cardinality of the set of all real numbers is 2°.

Problem 5.27. Given an > an+1 > 0 and a natural number p, such that a lim sup an < µ, n

aµn

prove that for all E > 0 there exist natural numbers N and no such that, for all n > no the following inequality holds: Nn

Eak <EEak. k=1

k=1=1

Solution. We have to find a natural number no such that for any E > 0 there exists a natural number N such that < E

k=1 ak EN

=1 ak

holds for all n > no, where condition (1) holds. Choose a real number v in the interval (limsup, (an/anµ,), t) . Define a natural number M such that an/anµ < v, for all n > M. Therefore, for all natural numbers l > 0 and n > M, the inequality an/v' < anµt holds.

3. SOLUTIONS TO THE PROBLEMS

516

First, we show that the series Ek=1 ak is divergent. Obviously, Mµ l

Mµl

Mµ1

am > Mp E aMµt >u1 v 1: ak > E k=1

k=1

= MaM µ v

Here M and am are fixed, so when 1 -> oo, then from v < p it follows that the right side of the inequality also goes to infinity, that is, we estimated the series Ek=1 ak from below by a divergent series. This divergence provides us with a number K, for which 1 ak < M+1 ak; so for any n, > K, En k=1 ak :5 2 k=M+1 ak. For given e, choose l such that 1/e < (p/v)1/2. Therefore, 1

2 (µ/v)l

>k=M+1 ak/v1 _

En 2

k=M+1 ak

< rkµMµl a(Lklv`J+1)pl < n

2 Ek=M+1 ak

Ek=M$L µ

2 Ek=M+1 ak

< Ek= ak n 2 Ek=M+1 ak Ek=1 ak n

a(Lk/vlj+1)/v'

3.10 TOPOLOGY

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3.10 TOPOLOGY

Problem T.1. Prove that any uncountable subset of the Euclidean nspace contains an uncountable subset with the property that the distances between different pairs of points are different (that is, for any points P1 # P2 and Q1 # Q2 of this subset, F1-P2 = Q1Q2 implies either P1 = Q1 and P2 = Q2, or P1 = Q2 and P2 = Q1). Show that a similar statement is not valid if the Euclidean n-space is replaced with a (separable) Hilbert space.

Solution. For the proof of the first statement of the problem, we say that a subset of the Euclidean n-space En has property T if the distances between all pairs of points of this subset are different. By induction on

n, we prove that if all subsets of a set H(C En) that have property T are countable, then H is countable. The case n = 0 is trivial. Suppose that the statement is true for n - 1(> 0). Let H(C_ En) be a set such that all subsets with property T are countable. By Tukey's lemma, there exists a maximal subset M of H with property T. By maximality of M, any point of H \ M either has equal distance from two distinct points of M or has a distance from a point of M that equals the distance between some pair of points in M. Therefore, H is covered by the perpendicular bisector hyperplanes of pairs of points of M, together with the spheres centered at and passing through points of M. The set of these hyperplanes F i and spheres S i is countable (i = 1, 2, ... ), since M is countable. Each H n S; is a countable set. Indeed, the preceding argument can be applied to H n S, instead of H (since H n Sj satisfies all assumptions made on H); therefore, the set of Si's and F's that replace the preceding Si's and Fi's and cover H n SS is countable; by the induction hypothesis H n F' and (since S3 n Sz is contained in a hyperplane of En) H n S; n Sz are countable

sets. Thus, H n S; is countable since it can be covered by a countable family of countable sets. Further, by the induction hypothesis, each H n F3 is a countable set. Therefore, H is a countable set since it is covered by a countable family of countable sets.

In order to prove the second statement of the problem, it suffices to construct an uncountable subset of the Hilbert space spanned by the orthonormal basis ei (i = 1, 2.... ), such that the set of distances between all pairs of points in this subset is countable. To this end, consider an uncountable set 2t of infinite sets of natural numbers with pairwise finite intersection. It is well known that such a set f2t exists. (For instance, the set of bounded, monotone sequences of rational numbers has this property.) Each set A(E 2t) determines a vector EÂ°O1(1/2i) eq; in the Hilbert space, where ai runs through A in increasing order. The set K of these vectors is uncountable. On the other hand, the distance between any pair of elements of K is the square root of a rational number, for if the vectors 1(1/2i) ear are determined by the sets a = E', (1/2') ea; and a' A and A' (E 2t), then

518

3. SOLUTIONS TO THE PROBLEMS

Ila-all2=11a112=I1a112-2(a,a')=3-2

12ij

a;=ai

where in the last term we have a finite sum since A n A' is finite.

Remark. Istvan Juhasz and Bela Bollobas pointed out that a similar argument proves the following generalization of the first statement of the problem: If m is a regular cardinal number, then any subset of cardinality m of the Euclidean n-space has a subset of cardinality m in which all distances between pairs of points are different; further, they considered generalizations of the statement to other metric spaces. Problem T.2. A sentence of the following type is often heard in Hungarian weather reports: "Last night's minimum temperatures took all values between -3 degrees and +5 degrees." Show that it would suffice to say, "Both -3 degrees and +5 degrees occurred among last night's minimum temperatures." (Assume that temperature as a two-variable function of place and time is continuous.) Remark. The formulation of the problem allows for various models. The proof is simplest when the country is assumed to be compact; not assuming compactness requires a different proof and yields a more general theorem. The argument needs some modification if the time interval is replaced with a (not necessarily metrizable) compact space. In the following solutions, the space is connected and the time is compact. The proof is carried out for metric spaces in the first one and for arbitrary topological spaces in the second one. Solution 1. Since any continuous image of a connected space is connected, and in the real line only the intervals are connected sets, it is sufficient to prove that the function that assigns to each point the minimum temperature attained there during the night is continuous.

Let E be a connected metric space, I be a compact metric space, R be the real line, and f : I --* R be a continuous function. For any fixed x E E, the function f (x, t) is continuous on the compact space I; therefore,

g(x) = mintEj f (x, t) exists. We show that g(x) is continuous. If this were not the case, then there would exist an e > 0 and a sequence {xk} (xk E E, k = 1, 2, ...) such that x, --* x, but I9(xk) - 9(x)I ? e. We consider two cases: (a) There is a subsequence of {x,,,} such that

9(xfk)

9(x) - E.

If f(xlk,tk) = g(xIIk), then the sequence {tk} has an accumulation point t E I, and we can assume that tk -> t. Then f(xnk,tk) = 9(xfk) C 9(x) - e C f(x,t) - E,

and f cannot be continuous.

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519

(b) There is a subsequence of {x,,,} such that g(xnk) ? g(x) + E. If f(x,t) = g(x),then f(xflk,t) >_ g(xnk) > g(x) + E > f(x,t) + E,

and f cannot be continuous. Since at least one of the cases must hold, we have a contradiction, which proves the statement.

Solution 2. Now let E be a connected topological space and I be a compact topological space. We keep other notations of Solution 1.

We prove again that g(x) is continuous on E. For e > 0, u c E, and v c I, let U(u, v) and V (u, v) be respective neighborhoods of u and v such that for all (x, t) E U(u, v) x V (u, v), we have

If(u,v)-f(x,t)I <E. For u fixed, the neighborhoods V (u, v) (v E I) cover I. Then, by its compactness, I is covered by a finite number of them:

V(u,vi),V(u,v2),...,V(u,vk). Then u = flk 1U(u, vi) is a neighborhood of u, and if x E U, then Ig(x) g(u) I < E, for if t c I, then t c V (u, vj) for some j, and then, since (x, t), (u, t) E U(u, vj) x V (u, vj), we have

if(x,t) - f(n,t)I < E. Remarks.

1. Assuming space and time to be compact topological spaces, Attila Mate considered temperature to take values in a metric space and proved, under these conditions, that the function g(x) is continuous. Compactness of space is not essential in this model. 2. Gyorgy Vesztergombi proved that the statement of the problem remains true if the definition of night depends on place (astronomical night), provided that the beginning and the end of the night are continuous functions of place.

3. Miklos Simonovits gave an example to show that it is essential to assume compactness of night. (If the night is not compact, then, of course, we have to speak of local infimum instead of local minimum.)

Let E = [-1. + 1], I = (-oo, +oo), f (x, t) = g(x)

and g(x) is not continuous.

' t) = E if (x { t

e-t2X2. Then

for x = 0,

forx # 0,

3. SOLUTIONS TO THE PROBLEMS

520

4. Juhasz and Posa gave examples that show that it is not sufficient to assume partial continuity of f (x, t). Juhasz's example is the following:

put E = I = [-1, +1], and

f(x,t)=

ifx<O,orx>Oandt>0,

1+1 ifx>0and-x<t<0, 1+t ifx>0and-1<t<-x.

In this case, 9(x)

for x < 0,

for x > O

is not continuous, and the statement of the problem is not true.

Problem T.3. Let A be a family of proper closed subspaces of the Hilbert space H = 12 totally ordered with respect to inclusion (that is, if L1, L2 E A, then either L1 C L2 or L2 C L1). Prove that there exists a vector x E H not contained in any of the subspaces L belonging to A. Solution 1. More generally, we prove the statement for separable Banach spaces. Let B be a separable Banach space, and let R be a system of subspaces L of B that satisfies the requirements of the problem. Suppose that ULERL = B. Consider a countable, everywhere-dense subset {XI, X2.... } of B. By recursion, define a countable increasing sequence of elements of R as follows. Let L1 be an element of B that contains x1. Suppose that the subspaces L1, ... , Ln (E R) have already been defined. Let L('+') be an element of R that contains xn+1. (Such an element exists by the assumptions.) One of the subspaces L 1, ... , L,a, L(n+1) contains all others; let this one be denoted by Ln+1. Then L. C _ L,,,+1 and x,a E Ln (n = 1, 2, ... ). Now, let L be an arbitrary element of R. Since L is a proper closed subspace of B, it cannot contain all elements of the dense set {x1, X2.... }. Suppose, say, that Xk V L. By definition, xk E Lk, so L C_ Lk since the system B is ordered with respect to inclusion. This means that U°__1Li = B. Now, for all natural numbers n, let f,a be a continuous linear functional on B for which IIfnII = n and fn (x) = 0 if x E L. (Such functionals exist. For example, take an arbitrary element yn in the complement of L. Since Ln is closed, the distance d of yn from the subspace Ln is positive. Consider now the linear subspace [yn] + Ln, where [yn] denotes the one-dimensional subspace generated by yn, and let .fn°) ()tyn - x) = nAd,

where x E Ln and A is a complex number. A ) is obviously linear on [yn] + Ln and fn°) (x) = 0 for x c Ln, and fn') II =sup

IAdl

IAynxll

sup

Ilyn

xIl

infxCL.Ilyn - xII = n.

3.10 TOPOLOGY

521

By the Hahn-Banach theorem, f (0) is extendable to a functional f,,, defined

on the whole B that has the required properties.) Now, if x E B, then f,,,(x) = 0, since, by U°_1Li = B, x is contained in some Li, and the sequence {L,,,} of subspaces is increasing with respect to inclusion. So, the sequence {f ,,, } of functionals is pointwise convergent; therefore, by the Banach-Steinhaus theorem, sup, 11f,,,11 < oo. But this contradicts the

choice off,,. Solution 2. We prove the following generalization of the problem. Generalization. Let M be a separable topological space of second Baire

category, and let 2t be a system of nowhere-dense closed subsets of M ordered by inclusion. Then f1LE%L # M. (It is obvious that the space 12 and the family of subspaces in the problem and, more generally, separable Hilbert and Banach spaces with a similar

family of subspaces all satisfy the hypotheses of this statement; indeed, these spaces equipped with the norm topology are separable topological spaces of second Baire category and, in any of these spaces, a proper closed subspace is nowhere dense.)

Proof. Suppose that, to the contrary, ULE21L = M, and consider a countable, everywhere-dense set R = {x1, X2.... } in M. Put

Mk=U{LE2t:xkEL} (k=1,2,...). Then Mk is a nonempty, nowhere-dense subset of M for all k. Therefore, U' 1 Mk is of first category, and 00

U Mk

k=1

On the other hand, if L E 2t, then L j4 M. L cannot contain R, that is, there is a natural number k such that xk V L. Since 2t is ordered with respect to inclusion, L C Mk follows from the definition of Mk. Thus, ULE21L C U

1Mk, a contradiction.

Remark. Several contestants remarked that the statement of the problem is not true for nonseparable Hilbert spaces. Attila Mate gave the following example. Let H be a nonseparable Hilbert space. Then H is isomorphic to a Hilbert space K ® L2 (W1 ), where K is a suitably chosen Hilbert space, and W1 is the set of countable ordinals considered as the discrete measure space in which the measure of all singletons is 1. Then

U (K ®L2( ) = K (D L2(Wl). Ewi

522

3. SOLUTIONS TO THE PROBLEMS

Problem T.4. Let K be a compact topological group, and let F be a set of continuous functions defined on K that has cardinality greater than continuum. Prove that there exist x0 E K and f g E F such that

f (xo) = 9(xo) = max f (x) = maxg(x). xEK

xEK

Solution. In the proof, we use the following theorem by Paul Erdos.

Theorem. If the edges of a complete graph of cardinality greater than continuum are labeled with natural numbers, then there exists an uncountable complete subgraph with all edges labeled with the same number. See,

for example, P. Erdds, A. Hajnal, and R. Radio, Partition Relations for Cardinal Numbers, Acta Math. Acad. Sci. Hung. 16 (1965), Theorem 1. If Fx denotes the system of functions in F whose maximum is x, then it is clear that the cardinality of Fx is greater than continuum for some real number x. Suppose that f-x fl g=x = 0 for all pairs of functions f g E Fx. Consider the complete graph with the functions in Fx as vertices. If f # g E Fx, then there exists a natural number n such that f>x-(l/n) fl 9>x_(1/n) = 0; otherwise, the closed sets { f>x-(1/n), g>x-(1/n) }n=1,2,... would form a family with the finite intersection property whose intersection,

by compactness, would be nonempty, but this intersection is contained in the set f=x fl g=x. By the theorem quoted above, there is an n and

an uncountable F' C Fx such that all edges of F' are labeled with n. Thus, the sets f.,>,/n, f E F' are nonempty, open, and pairwise disjoint. This contradicts the well-known fact that K admits a finite Haar measure, and therefore any family of pairwise disjoint, nonempty, open sets in K is countable. So, the statement of the problem is true for some pair of functions f 54 g E F. In fact, we proved the following theorem.

Theorem. The statement of the problem holds for a compact space K if K satisfies the following condition: any family of pairwise disjoint, nonempty, open sets in K is countable.

Problem T.5. Prove that two points in a compact metric space can be joined with a rectifiable arc if and only if there exists a positive number K such that, for any E > 0, these points can be connected with an E-chain not longer than K.

Solution. If A and B are two points of the metric space, then let t(A, B) denote their distance. By a rectifiable arc joining A and B we mean a homeomorphic image of the real interval [a, b], where a and b are mapped to A and B, and for any subdivision a = to < t1 < t2 < ... < t,,,=1 < t,,, = b, denoting the image of ti by Ti, we have n-1

1: t(Ti,Ti+1) < K i=o

3.10 TOPOLOGY

523

for some fixed K. The infimum of such numbers K is called the length of the arc. Therefore, the half of the problem stating that if two points can be joined with a rectifiable arc, then for any e > 0 they can be joined with an a-chain of length at most K, is obvious. Now suppose that, for any e > 0, A and B are joinable with an a-chain not longer than K. Let L = {H0,. .. , H,,, } be a sequence of points in the metric space. We use the following notation:

q(L) = max t(Hi-1, Hi) 1<i<n

and

K(L) _

>t(Hi-1,Hi).

i=1

For 0 < h < 1, let L(h) denote the Hk for which k+1

t(H1-1, H1) < h K(L) <: t(Hz-1, Hi), i=1

i=1

and let L(1) = Hn. Then, obviously,

t(L(H1)L(H2)) < Ihl - h2I . K(L) + 4(L) Now choose an indefinitely refining sequence So of chains connecting A and B, where the lengths of these chains do not exceed a fixed constant K. The

hypotheses in the problem guarantee the existence of such a sequence of chains. Arrange the rational points of the interval [0, 1] into a sequence. Then let Sk be an indefinitely refining sequence So of chains connecting A and B such that for k = 1, 2, ... , (a) Sk is a subsequence of Sk_1i and (b) L(h) is convergent in the first k rational points when the chains are taken from Sk and k is kept fixed. (More precisely, instead of L(h), we should write Ln,k(h), where this

denotes L(h) for the nth chain of Sk. In (b), k and h are fixed while n -> oo.) The sequence of chains Sk can be defined for all k by compactness of the space. Now define the function f (h) on the set of rational numbers in [0, 1] by the formula

f (h) = lim Lk,n(h) if h is one of the first k rational numbers. This definition is correct by (a). Let h1 and h2 be rational numbers in [0, 1]. Choose a sufficiently large k so that h1 and h2 occur among the first k rational numbers. Then

t(f(h1),f(h2))

K. Ihl -h21

Since a uniformly continuous function defined on a dense subset of the interval [0, 1] can always be continuously extended over the whole interval

524

3. SOLUTIONS TO THE PROBLEMS

[0, 1], provided the target space is compact, the function f (h) is extendable,

and the last estimate remains valid. The image of the interval [0, 1] is a continuous curve connecting the points A and B, and from the estimate it immediately follows that all approximating chains have length at most K. Therefore, the curve is rectifiable of length < K. In general, it is not true that we obtain an arc. But if we choose the above K to be the smallest constant such that, for any e > 0, A and B are joinable with an e-chain not longer than K, then we get an arc. Indeed, otherwise there would exist u and v in [0, 1] (u # v), with f (u) = f (v). Then we could delete the image of the open interval (u, v) from the curve constructed above, and we would get a rectifiable curve from A to B whose length K+ is less than K. Then, by our initial observations, for any e > 0, A and B would be joinable by an e-chain shorter than K+. This contradicts the minimality of K, proving the statement of the problem. Remarks. 1. Laszlo Lovasz proved that, instead of compactness, assuming only local compactness and completeness of the space, the statement remains valid. By deleting a chord from a disk in the plane, we obtain a locally compact space in which two points on different sides of the chord cannot be joined with an arc although they are joinable with an arbitrarily fine e-chain. Therefore, it does not suffice to assume only local compactness of the space. Completeness alone is not sufficient either. To show this, Lovasz gave the following example.

Example. Let eo, e1, .

, en, ... be an orthonormal basis in the Hilbert

space 12. Consider the segments connecting 0 with the endpoints of the

vectors el, e2.... , and similarly the segments connecting eo with the points el+eo,e2+eo, .... Divide the segment between en and en+eo into n equal parts, that is, consider the points en + (i/n) eo when i = 1, 2, ... , n - 1.

The segments and points just defined form a closed set in the Hilbert space; therefore, they define a complete metric space. In this space the points 0 and eo for any e > 0 are joinable with an a-chain not longer than

3, but they cannot be joined with an arc. This space can be made into a connected counterexample by connecting the points en + ((i - 1)/n) eo and en + (i/n) eo with suitably chosen, nonrectifiable arcs. 2. A sketch of the proof of Lovasz's generalization is the following.

Proof. We call a point accessible if, for some t, B and the point are joinable with an arc of length t and, for any e > 0, the point and A are joinable with an a-chain not longer than K - t, where K is the minimal constant used above. Of the accessible points x and y we say that x is finer than y if, in the definition of accessibility of x, the arc from B can be chosen through y. Then, using Zorn's lemma, we define a "finest" point x0, which we show to be necessarily A. Assuming the contrary, take a compact neighborhood U of x0 that does not contain A. Then define a point z on the boundary of U that is a limit point of certain points of chains from x0 to A in the definition of accessibility of xo. We show that z is accessible

3.10 TOPOLOGY

525

and finer than x0. To this end, it suffices to see that x0 and z are joinable with a sufficiently short arc. But we can apply the result already obtained for compact spaces to the points x0 and z, and this proves the statement.

Problem T.6. Let a neighborhood basis of a point x of the real line consist of all Lebesgue-measurable sets containing x whose density at x equals 1. Show that this requirement defines a topology that is regular but not normal.

Solution. (1) Let m(A) denote the measure of the Lebesgue-measurable set A. Let A and B be basis neighborhoods of the point x, we show that A fl B is as well. Since A fl B is measurable and x E A fl B, we only have to show that its density at the point x is 1, that is, m((A fl B) fl I)

m(I)

-4

when the interval I shrinks to x:

m(AflBfll) m(I)

-1

m(I) - m(I - (A f1 B)) m(I)

_ m((I-A)U(I-B)) m(I) m(I - A) + m(I - B) m(I) m(I) The last expression converges to 0 when I shrinks to x. So A fl B is indeed a neighborhood of x. In order to obtain a topology, we show that any neighborhood A of x contains a neighborhood B of x that is a neighborhood of all of its points. Let B be the set of points in A where the density of A is 1. Obviously, x E B, and by the Lebesgue density theorem, A - B has measure 0. Therefore, B is also measurable, and at points of B, B has the same density as A, that is, 1. So, we have a topology indeed. Let E denote this new topology, and call its open sets E-open while keeping the adjective "open" for open sets in the usual topology. Similarly, we distinguish closed and E-closed sets. It is not obvious but it is true that the E-open sets are measurable. We can prove this as follows.

Let H be an E-open set. We can assume that H is bounded; then the outer measure m(H) of H is finite. Now put a = sup{m(C) : C C H, C is measurable},

and for all n choose Cn so that C" C H and m(CC) > a - (1/n). Then S = UnCC is a subset of H with the property that every measurable

3. SOLUTIONS TO THE PROBLEMS

526

subset of H-S has measure 0. (S is usually called the measurable core of H.) If V C H is measurable and Q = H - S, then, by (V fl Q) U (V fl s) _ V, V fl Q is measurable, and therefore it has measure 0. So,

m(V) = m(V fl s). Q decomposes as the union of two sets Z and X, where the former has measure 0, and the density of Q at all points of the latter is 1. In order to prove that H is measurable, we show that Q has measure 0, and to this end, we need X to be empty. This will imply measurability of H.

Arguing by contradiction, we suppose that p E X. Let I denote an interval shrinking to p. By definition of X, m(Q fl I) = (1 + o(1))m(I). Now let U be an E-neighborhood of p contained in H. Then

m(U fl I n s) = m(U fl I) = (1 + o(1))m(I), so u fl S has density 1 at p. Therefore, by I - (U fl s) D Q fl I, we have

m(Q fl I) < m(I - (U n s)) = o(m(I)), which contradicts our assumption. (2) We show that the topology E is regular. Suppose that K is an E-closed set, and x V K. Then x has a neighborhood A that does not meet K. So, m(AfI) = (1+o(1))m(I), where the interval I shrinks to x. Thus, m(A fl I) = o(m(I)). (Here A denotes the complement of the set A.) K C A, so m(K fl I) < m(A fl I), and therefore

m(K fl I) m(I)

Put x,,, = x - (1/n) and yn = x + (1/n), n = 1, 2,... . It is well known that, for any measurable set X and e > 0, there exists an open set G such that X C G and m(G) < m(X) + e. Choose the open set Gn in the interval [xn_1i Xn+21 so that K fl [xn, xn+1] C Gn and

m(Gn) < m(K n [xn, xn+1]) +

hold. For n = 1, we require Gl C (-oo, x2]. Choose the sets Hn in a similar way for the intervals [yn+l, yn]. Put

B=(-oo,xl)UGIUG2U...UH1UH2U...U(yl,oo); then B is open and contains K. Since open sets are E-open (because they have density 1 at every point), it will suffice to prove that the density of B at x is 0. Then the complement of the E-closure of B and B will be disjoint E-neighborhoods of x and K, respectively. It is enough to show

that m(I fl B)/m(I) -> 0 for I = [x, x + k] (k -> 0), because a similar argument shows the corresponding estimate for the left-hand intervals,

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527

and these two together imply the same for arbitrary intervals I shrinking

to x. Assume y,,,+1 < x + k < Then m(I) = k > 1/(m + 1). We have m(B n I) < Em-2 m(H2), since H2 n [x, ym] = 0 for i < m - 2. Therefore cc

m(B n i) <

m(K n [yi + 1, y2]) +

i=m-2

i=m-2 1

= m(K n [x, yn - 2]) + and so

m((II) )

2,,,_s

< m(K n [X, ym-2]) + 2 1

m+1

If I shrinks to x, then m -+ oo, so both terms in the right-hand side of the last inequality converge to 0. This is obvious for the second term; for the

(3)

first one, use the already established convergence m(K n I)/m(I) -p 0 and the fact that (m + 1)/(m - 2) < 4 if m > 3. So, B indeed has density 0 at x, and the space is regular. We show that the space is not normal; namely, that an E-closed set K1 of second Baire category and an E-closed everywhere-dense set K2 cannot be separated by E-open sets. Since a countable set has measure 0, and thus is E-closed, the existence of such a set K2 is clear. The existence of such a K1 is also well known: for every n, there exists a nowhere-dense set of measure greater than 1 - 1/n in [0, 1]; the union of these is a first category set of measure 1; and the complement of this in [0, 1] is of second category and has measure 0 and is therefore E-closed. Since K1 is chosen to have measure 0, K2 can be chosen in the complement of K1, then K1 and K2 are disjoint. Suppose that there exist two disjoint E-open sets C1 and C2 with K1 C C1 and K2 C C2. Then, for any x E C1, there is a neighborhood A C C1 of x for which

M(An I) m(I)

--+ 1 ,

an d thus

m(C1 n I)

m(I)

-1

when I shrinks to x. Therefore, by K1 C C1, for any x E K1, there exists an no such that for n > no, m (C1 n [x - (1/n), x + (1/n)]) >

that is,

M /C1

- 2'

2/n 1 n X_ n,1 x + nJ I> 1

(n > no).

If we divide the elements of K1 into a countable number of subsets depending on whether the last inequality holds from a certain index,

3. SOLUTIONS TO THE PROBLEMS

528

then at least one of these subsets is dense in an interval [a, b]. Thus, there exists an no such that the last inequality holds from this no on an everywhere-dense subset of the interval (a, b). Since K2 is everywhere dense, we can choose an element y of K2 in (a, b). Since y E C2, there is an n > no with m (C2 fl [y - (1/n), y + (1/n)]) > 3

2/n

that is, 1

m C2 fl y - y n'

+ n) ?

4n K1 is dense in (a, b), so there exists an x in the interval (y, y + (1/8n)) for which

mICln[x n,x+nJ)>n holds. The length of the interval I = [y - (1/n), x + (1/n)] is at most (2/n) + (1/8n). C1 and C2 are disjoint and m(C1 f1 I) > n

and m(C2 fl I) > 2n

Therefore,

C2+8 n >m(I)>m(C1f1I)+m(C2nI)> (1+2)

a contradiction, which proves that the topology E is not normal. Remarks.

1. In order to solve the problem, it is not necessary to prove that E-open sets are measurable, but this simplifies the proof at various points. 2. Lajos Posa proved the nonnormality statement using a cardinality argument. The sketch of his proof is as follows. Proof. There exists a set of cardinality c = and of measure 0. All subsets of this set are E-closed; therefore it can be partitioned into two disjoint E-closed sets in 2` different ways. If the space were normal, then for each such partition we could find two disjoint E-open sets that separate the two parts. It can be proved that the pairs consisting of the E-closures of the separating E-open sets corresponding to different partitions are different.

It is also easy to see that the E-closure of every E-open set equals the E-closure of an F,-subset, which only differs by a set of measure 0. Since the cardinality of the set of F,-sets is only c, the cardinality of the set of all pairs of closures of F,-sets is also only c. Therefore, the cardinality of the set of pairs of separating E-open sets is only c, a contradiction. 3. Laszlo Lovasz proved complete regularity (which is stronger than reg-

ularity). He noticed that the above technique yields that if A and B are disjoint E-closed sets, then they can be separated by an open set containing B and an E-open set containing A. Then the proof of the Urysohn lemma applies.

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Problem T.7. Suppose that V is a locally compact topological space that admits no countable covering with compact sets. Let C denote the set of all compact subsets of the space V and U the set of open subsets that are not contained in any compact set. Let f be a function from U to C such that f (U) C U for all U EU. Prove that either (i) there exists a nonempty compact set C such that f (U) is not a proper subset of C whenever C C_ U EU, (ii) or for some compact set C, the set

f-1(C) = U{U E U: f(U) C C} is an element of U, that is, f -1(C) is not contained in any compact set.

Solution. The statement is trivial. Indeed, if f -1(0) = V, then (ii) holds.

If not, then for an arbitrary x E V - f-1(0), (i) holds with C = {x}. (We do not use that V is not or-compact, only noncompactness of V is necessary.)

Remark. LAszlo Babai proves that if, under the hypotheses of the problem, (ii) is not true, then (i) holds with a set C of two elements. The proof of this is fairly difficult.

Problem T.8. Let T1 and T2 be second-countable topologies on the set E. We would like to find a real function or defined on E x E such that 0 < a(x, y) < +oo, Q(x, z) < o (x, y) + a (y, z)

Q(x, x) = 0, (x, y, z E E),

and, for any p E E, the sets Vi (p, e) = {x : Q(x, p) < E}

(E > 0)

form a neighborhood base of p with respect to TI, and the sets

V2(p,E)={x:0'(p,x)<E} (E>0) form a neighborhood base of p with respect to T2. Prove that such a function or exists if and only if, for any p E E and Ti-open set G 3 p (i = 1, 2), there exist a Ti-open set G' and a T3_i-closed set F with p E

G'CFCG.

Solution. Suppose that there exists a function Q(x, y) with the required properties. Let p E E, G be open in T, p E G. Then there exists E > 0 with Vi(p, E) C G. Let 0 < 6 < E, G' = U (p, 6), and let F be the T3_iclosure of G'. Using the triangle inequality, it easily follows that G', F, G

3. SOLUTIONS TO THE PROBLEMS

530

indeed satisfy the hypotheses. (The assumption on second countability is not needed in this direction.) In what follows, G always denotes an open set and F a closed set. Superz scripts indicate which topology this means. For example, G(1) denotes the Tz-closure of the Tl-open set G(1). We prove that the hypothesis given in the problem is sufficient for the existence of Q. The proof is a modification of the well known-proof of the Urysohn metrization theorem. Tikhonov lemma. If F(1) nF(z) = 0, then there exist G(1) and G(z) such that G(1) i F(z), G(z) D F(1), and G(1) n G(z) = 0.

Proof. Let p E F(i). Then p E E - F(3-i), this set is T3_i-open, so by the assumption there exists a G' = (p) such that p E G' (3-i) (p) and G'(3-Z)

G'(3-i)(p)(Z)

C E - F(3-i), that is, G'(3-a)(p)(i) n F(3-i) = 0

To each such G'(j) (p), assign an element UW of the countable basis in T : p E UU) C G'(j) (p). We have countably many sets UU); these can be arranged as U(j), n = 1, 2, ... , (j = 1, 2). We also have oo

U(i) (3-i) nF(i)=0.

F (3-i) c u Un(z)

(1)

n=1

Put U() = U(x)

- J I U(31)(y) k I

k=1

These Un('*) are T-open, they also satisfy (1), and obviously U. n U(m3*i) 0. Therefore, the sets

G(i) = U Un

(i = 1, 2)

n=1

clearly satisfy the requirements of the lemma. Urysohn lemma. If F(1) C G(z), then there exists a real function p(x, y)

on E x E such that 0 < p(x,y) < 1,

p(x,x) = 0,

P(x, z) < P(x, y) + P(y, z)

(x, y, z E E);

(2)

the sets 01) (p, E) = {x : p(x, p) < E}

(E > 0)

(3)

(E > 0)

(4)

G(z).

(5)

are Tl-open; the sets Vn(z) (p, E) = {x : p(p, x) < E}

are Tz-open; an d

p(x, y) = 1

if x E F(1), Y

3.10 TOPOLOGY

531

FMMM, G12) = G(2), Proof. Introduce the notations Got) = 0, Fol) Fil) = E. Let D be a countable dense subset of the-closed interval [0, 1]: D = fro, rl, r2i ... }, where ro = 0, rl = 1.

By recursion on n, we define the sets

and F(1) so that Gn(2) C F(1)

and, for rn, < r,,,,, hold. These inclusions are true in the Th C GM) cases already defined (n, m = 0, 1). If the sets with subscripts less than n (n > 2) have already been defined, then find the neighbors of rn among ro, r1, ... , r,i_1. Let them be rk and r1: rk < rn, < rl, 0 < k, 1 < n - 1. Apply the Tychonoff lemma to the sets Fkl) and E - G12): by rk < rj and the induction hypothesis, Fk(1) n (E - G12)) = 0, so by the the Tychonoff lemma, there exist

and Gnl) with

G(2) j Fkl),

n Gn,:) = 0 and

G(,:) D E

that is, using the notation F(1) _ E - Gn :) Fn(l) C G(2)

and Gn2) n G,:)

0 means that G(2) C F(1).

It is clear that the validity of the induction hypothesis is inherited to these sets, that is, the recursive definition is correct. For x E E, put g(x) = inf({rn, : x E

U {1}),

f (x) = sup({rn : x V Fn(l) } U {0}),

(0 < f (x), g(x) < 1). It is clear that f I FM = gIF(1) = 0,

f J E - G(2) = gI E - G(2) - 1.

Further, if n > 2, then g(x) < rn,

x E G(2)

= x E F(1) = f (X) < rte,

so f (x) < g(x) for all x E E. But if f (x) < g(x) for some x E E, then there would exist rte,, rm E D such that

f(x) <rn <r,n. <g(x). Then x E Fn(l) and x V GMT, but F(1) C .f (x) = g(x) on E.

a contradiction. Therefore,

3. SOLUTIONS TO THE PROBLEMS

532

For x, y E E, put if f (x) >- f (y),

P(x, y) =

1 f (y) - f (x), if f (x) < f (y)

Obviously, 0 < p(x, y) < 1, p(x, x) = 0, and (2) and (5) hold. To prove (3) and (4), by (2) it suffices to show that p E int(i) V(a) (p, E) (p E E, e > 0) (i = 1, 2). For i = 1: Let e > 0, p E E. Then V(1) (p,

E) = {x : p(x, p) < E} = {x : f (p) - f (x) < E} D E - F(1) E) p

if n is chosen so that f (p) - E < r,, < f (p). In case f (p) = 0, V(1) (p, e) _ EE) p. E - F n(1) and E are Ti-open. For i = 2: VP(2) (p,

E) = {x : p(x, p) < E} = {x : f (X) - f (p) < E}

= {x : g(x) - g(p) < E} D

if 9(p) < rn < 9(p) + E. In case 9(p) = 1, This proves the Urysohn lemma.

V(2) (p, E)

= E.

We turn now to the proof of the theorem. Let P be the set of all pairs (U(2), V(ii), where UW and VW are elements of the countable basis for the topology T and U(-,

(3 -z)

C V(i)

(for i = 1, 2).

Then P is countable: P = {(U1,V1), (U2,V2),... I-

Now for k = 1, 2, ... we define a function pk(x, y). Put (Uk, Vk) = V(')) If i = 2, then apply the Urysohn lemma to

the pair of sets F'1) = U7k(2iGig' = V(2), and call pk the function p given

by the lemma. If i = 1, then put F(2) = GM = Vile, and apply the version of the Urysohn lemma where the superscripts (1) and (2) are interchanged and p(a, b) is replaced with p(b, a) everywhere. (Then (3) and (4) simply interchange and (2) remains the same.) Call Pk the function p thus obtained. Put 00

2k Pk(x, y) =E k=1 1

o (x, y)

3.10 TOPOLOGY

533

It is clear that 0 < o (x, y) < 1, o(x, x) = 0 and, by (2), u(x, z) < a (x, y) + o (y, z) holds for all x, y E E.

We prove that fore>0andpEE

(i=1,2).

p E int(z) U (p, E)

Let N be sufficiently large so that >k N

< 2. Then

i(P,E)

nv )(p,2) k=1

But, by (3) and (4), the right-hand side is a Ti-open set containing p. It only remains to show that if p E int(i) U, then p E V (p, e) C U for some e > 0. Since p E int(i)U, there is a G(i), then, by the assumptions, there is a G'(z) in the countable basis for Ti, such that p E Gt(z)

C G(i) C U.

Therefore, (G'('), G()) = (0), V(z)) E P for some k. So, using p E G'(') and (5), Vi

(p,

C p(k)(P,1) C G(i).

Problem T.9. Prove that there exists a topological space T containing the real line as a subset, such that the Lebesgue-measurable functions, and only those, extend continuously over T. Show that the real line cannot be an everywhere-dense subset of such a space T.

Solution. 1. Let { fi : i E I} be the set of all Lebesgue-measurable functions, Xi be a copy of the real line with the usual topology (i E I), and

T = fXi. iEI

If h : R -> T is the map defined by

pi(h(x)) = fi(x), where pi is the projection onto Xi, then h is injective and h(R) can be considered as a copy of the real line. We prove that T is as required. Let f be a measurable function, say, f = fi. Then f = piI h(R) and a continuous extension of this over T is pi. Conversely, if k : T -> R is continuous, then kIh(lR) is measurable. Indeed, by a well-known theorem

(see, for example, R. Engelking, Outline of General Topology, PWNPolish Sci. Publ, Warsaw, 1968, p. 98, Problem R), k only depends on

3. SOLUTIONS TO THE PROBLEMS

534

countably many coordinates, that is, k = k' o p, where p : T --> T' is the projection onto the product T' = 11iE1, Xi, where I' C_ I is countable and k' : T' -> R is continuous. Now,

{x : k(h(x)) > c} = {x : k'(p(h(x))) > c} = {x : p(h(x)) E U} where UU is open in V. The sets of the form H Yi,

iEl' where Yi = Xi with a finite number of exceptions when Yi is an interval

with rational endpoints, form a countable basis in V. Therefore, the set {x : k(h(x)) > c} can be written as a countable union of sets of the form {x : p(h(x)) E 11iEl' Yi}. But

x : p(h(x)) E fl Yi

= IX: fi(x) E Yi (i E I)},

iEl'

being a union of countably many measurable sets, is measurable. Thus, {x : k(h(x)) > c} is measurable. 2. Suppose that T has the required properties and R C T is dense. Since all measurable functions are continuous, it is obvious that the topology of T induces the discrete topology on R. Let p c T. Let f * denote the extension of the function f (x) = x over T (this is unique since R is dense in T). Further, consider the function

g(x) =

if x = f * (p), if x f*(p)

10, i

and let g* be its continuous extension over T. Put UP =

{q: Ig*(p) - g*(q)I < 1, If *(p) - f*(q)I <

g*(q) + 1

}

Let x E Up n R (since R is dense, such an x exists); we show that x = f * (p). Indeed, if x

f * (p), then

Ig(x)I = x _ f*(p) Ig*(p)I + 1, > I

which contradicts that I g* (p) - g* (x) I < 1. So Up n R = { f * (p)}. From this it also follows that f * (q) = f * (p) for every q E Up; indeed, UgnUpnR

is nonempty (since R is everywhere dense), but its only element must equal both f * (q) and f * (p). So f * (p) = f * (q). Now for an arbitrary, nonmeasurable, real function cp, the function 0* = f* (

is a continuous extension of cp, because this function, being constant on the neighborhood Up of any point p E T, is continuous. Thus, nonmeasurable functions also have continuous extensions over T, and this is a contradiction.

3.10 TOPOLOGY

535

Problem T.10. Let A be a closed and bounded set in the plane, and let C denote the set of points at a unit distance from A. Let p E C, and assume that the intersection of A with the unit circle K centered at p can be covered by an arc shorter than a semicircle of K. Prove that the intersection of C with a suitable neighborhood of p is a simple arc of which p is not an endpoint.

Solution. Let ab be the minimal arc of K containing K fl A (possibly a = b). Introduce Cartesian coordinates in the plane so that p is the origin, a and b lie in the left half-plane symmetrically with respect to the horizontal axis, and in case a 54 b, b lies in the upper half-plane. We claim that with a suitable 6 > 0, the intersection C1 of C with the upper half-plane and with the disc of radius 6 centered at p is a simple arc starting from p that only meets the horizontal axis at p. Then a similar statement is true for the lower half-plane, and this proves the theorem. In order to prove our claim, it suffices to show that, for any r < 6, the set C meets the upper semicircle Kr of radius r centered at p in a single interior point. In this case, assigning to points of C1 their distance from p is a topological map from the compact set C1 onto the interval [0, b], since it is continuous and bijective. Let T be a convex plane sector, containing a and b in its interior, with vertex at p, and with the horizontal coordinate axis as axis of symmetry. Then, obviously, p(p, A - T) = 1 + el > 1. Further,

there is an e2 > 0 such that if x E T, say, in the upper half-plane, and o(x, p) < E2, then the triangle pxb has an obtuse angle at p; Therefore, o(x, b) < p(p, b) = 1, and so o(x, A) < 1. Now let b < min(el, e2). If r < b, then K, fl Cl is nonempty. Indeed, let c and d be the endpoints of Kr in the left and right half-planes, respectively. Then o(d, A) < 1 by the above. On the other hand, if x E A - T, then o(x,c)

e(p,x) - o(p,c) > 1+6-6 > 1;

and if x E A fl T, then the triangle xpc has an obtuse angle at p, so o(c, x) > o(p, x) > 1.

Thus, o(c, A) > 1. By continuity of o, there is a point y on the arc Kr for which o(y, A) = 1, that is, y E C1. It also follows that c, d V C1.

On the other hand, suppose that y1, Y2 e C1 fl Kr. By the above, y1, y2 V T. Let yl be the one of y1i y2 that is closer to d, and let t E A be a point such that o(y2i t) = 1. Then t E T, because o(p, t) <- o(P, y2) + o(y2, t) < 1 + 6 < 1 + el.

Since o(yl, t) > 1 = o(y2i t), the point t is on the same side of the perpendicular bisector of the segment yl y2 (which goes through p) as y2. Therefore, the angle y2ptL, measured in positive orientation, is less than 180 degrees

and obviously is greater than 90 degrees. Then the triangle y2pt has an obtuse angle at p, and so o(y2i t) > p(p, t) > 1, a contradiction.

536

3. SOLUTIONS TO THE PROBLEMS

Problem T.11. Suppose that 'r is a metrizable topology on a set X of cardinality less than or equal to continuum. Prove that there exists a separable and metrizable topology on X that is coarser than T. Solution 1. We say that a set A is separated by a family of subsets if for any a, b E A, a # b, there exists a set in the family that contains a but does not contain b. If the cardinality of A is at most continuum, then A is always separated by a countable family of subsets. Indeed, embed A into R and take the subsets corresponding to intervals with rational endpoints. First, we show that there exists a sequence of closed subsets in X that separates X. It is well known that every metrizable space admits a adiscrete basis. Let B = Un__1Bn be a basis in X such that B,, consists of pairwise disjoint, nonempty sets. Then the cardinality of Bn is less than or equal to continuum, so we can take a family {Bn : i = 1, 2, ... } (a family of families of sets) that separates Bn. Let U, be the union of the elements of then U, is an open set with respect to the topology T. The family {Un} separates X. Indeed, let x, y E X, x # y. Since B is a basis, there is an n and V E Bn such that x E V and y V V. There may or may not exist a V' E Bn with y E V', but there is at most one such V', since Bn is a disjoint family. Choose a B; such that V E Bn and V' V Bn (or, an arbitrary B; containing V if there is no V'). Then, obviously, x E U, and y V U, . As the family {Un} separates X, so does the family consisting of the complements of the sets U, . Arrange these complements into a sequence F1, F2,... . Then {Fn} is a family that separates X and consists of closed sets. Let d be a bounded metric on X that induces the topology T. Consider the following map f : X -+ 11 :

f (x) =

Cd(Fi, x) d(F2, x) 21

Then f is injective, because if x # y there would exist an Fn with x E Fn and y V Fn, and then d(Fn, x) = 0 # d(Fn, y). The map f is continuous, for if D denotes the distance in 11, then 00

D(f (x), f (y)) = > 2-n I d(Fn, x) - d(Fn, y) I n=1

E 2-nd(x, y) = d(x, y) n=1

Consider now the topology on X determined by f as the inverse image topology. Since 11 is a separable metric space (and separability is a hereditary property of metric spaces) and f is injective, this topology is separable and metrizable and, since f is continuous, it is coarser than T.

Solution 2. For any cardinal number m > 0, we denote by J(m) the following metric space: the points of J(m) are the pairs (i, x), where i E I

3.10 TOPOLOGY

537

(I is an arbitrary set of cardinality m) and x E [0, 1], with the points of the form (i, 0) identified. The metric d of J(m) is defined as y))

x+y,

ifi

if i = j. In the proof, we use the theorem that asserts that every metric space of ix - YI,

weight m is topologically embeddable into the topological product of countably many copies of the space J(m). (See, for example, R. Engelking, Outline of General Topology, PWN-Polish Sci. Publ, Warsaw, 1968, p. 197, Theorem 7.) Let P denote the following property of topological spaces (X, T): there

exists on X a separable metrizable topology coarser than r. It is obvious that all subspaces of a space with property P and topological products of countable families of spaces with property P also have property P. It is also clear that metric spaces satisfying the hypotheses of the problem are of weight less than or equal to c (continuum). Since, by the theorem cited above, all metric spaces of weight < c are embeddable into the topological product of countably many copies of J(c), it suffices to show that J(c) has property P. It is obvious that J(c) is homeomorphic to the metric space whose underlying set is the unit disc and in which the distance between two points on the same radius is their usual distance, and the distance between two points on different radii is the sum of the norms of the points. The usual topology of the unit disc is coarser than this topology and is separable and metrizable.

Problem T.12. Suppose that all subspaces of cardinality at most R1 of a topological space are second-countable. Prove that the whole space is second-countable.

Solution. We argue by contradiction. Suppose that the topological space

X is not second-countable but that all subspaces of cardinality < R1 are. We construct a certain subspace Y of cardinality < R1, and secondcountability of Y will lead to a contradiction. The subspace Y is the union of subspaces Y« (a < w1) to be defined by transfinite recursion. Together with the Y«, we simultaneously define a family 9« of open sets in X so that 9« IY« is a basis for Y. (If f is a family of subsets in X, and Z C X, then 1-1I Z = {H f1 Z : H E f} is the trace of f in Z.) The recursion is as follows: Yo = go = 0. If the countable YY and 9g have already been defined for < a < w1, then define Y« and 9« the following way. The family UC<«9g, being countable, is not a basis for X, therefore there exists a point p E X and a neighborhood V of p such that, for any element G of Ug<«9g containing p, we have G 0 V. Choose a point PG E G \ V for each such G, and put

<«

U9e

<«

538

3. SOLUTIONS TO THE PROBLEMS

It is clear that Y,, is countable; therefore, by the assumptions, it is secondcountable. Let g, be a countable family of open sets in X such that 9aIY,, is a basis in Y. Consider the subspace Y = U,, <,,1 Ye, . It obviously has cardinality < 1Z 1, so it is second-countable. The crucial observation now is that Ua<,,,1 Ga IY is a basis for Y. Let us postpone the proof of this statement and show first how to complete the solution, assuming that this observation is valid. Since Y is second-countable, the basis U,,<,1 ga IY contains a countable is a basis basis. Therefore, for a sufficiently large a, the family

is a basis for Y,,. But this is a contradiction since u< ,,G I Y,,, does not contain a basis of neighborhoods for the point p used in the definition of Y. It only remains to show that U,<,1 Ga IY is a basis for Y. This statement is immediate from the following lemma. Let Y be a second-countable topological space, and let Y = U,<",1Ye,, where YQ C Y,3 for a < /3. For a < wl, let G,, be a family of open sets in Y such that GO IY,, is a basis for YQ. Then U,,<",1 g,, is a basis for Y.

for Y. Then, with this a,

The proof is again by contradiction. Suppose that U,,<1G is not a basis for Y. We construct a sequence {V,, : a < wl } of open sets with the property Vp \ UQ> VQ # 0 (/3 < wl ), but such a sequence cannot exist in a second-countable space. The sets Vim, together with a sequence of points rQ E Vim, are defined by transfinite recursion as follows. Since U,<,1g,, is not a basis for Y,

there exists a point q E Y and a neighborhood U of q such that G 0 U whenever q E G E U,,,<L,19,,,. Put ro = q and Vo = Y. If YY and rg have already been defined for < a < wl, then first choose a p < wl with Yp D {rg : < a}. Since 9pjYP is a basis for Yp, there exists a set V,, E c such that q E Yp n V0, C U. On the other hand, V,,, 0 U; therefore we can choose a point rQ E Vc, \ U. Thus, the sequences { VQ }, {rte } are defined. Notice that if /3 < a, then r,3 V. Indeed, r,3 E Yp \ U while V, n Y,, C

U. Thus, rp E V,3 \ U,,,>,3V,, (3 < wl). But this contradicts that Y admits a countable basis {B1, B2, ... }. Indeed, for each a choose a Bna with r,,, E B,,,a C V. Since there are only countably many B,,,, there exists a > /3 with n, = n,3 = n. But then rp E Bn and r,3 # Vc D Bn, which is impossible.

Remark. Unfortunately, the phrase "at most" was missing from the text of the problem when it was posed for the competition. The statement is not true in that version; a counterexample is provided by the set of natural numbers endowed with the topology in which the nonempty, open sets are the sequences of density 1. This is not a second-countable space, but the hypothesis is vacuously true. The contestants usually noticed this defect and considered the correctly modified version of the problem.

Seven solutions were submitted. Best results were obtained by Emil Kiss. He proved the statement for T3-spaces among other arguments that contained all the ideas necessary to prove the statement in full. Nandor Simanyi proved the statement assuming the continuum hypothesis and reg-

3.10 TOPOLOGY

539

ularity of the space. Vilmos Totik used the hypothesis that every point has a basis of neighborhoods of cardinality N1, and Zoltan Szabo assumed firstcountability of the space.

Problem T.13. Suppose that the T3-space X has no isolated points and that in X any family of pairwise disjoint, nonempty, open sets is countable. Prove that X can be covered by at most continuum many nowhere-dense sets.

Solution 1. By transfinite recursion on a < wl, we define a family of open sets in X indexed by sequences of length a consisting of natural numbers,

that is, by functions f : a --+ w. For a = 0, the only such function is the empty set; put Go = X. If a is a limit ordinal and f : a -4 w, then put Gf = int (lgGg), where g ranges over restrictions of f to all smaller ordinals. If a has the form a = Q + 1 and G f = 0 for some f : ,3 -* w, then we define Gg = 0 for all extensions g : 03 - w of f. If Gf # 0, then we can define a sequence {Hl, H2, ... } of open sets such that H,,CGf,H,,,f1H 01,H,,0 0 foralln54 m, and Hl, H2,... is a maximal system of nonempty, open sets in Gf. (Here we used regularity of X and the property formulated in the problem. It is easy to avoid that this

system is finite.) Let the sets Hl, H2i ... be indexed with the extensions 91,92.... of f over 0 + 1; the cardinality of these is also count ably infinite: {Hl, H2,. .. } = {Ggl, Gg2, ... }. So the sets Gf are defined for all functions

f :a-+ w (a<wl). It is obvious that GfflGg=0if f,g:a-*wand

g, and that Gf 2 Gg if f C g. First, we prove that there is no p E X that is contained in Gf with some f : a --> w for all a < wl. Indeed, let us suppose the contrary. Then, by our preceding remarks, for all a < wl there exists precisely one function fa : a -4 w with p E C1, and these functions are restrictions of a single function F : wl -+ w. Then none of the sets G f. is empty, and

if we define the functions ga : a + 1 :-+ w by ga(d) = fa(d) ( < a) and g. (a) = fa+l (a) + 1, then the sets Gga are pairwise disjoint, nonempty, open sets, which contradicts the assumptions. Thus, each point p "dies out" at a certain level below w1. Therefore, the space X can be covered as follows:

U Gg Ua<wl U f:a-w U (flCCf). X=a<wl U f:a-.w U Gf-g:a+1-.w gj f

limit

9vf

It is immediate that this is a covering by 2Â° sets. It remains to show that the summands are nowhere-dense sets. The set

G f - U Gg g:a+1-.w

gjf

540

3. SOLUTIONS TO THE PROBLEMS

is nowhere dense because otherwise it would have a nonempty interior, and

the family of the sets G9 would not be maximal. For the other type of summands, denoting the restriction of f to Q by g,3 for Q < a, we have

n Gsv g n

,3<a

G97o,

'3+1<a

C I I Gsv R<a

that is, the summand is a closed set minus its interior, which, being the boundary of a closed set, is nowhere dense.

Solution 2. Let {Ga : a < Al be a well-ordering of the set of nonempty, open sets of X. Since there are no isolated points, for each a < A there exists a pair GT), G,(1) of nonempty, open sets with GT) n G,(1) = 0 and Gq째) U GT C_ G. Let p c X be arbitrary, and define the set' H(a, p) by transfinite recursion on a as follows. Suppose that the sets H(0, p) are already defined for Q < a. If (U0<aH(,3, p)) n Ga # 0, put H(a,p) = 0. If (U0<aH(,3, p)) n Ga = 0 and p E G7째), put H(a,p) = G(l); otherwise put H(a, p) = Gc(째i. Finally, let Fp = X - Ua<AH(a, p). It is obvious that Fp is a closed set containing p; moreover, it is nowhere dense, because if it contained a nonempty, open set, then it would contain another nonempty, open set, say, Ga, which would not contain p, but then, by the construction, G,(째) would not be a subset of Fp. It remains to prove that the number of different FP's cannot be greater than continuum. For each p there can only be countably many a with H(a, p) 0, since

the sets H(a, p) are pairwise disjoint. Let the set of these be jag(p) : < }, where cp = cp(p) < w1. We claim that if p and q are points with ap(p) = cp(q) such that, for every < cp(p), H(ag(p), p) = G(1e(p) implies H(ag(q),q) = GaaE(gI (that is, for each < cp(p), the points p and q "ramify in the same direction"), then Fp = F9. Taking this statement for granted, we can complete the proof by the observation that the cardinality of 0-1 sequences indexed by countable ordinals is continuum; therefore the number of different Fp's is at most continuum. It remains to prove our claim. It will suffice to show, by transfinite induction, that ag (p) = aC (q). If a((p) = a((q) holds for all ( < C, then obviously H(a((p),p) = H(a((q),q), and ac(p) is the first ordinal

a > sup{a( : ( < } with Ga n p)) = 0; similarly, aC(q) is the first a with Ga n (U(<CH(a(, q)) = 0, and therefore ap(p) = ag(q). Remark. Janos Kollar generalized the statement of the problem; by a modified version of Solution 1, he proved the statement for T2-spaces. In fact, Solution 2 does not use regularity, but this was not noticed by the author.

Problem T.14. Construct an uncountable Hausdorff space in which the complement of the closure of any nonempty, open set is countable.

3.10 TOPOLOGY

541

Solution 1. Let X be an arbitrary uncountable set. It is easy to see that there exists a topology on X with the required properties if and only if there exists a family A of subsets of X that satisfies the following conditions:

(1) JAS=worA=OforallAEA; (2) A is closed under finite intersections;

(3) for every pair a, b c A, a # b, there exist A, B E A such that a E A, (4) if A' C A and A E A\ {O}, then (UA') fl A = O implies I U A'I < w. Indeed, if A has properties (1)-(4), then X endowed with the topology generated by the basis A satisfies the requirements of the problem. Conversely, if r is a topology required by the problem, then the family A of all countable T-open sets satisfies the conditions (1)-(4). Now choose X to be the set wl of countable ordinals. Let {(b0,, c,,)

a E wl } be a sequence of all pairs of elements of wl with b0 # c,,,. By transfinite recursion, we construct a sequence of families A,,, : a E wl of subsets of wl so that for all a E wl, we have (10,) JAI=w orA=O for allAeA0,, (20,) UQ.<0,Ap C A.,

(30,) A, is closed under finite intersections, (40,) IA0I < w,

(50,) there exist B, C E A, with bc. E B, c0, c C, and B fl C = 0, (60,) if A E A, \ Up<0,Ap and A' E Up<0,A0, then JA fl A'I = w. Assuming that we have constructed such a sequence, we show that A = U0,<,IA0, satisfies conditions (1)-(4). (1), (2), and (3) are obvious. In order to check (4), let A' C A and A E A \ {0}, with (UA') fl A = 0. Let

a E wl be the smallest ordinal such that A E Ate. We shall prove that UA' c UA0,. To this end, it is sufficient to see that for each A' E A', there

exists B E A,, with A' C B. Let A' E A' be arbitrary, and let Q E wl denote the smallest ordinal B E Ap such that B D A' and B fl A = 0. (The definition of /3 is correct since there exists B E A with B D A' and B n A = 0; for example, B = A' is such.) Then Q > a would contradict

(60,), so3<a,that is,BEApcA,. In order to construct the sequence {A,, : a c wl }, we need the following simple lemma.

Lemma. If JEl = w and E is a nonempty, countable family of subsets

of E, then there exist subsets B and C of E such that B fl C = 0 and

IBfAI=ICflAl=w for allAEe. Proof. Indeed, put e = {An : n E w} (possibly with multiple appearances), and for n c w, let An = UkEWAnk be a partition of An, into pairwise disjoint infinite sets. Arrange the sets Ask (n, k c w) into a sequence {A' : n c w}. Since each A' is infinite, by recursion we can define the elements bn,, cam, E w such that bn,, cn, E A' \ ({bk : k < n} U {ck : k < n}) for

all n c w. Then B = {bn : n E w} and C = {cn, : n E w} are as required.

Turning now to the construction of the sequence {A,, : a < wl }, let B0, Co C wl be two arbitrary countably infinite sets such that bo E Bo,

3. SOLUTIONS TO THE PROBLEMS

542

co E Co and Bo n Co = 0. Then Ao = {Bo, Co, 0} satisfies conditions (l0)-(6o). Suppose that for some ordinal a E w1, a > 0, we already have defined the sequence {A0 :,3 < a} so that conditions (1Q)-(60) are fulfilled

for all 0 < a. Then, by applying the lemma to E = U0<a(UA,3) and e = U0<aA0, there exist countably infinite sets B, C C Uj<a(UAO) such that jBnAj = ICnAI = w for all A E Up<aAp. If Aa is defined to consist of finite intersections from the family UQ<aApU{B U {ba}, C U {cam}}, then it is routine to check that conditions are fulfilled, which completes the proof.

Solution 2. Let D be a countable, dense subspace of the product space 2"1, which exists by the Marczewski-Pondiczery theorem. Let F = { fa

aEw1}C2"1 \D such that f, I (wl \ 'Y)

fR I (wl \ 7)

and f I (wl \ 'Y)

fa I (wl \ 7)

for all a, Q, y E wl, a 5 4,3, and f E D. Choose D U F to be the underlying set of the space X to be defined; let D be open in X, and let its subspace topology coincide with the topology inherited from 2"1. It remains to define basis neighborhoods for fa (a E w1). To this end, let B = {B,,, : a E wl } be the family of elementary open sets in the product

space 2", let the finite sets Ta (a E w1) be the supports of the Ba, and put A. = sup(Uj< T0) + 1. Then let the neighborhoods of fc. be the sets of the form If, I U (B n D), where B ranges over the elements of B that contain f0, and are supported in w1 \ A. It is easy to check that we have defined a Hausdorff topology. By the construction, for all B,,, E B, we have

clx(BanD)D{f,3 :i3Ew1\(a+1)}, and thus the closure of any nonempty, open set in X has a countable complement.

Remark. It can be shown that the maximum cardinality of spaces satisfying the hypotheses of the problem is 210.

Problem T.15. Let W be a dense, open subset of the real line R. Show that the following two statements are equivalent:

(1) Every function f : R --+ R continuous at all points of R \ W and nondecreasing on every open interval contained in W is nondecreasing on the whole (2) R \ W is countable.

Solution. Suppose first that F = R\W is uncountable. Then we construct a continuous function f : ][8 -* R that is constant on every subinterval of W and that is nonconstant and decreasing on the whole We may assume that F has no isolated points (otherwise we pass to the set of its points of condensation). Let W = UV, where V is a countable

3.10 TOPOLOGY

543

family of pairwise disjoint, open intervals. Consider the natural ordering of V inherited from R. Since F is nowhere dense and perfect, this ordering of V is dense (that is, between any two elements of V, there exists a further element of V); therefore, there exists an order-preserving map q --+ VQ from

the set Q of rational points of the interval [0, 1] into V. For x E R, put

f(x)=1-sup{qEQ: (-oc,x]f1V954 0}. The function f is obviously constant on elements of V, f (x) = 1 for x E Vo, f (x) = 0 for x E Vi, and f is monotone nonincreasing. Moreover, f takes all values t E [0,1], since f (x) = t for x = sup U{V. : q E Q fl [0,1 - t]}. Therefore, f has no jumps and f is continuous.

Suppose now that JR \ W = {an : n E N} is countable. Let e > 0 be arbitrary; we show that for any pair x, y E R, x 54 y, we have f (x) < f(y) + E.

Since f is continuous at the points an (n E N), for each n E N there exists an open neighborhood Un of an such that for all u, v E Un the inequality f (u) - f (v) < e/2n+1 holds. Let W = UV, where V is a countable family of open intervals such that the restriction of f to any element of V is nondecreasing. The family U = V U {Un : n E N} is an open cover of the real line R; consider a finite subfamily U' = V U {Un : n < no} of U such that V C V and U' covers the interval [x, y]. Let L be the set of points y' in [x, y] for which there exists a finite sequence x = zo < zi < . . < zk = y' such that [zi_1, zi] C U(i) with some U(i) E U' for all 1 < i < k. It is easy

to see that L is an open-and-closed subset of [x, y] containing x, and so L = [x, y]. Therefore, there exists such a sequence z 0 , . . . , zk with zk = y. Then no

(f ('zi-1) - f (zi)) < E

P X) - f (y) = i=1

9n+1

< e.

n=O

Remark. Statement (2)x(1) of the problem can also be proved by transfinite induction on isolated points of R \ W. In connection with a few incorrect solutions, it seems worth noting that such an induction does not necessarily terminate at the first infinite ordinal.

Problem T.16. Let n > 2 be an integer, and let X be a connected Hausdorff space such that every point of X has a neighborhood homeomorphic to the Euclidean space Rn. Suppose that any discrete (not necessarily closed) subspace D of X can be covered by a family of pairwise disjoint, open sets of X so that each of these open sets contains precisely one element of D. Prove that X is a union of at most R1 compact subspaces.

Solution. Let X be a space that satisfies the hypotheses of the problem. First, we prove the following.

544

3. SOLUTIONS TO THE PROBLEMS

Statement (*). If F is an arbitrary subspace of X, then there exists a family G of open subsets of X homeomorphic to Rn such that (ug) n F is dense in the subspace F, and the family G is o-disjoint, that is, g is a union of countably many families consisting of pairwise disjoint sets.

If F = 0, then this statement is obvious. If F 54 0, then consider a maximal family U of pairwise disjoint, nonempty, separable, relative open

subsets of the subspace F. (Such a family exists by Zorn's lemma.) For each U E U, let S(U) = {xn,(U) : n = 0, 1.... } be a countable, dense subset of U. For each n, the set S,,, _ {xn(U) : U E U} is a discrete subspace of F and, thus, of X. By the hypothesis of the problem, there exists a family Gn = {Gn(U) : U E U} of pairwise disjoint, open subsets of X, such that xn(U) E Gn(U) for all U E U. Since X is locally Euclidean, we may assume that Gn consists of sets homeomorphic to Rn. Then the family G = U째째_oGn satisfies the requirements in (*), since the closure of

(ug) n F i U째째_0Sn in the subspace F contains the closure of UU in F, and, by maximality of U, the set UU is dense in F. We define a sequence { (Fe, G,,) : a E w1 } by transfinite recursion as follows. (As usual, w1 denotes the set of all countable ordinals.) Let Fo =

X and go be a maximal family of pairwise disjoint, open subsets of X homeomorphic to Rn. If a E wl, a > 0 and {(FQ,90) : Q < a} has already been defined, then put Fa = X \ Ups,,(UGp), and let G,, be a v-disjoint family of open sets of X homeomorphic to Rn such that (UGa) nF,, is dense in Fa. (The existence of such a family G,,, is guaranteed by (*).)

We show that X = U,1 (UG,,). Arguing by contradiction, assume that there exists a point x E X \ U-E- (UGa), and consider a neighborhood V of x that is homeomorphic to Rn. Since (UGa) n F,, is dense in Fa _ X \ Up<a(UG,), it follows that the sets V,, _ (Up<a (UGQ)) n V (a E wl) form a strictly monotone increasing sequence of order type w1 of open sets in V; this is impossible in a space homeomorphic to Rn. Therefore, the family G* = UaEll, Ga is a covering of X by open sets

homeomorphic to Rn, and G* is a union of at most l1 disjoint families. Since a subspace homeomorphic to Rn can only meet countably many disjoint open sets, each G E G* meets at most l1 members of G*. Finally, consider the equivalence relation on g* in which G, G' E G* are equivalent if and only if there exists a finite sequence Go, ... , Gk E G* such

that Go = G, Gk = G', and Gi n Gi+l # 0 (i = 0,1, ... , k - 1). Using the result of the last paragraph, induction on k shows that every equivalence class contains at most R1 sets from g*. On the other hand, if G C G* is an equivalence class, then UG is open and closed in X, and then g = g* by connectedness of X. Therefore, the cardinality of g* is at most 1Z1.

Thus, X is a union of at most R1 subsets homeomorphic to Rn, and therefore it is a union of at most l1 compact sets.

3.10 TOPOLOGY

545

Remarks. 1. The most well-known nonmetrizable topological manifold, the long line, is an example of a space that satisfies the hypotheses of the problem but is not a union of count ably many compact subspaces. 2. Several contestants proved the statement of the problem using the assumption that all separable subspaces of X are second-countable. Gabor Moussong showed that there exists a model of ZFC where this assump-

tion holds. Note, however, that by a construction of M. E. Rudin and P. H. Zenor in some other models of ZFC there exist separable and not second-countable topological manifolds that satisfy the hypotheses of the problem. 3. Gabor Moussong noticed that the statement of the problem follows from the continuum hypothesis, since the cardinality of a connected topological manifold of dimension > 1 is continuum.

Problem T.17. A map F : P(X) -> P(X), where P(X) denotes the set of all subsets of X, is called a closure operation on X if for arbitrary A, B C X, the following conditions hold: (i) A C F(A);

(ii) A C B = F(A) C F(B); (iii) F(F(A)) = F(A). The cardinal number min{ CAI : A C X, F(A) = X} is called the density ofF and is denoted by d(F). A set H C X is called discrete with respect

to F if u

F(H - {u}) holds for all u E H. Prove that if the density of the closure operation F is a singular cardinal number, then for any nonnegative integer n, there exists a set of size n that is discrete with respect to F. Show that the statement is not true when the existence of an infinite discrete subset is required, even if F is the closure operation of a topological space satisfying the T1 separation axiom.

Solution. (a) We prove the statement by induction on n. For n = 0, the statement is obvious; assume that it is true for n. Let the density of the closure

operation F on X be A with cf (A) = n < A. Let JAI = A, with F(A) = X, and well-order A in order type A. By recursion, define a sequence B = {x : < Al by x£ = min (A - F ({x,, : q < l;})). Then F(B) D F(A) = X. Let C C B such that ICI = , and C is cofinal in B, and for Y C B put F'(Y) = F(Y U C) fl B. It is easy to see that F' is a closure operation on B and that d(F') _ A. By the induction hypothesis, there exists a set {x£; : i < n} that is discrete with respect to F'. Let Cn > i (i < n) be such that xC E C. Then the set {xC, i < n} is discrete since xC V F ({x£; : i < n}) C F ({xn : q < En}), and x£i V F ({x£, : i # j, j < n}) C :

F'({xC3 :i# j, j <n}) for i <n. (b) Let A be a singular, strong-limit cardinal number, and let f C P(A) be a maximal, almost disjoint family of countable sets. (Such a family

546

3. SOLUTIONS TO THE PROBLEMS

exists by Zorn's lemma.) Call F C A closed if, for each H E N, IF n HI = w implies F D H. It is easy to see that by this a T1 topological space is defined in which there are no infinite discrete subsets. It remains to prove that the density is A. Let A0 C A, with IAoI < A. For < wl, we define a sequence A by the following recursion: if EN:IHnA,.I =w}, and if is a limit =71+1, then ordinal, then put A£ = U{Aq :77 < }. Since IA,7+11 _< IAqIw, it is easy to see that IACI < IAoIw < A for < wl. On the other hand, A,,,, is obviously closed, and therefore A0 is not dense.

Problem T.18. Suppose that K is a compact Hausdorff space and K = U'--OA,,, where An is metrizable and An C A,,,, for n < m. Prove that K is metrizable. Solution. First, we prove the following lemma. Lemma. If a subspace H of a metric space A is not separable, then there exists an uncountable, discrete, closed subspace Z C H.

Proof. Let Zn be a maximal subset of A such that d(x, y) > 1/n for all x, y E Zn, x # y. If a E A, then d(a, x) < 1/2n can hold for at most one point of Zn, so Zn is discrete. On the other hand, by maximality of Zn, for each h E H there exists x E Zn such that d(h, x) < 1/n, so U°_IZ, is dense. If H is not separable, then this implies that Zn is uncountable for some n. Now we prove that An is separable for all n. Indeed, assuming that it is not separable, there is an uncountable discrete closed subset Zn in An; this set is dicrete in An+1 and is not separable, so there exists an uncountable,

discrete, closed subset Zn+l in Zn. Thus, we obtain a sequence Z,n such that Z,,,, 3 Z,,,, for ml < m2, and Z,n is an uncountable, discrete, closed subset of A,,,,. Since K is compact, the set Z;,, of accumulation points of Z,n is nonempty; therefore n°O=nZm 0. On the other hand, Z;,, nA,n = 0 since Z.. is discrete and closed in An, and therefore 0, and which is a contradiction. Fix k and consider the set A n n Ak (k < n), which is a separable metric space and therefore has a countable basis {Bn,,,, : m E w}. Let Gn,m be an open set in Ak such that Bn,m = Gn,m n An. The family {Gn,m : n > k, m E w} is a subbasis for a Hausdorff topology on Ak, which is coarser than the compact topology of Ak; therefore, these two topologies are equal. This means that Ak is a second-countable, compact topological space. Let {fn,k : n E w} be a family of continuous functions on Ak that separates the points. Let Fn,k be a continuous function on K for which Fn,klAk = fn,k

(Such functions Fn,k exist by the Tietze theorem.) The functions Fn,k separate the points of K; therefore the metric defined by these functions induces the topology of K.

3.10 TOPOLOGY

547

Problem T.19. Let U denote the set If E C[0,1] : If (x) I < 1 for all x E [0, 1]}. Prove that there is no topology on C[0,1] that, together with the linear structure of C[0, 1], makes C[0,1] into a topological vector space in which the set U is compact. Solution. We assume that topological vector spaces are Hausdorff spaces. Arguing by contradiction, assume that C[0,1] is a topological vector space in which U is compact. Then U is closed since the topology is Hausdorff. For each f E C[0, 1] and 6 > 0, the set Uf,6 = {g E C[0,1] : lg(x) - f (x) 1 < b for all x E [0, 1]}

is also compact and closed. Define the functions fn, gn E C[0,1] for each natural number n > 2 as

if0<x<2-n,

10 f,, (x) =

nx + 1 - 2

if 2 - n < x <

if1<x<1,

10 gn(x) = nx - 2 1

if0<x< 2, if 2 < x < 2 + n, if

2+n

<x<1.

Put Kn = {g E C[0,1] : gn(x) < g(x) < fn(x) for all x E C[0,1]} (n > 2). fl Then K2 D K3 3 , and Kn is compact and closed since Kn = Ugn_1,1. So nn'=2Kn # 0. Let f E nn'=_2Kn. Then f (x) = 0 for x < 1/2, and f (x) = 1 for x > 1/2, which is impossible, since f is continuous. This contradiction shows that U cannot be compact. Problem T.20. Let 4 be a family of real functions defined on a set X such that k o h E 4 whenever fi c 4 (i E I) and h : X - 1[81 is defined by the formula h(x)i = fi(x), and (1) k: h(X) ---> 1[8 is continuous with respect to the topology inherited from the product topology of R1. Show that f = sup{gj : j E J, gj E implies f E 4. Does this statement m E M, hn,, E remain true if (1) is replaced with the following condition? (2) k: h(X) -> 1[8 is continuous on the closure of h(X) in the product topology.

Solution. We may obviously assume that J and M are disjoint. Define p : X -+ JJUM by p(x)j = gj (x), for j E J, and p(x),,+. = h,,, (x), form E M. Then, for ally E p(X), we have sup{yj : j E J} = inf{y,n : m E M}. Define k : p(X) -> JR by k(y) = sup{yj : j E J} = inf{ym : m E M}. Then f = k o p, so, in view of (1), it suffices to show that k is continuous on p(X). Let y E p(X) and E > 0. We shall define a neighborhood S of y in

p(X) such that Ik(z) - k(y)l < E for all z E S. Choose jo c J with

548

3. SOLUTIONS TO THE PROBLEMS

gjo(y) > k(y) - (e/2) = sup{gj(y) : j E J} - (E/2), and choose mo E M with h,,,.o(y) < k(y) + (E/2) = inf{hm(y) : m c M} + (e/2). Such jo and mo exist by the definition of supremum and infimum. Put

S = {z ep(X) : xjo -Yjo1 < 2 and Izmo - ymoI < 2 Then, for z E S, we have

k(z)=sup{zj:jEJ}>zjo>yjo-2>k(y)-e and

that is, k(z) - k(y)I < E.

Therefore, for every y E p(X) and e > 0, there exists a neighborhood S

of y such that Ik(z) - k(y)I < e for all z E S, that is, k is continuous on p(X). This solves the first part of the problem. If condition (1) is replaced with (2), then the statement is false. Indeed, let X = N be the set of natural numbers, and let 4) consist of all functions f : N -p ]R such that the sequence { f (n)}n 1 is convergent. We claim that this is a counterexample. Suppose that fi E 4) (i E I), that h : N R is defined by h(n)i = fi(n)

(i E I), and that k : h(X) -> R is continuous. Let b be the element of R1 such that bi = lim fi. Then, by the definition of h, h(n)i -+ bi for all i E I; therefore, h(n) -4 b in the product topology. This implies that b is contained in the closure of h(X) since it is the limit of a sequence in h(X). By continuity of k on h(X), we have that k(h(n)) -+ k(limh(n)) = k(b), that is, the sequence k(h(n)) is convergent, and therefore k o h E This shows that the hypotheses of the problem are satisfied. Let J = M = N; let gj (n) = 1, for n < j and n even, gj (n) = 0 otherwise; and let hm(n) = 0, for n < m and n odd, hm(n) = 1 otherwise. These functions are indeed in (P, since gj(n) --+ 0 and hm(n) -+ 1 for all j E J and

m E M. On the other hand, sup{gj(n) : j E J} = inf{h,.. (n) : m E M} equals 1 if n is even and equals 0 if n is odd. This sequence is not convergent;

therefore f = sup{gj(n) : j E J} = inf{h,,,(n) : m E M} does not belong to

Problem T.21. Characterize the sets A C R for which

A+B={a+b:aeA, bEB} is nowhere-dense whenever B C R is a nowhere dense set.

Solution. We prove that the sets with this property are the bounded sets with countable closure. More precisely, we show that for a set A C R the following are equivalent:

3.10 TOPOLOGY

549

(i) If B C R is nowhere dense, then so is A + B; (ii) For any sequence an (n = 1, 2, ...) of positive numbers, there exists a finite family of intervals II (j = 1, ... , k) that covers A and such that the length of Ij is II, I = aj (j = 1, 2, ... , k); (iii) A is bounded and its closure is countable. (i) = (ii): Suppose that (ii) does not hold. Then there exists a sequence consisting of positive numbers an (n = 1, 2, ...) such that for any finite family of intervals Ij, jIj I = aj (j = 1,2,... , k) we have A Â˘ Uj=1Ij. Using this, we shall construct a nowhere-dense set B such that A + B contains all rational numbers, which contradicts (i). Arrange the rational numbers into a sequence rn (n = 1, 2.... ). Now we define the numbers xn and yn by induction on n. Let x1 E A be arbitrary, and y1 = r1 - x1. Suppose that k > 1 and the numbers y1, y2, , yk-1 have already been defined. Then, by our assumption, the intervals

h- (rk - yJ

rk - yJ +

2I' j

=1,2,...,k-1

do not cover A. Therefore, we can choose an element xk E A \ Finally, put Yk = rk - xk. Then, with B = {yk : k E N}, the set A + B contains all rational numbers since rk = xk + yk E A + B. We show that

every point of B is isolated. Indeed, for j < k, we have xj (rk - yj aj /2, rk - yj + aj /2). Therefore, yk = rk - xk V (yj - a3/2, yj + a3/2). This means that the set B contains at most j points in the neighborhood of radius aj/2 of the point yj, and thus yj is an isolated point of B. This implies that B is nowhere dense. (ii) (iii): If (ii) is true; then A is obviously bounded. Arguing by contradiction, suppose that (iii) does not hold. Then the closure of A is

uncountable and, by the Cantor-Bendixson theorem, contains a nonempty,

bounded, and perfect set P. Let us call a sequence (a1i a2, ...) of positive numbers insufficient if for any family of closed intervals Ij, I3 I = a3 (j = 1, 2, ... , k), we have P Â˘ In order to arrive at a contradiction, by induction we shall define an infinite sequence of positive numbers (a1i a2, ...) whose all finite initial segments are insufficient. Let a1 be a

positive number that is smaller than the diameter of P; then (a1) is obviously insufficient. Suppose that n > 1 and we have already defined an insufficient sequence (a1i a2.... , an). We show that there exists a number an+1 such that (a1i a2, ... , an+1) is insufficient. Suppose this is not true; then for every k E N and an+1 = 1/k there exists a sequence of intervals Ij,k, JIj,k = aj (j = 1, 2, ... , n+ 1) such that P C U +1 Ij,k holds. We may assume that the intervals Ii,k all meet the set P, and thus they are all contained in a bounded set. By passing to a suitable subsequence, we may also assume that for each fixed j = 1, 2, ... , n + 1, the sequence of intervals Ij,k converges to an interval I j when k --f oo. Then II; = aj, for j = 1, 2, ... , n, and In+1 consists of a single point. Since (a1, a2, ... , an) is insufficient, the set P is not contained in U 1I3 . Moreover, being perfect, P has infinitely many points in the complement of U 1II. So we can choose a point

550

3. SOLUTIONS TO THE PROBLEMS

x E P \ UÂąi

But then, for sufficiently large k, x 0 U +1 Ij,k, which . contradicts the choice of the intervals Ii,k. This proves that the sequence (ai, a2i ... ,1/k) is insufficient, completing the induction. (iii) = (i): Suppose that, to the contrary of (i), with a nowhere-dense set B, the set A + B is dense in some interval I. Let A0 denote the closure of A; then A0 + B is also dense in I. Define the sets A,,, for all countable ordinals as follows: If /3 > 0 is an ordinal such that A,, has already been defined for all a < 3, then let Ap = n,<,3A0, if 0 is a limit ordinal, and

let A0 be the set of accumulation points of A,, if 0 = a + 1. Since A0 is countable, there exists a countable ordinal ry such that A., = 0. Then obviously A.1 + B = 0 and A., + B is not dense in I. Let ,3 denote the smallest ordinal for which Ap + B is not dense in I; then ,3 > 0. Let J be a subinterval of I with (A0 + B) n J = 0, and let 0 < E < IJ1/3. Let U(H, S) denote the neighborhood of radius S of the set H. It is easy to see that U(H, S) + B C U(H + B, S) holds for every set H and b > 0. Thus, U(A0, e) + B C U(A0 + B, E), and therefore U(A0, E) + B does not meet the middle third K of J. Suppose that 0 is a limit ordinal. Since {A,,, : a < 3} is a nested sequence

of compact sets whose intersection is A0, there exists a < 0 with A, C U(A0, E). Then A,, + B does not meet K. But this is impossible, since a < ,3 and A,,, + B is dense in I. Finally, suppose that 3 = a + 1. Then V = A, \ U(Ao, E) is a finite set, since all accumulation points of A,, are in A0. Then the set V + B is nowhere dense. Therefore, from (A,, + B) fl K = (V + B) fl K, it follows that (A, + B) fl K is nowhere dense. But again this contradicts that a < /3 and that A,, + B is dense in I. This completes the proof. Problem T.22. Prove that if all subspaces of a Hausdorff space X are or-compact, then X is countable.

Solution. Since X itself is a union of countably many compact sets, it is sufficient to prove the statement for compact spaces X. Suppose that X has no subset that is dense in itself. Then there is a well-ordering on X in which all initial segments are open. Indeed, suppose that the points pp E X have already been defined for all ,3 < a and that

Xa = {pp : /3 < a} # X. Then there exists a point pa E X \ X0. that is not an accumulation point of X \ X0,, and so p0. E G. C_ X. U {p.} with a suitable open neighborhood G0, of p,,. This recursion defines a wellordering of X, and UQ<0,GQ C X. = U0<0,{p0} C_ Up<0,Gp shows that X4 = Up<0.Gp is an open set for each a. An initial segment of order type

wi in X cannot be or-compact since a compact subset in X must have a greatest element. Therefore, X is countable. Thus, it suffices to prove that X does not contain a subset dense in itself. Arguing by contradiction, assume that there exists such a subset Y. Then Y is dense in itself and closed. We define a continuous function f : Y -+ [0, 1] as follows. Since Y is a compact Hausdorff space, for any two distinct points p and q there exist open sets Gp, Gq and closed sets FP,

3.10 TOPOLOGY

551

Fgsuch that pEGp9Fp,gEGg9Fq,andFpnFq=0. So, for all finite 0-1 sequences lei}, we can define closed sets AE1i... En C Y so that AEI,...,e.,o n AEI,...,En,l =0

and

AE1....,e, ,O U AEl ...,e, ,l C A,, ---,En

hold for all natural numbers n and el, ... , en E 10, 11. For any infinite 0-1 sequence e1,E2,...,the set 00

A(El, E2....) = n A,

62,--En

n=1

is nonempty and closed. If x E A(E1, E2.... ),then put °O

f (x)

Ei 2i

i=1

The function f maps A(El, E2, ... ), which is a closed set since it is an intersection of closed sets, continuously onto the interval [0, 1]. Then f extends continuously over V with Im f = [0, 1]. The interval [0, 1] contains 2x° or-compact sets (that is, Fe-sets), since the number of closed sets in [0, 1] is 2x°. Therefore, there exists a set Z C [0, 1] that is not or-compact; then its inverse image f -1(Z) C V cannot be a-compact, which contradicts the hypothesis of the problem. This proves that X contains no subset that is dense in itself.

3. SOLUTIONS TO THE PROBLEMS

552

3.11 SET THEORY Problem K.1. Does there exist a function f (x, y) of two real variables that takes natural numbers as its values and for which f (x, y) = f (y, z) implies x = y = z ?

Solution 1. First, we show that it is sufficient to construct for each real number c, two sets Ac and Bc of natural numbers that satisfy the following conditions: (1) 1 V Ac, 1 V Bc for all c,

(2) A,nB,=O forallc, (3) A.nBd54 0forc#d. Indeed, put f (c, c) = 1, and for c # d, let f (c, d) be an arbitrary element of the nonempty set Ac n Bd. Now, if

f(x,y)=f(y,z)=m01, then

mEAxnBy

and

mEAynB,,

that is,

mEA,nB.nAynBZ cAynBy=O, which is a contradiction. But m = 1 is only possible when x = y = z.

In order to construct Ac and B,, map the set R of rational numbers bijectively onto the set N = 12,3.... }; let n(r) denote the image of r E R. Assign to each real number c a sequence {rk(c)}- 1 that converges to c and consists of rational numbers different from c. Let A, = {n(rk(c)) : k = 1, 2, ... } c N,

B,=N-Ac. It is easy to see that these sets Ac and Bc satisfy conditions (1), (2), and (3). Thus, we have constructed a function with the required property.

Solution 2. Arrange the set of rational numbers into a sequence r1, r2, . Define the function f (x, y) the following way: If x = y, put f (x, y) = 1; If x < y, put f (x, y) = 2i, where i is smallest possible with x < ri < y; If x > y, put f (x, y) = 2i+1, where i is smallest possible with x > ri > y. We show that f (x, y) is as required. It is obviously defined on the reals, and its range is the set of natural numbers. Suppose that

f(x,y)=f(y,z)=m. If m = 1, then x = y = z is the only possibility. If m > 1 and m = 2i, then

x<ri<y

and

y<ri<z,

3.11 SET THEORY

553

which is a contradiction. Similarly, if m > 1 and m = 2i + 1, then

x>ri>y

and

y>ri>z,

which is again a contradiction. Therefore, x = y = z follows.

Solution 3. We will decompose the plane into a countable family of pairwise disjoint subsets with the following property: if one of the lines parallel to the coordinate axes through the point (y, y) intersects one of the subsets, then the other does not, except when the subset is the line x = y, which will be one of the subsets. It suffices to construct such a partition of the open half-plane below the line x = y, because by reflecting in the line x = y and adding the line x = y, we obtain the required partition of the whole plane.

Put

Ao={(x,y):x>0, y<0},

Ak={(x,y):x>k, k-1<y<k}, AÂ°k={(x,y):-k<x<-k+l, y<-k}, k

An = { (x, y) : An k =

{

2k-1 2n

< x < 2n ,

2k-2 2n

-2k+1 <X< -2k+2 )n , 2n

2k-1 2n

-2k -2k + 11 2n < y C 2n j

where k, n = 1, 2, .... It is obvious that the union of these sets is the open half-plane below the line x = y. This defines a partition of the whole plane, as we said above. Label the sets of this partition by the natural numbers, and assign to the point (x, y) the label of the set containing (x, y) as the value of the function f (x, y). If x, y, z are real numbers with f (x, y) = f (y, z), then the points (x, y) and (y, z) are in the same subsets, but then (x, y) (and (y, z)) is on the line through the point (y, y) parallel to the x-axis (y-axis, respectively), and any of the subsets can only meet one of these, except for the subset

{(x,y):x=y}, but then x=y=z. Remarks.

1. Obviously, the existence of such a function is a question of cardinality. In general, it makes sense to ask whether there exists a function f (x, y)

defined on a set A of cardinality m that takes its values in a set B of cardinality n (both m and n are infinite) and for which f (x, y) = f (y, z) implies x = y = z. J. Gerlits, L. Lovasz, L. Posa, and M. Simonovits proved that such a function exists if and only if m < 2Â°. Here we present the proof by Posa.

3. SOLUTIONS TO THE PROBLEMS

554

Let m < 2". It suffices to give the construction for a set A of cardinality 2". Let a be the smallest ordinal of cardinality n. We may assume that A is the set of sequences of the numbers 0 and 1 of order type a. Define the function f (x, y) as follows:

If x = y, put f (x, y) =0; If x # y, then let 13 be the smallest ordinal where the elements in x and y are different, let j be the 13th element of x (j = 0 or 1), and put f (x, y) = (13, j). The range off is the set of the symbol 0 and the ordered pairs (13, j), where 13 < a and j = 0 or 1. Therefore, it has cardinality n and can be mapped bijectively onto B. f (x, y) = f (y, z) = (13, j) means that all elements before 13 agree both

in x and y, and in y and z, and the 13th element is j in x and y. But then the 13th elements of x and y are equal, which is a contradiction. Therefore, f (x, y) = f (y, z) = 0, which implies that x = y = z. (This part of the statement is also proved by Bollobas and Juhasz.)

Suppose that m > 2", and there exists a function required in the problem. Consider the range of f (x, y) when x is kept fixed. This is a subset of B, and since the cardinality of all possible x's is greater than the cardinality of the set of all subsets of B, for some x1 $ x2 the range of f (xii y) equals that of f (x2, y). So, the value f (x1, x2) is contained in the range of f (x2i y), that is, for some yo, f (x1, x2) = f (x2i yo), which is a contradiction. 2. Bela Bollobas and Miklos Simonovits notice that there exists a function F(xl,... , xn) defined on the real numbers that takes its values in the natural numbers, and for which F(a, x2i ... , xn) = F(yi, a, y3, ... , yn) = ... = F(u1,... , un-1, a)

implies a = xi = yi =

= ui (i = 1, 2, ... , n). For example, n

F(x1i... , xn) _ 11

pj(ijxi,xi)

i,j=1

is such a function, where the pij (i, j = 1, ... , n) are different primes, and f (x, y) is the function defined in either solution of the problem.

Problem 1t.2. Prove that there exists an ordered set in which every uncountable subset contains an uncountable, well-ordered subset and that cannot be represented as a union of a countable family of well-ordered subsets. Solution. Let R be the set of all limit ordinals less than w1i and for a E R, let fa(n) be a monotone increasing sequence (of order type w) of ordinals converging to a. Order R as follows: for a,13 E R, a -< 13 if f, (n) < fp(n) holds for the first n with fa(n) 54 f1i(n).

3.11 SET THEORY

555

Any uncountable subset X of R contains a well-ordered subset of order type w1 with respect to the ordering -. Indeed, let

XIn = { (ao, al, ... , an) : there exists a E X, such that f, ,(m) = a(m) for all m < n}. For n sufficiently large, X In is uncountable, since for large enough n the ordinal

ryn=sup{fa(n): aEX} equals w1, since, by f,, (n) - a,

sup{ryn:n<w}=sup{a: aEX}=w1. Order XIn lexicographically, that is,

(ao,...,an) -<' (00,

On)

if am <,3m for the smallest m with am # /3m. Then X In is well-ordered and, if n is large enough for X In to be uncountable, it contains a wellordered subset of order type w1. Let X' be a subset of X such that for any

(ao,...,an) E XIn, there exists a unique a E X' with f.(0) = ao,...,fa(n) = an.

Then the ordered set (X', -<) is isorr. )rphic to (X In, -<'). Since we have seen that the latter contains a well-ordered subset of order type w1i the same is true of the former. This is what we wanted to prove. Now we show that R is not a union of a countable family of well-ordered subsets. First we make a digression. A function that maps a set of ordinals into the ordinals is called regressive if f holds for all ordinals # 0 in the domain of f . If the domain of f is a subset of w1 and the set

{ : f() < Al is not cofinal with Wl for any p < w1, then f is called divergent. X C wl is called thin if there exists a regressive divergent function defined on X; otherwise, it is called stationary. The set wl is stationary. Indeed, assuming that f is a regressive function on w1, the sequence

S'f(01 f(f(0),... , being a descending sequence of ordinals, can contain only a finite number of elements different from zero. Therefore,

W1 = U Xn n<w

556

3. SOLUTIONS TO THE PROBLEMS

where

Xo = {0}, and

X.+1

= {e : f (e) E Xn}. Assuming further that f is divergent, induction on n shows that sup Xn < w1 for all n, which obviously is a contradiction. From this it easily follows that the set R of limit ordinals less than w1 is also stationary. It quickly follows from the definition that the union of a countable family

of thin sets is also thin. Further, if X is stationary and f is a regressive function on X, then for some p < w1 the set

is stationary (this is a special case of Fodor's theorem). Indeed, assume that all Xµ are thin, and let fµ be a regressive divergent function defined on X. Then the function g defined on X by max(p, f,, (C))

(C E Xµ, p < wi)

is regressive. It is divergent, too, because for all v < w1 the set

{C:g(c) <v}C U{c: A -W <v} µ<v

is not cofinal with wl, since neither one of the summands in the righthand side is, because the functions fµ are divergent. This contradicts the assumption that X is stationary, which proves our statement. Let X C R be stationary. We show that X is not well ordered with respect to the ordering -<. Put Y_1 = X and

Yn={aEYn_1 : fa(n)=6n}, where 6n is the smallest ordinal for which Yn is stationary (such 6n exists by the Fodor theorem proved above).

Put

X'=X- UYn, n<W

where

Y,'={aEYn-1:fa(n)<6n}. By the choice of 6n, Y,' is not stationary, and neither is X - X'. Put 6 = sup{6n : n < w}.

It is obvious that 6 < w1. Let a E X' be greater than 6. (X' is stationary and so it is cofinal with w1.) Since fa(n) -p a, for some n < w we have f, (n) > 6 > 6n,

3.11 SET THEORY

557

so f, (m) = 6,,,, cannot hold for all m. It follows from a E X' that fa (m) >

8,,, for the smallest m with f, (m) # 6m. Thus, for this m, a is greater than all elements of Y,,, with respect to the ordering -<. Put X0 = X, ao = a, and Xl = Y,,,,. Since Y,,,, is stationary, we can repeat the above argument with Xl instead of X0. We obtain a1 and X2,

then a2 and X3, and so on. Since ao >- a1 >- a2 >- ..., X is not wellordered with respect to the ordering -<, and this is what we wanted to prove.

Now, if R = Un< R,,, then some R is stationary, and this R, is not well-ordered with respect to -<. This proves the second statement of the problem.

Remark. Laszlo Babai showed that for each cardinal number is > w, there exists an ordered set B of cardinality is such that all subsets of cardinality r. contain a well-ordered subset of cardinality K and B is not a union of a family of cardinality less than K of well-ordered subsets.

Problem R3. Let < be a reflexive, antisymmetric relation on a finite set A. Show that this relation can be extended to an appropriate finite superset B of A such that < on B remains reflexive, antisymmetric, and any two elements of B have a least upper bound as well as a greatest lower bound. (The relation < is extended to B if for x, y E A, x < y holds in A if and only if it holds in B.)

Solution. B will consist of all the elements and subsets of A without identifying the elements with the corresponding singletons. In what follows, lowercase letters denote elements, whine capitals mean subsets. Define < as follows:

a < b,

as given,

a<P, ifaEP, P<a, ifaÂ˘P, P<Q, ifPCQ. This is obviously a reflexive and antisymmetric relation. To show the other property, it suffices to treat incomparable elements, that is, two incomparable elements in A, or two subsets of A, none containing the other, as a subset. If a, b E A are incomparable, {a, b}, A \ {a, b} are the least upper (resp. greatest lower) bounds. If P, Q C_ A, then P U Q and P n Q will serve as the least upper and greatest lower bounds.

Problem 1t.4. Let .F be a family of subsets of a ground set X such that UFE.FF = X, and (a) if A, B E.F, then A U B C C for some C E .F;

(b) if A E F (n = 0,1, ... ), B E F, and Ao C Al C .... then, for some

k>0,A,,,nB=AknBforalln>k.

3. SOLUTIONS TO THE PROBLEMS

558

Show that there exist pairwise disjoint sets X., (7 E F), with X = U{X.y : 7 E F}, such that every X y is contained in some member of .F, and every element of.F is contained in the union of finitely many X.y's.

Solution. Let < be a well-ordering of F. Let F be the set of all finite subsets of F. Well order F as follows:

71,72EF,7154 72,71={A1i...,A,}, 72={B1,...IBm}1

An <... <A1i B71. <...B1i m<n. Put 7i > 72 if Ai = Bi for 1 72 if Ai < Bi or Ai > Bi. A straightforward checking shows that this well orders F. We notice that 71 < 72 implies 71 < 72. For 7 E F, we define Y.y E F as follows. If y = {A}, put Y.y = A. Assume that Y6 is defined for b < 7. Then let Yy be a member of F, covering every Yp (,Q < 7). This is possible by (a). Now put X.y = Yy \ U6<.yY6. The only nontrivial thing to prove is that

every A E F is covered by finitely many Xp. We prove by transfinite induction on -y that Yy is covered by finitely many X6's. If this first fails

for 7, there exists /31 < 02 < ... < 7 such that Y.y n Xp # 0. Let Y y = {Al,... , Ak}, A ,>. . .> Ak. For every n there is a 1 < j < k such that

A1i...,Aj_1E/n,

A.7Oi3

We can assume, by shrinking, that j is the same for every n. Put ijn = max(01, ... , Qn) < 7. As 7 is minimal, Yp meets finitely many X. As Yn, C Yn2 C

..., by (b) there is an m such that Y,,m nYy = Ynm+, nYy = ....

Xnm+: n Yy # 0 (i = 0, 1, ...) that is, Y1m meets X'7_' X,7-+,, ... , a contradiction. so Y,,m

Ynm+: n Yy

Problem R.5. Show that there exists a tournament (T, -4) of cardinality N1 containing no transitive subtournament of size N1. (A structure

(T, -) is a tournament if - is a binary, irreflexive, asymmetric, and trichotomic relation. The tournament (T, -4) is transitive if - is transitive, that is, if it orders T.) Solution. We use the existence of,a Specker type, that is, an ordered set (A, <) of size ail, not containing subsets similar to w1, wi, any uncountable subset of the reals. See (P. Erdos, A. Hajnal, A. Mate, R. Rado, Combinatorial set theory: Partition Relations for Cardinals, Akademiai Kiado, Budapest, 1984, p. 326.) Enumerate A as {a(a) : a < w1 }, and let {x(a) : a < w1} be different real numbers in [0, 1]. If a < /3, put a +- ,Q iff either x(a) < x(/3) and a(a) < a(/3) or x(a) > x(/3) and a(a) > a(,3).

3.11 SET THEORY

559

Let B C A be uncountable. Let X = {a < w1: a(a) E B}. Claim 1. There is an a E X such that {/3 E X : a(a) < a(0)

and

x(a) < x(,3)}

is uncountable.

Proof. For a E B, let f (a) be the least t E [0,11 such that x(3) < t for all but countably many 3 E X, a(,3) > a. Since f is a non-increasing real-valued function on (A, <), it can only have countably many different values; otherwise, there would be an uncountable subset of A, similar to a set of reals. Hence f is constant on an uncountable B0 C B. Define X0 analogously to X. We can choose ao E X0 such that {,Q E Xo: x(/3) > x(ao)} is uncountable; otherwise (A, <) would contain a subset of type wi.

Now we can choose a E Xo such that a(a) > a(ao) and x(a) < x(ao). Since g(a(a)) = g(a(ao)) > x(a), there are uncountably many /3 E X such

that a(3) > a(a) and x(Q) > x(a).

Claim 2. There are uncountable X0, X1 C X such that if a E X0, Q E X1, then a(a) < a(,3) and x(a) < x(/3).

Proof. Let U be the set of those a E X such that {/3 E X : a(a) < a(0)

and x(a) < x(,0)}

is countable, and let L be the set of those a E X such that {,Q E X : a(,3) < a(a)

and x(,3) < x(a)}

is countable. If U or L is uncountable, we get a contradiction by Claim 1, so we can select a E X \ U \ L, and put Xo = {,Q E X : a(,3) < a(a)

X1 = {/3 E X : a(a) < a(/3)

and x(/3) < x(a)}, and x(a) < x(,3)} .

By an application of Claim 2 to X0 and the Specker type (A, >), one can find Yo, Y1 C_ Xo such that a E Yo, ,0 E Y1 imply a(a) < a(,3), and x(a) > x(Q). Now select a E Yo, 3 E X1, -y E Y1 with a < Q < y. Then a <- /3 +- y and -y +- a, so our tournament is not transitive on B. Remark. The result in the problem was first proved by R. Laver.

Problem R.6. Assume that R, a recursive, binary relation on N (the set of natural numbers), orders N into type w. Show that if f (n) is the nth element of this order, then f is not necessarily recursive.

Solution. We use the following well-known facts. There is a set A C N that is recursively enumerable, that is, the range of a recursive function, but

560

3. SOLUTIONS TO THE PROBLEMS

is not recursive, namely, its characteristic function is not recursive. For every infinite r.e. set A there is a recursive function enumerating A's elements in a one-to-one manner. An infinite set possesses an increasing such enumeration if and only if it is recursive. Take a (necessarily infinite) r.e. but not recursive set A, and a function g enumerating A without repetition. Put aRb if g(a) < g(b). Clearly, R is recursive and orders N into type w. If f (n)=the nth element by this order, and f is recursive, then the recursive h(n) = g(f (n)) would enumerate A's elements in increasing order, which is impossible.

Problem R.7. Let H be the class of all graphs with at most 2K0 vertices not containing a complete subgraph of size 1Z1. Show that there is no graph

H E H such that every graph in H is a subgraph of H.

Solution. We have to show that if (V, G), a graph with IVI = 2s0, does not contain a complete graph on R1 vertices, then there is a graph (W, H) with these properties such that (W, H) cannot be embedded into (V, G). Let W be the set of those functions injecting a countable ordinal (possibly 0 # 0) into a complete subgraph of G. To define H, join two such functions if one extends the other. Clearly, Ja->VI

JWJ= a<wl

First, we show that (W, H) does not contain a complete R1-gon. Assume

that If,,: a < wl } are pairwise joined, and that the ordinals Dom (fe) are in increasing order. By the construction of H, for,3 < a, f,, extends fQ, so the union of the fa's gives a function f : w1 --+ V onto a complete R1-gon in (V, G), a contradiction. Next we prove that (W, H) may not be embedded into (V, G). Assume that h: W -4V is such an embedding. By transfinite recursion on a < w1, we define fa : a -4V, fc. E W, and x,, = h(fa) E V in such a way that fa extends fp for i < a. Put fo = 0, fo E W. If fp (i3 < a) are defined, then

{xo:,3<a}=h{fp: i3<a} is a complete subgraph, so we can define f,, (03) = x0. This clearly satisfies

the requirements. But then {x,,: a < W11 is a complete subgraph in G, a contradiction. Remark. This is an unpublished result of R. Laver. Problem N.8. For which cardinalities ,c do antimetric spaces of cardinality is exist? (X, o) is called an antimetric space if X is a nonempty set, o : X2 -+ [0, oo) is a symmetric map, o(x, y) = 0 holds iff x = y, and for any three-element subset {a1, a2, a3} of X

o(alf,a2f) + o(a2f,a3f) < o(aif,a3f) holds for some permutation f of {1, 2, 3}.

3.11 SET THEORY

561

Solution. For 0 < ,c < 2H0. First, if X C_ R is nonempty, then (X, o) is antimetric, where o(x, y) = (x - y)2 and clearly I X I can take any value in the given interval. For the other direction, let (X, o) be antimetric and IXI > 2K0. Color the complete graph on X by the integers as follows. Let {x, y} get color k if 21c < o(x, y) < 2k+1 By the Erdos-Rado theorem (see P. Erdds, A. Hajnal, A. Mate, R. Rado, Combinatorial Set Theory: Partition Relations for

Cardinals, Akademiai Kiado, Budapest, 1984, p. 98), there exist x, y, z such that {x, y}, {x, z}, {y, z} get the same color. But then, x, y, z do not meet the requirement on antimetricity.

Problem 3t.9. If (A, <) is a partially ordered set, its dimension, dim (A, <), is the least cardinal K such that there exist K total orderings {<a: a < ic} on A with < = na<,, <a. Show that if dim(A, <) > Rio, then there exist disjoint Ao, Al C A with dim(Ao, <), dim(Al, <) > No. Solution. Assume indirectly that (A, <) is a counterexample of minimal cardinality A. Without loss of generality, A = A. Put

I={BC A:dim(B,<) <Ro}. Clearly, if B has I B I< A, then B E I.

Claim 1. If B,,, E I (n = 0,1, ...) and B is such that every pair of B is covered by some Bn, then B E I. Proof. Straightforward from the definition.

Claim 2. There are Bo, B1 E I such that Bo n B1 = 0 and Bo U B1 ¢ I.

Proof. If not, then I is a a-complete primideal containing every subset of cardinality < A. For a < A, let <a,n be a total order on a establishing dim (a, <) < No. If, for /3 < 'y < A we put /3 <n ry if {a < A: a <a,n y} ¢ I, then <n establish that dim (A, <) < Rio, a contradiction. We can assume that Bo, B1 as above are selected with is = IB1 I minimal.

Put

I1={CC B1:BoUCEI}. Then B1 ¢ Il and every subset of B1 of size < is is in I,. Repeating the argument of Claim 2, we get that there exist Co, C1 E Il such that Co n C1 = 0 and Bo U Co, Bo U Cl ¢ I. Repeating the above argument for Co, Bo this time, one gets Do, D1 C Bo, Do n Dl = 0 such that Do U Co, D1 U Co E I, Do U D1 U Co ¢ I. As every pair of Do U Dl U C1 is contained either in Bo or in Do U C1 or in D1 U C1, by Claim 1, either Do U C1 or D1 U C1 is not in I. In the former case, Do U C1 and D1 U Co are two disjoint sets not in I, and we conclude similarly in the other case, too.

3. SOLUTIONS TO THE PROBLEMS

562

Problem 12.10. A binary relation -< is called a quasi-order if it is reflexive and transitive. The infimum of the quasi-order (Q, -<) is the greatest subset J C_ Q such that (i) for every B E Q there is an A E J with A - B, and (ii) A -< B, A, B E J imply B -< A. Let X be a finite, nonempty alphabet, let X* be the set of all finite

words from X, and let P be the set of infinite subsets of X*. For A, B E P, let A - B if every element of A is a (connected) subword of some element of B. Show that (P, -<) has an infimum, and characterize its elements.

Solution. As follows, subword will always mean connected subword. Let a < b denote that a is a subword of b. Call A E P minimal, if for all w E A, all but finitely many subwords of the elements of A contain w, as subword. Claim 1. If A is minimal, B

A, then A

Proof. Assume that w E A and k is so large that every subword in A longer than k, as B -< A, is a subword, so w < b. Claim 2. If A E P is not minimal, then there exists a B E P such that B

Abut A.AB. Proof. Let w be a subword of A such that

H={h: there is an aEA,h<a,w:9 h} is infinite. Then H -< A, but A - H is established by w. Claim 3. For every A E P there is a minimal B -< A.

Proof. Such a B is given by the following algorithm. Step 1: Let X = {al, ... , an}. If, among the subwords of A, infinitely many do not contain a1 i let Hl be their set. H1 E P, H1 -< A, and it does not contain al. Then get successively H2, H3, ... using a2, a3, .... If Hn_1 is defined, it may only contain an, so it is minimal. We can therefore assume the existence of Hi_1 such that all but finitely many subwords of Hi_1 contain ai. Put b1 = ai, B1 = Hi_1 -< A, and there follows Step 2. Here Step k: If bk_1i Bk_1 are given, and b1 < ... < bk_1i Bk_1 -< B1 - A are given, and all but fi-itely many subwords of Bt contain bt, consider the word bk_1a1. If infinitely many subwords of Bk_1 omit it, let Hi -1 be this set. bk_1a1 therefore is not a subword of Hi-1. Repeating this to bk_1a2i we get H2 -1, etc. If we can reach Hn=i, then there exists a natural number in, such that in every subword of length 2m a bk_1 can be found in the first m positions, but it must be followed by an. So we can select Hk-1 as Bk and bk_1ai as bk. By this process, we get b1 < b2 < .... B1 >- B2 ... . Put L = {b1i b2, ... }. Then L A and L is minimal. In fact, for bi c L , E Bi, so only finitely many subwords of them omit bi. bi+i, bi+2, We get that the minimal members constitute the infimum. Claim 3 gives (i), Claim 1 (ii), and the maximality holds as if A is nonminimal, B -< A

3.11 SET THEORY

is such that A 7k B, C is minimal, C - B, then C -< A, A

563

C, so the

addition of C would violate (ii).

Problem N.11. Define a partial order on all functions f : R -* R by the relation f -< g if f (x) < g(x) for all x E R. Show that this partially ordered set contains a totally ordered subset of size greater than 2s0 but that the latter subset cannot be well-ordered.

Solution. For the first part of the problem, it suffices to find a family of more than 20 subsets of R totally ordered by c, as then the characteristic function will do the job. In order to show that if is is an infinite cardinal, then there are #c+ subsets of a set of size K, ordered by C, let A be the least

cardinal such that 2A > ic. By Cantor's theorem, .\ exists and is < K. If we find a totally ordered set of size f+, with a dense subset of cardinality < #c, we are done by the Dedekind cuts. For such a set, take the .\ -> {0, 1}

functions, ordered lexicographically. A dense subset is formed by those functions that take 0 from a place onward. Their number is

1: 21`' <a Ck<a

For the second part, assume that f0, < fp for a < 0 < (2"0)+. Let x0, E R satisfy f,, (x,,) < f,,+1(xa), for (2k0)+ many a, xa = x, and so the f,, (x) are (2'0) + different elements of IR, which is impossible.

Index of Names A

Abel, N. H., 489 Aczel, J., 192 Adam, A., 16 Adian, S. I., 82 Ajtai, M., 30

Cayley, A., 421 Clifford, A. H., 85 Corradi, K., 10 Coxeter, H. S. M., 91 Csakany, B., 8, 22, 28, 46 Csaszar, A., 2, 4, 7, 12, 13, 18, 21, 30, 42 Csikos, B., vii, 46, 52 Csorgo, M., 443 Csorgo, S., 35, 42, 48 Czach, L., 12 Czedli, G., 46, 51 Czipszer, J., 2

B Babai, L., 25, 32, 50, 51, 282, 491, 529, 557 Baer, R., 63, 81 Bajmoczi, E., 51 Balogh, A., 39 Balogh, Z., 32, 33, 51 Bara, T., 51 Beck, J., 51 Beleznay, F., 52 Benczur, A., 53 Berlekamp, E. R., 374 Bernstein, I. N., 332 Bernstein, S., 158 Birkas, Gy., 53 Biro, A., 53 Bod6, Z., 52 Bognar, J., vii Bognar, M., 13, 20, 21, 27 Bognarne, K., 425 Bohus, G., 52 Bolla, M., vi Bollobas, B., 50, 263, 518, 554 Boltjainskii, V. I., 184 Borges, R., 3 Boros, E., 51 Bosznay, A., 32 Brindza, B., 298 Buczolich, Z., 52 Burnside, W., 96, 118

Daroczy, Z., 12, 20, 25, 38, 192, 513

Deak, J., 51 Denes, J., 33 Deny, J., 207 Dini, U., 489 Domokos, M., 53 Domosi, P., 44 Drasny, G., 53 Druszt, E., 34

E Elbert, A., 50 Elek, G., 52 Engelking, R., 533, 537 Eotvos L., v Erdelyi, A., 183 Erdos, J., 11, 14, 19 Erdos, L., 52, 53 ror

INDEX OF NAMES

566

Erdos, P., 7, 11, 12, 15, 16, 19, 23, 26, 33, 36, 38, 43, 44, 145, 522, 558, 561

F Fazekas, I., 39, 45 Fejer, L., v, 151, 479 Fejes-Toth, G., 30, 40 Fejes-Toth, L., 4, 7, 9, 11, 13, 18, 20, 24, 27, 30, 36, 40 Feller, W., 406 Fischer, E., 471 Fleiner, T., 53 Foiaยง, C., 396 Frankl, P., 51 Freud, R., 50 Fried, E., 2, 7, 9, 17, 23, 27, 41 Fritz, J., 50 Fuchs, L., 2 Fi redi, Z., 51

Hajos, Gy., 3 Halasz, G., 18, 21, 24, 30, 33, 36, 40, 46, 50, 465, 467 Halmos, P. R., 348 Harcos, G., 53 Hardy, G. H., 217, 470, 484 Hatvani, L., 23, 29, 41, 48, 190 Hausel, T., 53 Hayman, W. K., 182 Heppes, A., 7 Hetyei, G., 52 Hewitt, E., 440 Horvath, L., 35 Hosszu, M., 3 Huhn, A., 31, 35 Huppert, B., 107

I Ivanyos, G., 52

J G

Gacs, P., 50, 51, 473, 477, 479 Galvin, F., 17, 30 Garay, B., 44 Geher, L., 9 Gelfand, I. M., 459 Gereb, M., 51, 52 Gerencser, L., 50 Gerlits, J., 553 Gesztelyi, E., 12, 26, 32, 37 Gondocs, F., 51 Gorenstein, D., 104 Greenleaf, F. R., 194 Gyires, B., 12, 20 Gyori, E., vii, 51 Gyori, I., 15 Gydry, K., 11, 25, 38

H Haar, A., v Hajdu, G., 53 Hajnal, A., 7, 10, 14, 20, 21, 23, 27, 32, 35, 39, 522, 558, 561

Hajnal, P., 52

Jaglom, I. M., 184 Joo, I., 36, 43 Juhasz, I., 9, 13, 24, 27, 31, 46, 50, 518, 520, 554

K Kalina, J., 27 Karman, T., v Karolyi, Gy., 52 Karteszi, F., 25 Katai, I., 10, 38, 513 Katona, Gy., 23, 39, 50 Keleti, T., 53 Kelly, D., 39

Kemperman, J. B. H., 336 Kennedy, P. B., 182 Kerchy, L., 29, 41, 48 Kertesz, A., 19, 84 Kery, G., 50 Kiss, E., 51, 52, 538 Knuth, E., 50 Koljada, K. I., 35 Kollar, J., 51, 56, 540 Komjath, P., vii, 42, 51, 52

INDEX OF NAMES

Komlos, J., 21, 50 Komornik, V., 51, 297 Kos, G., 53 Kovics, B., 38 Kovics, I., 9 Kovics, S., 52, 53 Krisztin, T., 34, 47, 48, 51, 52, 242

Laczkovich, M., 17, 24, 27, 33, 36, 39, 43, 44, 45, 50, 51 Langberg, N. A., 443 Laver, R., 559, 560 Leindler, L., 8, 15, 16, 22, 47, 491 Lempert, L., 24, 27, 43, 51 Leon, R. V., 443 Levi, B., 439 Linnik, Yu. V., 107 Littlewood, J. E., 470 Losonczi, L., 12, 26 Lovasz, L., 20, 22, 23, 28, 29, 36, 50, 51, 263, 415, 473, 477, 479, 524, 553

Lukics, E., 42

567

Montagh, B., 113 Moor, A., 9 Mori, T. F., 32, 33, 37, 43, 46, 51 Moricz, F., 16, 29, 47, 438 Moussong, G., vii, 52, 545

N Nagy, P., 29 Nagy, Zs., 51 Neumann, J., v Novikov, S. P., 82 Natanson, I. P., 168, 479 Nemetz, T., 465

0 Odor, T., 52

P Pach, J., 43 Pales, Zs., 31, 38, 52 Palfy, P. P., 31, 36, 39, 51, 52, 297 Pap, Gy., 37 Pelikan, J., vii, 17, 20, 30, 50, 51, 81

M Magyar, A., 52, 53 Magyar, Z., 51, 52 Major, P., 21, 450, 459 Majoros, L., 53 Makai, E., 18, 24, 50, 51, 477 Makay, G., 53 Makkai, M., 10, 13 Maksa, Gy., 25, 38 Marki, L., 28 Mate, A., 11, 15, 50, 491, 519, 521, 558, 561

Mate, E., 50 McLean, R. P., 85 Medgyessy, P., 19 Megyesi, L., 22 Michaletzki, Gy., 40 Miklos, D., vi, 52 Mocsy, M., 52, 53 Mogyorodi, J., 425

Petruska, Gy., 17, 50 Pinkus, A., 36, 209 Pinter, L., 29 Pintz, J., 51 Pollak, Gy., 3, 41 Polya, Gy., 180, 462, 470, 492 Pontriagin, L. S., 189 P6sa, L., 23, 28, 50, 51, 479, 520, 528, 553

Prachar, K., 107 Prekopa, A., 425 Preston, G. B., 85 Proschan, F., 443 Pyber, L., 42, 45

R Rado, R., 145, 522, 558, 561 Rado, T., v Ramsey, F. P., 122, 134

Ritz, L., v Redei, L., 3, 8, 13, 25, 103

INDEX OF NAMES

568

Renyi, A., 3, 4, 8, 11, 425, 433, 443, 453, 461 Revesz, P., 14, 19, 24, 37, 443 Revesz, Sz. Gy., 41 Reviczky, J., 51 Riesz, F., 394 Riesz, M., v

Riman, J., 38 Rudin, M. E., 545 Rudin, W., 505 Ruzsa, I. Z., vii, 27, 39, 45, 50, 51, 440, 445, 477 S

Saks, S., 330, 336 Salem, R., 479 Schrijver, A. A., 28 Schottky, W., 228 Schur, I., 382 Schweitzer, M., v, vi Seb6, A., 51 Seress, A., 52 Shapiro, H. S., 27 Shilow, G. E., 459 Shockey, H., 207 Sidon, S., 135 Sigray, I., 52, 53 Simanyi, N., 51, 52, 290, 538 Simonovits, A., 477, 484 Simonovits, M., 50, 470, 519, 553, 554

Somorjai, G., 27, 51, 501 Sos, V. T., 30, 33, 34 Stromberg, K., 440 Suranyi, J., 2, 9, 13, 18 Sylvester, J. J., 382 Szabados, J., 22, 24, 30, 40, 45 Szabo, E., 52 Szabo, Gy., 38, 44 Szab6, L. I., 46, 52 Szabo, T., 53, 513 Szab6, Z., 28, 34, 35, 52, 53, 539 Szasz, D., 50, 425 Szegedy, M., 52 Szeg6, G., v, 180, 492 Szekely, G. J., vi, 32, 440, 445

Szekely, L., 34, 36 Szekelyhidi, L., 25, 44 Szekeres, Gy., 7, 30, 33, 40 Szemeredi, E., 27 Szendrei, A., 41, 47 Szendrei, M., 51 Szenes, A., 52 Szenthe, J., 16 Szep, G., vii Szigeti, F., 50 Szilard, L., v Szilasi, J., 39, 45 Szokefalvi-Nagy, B., 4, 9, 22, 394, 396

Szucs, J., 22, 23, 31, 35, 50

T Tamassy, L., 12, 25 Tandori, K., 4 Tardos, G., 52 Teller, E., v Terjeki, J., 34 Tomko, J., 26 Torocsik, J., vii, 52, 53 Totik, V., vii, 29, 34, 36, 39, 41, 42, 46, 47, 48, 51, 209, 297, 539

Turin, P., 2, 3, 8, 10, 14, 18, 158, 431

Tusnady, G., 14 Tuza, Zs., 51, 297

V Vamos, P., 50 Varga, J., 52 Varlet, J., 27 Veres, S., 51 Vesztergombi, Gy., 50, 519 Vitali, G., 336, 510 Vu, H. V., 53

W Weiss, B., 39 Wiegandt, R., 34, 43 Wielandt, H., 96, 97

Wigner, E., v

Zempleni, A., 52 Zenor, P. H., 545 Zorn, M., 86, 345

Problem Books in Mathematics

(continued)

Demography Through Problems by Nathan Keyfitz and John A. Beekman

Theorems and Problems in Functional Analysis by A.A. Kirillov and A.D. Gvishiani

Exercises in Classical Ring Theory by T. Y. Lam

Problem-Solving Through Problems by Loren C. Larson

A Problem Seminar by Donald J. Newman

Exercises in Number Theory by D.P. Parent

Contests in Higher Mathematics: Miklos Schweitzer Competitions 1962-1991 by Gabor J. Szekely (editor)