Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 1, Issue No. 2, 130 - 134
Fixed Points Theorems In Three Metric Spaces ( FPTTMS) Manish Kumar Mishra and Deo Brat Ojha
mkm2781@rediffmail.com, deobratojha@rediffmail.com
Department of Mathematics R.K.G.Institute of Technology Ghaziabad,U.P.,INDIA
satisfying integral type inequality.
for all x in X, y in Y and z in Z, where 0≤ c < 1. Then RST has
Mathematics Subject Classification: 54H25
a unique fixed point u in X, TRS has a unique fixed point v in Y and STR has a unique fixed point w in Z. Further Tu = v,
ES
Keywords- Three metric space, fixed point, integral type inequality. .
d ( RSTx, RSy ) c max{d ( x, RSy ), d ( x, RSTx), ( y, Tx), ( Sy, STx)} (TRSy, STz ) c max{ ( y, TRz ), ( x, TRSy), ( z, Sy), d ( Rz, RSy)} ( STRz, STx) c max{ ( z, STx), ( z, STRz), d ( x, Rz), (Tx, TRz)}
T
Abstract— We obtained fixed point theorem on three metric spaces
Sv = w and Rw = u. The next theorem was proved in [3]. Theorem 1.3 : Let ( X , d ) , (Y , ) and ( Z , ) be complete
I.
INTRODUCTION
metric spaces and suppose T is a continuous mapping of X
The following fixed point theorem was proved by Fisher [1].
into Y, S is a continuous mapping of Y into Z and R is a continuous mapping of Z into X satisfying the inequalities:
complete metric spaces. If S is a continuous
mapping of X into Z, and R is a continuous
d ( RSTx, RSTx ') c max{d ( x, x '), d ( x, RSTx), d ( x ', RSTx '), (Tx, Tx '), STx, STx ' }
IJ A
Theorem 1.1: Let ( X , d ) and ( Z , ) be
mapping
of
Z
into
X
satisfying
the
inequalities:
d ( RSx, RSx ') c max{d ( x, x '), d ( x, RSx), d ( x ' RSx '), d ( Sx, Sx ')} (SRz, SRz ') c max{ ( z, z '), ( z, SRz), ( z ' SRz '), ( Rz, Rz ')}
(TRSy, TRSy ') c max{ ( y, y '), ( y, TRSy), ( y ', TRSy '), ( Sy, Sy '), d RSy, RSy ' }
(STRz, STRz ') c max{ ( z, z '), ( z, STRz), ( z ', STRz '), d Rz, Rz ' (TRz, TRz ')}
for all x, x in X, y, y in Y and z, z in Z where 0 c <1. Then RST has a unique fixed point u in X, TRS
for all x, x' in X, and z, z' in Z, where 0≤ c < 1, then
has a unique fixed point v in Y and STR has a unique
RS has a unique fixed point u in X and RS has a
fixed point w in Z. Further, Tu = v, Sv = w and Rw = u. In
unique fixed point w in Z. Further Su = w and Rw =
recently Ansari , Sharma[6] generate Related Fixed Points
u. The next theorem was proved in [2].
Theorems on Three Metric Spaces. Now We obtained related
Theorem 1.2: Let
( X , d ) , (Y , ) and ( Z , ) be complete
metric spaces and suppose T is a continuous mapping of X into Y, S is a continuous mapping of Y into Z and R is a continuous mapping of Z into X satisfying the inequalities:
ISSN: 2230-7818
fixed point theorem on three metric spaces satisfying integral type inequality. Let
(X,d ) be
[0,1], f : X X
@ 2010 http://www.ijaest.iserp.org. All rights Reserved.
a
complete
metric
space,
a mapping such that for each
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Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 1, Issue No. 2, 130 - 134 d ( fx , fy )
x , y X ,
(t )dt
d ( x, y )
0
; R R summable,
(t )dt ,
Where
such
that,
for
each
0, (t )dt 0 .
d 2p (TSz ,TSRy )
d3p
0
is a lebesgue integrable mapping which is and
d1p ( SRy , SRTx )
0
0
nonnegative
Then f has a unique common fixed
max M 1 ( x , y )
max M 2 ( y , z )
0
(t ) dt
max M 3 ( z , x )
0
Proof. Let
condition by the following
sequences
(t )dt
0
and
inequality
J P ( Fx , Fy )
(t ) dt
0
IJ A
0
ad ( fx , fy ) d P 1 ( fx , Fx ), ad ( fx , fy ) d P 1 ( fy , Fy ), P 1 P 1 ad ( fx , Fx ) d ( fy , Fy ), cd ( fx , Fy ) d ( fy , Fx ) (t ) dt
for all
x , y in X ,where p 2 is an integer a 0 and
0 c 1 then f and F have unique common fixed point
x, y
in
0 c 1 then f X
X
,where
and
F
p2
is an integer
a0
all (t )dt n N
.
We
will
Taking
spaces and
mappings satisfying the following inequalities:
(t ) dt
and
.
Similarly, if
in (1) and (4), we obtain:
{ d1 ( xn , SRy n ), d1 ( xn , SRTxn ), d 2 ( y n , Txn )} p
p
p
0
{ d1p
( xn , xn ), d1p
( xn , xn 1 ), d 2p
{0, d1p ( xn , xn 1 ), d 2p ( yn , yn 1 )}
( xn , xn 1 )
0
(t )dt
d1p
( SRyn , SRTxn )
0
( yn , yn 1 )}
max{0, d1p ( xn , xn 1 ), d 2p ( yn , yn 1 )}
0
(t )dt
( yn , yn 1 )
since, if
d1p ( xn , xn1 )
0
(t )dt
(t )dt
F 0
0
d 2p
(t )dt
(t ) dt
0
0
(t ) dt
(t )dt
F min M 1 ( xn , yn )
max M1 ( xn , yn )
(t ) dt
(t )dt
max M1 ( xn , yn )
0
(t )dt
(t ) dt
d1p ( xn , xn1 )
0
d1 ( xn , xn 1 ) p
0
(t ) dt
then by the inequality
(t )dt
it follows xn 1 xn since 0 1 .
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If
zn zn1 ,
Thus
ISSN: 2230-7818
and we
xn , yn and
0
d1p
0
are three
for all n. Otherwise, if
the latter equality implies that
0
MAIN RESULTS
T : X Y, R :Y Z, S : Z X
that
xn xn1 .
x xn y yn
M1 ( xn , yn )
0
be three metric
assume
zn are Cauchy sequences.
have unique common fixed point in
( X , d1 ) , (Y , d 2 ) and ( Z , d3 )
. Further,
xn a, yn1 b and zn1 c
First, we prove that the sequences
Now, we will give and prove our theorem as follows: Theorem 2.1 Let
cZ
yn1 yn2 , zn1 zn2
for some n, then
put
then again
.
II.
SRT
be an arbitrary point. We define three
yn yn1 , then zn zn1 and
in X .
for all
then
has a unique fixed point
SRTxn1 SRTxn , i.e. xn xn1 .
max
6
n
could
commutative (OWC) and satisfy the
(t )dt
xn , yn and zn with X, Y, Z respectively as
xn xn1
a
multi-valued map respectively, suppose that f and F are occasionally weakly
x X
ES
f : X X , F : X CB( X ) be a single
5
0 1 ,
xn xn1 , yn yn1 and zn zn1
Let ( X , d ) be a metric space and let
Ojha (2010) [5]
(t )dt
T
F min M 3 ( z , x )
0
4
xn SRT x0 , yn Txn1 , zn Ryn
follows:
1 max{d ( x , y ), d ( x , fx ), d ( y , fy ), [ d ( x , fy ) d ( y , fx )] 2 for 0
(t )dt
has a unique fixed point
Rhoades(2003)[4], extended this result by replacing the above
d ( fx , fy )
0
, where
Ta b, Rb c and Sc a.
n
F min M 2 ( y , z )
(t ) dt
z X such that for each x X , lim f x z. n
F min M1 ( x , y )
0
a X , TSR
has a unique fixed point
b Y and RTS
0
(t ) dt
x X , y Y , z Z
for all
(t ) dt
0
(t ) dt
( RTx , RTSz )
0
(t ) dt
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Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 1, Issue No. 2, 130 - 134
d1p ( xn , xn1 )
(t )dt
0
d2p ( yn , yn1 )
0
Taking
y yn , z zn1
M 2 ( yn , zn 1 )
0
(t ) dt
p
p
0
{ d 2p ( yn , yn ), d 2p ( yn , yn 1 ), d 3p ( zn 1 , zn )}
{0, d 2p
( yn , yn 1 ), d3p
(t )dt
( zn 1 , zn )}
d 2p (TSzn 1 ,TSRyn )
0
max{0, d 2p ( yn , yn 1 ), d3p ( zn 1 , zn )}
d3p ( zn 1 , zn )
0
(t )dt
(t )dt
0
max M1 ( xn , b )
0
(t )dt
F 0
M1 ( xn , b )
0
0
0
0
(t ) dt
d 2 ( yn , yn 1 ) p
0
( yn , yn1 )
(t ) dt
then by the
d1p ( SRb , a )
(t )dt
d3p ( zn1 , zn )
0
8
(t )dt
0
(t ) dt
{ d 3 ( z n , RTxn 1 ), d 3 ( z n , RTSz n ), d1 ( xn 1 , Sz n )} p
p
p
0
{ d3p
( zn , zn ), d3p
( zn , zn 1 ), d1p
{0, d3p ( zn , zn 1 ), d1p ( xn 1 , xn )}
( xn 1 , Szn )}
0
( zn , zn 1 )
0
(t )dt
d3p
( RTxn 1 , RTSzn )
0
max{0, d3p ( zn , zn 1 ), d1p ( xn 1 , xn )}
0
(t )dt
0
0
( xn 1 , Szn )
0
Replacing
(t )dt
d3p ( zn1 , zn )
n
d1p ( xn2 , xn1 )
0
(t )dt
9
(t )dt
Using (7), (8) and (9) we get
d1p ( xn , xn 1 )
0
(t ) dt
d 2p ( yn , yn 1 )
0
3 d1p ( xn 2 , xn 1 )
0
(t ) dt
2 d3p ( zn 1 , zn )
0
(t )dt
4 d1p ( xn 4 , xn 3 )
0
(t )dt 0 ......... 3 k p d1 ( x0 , x1 ) (t ) dt 0 3 k d1p ( x1 , x2 )
M 1 ( a , yn )
(t )dt
(t )dt
{ d1p ( a , SRyn ), d1p ( a , SRTa ), d 2p ( yn ,Ta )}
(t )dt
0
{ d1p
( a , xn ), d1p
( a , SRTa ), d 2p
( yn ,Ta )}
(t )dt
d1p ( a , SRTa )
0
(t )dt
max{d1p ( a , a ), d1p ( a , SRTa ), d 2p ( b ,Ta )}
0
max{d1p ( a , SRTa ), d 2p ( b ,Ta )}
0
(t )dt
(t )dt
from which it follows or
d1p ( a , SRTa )
0
or
(t )dt
d1p ( a , SRTa )
0
d1p ( a , SRTa )
0
(t )dt
(t )dt SRTa a
d 2p ( b ,Ta )
0
(t )dt
which can be also written in the following form
d1p ( a , Sc )
0
(t )dt
d2p ( b ,Ta )
0
(t )dt
11
since RT a = c.
(t ) dt
Taking,
(t )dt
z zn , y b
above we
for n 2k
obtain:
xn , yn and zn
(t )dt
Letting n tend to infinity we get
for n 2k 1
Since 0 1 , the sequences
F min M1 ( a , yn )
0
n 1 we obtain:
(t )dt
0
(t )dt
(t )dt with
d1p ( SRyn , SRTa )
0
(t ) dt
F 0
x a , y yn we get
(t )dt
max M1 ( a , yn )
0
(t )dt
0
d1p
SRb a .
where
(t )dt
F min M 3 ( zn , xn 1 )
max M 3 ( zn , xn 1 )
(t )dt
0
0
(t ) dt
(t )dt
d1p ( xn , SRTa )
IJ A
0
d3p
d1p ( a , SRb )
0
0
Taking x xn 1 z zn in (3) and (6) we obtain: M 3 ( zn , xn 1 )
(t )dt
Using (4), if we take
Thus 0
(t )dt
In the same way it can be shown that T Sc = b and RT a = c.
(t )dt
it follows yn yn 1 since 0 1 .
(t )dt
Letting n tend to infinity in the inequality (10) and by the fact that F
from which it follows d2p
0
d2p ( yn , yn1 )
{d1p ( xn , SRb ), d1p ( xn , xn 1 ), d 2p ( b , yn 1 )}
is continuous in 0 we get
max{d 2p ( yn , yn 1 ), d3p ( zn 1 , zn )}
(t )dt
10
(t )dt
{ d1p ( xn , SRb ), d1p ( xn , SRTxn ), d 2p ( b ,Txn )}
ES
( yn , yn1 )
0
(t )dt
F min M1 ( xn , b )
(t )dt
(t )dt
(t )dt
inequality
d1p ( SRb , SRTxn )
0
0
d2p
(t )dt
0
(t )dt
since, if
d1p ( SRb , xn1 )
y b in the inequality (4) we obtain
and
where
0
0
(t )dt
(t )dt
0
0
(t ) dt
(t )dt
F min M 2 ( yn , zn 1 )
max M 2 ( yn , zn 1 )
n
x xn
Taking
p
0
0
n
in (2) and (5), we obtain:
0
d 2p ( yn , yn 1 )
lim xn a X , lim yn b Y , lim zn c Z
n
{ d 2 ( yn , TSz n 1 ), d 2 ( yn , TSRyn ), d 3 ( zn 1 , Ryn )}
have
7
(t )dt
T
are Cauchy
d2p ( b ,Ta )
0
(t )dt
d3p ( c , Rb )
0
in the inequality (5) in the same way as
(t )dt
12
In the same way, we obtain:
sequences. Since ( X , d1 ) , (Y , d 2 ) and ( Z , d3 ) are complete metric spaces we
ISSN: 2230-7818
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Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 1, Issue No. 2, 130 - 134
d3p ( c , Rb )
d1p ( a , Sc )
(t )dt
0
0
13
(t )dt
( a , Sc )
(t )dt
0
3
d1p
( a , Sc )
0
Thus, again
d1p ( a , SRTa )
0
d3p ( RTa ', c )
(t )dt
0
d3p ( RTa ', c )
0
(t )dt Sc a.
(t )dt
d1p ( a , a ')
(t )dt
16
(t )dt
0
or
By (11), (12), (13) it follows: d1p
d3p ( RTa ', c )
0
17
(t )dt
By (17) it follows that RTa ' c since 0 1 .
d1p ( a , Sc )
0
(t )dt 0
By (14), (15) and (16) we obtain
d1p ( a , a ')
0
or SRTa a
0
and by
So, we proved that a is a fixed point of SRT .
d2p (Ta ',b )
(t )dt
d1p ( a , a ')
(t )dt
(t )dt
0
2 d3p ( RTa ', c )
0
3 d1p ( a , a ')
0
(t )dt
3 d1p ( a , a ')
0
(t )dt
(t )dt where 0 1 , it follows
a' a .
In the same way it can be shown that b is a fixed point of T SR and c
that
is a fixed point of RT S.
Thus we proved that a is the unique fixed point of SRT .
Further, we showed that T a = b, Rb = c, Sc = a.
In the same way, it can be shown that b is the unique fixed point of T
is another fixed point of SRT ,
x a, y Ta ' , we get
different from a. Using (4), if we take
d1p ( a , a ')
0
(t )dt
d1p ( SRTa ', SRTa )
0
max M1 ( a ,Ta ')
0
(t )dt
(t )dt
F min M1 ( a ,Ta ')
0
(t )dt
0
{ d1p ( a , SRTa '), d1p ( a , SRTa ), d 2p (Ta ',Ta )}
0
{ d1p ( a , a '), d1p ( a , a ), d 2p (Ta ', b )}
0
Z Y , d1 d2 ), and the mapping
( a , a ')
0
(t )dt
(Ta ', b )
(t )dt
(t )dt
14
in the inequality (5) we obtain:
d2p (TSc ,TSRTa ')
0
(t )dt
max M1 (Ta ', c )
0
F min M 2 (Ta ', c )
(t )dt
0
(t )dt
where
M 2 (Ta ', c )
0
(t )dt
{ d 2p (Ta ',TSc ), d 2p (Ta ',TSRTa '), d3p ( c , RTa ')}
(t )dt
0
{ d 2p
(Ta ', b ), d 2p
(Ta ',Ta '), d3p
( c , RTa ')}
0
(t )dt
Then
d2p ( b ,Ta ')
0
d3p ( c , RTa ')
0
Taking
(t )dt
z c, y Ta '
d3p ( RTa ',c )
0
(t )dt
max M 3 ( c , a ')
0
(t )dt
15
(t )dt
in the inequality (6) we obtain:
d3p ( RTa ', RTSc )
0
F min M 2 ( c , a ')
0
(t )dt
M 3 ( c , a ')
0
(t )dt
{d3p ( c , RTa '), d 2p ( c , RTSc ), d1p ( a ', Sc )}
0
{ d3p ( c , RTa '),0, d1p ( a ', a )}
0
d1p ( Sy , STx )
0
(t )dt
(t ) dt
M1 ( x , y )
0
d2p (TSy ,TSy )
0
apply
the
inequalities
(4)(5)
and
(6)
for
as the identity mapping.
R
max M 1 ( x , y )
0
(t ) dt
p { d1
(t ) dt
p ( x , Sy ), d1
F min M1 ( x , y )
0
p ( x , STx ), d 2
( y , Tx )}
0
(t ) dt
max M 2 ( y , y )
0
(t ) dt
(t )dt
(t ) dt
F min M 2 ( y , y )
0
(t )dt
The inequality (6) takes the form:
d2p (Tx ,TSy )
(t ) dt
0
max M 3 ( y , x )
0
(t ) dt
F min M 3 ( y , x )
0
(t )dt
where
M3 ( y , x )
0
(t ) dt
{ d 2 ( y , Tx ), d 2 ( y , TSy ), d1 ( x , Sy )} p
p
p
0
x X , y Y
for all
(t ) dt
. In sequel,
is denoted with
M 3 ( y, x)
M 2 ( y, x) .
Thus, the following theorem (Theorem 1[1]) is obtained: ( X , d1 )
,
(Y , d 2 )
be
T : X Y, S :Y X (t )dt
which holds always since the left hand is zero.
two
metric
spaces
d1p ( Sy , STx )
0
(t ) dt
M1 ( x , y )
0
max M 1 ( x , y )
0
(t ) dt
p { d1
0
(t ) dt
p ( x , Sy ), d1
F min M1 ( x , y )
0
p ( x , STx ), d 2
( y , Tx )}
(t )dt
(t ) dt
and
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and
are two mappings satisfying the
following inequalities:
where
Then
where
Let
(t ) dt
where
We
The inequality (5) takes the form:
IJ A
0
Proof.
Z Y , d3 d2 , z y and
(t )dt
z c, y Ta '
d2p ( b ,Ta ')
as the identity mapping of
The inequality (4) takes the form:
0
Taking
d2p
R
Y Ry y, y Y , then we obtain Theorem 1[1].
(t )dt
form which it follows d1p
the same with the metric space (Y , d 2 ) , (that is
ES
(t )dt
Corollary 2.2 If we consider in Theorem 2.1 the metric space and
( Z , d3 )
where M1 ( a ,Ta ')
SR and c is the unique fixed point of RT S.
T
a ' X
Let assume now that
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Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 1, Issue No. 2, 130 - 134
d2p (Tx ,TSy )
0
(t ) dt
max M 2 ( y , x )
0
(t ) dt
F min M 2 ( y , x )
0
(t )dt
where
M2 ( y, x)
0
(t ) dt
{ d 2 ( y , Tx ), d 2 ( y , TSy ), d1 ( x , Sy )} p
p
0
p
(t ) dt
,
for all x X , y Y and 0 1 , then ST has a unique fixed point
a X
and T S has a unique fixed point
b Y
. Further,
T a = b and Sb = a. It is clear that in the Theorem 1[1] the functions
F1
and
replaced by F such that F t max F1 t , F2 t and
can be
c1 , c2
can be
c max c1 , c2 .
Corollary 2.3 For
p 1 and F t 0 for all t R , by Theorem
2.1 we obtain a theorem which extends the result of Theorem 1([2]) to three metric spaces. Corollary 2.4 For
p2
and F t 0 for all
t R , we obtain a
REFERENCES [1]
S. C.Nesi, “A fixed point theorem in two metric spaces”, Bull. Math. Soc. Sci. Math. Roumanie (N.S.) 44, No. 92 (2001), 253257. B. Fisher, “Fixed points on two metric spaces”, Glasnik Mat.16, No. 36(1981), 333-337. V. Popa, Fixed points on two complete metric spaces, Zb. Rad. Prirod.-Mat. Fak. (N.S.) Ser. mat. 21, No. 1 (1991), 83-93. R. K. Jain, H. K. Sahu, B. Fisher,Related fixed point theorems for three metric spaces, Novi Sad J. Math. 26, No.1 (1996), 1117. Luljeta Kikina, “Fixed Points Theorems on Three Metric Spaces”, Int. Journal of Math. Analysis, Vol. 3, 2009, no. 13, 619 – 626. Deo Brat Ojha, Manish Kumar Mishra and Udayana Katoch,A Common Fixed Point Theorem Satisfying Integral Type for Occasionally Weakly Compatible Maps, Research Journal of Applied Sciences, Engineering and Technology 2(3): 239-244, 2010. Rhoades, B.E.,Two fixed point theorem for mapping satisfying a general contractiv condition of integral type. Int. J. Math. Sci., 3: 2003 4007-4013.
IJ A
[2]
ES
generalization of Theorem 2[3], extended to three metric spaces.
T
replaced by
F2
[3] [4]
[5]
[6]
[7]
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Page 134