Ijeart06524

International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015

A Common Fixed Point Theorem in Cone Metric Spaces Manoranjan Singha, Kousik Das 

(iii)Suppose x = ( x1 , y1 ) then  x = ( x1 , y1 ) .

If x = ( x1 , y1 )  P then x1  0 and y1  0.

Again if  x = ( x1 , y1 )  P then x1  0 and y1  0 , i.e., x1  0 and y1  0 .

Combining above we have x1 = 0, y1 = 0 , i.e., x =  .

Abstract— The purpose of this paper is to translate a set of generalized contractive conditions for a couple of self-mappings to have a unique common fixed point in the language of cone metric spaces.

Index Terms— Cone, cone metric space, complete cone metric space, totally ordered cone, contraction, fixed point.

Let us prove d is a metric on X. (i) | x  y | 0 and  | x  y | 0 (Since   0 )  d ( x, y) = (| x  y |, | x  y |)  (0,0) =  , x, y  X .

Now, d ( x, y) =   (| x  y |, | x  y |) = (0,0) | x  y |= 0, | x  y |= 0 | x  y |=  (Since   0 )  x = y. (ii) d ( x, y) = (| x  y |, | x  y |) = (| y  x |, | y  x |) = d ( y, x)  x, y  X . (iii) | x  y |=| x  z  z  y || x  z |  | z  y | and  | x  y |  | x  z |  | z  y | (Since   0 ).

Therefore, d ( x, y ) = (| x  y |,  | x  y |)  (| x  z |  | z  y |,

I. INTRODUCTION Let’s begin with some basic definitions and results which will be used later in the sequel: Let B be a real Banach space and P be a subset of B . By  we denote the zero element of B and by IntP the interior of P . P is called a cone in B if (i) P is closed, nonempty and P  { } ; (ii) a, b R, a, b  0, x, y  P  ax  by  P and (iii) x, x  P  x =  ,i.e., P  ( P) = { } .

For a given cone P in a Banach space B we define a partial ordering  with respect to B by x  y if and only if y  x  P; x < y implies x  y but x  y, while x << y will stand for y  x  IntP . If x , y , z  P so that x  y  z, then x  z . A cone P in a Banach space B is called totally ordered if for any x , y  B either x  y  P or y  x  P i.e., either y  x (in this case we write max{x, y} = x ) or x  y .

Let X be a nonempty set. Suppose the mapping d : X  X  B satisfies (i)   d ( x, y)  x , y  X and d ( x, y) =  iff x = y , (ii) d ( x, y) = d ( y, x)  x , y  X , (iii) d ( x, y)  d ( x, z)  d ( z, y)  x, y, z  X .

 | x  z |  | z  y |) = (| x  z |,  | x  z |)  (| z  y |,  | z  y |) = d ( x, z )  d ( z , y ),  x, y, z  X .

Therefore (X , d ) is a cone metric space.

A sequence {xn } in the cone metric space (X , d ) is said to converge to x  X if for any c  B with  << c,  N  N such that d ( xn , x) << c,  n  N .

A sequence {xn } in the cone metric space ( X , d ) is said to be a Cauchy sequence if for any c  B with  << c ,  N  N such that d ( xn , xm ) << c,  n, m  N. If every Cauchy sequence is convergent in ( X , d ) , then ( X , d ) is called a complete cone metric space.

Then d is called a cone metric on X , and the ordered pair ( X , d ) is called a cone metric space. As an example, let

Throughout this paper, we always suppose B is a Banach space, P is a totally ordered cone in B with IntP   and  is the partial ordering with respect to P and we denote by CardA the cardinality of A. The whole work of this paper is a translation of the Theorem 2.1 ([1]) in the language of cone metric spaces. This has been motivated by [2], [3], [4], [5], [6], [7], [8]. Now we prove the main theorem proposed in the abstract of this paper followed by some lemmas which are required to prove it.

B = R2 , P = {( x, y)  B : x, y  0}  R 2 , X = R 2 , d : X  X  B such that d ( x, y) = (| x  y |, | x  y |) , where   0 is a constant. Then ( X , d ) is a cone metric space. For, R 2 is a Banach space and clearly P is a cone.

Since, (i) P is a closed subset of R 2 . (ii)for any a, b R with a, b  0 and x = ( x1, x2 ), y = ( y1 , y2 )  P ax1  by1  0 and ax2  by 2  0.  (ax1  by1 , ax2  by 2 )  P  a( x1 , x2 )  b( y1 , y 2 )  P for all ( x1 , x2 ), ( y1 , y 2 )  P.

64

www.ijeart.com

A Common Fixed Point Theorem in Cone Metric Spaces

II. MAIN RESULTS Lemma 1: Let ( X , d ) be a totally ordered cone metric space. For any x, y, z  X ,

d ( x, z ) }  max{d ( x, y ), 2 . d ( x, y )  d ( y , z ) d ( y, z ), } 2 Proof. max{d ( x, y ), d ( y, z ),

Let us suppose that d ( x, y ) = max{d ( x, y ), d ( y, z ), Then d ( x, y)  d ( y, z) and d ( x, y ) 

d ( x, z ) }.

2

d ( x, z ) , 2

implies and d ( x, y)  d ( y, z)  P d ( x, z ) d ( x, y )  P .

2 d ( x, y )  d ( y , z ) d ( x, y )  d ( y , z ) d ( x, y )  = P Now, 2 2 d ( x, y )  d ( y , z )  d ( x, y )  .

2 d ( x, y )  d ( y , z ) } = d ( x, y ).

Therefore, max{d ( x, y ), d ( y, z ), 2 Similarly, we can check for d ( y, z) . d ( x, z ) d ( x, z ) = max{d ( x, y ), d ( y, z ), }, Again if 2 2 d ( x, z ) d ( x, z )  d ( x, y ) and  d ( y , z ) . ............(1) then 2 2 Now by using triangle inequality we have, d ( x, z)  d ( x, y)  d ( y, z) for all x, y, z  X  d ( x, y)  d ( y, z)  d ( x, z)  P d ( x , y )  d ( y , z ) d ( x, z )   P 2 2 d ( x, z ) d ( x , y )  d ( y , z ) i.e.,  ............(2) 2 2 By using (1),(2) and applying transitivity we have, d ( x, y )  d ( y , z ) d ( x, y )  d ( y , z ) d ( x, y )  and d ( y, z )  .

2 2 Therefore, d ( x, y )  d ( y , z ) d ( x, y )  d ( y , z ) max{d ( x, y ), d ( y, z ), }= .

2 2 Hence, we conclude that, which

d ( x, y )  d ( y , z ) d ( x, y )  d ( y , z )  d ( x, y ) and  d ( y, z ) .

2 2 d ( x, y )  d ( y , z ) But if is equal with one of d ( x, y) and 2 d ( y, z) , then the equality holds.

So we consider d ( x, y )  d ( y , z ) d ( x, y )  d ( y , z ) > d ( x, y ) and > d ( y, z ) 2 2 d ( x, y )  d ( y , z )   d ( x, y )  P and 2 d ( x, y )  d ( y , z )  d ( y, z )  P with d ( x, y )  d ( y, z ).

2 d ( y , z )  d ( x, y )  P 2 d ( x, y )  d ( y , z )  P with d ( x, y )  d ( y, z ) . and 2 d ( y , z )  d ( x, y ) d ( y , z )  d ( x, y )  P and  P i.e., 2 2 with d ( x, y )  d ( y, z ) d ( y , z )  d ( x, y )  = 2  d ( x, y) = d ( y, z), which contradicts the fact that

d ( x, y)  d ( y, z).

Therefore, max{d ( x, y ), d ( y, z )} = max{d ( x, y ), d ( y, z ),

d ( x, y )  d ( y , z ) }.

2 Lemma 3: Let ( X , d ) be a cone metric space and {xn } and { y n } be two convergent sequence in B such that xn  y n , n  N , then lim n xn  lim n y n .

Proof. Suppose lim n xn = x and lim n yn = y.

Now we have, xn  yn for all n  N  yn  xn  P for all n N  y  x  P (Since xn  x, yn  y as n   and P is closed.)  x  y. i.e., lim n xn  lim n yn .

Lemma 4: Let ( X , d ) be a cone metric space and f : X  X be continuous, then for every sequence d ( x, z ) d ( x, y )  d ( y , z ) max{d ( x, y ), d ( y, z ), }  max{d ( x, y ), d ( y, z ), {xn } in X converging to x , f ( xn )  f ( x).

2 2 .

Proof. Since {xn } converges to x in X , then for every Lemma 2: Let ( X , d ) be a totally ordered cone metric  N N  B  <<  with such space. For any x, y, z  X , that, n  N , d ( xn , x) <<  . d ( x, y )  d ( y , z ) max{d ( x, y ), d ( y, z )} = max{d ( x, y ), d ( y, z ), }. Again f is continuous on X and x  X , then for every 2 c  B with c >>  , there exists  >>  such that, d ( xn , x) <<   d ( f ( xn ), f ( x)) << c. Hence we conclude Proof. If d ( x, y )  d ( y , z ) d ( x, y )  d ( y , z ) that,  n  N d ( f ( xn ), f ( x)) << c  f ( xn )  f ( x) . max{d ( x, y ), d ( y, z ), }= , 2 2 i.e., lim n f ( xn ) = f ( x) = f (lim n xn ). then Theorem 1: Let ( X , d ) be a totally ordered complete cone metric space. Let S and T be mappings on X satisfying the following:

65

www.ijeart.com

International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015 implies (1 r)d (SIx, SIIx)  P .

Hence we have, (1  r )d (SIx, SIIx) =   d (SIx, SIIx) =  , (Since (1  r ) > 0 ), which implies d (Sx, SIx) <  , which contradicts the fact that d (Sx, SIx)   .

So we have, max{d (Sx, SIx), d (SIx, SIIx)} = d (Sx, SIx).

Thus d (SIx, SIIx)  rd (Sx, SIx),  x X .............(1)

(a) S is continuous; (b) T (X)  S (X); (c) S and T commute.

Let us define a non-increasing function

1  : [0,1)  ( ,1] 2

by,  (r ) = 1, if 0  r 

Let u  X and put u 0 = u . Now we consider u n = I n u , n  N . Then we find,

5 1 1 r ; 2 2 r 2 1 1 = , if  r < 1.

1 r 2 Suppose that there exists r [0,1) such that, =

1 r

5 1 ; 2

, if

u n1 = I n1u

= I ( I n u) = Iun and Su n1 = SIun = Tun ,  n  N. Now we will prove {Su n } is a Cauchy sequence. For this we have, d (Su n , Su n1 ) = d (SIun1 , SIIun1 )  rd (Sun1 , SIun1 ) [Using (1)] ..............(B) = rd (Su n1 , Su n )  r.rd ( Su n2 , Su n 1 )

(r)d (Sx,Tx)  d (Sx, Sy) implies d (Tx, Ty)  rM S ,T ( x, y ) for all x, y  X , where M S ,T ( x, y ) = max{d ( Sx, Sy ), d ( Sx, Ty)  d ( Sy, Tx) }...........(A).

2 Then there exists a unique common fixed point of S and T . d ( Sx, Tx), d ( Sy, Ty),

Proof. Using (b) we define a map I : X  X such that S ( I ( X )) = T ( X ) , i.e., SIx = Tx, x  X .

Since we are going to find the common fixed point of S and T so we have to deal with their common range. So we have introduced the mapping I on X. Since 1  (r )  0 and d (Sx, Tx)  P so we have,

. .  r n d ( Su 0 , Su1 ) (Using (1) repeatedly ).............(C ) If m > n then as in general case, d ( Su n , Su m )  d ( Su n , Su n 1 )  d ( Su n 1 , Su n  2 )  ...

 d (Sx,Tx)  (r)d (Sx,Tx)  P  d (Sx, SIx)  (r)d (Sx,Tx)  P (Since SIx = Tx,  x  X )  (r)d (Sx,Tx)  d (Sx, SIx).

Then by using (A) we have, d (SIx, SIIx) = d (Tx,TIx)  rM S ,T ( x, Ix) (1  (r ))d (Sx, Tx)  P

= rmax{d ( Sx, SIx), d ( Sx, Tx), d ( SIx, TIx),

 d ( Su m 1 , Su m )( By repeated use of triangle inequality )

 r n d (Su0 , Su1 )  r n1d (Su0 , Su1 )  ...  r m1d (Su0 , Su1 ) [Using (C )] = (r n  r n1  ...  r m1 )d (Su0 , Su1 )

mn d ( Sx, TIx)  d ( SIx, Tx) }= r n ( 1  r )d ( Su 0 , Su1 ) [ Since r < 1] 2 1 r

= rmax{d ( Sx, SIx), d ( Sx, SIx),

rn 1  r mn  rn ) > 0 and d (Su0 , Su1 )  P , Since ( d ( Sx, SIIx)  d ( SIx, SIx) 1 r 1 r d ( SIx, SIIx), }(Since Tx = SIx,  x  X ) 2 rn 1  r mn d ( Sx, SIIx)  rn )d ( Su 0 , Su1 )  P and so ( = rmax{d ( Sx, SIx), d ( SIx, SIIx), } 1 r 1 r 2 1  r mn rn )d ( Su 0 , Su1 )  ( )d ( Su 0 , Su1 ) therefore, r n ( 1 r 1 r d ( Sx, SIx)  d ( SIx, SIIx)  rmax{d ( Sx, SIx), d ( SIx, SIIx), } Hence by using transitivity we have, 2

.

rn ) d ( Su 0 , Su1 ) ..............(D) 1 r Let  << c and choose a  > 0 so that c  N ( )  P,

(Using Lemma 1)

d ( Su n , Su m )  (

= rmax{d (Sx, SIx), d (SIx, SIIx)}(Using Lemma 2) Now if max{d (Sx, SIx), d (SIx, SIIx)} = d (SIx, SIIx) , then d (SIx, SIIx)  d (Sx, SIx) and as a result we have, d (SIx, SIIx)  rd (SIx, SIIx) .

Again, if d (SIx, SIIx) = d (Sx, SIx) then we deduce that

where N ( ) := { y  B : y <  } . Again choose a natural number N such that, (

n  N . So,  (

d (SIx, SIIx)  rd (Sx, SIx).

So we consider, d (SIx, SIIx) > d (Sx, SIx).

Now, d (SIx, SIIx)  rd (SIx, SIIx)  (r  1)d (SIx, SIIx)  P i.e.,(1  r )d (SIx, SIIx)  P. Again (1  r ) > 0 and d (SIx, SIIx)  P

implies all n  N .

66

rn )d ( Su 0 , Su1 )  N  ( ) for all 1 r

rn )d ( Su 0 , Su1 )  N  ( ) for all n  N , that 1 r

c(

rn ) d ( Su 0 , Su1 )  c  N  ( )  P 1 r

for

www.ijeart.com

A Common Fixed Point Theorem in Cone Metric Spaces

Thus for all n  N , c  (

rn )d ( Su 0 , Su1 )  IntP 1 r

d ( z, Tx)  d ( Sx, z ) d ( z, Tx)  d ( Sx, z )  d ( Sx, z ) with  d ( Sx, z ); 2 2 d ( z, Tx)  d ( Sx, z )   d ( Sx, z )  P with d ( z, Tx)  d ( Sx, z ); 2 d ( z, Tx)  d ( Sx, z )   P with d ( z, Tx)  d ( Sx, z ); 2



rn ) d ( Su 0 , Su1 ) << c ............(E) 1 r Thus (D) and (E) together imply that {Su n } is a Cauchy sequence.

Since X is a complete cone metric space, then there exists a point z  X such that Su n  z .

We also have Tun = Su n1  z. Next we show that, for all d (Tx, z)  rmax{d ( z, Sx), d (Sx,Tx)} ............(2) x  X with Sx  z . Since Sun  z, Tun  z as n   (

and Sx  z then d (Sun , Tun )  d ( z, z) =  and d (Sun , Sx)  d (Sx, z) >  as n  .

Hence, there exists N1  N such that, (r )d (Sun , Tun )  d (Sun , Sx),  n  N1. For, if possible let (r )d (Sun , Tun ) > d (Sun , Sx),  n  N1 , which implies,  (r )d ( z , z )  d ( Sx, z ) [ Since Su n  z , Tu n  z as n  , using Lemma 3]

i.e.,   d (Sx, z)  d (Sx, z)  P. Also we have, d (Sx, z)  P. Therefore, combining these two we have, d (Sx, z) =   Sx = z, which is a contradiction to the fact that Sx  z.

Hence by (A) we have, d (Tun , Tx)  rM S ,T (u n , x) for n  N with n  N1. i.e., d (Tu n , Tx)  rmax{d ( Su n , Sx), d ( Su n , Tu n ),

d ( Su n , Tx)  d ( Sx, Tu n ) } 2  rmax{d ( Su n , Sx), d ( Su n , Tu n ), d ( Sx, Tx),

 d ( z,Tx)  d (Sx, z)  P with d ( z,Tx)  d (Sx, z);  d ( z, Tx)  d (Sx, z) with d ( z, Tx)  d (Sx, z); i.e., d ( z, Tx) > d (Sx, z).

Since max{d ( z , Sx), d ( Sx, Tx),

d ( z , Tx)  d ( Sx, z ) d ( z, Tx)  d ( Sx, z ) }= , 2 2

then from (3) we have, d ( z, Tx)  d ( Sx, z ) d ( z, Tx)  d ( z, Tx) d ( z, Tx)  r <r 2 2 [Since d ( z, Tx) > d (Sx, z) ] d ( z, Tx)  d ( z, Tx)  d ( z, Tx) < r [Using transitivity] 2 i.e., d ( z, Tx) < rd ( z, Tx).

Again (1  r ) > 0 and d ( z, Tx)  P implies (1 r)d ( z,Tx)  P  rd ( z, Tx)  d ( z, Tx) , which is a contradiction to the fact that d (Tx, z ) < rd (Tx, z ).

Therefore, d (Tx, z)  rmax{d ( z, Sx), d (Sx,Tx)} . Hence (2) holds for all x  X with Sx  z . Now we will prove z is a fixed point of S. Let us consider the case when Card {n : d (Sun , Tun ) > d (Sun , SSun )} = , then there exists a subsequence {u n } of {un } such j

d ( Sx, Tx),

that d ( Su n , Tun ) > d (Su n , SSu n ) j

j

j

j

 d ( Su n , Tun )  d (Su n , SSu n )  P with j j j j d ( Su n , Tx)  d ( Sx, Tu n ) }  d (Tu n , Tx)  P d ( Su n , Tu n )  d ( Su n , SSu n ) ............(4) j j j j 2  rmax{d ( z , Sx), d ( z , z ), d ( Sx, Tx), Now by using triangle inequality we have, d ( SSu n , z )  d ( SSu n , Su n )  d ( Su n , z ) d ( z , Tx)  d ( Sx, z ) j j j j }  d ( z , Tx)  P 2  d ( SSun , Su n )  d (Su n , z )  d (SSu n , z )  P ........(5) j j j j (Since, Su n  z, Tu n  z as n   and using Lemma 3) Therefore from (4) and (5) we have, d ( z , Tx)  d ( Sx, z )  rmax{d ( z , Sx), d ( Sx, Tx), }  d ( z , Tx)  P d ( Su n , Tu n )  d ( Su n , SSu n )  d ( SSu n , Su n ) j j j j j j 2  d ( Su n , z )  d ( SSu n , z )  P with j j (Since d (Sx, z) >  so we drop the term d ( z, z) =  .) d ( Su n , Tu n )  d ( Su n , SSu n ). i.e., j j j j d ( z , Tx)  d ( Sx, z ) d ( z , Tx)  rmax{d ( z , Sx), d ( Sx, Tx), }...........(3) d ( Su n j , Tun j )  d ( Su n j , z )  d ( SSu n j , z )  P 2 ,  d ( z, z)  d ( z, z)  d (Sz, z)  P [Since Su n  z Let j d ( z , Tx)  d ( Sx, z ) d ( z , Tx)  d ( Sx, z ) Tu  z , as n   and using Lemma 3] = max{d ( z , Sx), d ( Sx, Tx), } nj j 2 2    d (Sz, z )  P , d ( z, Tx)  d ( Sx, z )  d ( Sx, z ) . then i.e., d (Sz, z)   ,  d (Sz, z) =  , i.e., Sz = z.

2 d ( z , Tx)  d ( Sx, z ) Therefore z is a fixed point of S in this case. = d ( Sx, z ) then Now if But if Card {n : d (Sun , Tun ) > d (Sun , SSun )} < , 2 d ( z,Tx)  rmax{d ( z, Sx), d (Sx,Tx)}. there exists N 4  N such that d (Sun , Tun )  d (Sun , SSu n ) d ( z , Tx)  d ( Sx, z ) for all n  N 4 . > d ( Sx, z ) But if 2 Now (1  (r ))  0 and d (Su n , Tun )  P implies

67

www.ijeart.com

International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015 (1 (r ))d (Sun , Tun )  P

= rmax{d (T n1 z, T n z ), d (T n1 z, T n z ), d (T n z, T n1 z ),

 (r )d (Su n , Tu n )  d (Su n , Tu n ).

Therefore, using transitivity we have, (r )d (Sun , Tun )  d (Sun , SSun ),  n  N 4 .

So, by (A) we have, d (Tu n , TSu n )  rM S ,T (u n , Su n ),  n  N 4 . .............(6)

d (T n1 z, T n1 z )  d (T n z, T n z ) } [ Since Sz = z ] 2 = rmax{d (T n 1 z , T n z ), d (T n z , T n 1 z ),

Since S is continuous and S , T commute then TSun = STun  Sz as n   (Using Lemma 4).

Therefore, from (6) we have, rM S ,T (u n , Su n )  d (Tu n , TSu n )  P,  n  N 4 .

 rmax{d (T n1 z, T n z), d (T n z, T n1 z), d (T n 1 z , T n z )  d (T n z , T n 1 z ) } 2 (Using Lemma 1)

i.e., rmax{d ( Su n , SSu n ), d ( Su n , Tu n ), d ( SSu n , TSu n ),

= rmax{d (T n1 z, T n z ), d (T n z, T n1 z )} ............(F )

d ( Su n , TSu n )  d ( SSu n , Tu n ) }  d (Tu n , TSu n )  P 2  rmax{d ( Su n , SSu n ), d ( Su n , Tu n ), d ( SSu n , STu n ),

(Using Lemma 2) Now if max{d (T n1 z, T n z), d (T n z, T n1 z)} = d (T n z, T n1 z) , then, d (T n z, T n1 z)  d (T n1 z, T n z).

d ( Su n , STu n )  d ( SSu n , Tu n ) }  d (Tu n , STu n )  P 2 [Since S and T commute.]  rmax{d ( z , Sz ), d ( z , z ), d ( Sz , Sz ),

If d (T n z, T n1 z) = d (T n1 z, T n z), then (7) is proved.

So we consider d (T n z, T n1 z) > d (T n1 z, T n z) ...........( F 1 ) Hence from ( F ) we have, d (T n z, T n1 z)  rd (T n z, T n1 z)

d ( z , Sz )  d ( Sz , z ) }  d ( z , Sz )  P 2 [Since Su n  z, Tun  z as n   and using Lemma 3] i.e., rmax{d ( z, Sz), d ( z, z), d (Sz, Sz), d ( z, Sz)}  d ( z, Sz)  P

 rd (T n z, T n1 z)  d (T n z, T n1 z)  P

 (1 r )d (T n z,T n1 z)  P Again (1  r ) > 0 and d (T n z, T n1 z)  P implies

 rd ( z, Sz)  d ( z, Sz)  P [Since d (Sz, Sz) =  = d ( z, z) and d (Sz, z)   ]

(1  r )d (T n z, T n1 z)  P. Hence (1  r )d (T n z, T n1 z) = 

 (1  r )d (Sz, z)  P Again (1  r ) > 0 and d (Sz, z)  P  (1  r )d ( z, Sz )  P. Hence (1  r )d ( z, Sz ) =   d ( z, Sz) =  [Since (1 r ) > 0] i.e., Sz = z.

Therefore z is a fixed point of S in the both cases.

Now we prove that,

 d (T n z, T n1 z) =  [Since (1  r ) > 0 ] Thus from ( F 1 ) we have, d (T n z, T n1 z) <  , which is a contradiction. Thus we conclude that max{d (T n1 z, T n z), d (T n z, T n1 z)} = d (T n1 z, T n z) Therefore, d (T n z, T n1 z)  rd (T n1 z, T n z).

Thus (7) holds for n  N. So, d (T n z, T n1 z)  rd (T n1 z, T n z)  r.rd (T n2 z, T n1 z)

d (T n z, T n1 z )  rd (T n 1 z, T n z ) for n  N, where T 0 z = z. ...........(7) Now, (1  (r ))  0 and d (ST n1 z, TT n1 z)  P implies

. . .

(1 (r ))d (ST n1 z,TT n1 z)  P.  (r )d (ST n1 z, TT n1 z)  d (ST n1 z, TT n1 z)

 r n d (T 0 z,Tz) [Using (7) repeatedly ]

= d (ST n1 z,T n Sz) [Since Sz = z.]

= r n d ( z,Tz) [Since T 0 z = z]

= d (ST n1 z, ST n z) [Since S and T commute.]

Therefore by (A) we have, d (TT n1 z, TT n z) = d (T n z, T n1 z)

i.e., d (T n z, T n1 z)  r n d ( z, Tz) for all n  N …….........(8) [Using transitivity repeatedly] We next show by induction that, d (T n z, z)  d ( z,Tz) for all n N. .............(9) For n=1, (9) is obvious.

We suppose that d (T n z, z)  d ( z,Tz) for some n N.

 rM S ,T (T n1 z, T n z ) = rmax{d ( ST n1 z, ST n z ), d ( ST n1 z, TT n1 z ) d ( ST n z, TT n z ),

= rmax{d (T

n 1

d ( ST n1 z , TT n z )  d ( ST n z, TT n1 z ) } 2 n

Sz, T Sz ), d (T

d (T n Sz, T n1 z ),

n 1

n 1

n 1

If T n z = z then T n1 z = T (T n z) = Tz and

d (T n1 z, z) = d (Tz, z)  d (Tz, z), then (9) is true.

n

Sz, T z ),

Now if T n z  z , then ST n z = T n Sz = T n z  z. [Since S and T commute.]

So we have by (2) that, d (TT n z, z)  rmax{d ( z, ST n z), d (ST n z,TT n z)} ; i.e.,

z )  d (T Sz, T z ) } 2 [Since S and T commute.]

d (T

Sz, T

d (T n 1 z , T n 1 z ) } 2

n

n

68

www.ijeart.com

A Common Fixed Point Theorem in Cone Metric Spaces

d (T n1 z, z )  rmax{d ( z, T n Sz ), d (T n Sz, T n1 z )}

 (r )d (T n z, T n1 z )  (

[ Since S and T commute.]

1 r

1 r

1 r r2

)d (T n z, T n1 z ) ...............(L)

for all n  N with n  2 .[Since r [0,1) ] r r2  rmax{d ( z, T n z), r n d ( z, Tz)} [Using (8)] 1 r 1 r i.e., n  2  0 for all n  N with n  2 . n  rmax{d ( z, Tz), r d (Tz, z)} [Using our assumption ] ..........(G) r r Therefore, 1 r 1 r As 0  r < 1, then r n < 1 for all n N  1  r n > 0 ( n  2 )d (T n z, T n1 z )  P r r for all n  N. Also d ( z,Tz)  P .

1  r 1 r  ( 2 )d (T n z, T n1 z )  ( n )d (T n z, T n1 z ) ............(M) Therefore, (1 r n )d (Tz, z)  P r r  r n d (Tz, z)  d (Tz, z) By (8) we have, Therefore, from (G) we have, d (T n z, T n1 z)  r n d (Tz, z) n1 d (T z, z)  rd ( z,Tz) ...............(H ).  r n d (Tz, z)  d (T n z,T n1 z)  P As (1  r ) > 0 and d (Tz, z)  P , then (1  r )d (Tz, z )  P .

1 r 1 r  ( n )r n d (Tz, z )  ( n )d (T n z, T n1 z )  P  d (Tz, z)  rd (Tz, z)  P . r r i.e., rd (Tz, z )  d (Tz, z ) ............(I) 1 r [Since n  0 ] Now using transitivity on (H) and (I) we have, r d (T n1 z, z)  d (Tz, z).

1 r  (1  r )d (Tz, z )  ( n )d (T n z, T n1 z )  P Hence by induction, (9) holds for n  N. r Now we will prove z is fixed point of T. We will prove this in 1 r  ( n )d (T n z, T n1 z )  (1  r )d (Tz, z ) ..............(N) two cases: r 1 By (L),(M),(N) and using transitivity we have, Case I: Now we consider the case, when 0  r  , then n n1 2 (r )d (T z,T z)  (1 r )d (Tz, z) ................(O) 1 r By (11) we have, (r )  2 .

1 r d ( z , T n z )  d (Tz, z )  P We first prove that, 1 r d (T n z,Tz)  rd (Tz, z) for all n N ............(10)  d ( z,T n z)  (1 r )d (Tz, z)  P [Since (1  r ) > 0 ] For n = 1 , (10) is obvious. n i.e., (1  r )d (Tz , z )  d ( z , T z ) ...............(Q) For n = 2 , d (T 2 z,Tz)  rd (Tz, z) [Putting n = 1 in (8)] By applying transitivity on (O) and (Q) we have, So, (10) is true for n = 2. (r )d (T n z,T n1 z)  d ( z,T n z) Let us suppose that, (10) is true for some n  N with n  2 , = d (Sz,T n Sz) [Since Sz = z] i.e., d (T n z,Tz)  rd (Tz, z) = d (Sz, ST n z) [Since S and T commute]  rd (Tz, z)  d (T n z,Tz)  P ...............(J) i.e., (r )d (T n z,T n1 z)  d (ST n z, Sz) Using triangle inequality we have, or, (r )d (T n Sz,T n1 z)  d (ST n z, Sz) [Since Sz=z] d ( z,Tz)  d ( z,T n z)  d (T n z,Tz)



= rmax{d ( z, T n z), d (T n z, T n1 z)} [Since Sz = z.]

Again,

 d ( z,T n z)  d (T n z,Tz)  d ( z,Tz)  P ............(K ) From (J) and (K) we have, d ( z,T n z)  d (T n z,Tz)  d ( z,Tz)  rd (Tz, z)  d (T n z,Tz)  P

 (r )d (ST n z,TT n z)  d (ST n z, Sz) [Since S, T commute] Therefore, by (A) we have, d (TT n z, Tz)  rM S ,T (T n z, z)

 d ( z,T n z)  (1 r )d (Tz, z)  P

1 1 d ( z, T n z )  d (Tz, z )  P [Since > 1 r 1 r 1 d ( z, T n z ) ..............(11) i.e., d (Tz, z )  1 r 1 r 1 r Since (r )  2 , i.e., 2  (r )  0 and r r d (T n z, T n1 z)  P then we have, 

(

1 r r

2

(

= rmax{d (ST n z, Sz), d (ST n z,TT n z), d (Sz,Tz), d ( ST n z , Tz )  d ( Sz , TT n z ) } 2 = rmax{d (T n Sz, Sz), d (T n Sz,TT n z), d (Sz,Tz),

0]

d (T n Sz , Tz )  d ( Sz , TT n z ) } 2 [Since S and T commute]

= rmax{d (T n z, z), d (T n z,T n1 z), d ( z,Tz),

 (r ))d (T n z, T n1 z )  P

1 r r2

n

d (T n z , Tz )  d ( z , T n 1 z ) }[Since Sz = z ] 2

)d (T n z, T n1 z )  (r )d (T n z, T n1 z )  P

By (8) we have, r n d ( z,Tz)  d (T n z,T n1 z)  P for all n N. ...............(R) By (9) we have,

69

www.ijeart.com

International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015

d (Tz, z)  d (T n z, z)  P for all n N .................(S )

r r r2 r d (Tz, z )  d (T n1 z, z )  d (Tz, z )  d (T n z, Tz ) 2 2 2 2 r r n 1 n n 1  d ( z, T z )  d (T z, Tz )  d (T z, Tz )  P 2 2 2 r r  (  )d (Tz, z )  d (T n1 z , Tz )  P ...............(A1) 2 2

Also, d (Tz, z)  d (T n1 z, z)  P ....................(T) Now, by our assumption we have, d (T n z,Tz)  rd (Tz, z) for some n N

i.e., rd (Tz, z)  d (T n z, Tz)  P for some n  N ...........(U )

r r2  0 2 2

If d ( z , T z ) = max{d ( z , T z ), d ( z , Tz ),

Again since r [0,1) then,

d ( z , T n 1 z )  d (T n z , Tz ) } 2 then, d (T n1 z,Tz)  rd ( z,T n z)

r r2  (  )d (Tz, z )  P for d (Tz, z )  P ...............(A2) 2 2 Therefore from (A1) and (A2) we have,

n

n

d (T n z , T n 1 z ),

 rd ( z,T n z)  d (T n1 z,Tz)  P .............(V) Therefore, from (S) and (V) we have, rd (Tz, z)  rd (T n z, z)  rd ( z,T n z)  d (T n1 z,Tz)  P [Since r [0,1) ]

r2 r r r2  )d (Tz, z )  d (T n1 z , Tz )  (  )d (Tz, z )  P 2 2 2 2 n1  rd (Tz, z)  d (T z,Tz)  P (

 d (T n1 z,Tz)  rd (Tz, z) Now if d (Tz, z) = max{d ( z,T n z), d ( z,Tz), d (T n z,T n1 z),

 rd (Tz, z)  d (T n1 z,Tz)  P i.e., d (T n1 z,Tz)  rd (Tz, z).

Again if d (T n z,T n1 z) = max{d ( z,T n z), d ( z, Tz), d (T n z,T n1 z),

d ( z, T n 1 z )  d (T n z, Tz ) } 2 then d (T n1 z, Tz)  rd (Tz, z) .

Thus in any case, d (T n1 z, Tz)  rd (Tz, z) Therefore, by induction, (10) holds for all n  N .

Now, arguing by contradiction we assume that, Tz  z,

d ( z , T n 1 z )  d (T n z , Tz ) } 2 then d (T n1 z, Tz)  rd (T n z, T n1 z)

 rd (T n z, T n1 z)  d (T n1 z, Tz)  P ................(W) From (R) and (W) we have, Then by (10) we have, T n z  z. r n1d ( z, Tz)  rd (T n z, T n1 z)  rd (T n z, T n1 z)  d (T n1 z, Tz)  P For, if T n z = z , then by (10) we have, d ( z, Tz)  rd (Tz, z )  (r 1)d (Tz, z)  P. i.e., (1 r)d (Tz, z)  P.  r n1d ( z,Tz)  d (T n1 z,Tz)  P Again, (1  r ) > 0  (1  r )d (Tz, z)  P. i.e., d (T n1 z, Tz)  r n1d ( z, Tz) .............(X) Hence, (1 r )d (Tz, z) =  Since r [0,1),(r  r n1 )  0 and also d (Tz, z)  P,  d (Tz, z) =  [Since (1  r ) > 0] . then (r  r n1 )d (Tz, z)  P i.e., Tz = z, which is a contradiction to our assumption Tz  z . or, rd (Tz, z)  r n1d (Tz, z)  P  r n1d (Tz, z)  rd (Tz, z) .............(Y) Hence using transitivity on (X) and (Y) we have, d (T n1 z,Tz)  rd (Tz, z). Now if

So, ST n z = T n Sz = T n z  z [Since S and T commute.]

Thus by (2) we have, d (TT n z, z) = d (T n1 z, z)

 rmax{d (ST n z, z), d (ST n z,TT n z)} d ( z , T n1 z )  d (T n z , Tz ) = max{d ( z , T n z ), d ( z , Tz ), d (T n z , T n1 z ), 2 = rmax{d (T n Sz, z), d (T n Sz,T n1 z)} [Since S and T commute.] n 1 n d ( z , T z )  d (T z , Tz ) = rmax{d (T n z, z), d (T n z,T n1 z)} [Since Sz = z.] } 2  rmax{d (T n z, z), r n d ( z,Tz)} [ By using (8).] n 1 n d ( z , T z )  d ( T z , Tz ) then, d (T n 1 z , Tz )  r i.e., d (T n1 z, z)  rmax{d (T n z, z), r n d ( z, Tz)} .............(12) 2 By triangle inequality, d ( z , T n1 z )  d (T n z , Tz ) r  d (T n1 z , Tz )  P ...............(Z) d (Tz, z)  d (Tz,T n z)  d (T n z, z) 2  d (Tz,T n z)  d (T n z, z)  d (Tz, z)  P .............(A3).

Now from (T) and (U) we have, r r By using (10) we have, d (Tz, z )  d (T n1 z , z )  P and 2 2 rd (Tz, z)  d (T n z,Tz)  P .................(A4).

2 r r Therefore, from (A3) and (A4) we have, d (Tz, z )  d (T n z , Tz )  P [ Since r > 0].

2 2 d (Tz,T n z)  d (T n z, z)  d (Tz, z)  rd (Tz, z)  d (T n z,Tz)  P Therefore, from the above two and (Z) we have,  d (T n z, z)  (1 r )d (Tz, z)  P  (1 r )d (Tz, z)  d (T n z, z) ............(A5)

70

www.ijeart.com

A Common Fixed Point Theorem in Cone Metric Spaces

Also there exists a N 5  N such that,  n  N 5

(1 r )  r n  0 and also d (Tz, z)  P

,

 ((1 r )  r )d (Tz, z)  P n

 (1 r )d (Tz, z)  r n d (Tz, z)  P . i.e., r n d (Tz, z)  (1 r )d (Tz, z) .............(A6) Therefore, from (A5)and(A6) and by using transitivity we have, r n d (Tz, z)  d (T n z, z) for all n  N5 Then by (12) we have, for all n  N 5

d (T n1 z, z)  rmax{d (T n z, z), r n d (Tz, z)}  rmax{d (T n z, z), d (T n z, z)} [ By Lemma 1]

= rd (T n z, z) i.e., d (T n1 z, z)  rd (T n z, z)

 r.rd (T n1 z, z)

. . . r

n N5 1

d (T

n N 1

N5 z, z )

i.e., d (T n1 z, z)  r 5 d (T 5 z, z) ............(A7) Now, let  << c and choose a  > 0 so that c  N ( )  P, N

1 )[d ( Su n1 , Su n )  d ( Su n , Su n1 )] 1 r ...........(A8) Now by triangle inequality we have, d (Sun1 , Sun )  d (Sun1 , z)  d (Sun , z) 1  d ( Su n1 , Su n ) < ( )[d ( Su n1 , Su n )  d ( Su n , Su n 1 )] 1 r [By (A8) and applying transitivity.] ............(A9) By (13) we have, rd (Su n1 , Su n )  d (Su n , Su n1 )  P r 1 1 ( )d ( Su n1 , Su n )  ( )d ( Su n , Su n1 )  P [ Since > 0.]

1 r 1 r 1 r r 1  [( )d ( Su n1 , Su n )  ( )d ( Su n 1 , Su n )] 1 r 1 r 1 1  [( )d ( Su n , Su n1 )  ( )d ( Su n1 , Su n )]  P. 1 r 1 r 1 1 r ( )[d ( Su n , Su n1 )  d ( Su n1 , Su n )]  ( )d ( Su n1 , Su n ) 1 r 1 r

where N  ( ) := { y  B : y <  }.

 [d ( Su n1 , z )  d ( Su n , z )] < (

Again choose a natural number N 6 such that,

r

n N5 1

d (T

N5 z, z )  N

n N5 1

So,  r This implies,

N d (T 5 z, z)  N



 ( )

n N 1

N

n N5 1

N5 z, z)  IntP .

n N 1

N

( ) for all n  N 6 . for all n  N 6 .

c  r 5 d (T 5 z, z)  c  N ( )  P for all n  N 6 .

Thus, for all n  N 6 , cr

d (T

 r 5 d (T 5 z, z) << c Therefore, by using (A7) and applying transitivity we have, d (T n1 z, z) << c for all n  N 6 ,  T n1 z  z as n  

i.e., T n z  z as n   .

By using (10) we have, rd (Tz, z)  d (T n z,Tz)  P

 rd (Tz, z )  d ( z, Tz)  P [ Since T n z  z as n   and P is closed ]  (1  r )d (Tz, z)  P Again (1  r ) > 0 and d (Tz, z)  P ,  (1  r )d (Tz, z)  P Hence, (1 r )d (Tz, z) =   d (Tz, z) =  [Since (1 r ) > 0]  Tz = z, which contradicts to our assumption that Tz  z .

Therefore, we obtain Tz = z .

Hence, z is a fixed point of T in this case.

1  r < 1. We Case II: Now we consider the case, when 2 will show that there exists a subsequence {n j } of {n} such

that, (r )d ( Su n , Su n j

j 1

We have, d (Su n , Su n1 ) = d (SIun1 , SIIun1 )  rd (Su n1 , SIun1 ) [Using (1).] = rd (Su n1 , Su n ) i.e., d (Su n , Su n1 )  rd (Su n1 , Su n ) ..............(13) Now we assume that, 1 ( )d ( Su n 1 , Su n ) > d ( Sun 1 , z ) and 1 r 1 ( )d ( Su n , Su n1 ) > d ( Su n , z ) ............(14) 1 r 1 ( )d ( Su n 1 , Su n )  d ( Su n1 , z )  P with 1 r 1 d ( Su n 1 , z )  ( )d ( Su n 1 , Su n ), and 1 r 1 ( )d ( Su n , Su n1 )  d ( Su n , z )  P with 1 r 1 d ( Su n , z )  ( )d (Su n , Su n 1 ).

1 r Therefore, 1 ( )[d ( Su n 1 , Su n )  d ( Su n , Su n 1 )]  [d ( Su n 1 , z )  d ( Su n , z )]  P 1 r with 1 ( )[d ( Su n1 , Su n )  d ( Su n , Su n1 )]  [d ( Su n1 , z )  d ( Su n , z )].

1 r

)  d ( Su n , z ) for j  N. j

71

= d (Sun1 , Sun ) i.e., 1 ( )[d ( Su n , Su n 1 )  d ( Su n 1 , Su n )]  d ( Su n 1 , Su n ) .........(A10) 1 r Hence applying transitivity on (A9) and (A10) we have, d (Sun1 , Sun ) < d (Sun1 , Sun ) , which is a contradiction.

Therefore the contrapositive of (14) is true. i.e., 1 )d ( Su n1 , Su n )  d ( Su n1 , z ) either ( 1 r 1 )d ( Su n , Su n1 )  d ( Su n , z ), for n  N or ( 1 r  either (r )d (Su2n1 , Su2n )  d (Su2n1 , z) or (r )d (Su 2n , Su 2n1 )  d (Su 2n , z) holds for n  N.

www.ijeart.com

International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015 REFERENCES

Thus there exists a subsequence {n j } of {n} such that, (r )d ( Su n , Su n

j 1

j

)  d ( Su n , z ) for j  N.

[1]

j

i.e., (r )d ( Su n , Tun )  d ( Sun , Sz ) for j  N. [ Since Sz = z and j

Su n

j 1

j

[3]

= SIun = Tun .] j

j

Therefore, by (A) we have, d (Tun , Tz)  rM S ,T (u n , z ) j

[4]

j

[5]

 rM S ,T (u n , z )  d (Tun , Tz)  P j

[6]

j

 rmax{d ( Su n , Sz ), d ( Su n , Tu n ), j

j

j

[7]

j

d ( Su n , Tz )  d ( Sz , Tu n ) d ( Sz , Tz ),

[2]

j

j

2  rmax{d ( z , Sz ), d ( z , z ), d ( Sz , Tz ),

[8]

}  d (Tu n , Tz )  P j

d ( z , Tz )  d ( Sz , z ) }  d ( z , Tz )  P [Since Su n  z, Tu n  z j j 2 as n   , and also using Lemma 3] d ( z , Tz )  d ( z , z )  rmax{d ( z , z ), d ( z , z ), d ( z , Tz ), } 2  d ( z , Tz )  P [ Since Sz = z.] d ( z , Tz )  rmax{d ( z , Tz ), }  d ( z , Tz )  P 2 [ Since Tz  z , i.e., d (Tz, z ) >  ]  rd (Tz, z)  d (Tz, z)  P,  (1  r )d (Tz, z )  P. Again (1  r ) > 0 and d (Tz, z)  P

O. Popescu, Two fixed point theorems for generalized contractions with constants in complete metric space, Cent. Eur. J. Math., 7(3), 2009, 529-538.

M. Singha, Two Fixed Point Theorems in the Language of Cone Metric Spaces, Journal of Mathematics, IV(1)(2012), 67-76.

C. D. Bari, P. Vetro,

 -pairs and common fixed points in cone metric

spaces, Rend. del Cir. Mat. di Pal 57, 279-285(2008).

S. K. Chatterjea, Fixed point theorems, comptes rendus de l’Academie Bulgare des Sciences, 25, (1972), 727-730.

Lj. B. Ciric, Generalized contractions and fixed point theorem, publ.

Inst. Math., 12, (1971), 19-26.

R. Kannan, Some results on Fixed points, Bull. Calcutta Math. Soc., 60, (1968), 71-76.

O. Popescu, Fixed point theorems in metric spaces, Bull. of Transilvania Univ., 2008, 50, 479-482.

T. Suzuki, A generalized Banach contraction principle which characterizes metric completeness, Proc. Amer. Math. Soc., 2008, 136, 1861-1869.

Manoranjan Singha, M.Sc, Ph.D. Assistant Professor, Department of Mathematics, University of North Bengal, India.. His present area of research interest are Topology, Functional Analysis and Algebra. E-mail: manoranjan_ singha@rediffmail.com Kousik Das, M.Sc, Research Scholar, Department of Mathematics, University of North Bengal, India.. E-mail: das.kousik1991@gmail.com

implies (1 r )d (Tz, z)  P. Hence, (1 r )d (Tz, z) =  ,  d (Tz, z) =  [Since (1 r ) > 0]; i.e., Tz = z.

Thus, we have shown that, Tz = z in both of the cases.

Hence, z is a common fixed point of both S and T .

Uniqueness of the common fixed point: Let z1 and z 2 be two distinct common fixed points of both S and T . Then, Sz1 = z1 , Tz1 = z1 and Sz 2 = z 2 , Tz2 = z 2 .

Now by using (2) we have, d (Tx, z1 )  rmax{d ( z1 , Sx), d (Sx, Tx)} for all those x  X satisfying Sx  z1 . Now putting x = z 2 in the above, we have, d (Tz2 , z1 )  rmax{d ( z1 , Sz 2 ), d (Sz 2 , Tz2 )} ; i.e., d ( z2 , z1 )  rmax{d ( z1 , z 2 ), d ( z 2 , z 2 )}

 d ( z1 , z2 )  rd ( z1 , z2 ) [Since d ( z2 , z 2 ) =  and d ( z1 , z 2 ) >  ].  rd ( z1 , z2 )  d ( z1 , z2 )  P, i.e., (1 r )d ( z1 , z2 )  P. Again (1  r ) > 0 and d ( z1 , z 2 )  P implies (1  r )d ( z1 , z 2 )  P. Henc, (1  r )d ( z1 , z 2 ) =   d ( z1 , z 2 ) =  [Since (1  r ) > 0] i.e., z1 = z 2 . This contradicts the distinctness of z1 and z 2 .Hence, there exists a unique common fixed point of S and T .

72

www.ijeart.com