International Journal of Mathematics and Statistics Invention (IJMSI) E-ISSN: 2321 – 4767 P-ISSN: 2321 - 4759 www.ijmsi.org Volume 2 Issue 4 || April. 2014 || PP-33-39

Coupled Fixed Point Theorems in S-metric Spaces Ajay Singh1, Nawneet Hooda2 1, 2,

(Department of Mathematics, DCRUST, Murthal, Sonepat, Haryana, India

ABSTRACT : The aim of this paper is to establish some coupled fixed point results in S-metric space which are the generlaizations of some fixed point theorems in metric spaces [6]. MSC: 47H10, 54H25

KEYWORDS - Coupled fixed point, mixed monotone property, S-metric space, ICS property. I.

INTRODUCTION

In 1922, the Polish mathematicians, Banach, proved a theorem which ensures, under approiate conditions, the existence and uniqueness of fixed point. This result is called Banach’s fixed point theorem or Banach contraction principle. Many authors extend this principle in different ways and imporoved it. Bahaskar and Lakshikanthamin [1] introduced the concept of coupled fixed point of a mapping and proved some coupled fixed point results in partially ordered metric spaces. After that many result are obtained by several authors. Mustafa and Sims [4] introduced the concept of generalized metric space called G-metric space. Now, recently Sedghi et al. [3] have introduced the notion of S-metric spaces as the generalization of G-metric and D* -metric spaces. After this they proved some fixed point theorems in S-metric spaces. After that several authors obtained many result on S-metric space.

II.

PRELIMINARIES

Definition 1 ([3]). Let X be a nonempty set. An S-metric on X is a function S : X 3  [0, ) that satisfies the following conditions, for each x, y, z, a  X .

S ( x, y, z)  0 (ii) S ( x, y, z)  0 if and only if x  y  z (iii) S ( x, y, z)  S ( x, x, a)  S ( y, y, a)  S ( z, z, a) Then the pair ( X , S ) is called an S-metric space. (i)

Definition 2 ([1]). Let ( X , ) be a partially ordered set and F : X  X  X be a mapping. F is said to have the mixed monotone property if F ( x, y) is monotone non-decreasing in x and is monotone nonincreasing in y , that is, for any x, y  X ,

x1  x2  F ( x1 , y)  F ( x2 , y) , for x1 , x2  X and y1  y2  F ( x, y2 )  F ( x, y1 ) , for y1 , y2  X . Definition 3 ([1]). An element ( x, y)  X  X

is said to be a coupled fixed point of the mapping

F : X  X  X if F ( x, y)  x and F ( y, x)  y . For further definition see [7-10]. Definition 4. Let ( X , d ) be a metric space. A mapping T : X  X is said to be ICS if T is injective, continuous and has the property: for every sequence {xn } in X , if {Txn } is convergent then {xn } is also convergent. Let  be the set of all functions  : [0,1)  [0,1) such that

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Coupled Fixed Point Theorems in S-metric Spaces (i) (ii) (iii)

 is non-decreasing,  (t )  t for all t  0 , lim  (r )  t for all t  0 . r t

Definition 5 ([1]). Let ( X , ) be a partially ordered set equipped with a metric S such that ( X , S ) is a metric space. Further, equip the product space X  X with the following partial ordering: for ( x, y),(u, v)  X  X , define (u, v)  ( x, y)  x  u, y  v . Theorem 1. Let ( X , ) be a partially ordered set and suppose there is a metric S on X such that ( X , S ) is complete S-metric space. Suppose T : X  X is an ICS mapping and F : X  X  X is such that F has the mixed monotone property. Assume that there exists    such that

S (TF ( x, y), TF ( x, y), TF (u, v))   (max{S (Tx, Tx, Tu), S (Ty, Ty, Tv)}) for any x, y, u, v  X for which x  u , v  y . Suppose either (a) F is continuous or (b) X has the following property: (i) if non-decreasing sequence xn  x , then xn  x for all n . (ii) if non-increasing sequence yn  y , then yn  y for all n . If there exist x0 , y0  X such that x0  F ( x0 , y0 ) , y0  F ( y0 , x0 ) , then there exist x, y  X such that F ( x, y)  x , F ( y, x)  y that is F has a coupled fixed point. Proof. Let x0 , y0  X such that x0  F ( x0 , y0 ) and y0  F ( y0 , x0 ) . Define x1  F ( x0 , y0 ) , y1  F ( y0 , x0 ) Like this, we can construct sequences {xn } and { yn } in X such that

xn1  F ( xn , yn ) , and yn1  F ( yn , xn ) Since F has the mixed monotone property, so it is easy that xn  xn1 , yn1  yn for n  0,1,2,... Let for some n  N , we take xn  xn1 or yn  yn1 then form (3), F has the coupled fixed point. Let if possible, for any n  N , xn  xn1 or yn  yn1 . Since T is injective, then for any n  N , by (4) 0  max{S (Txn , Txn , Txn1 ), S (Tyn , Tyn , Tyn1 )} Using (1) and (3), we get S (Txn , Txn , Txn1 )  S (TF ( xn1, yn1 ), TF ( xn1, yn1 ), TF ( xn , yn )} .    max{S (Txn1, Txn1, Txn ), S (Tyn1, Tyn1, Tyn )} 

(1)

(2) (3)

(4) (5)

(6)

Also

S (Tyn , Tyn , Tyn 1 )  S (Tyn 1 , Tyn 1 , Tyn )  S (TF ( yn , xn ), TF ( yn , xn ), TF ( yn 1 , xn 1 )}

.

  max{S (Tyn , Tyn , Tyn 1 ), S (Txn , Txn , Txn 1 )} Also, we are given that  (t )  t for all t  0 , so from (6) and (7), we set 0   max{S (Txn , Txn , Txn 1 ), S (Tyn , Tyn , Tyn 1 )}     max{S (Txn 1 , Txn 1 , Txn ), S (Tyn 1 , Tyn 1 , Tyn )}    max{S (Txn 1 , Txn 1 , Txn ), S (Tyn 1 , Tyn 1 , Tyn )} 

(8)

This shows that

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Coupled Fixed Point Theorems in S-metric Spaces max{S (Txn , Txn , Txn1 ), S (Tyn , Tyn , Tyn1 )}  max{S (Txn1, Txn1, Txn ), S (Tyn1, Tyn1, Tyn )} Thus  max{S (Txn , Txn , Txn1 ), S (Tyn , yn , yn1 )}  is a positive decreasing sequence, so there exist r  0 such that lim max{S (Txn , Txn , Txn 1 ), S (Tyn , yn , yn 1 )}  r n 

Assume that r  0 . Letting n   in (8), we get 0  r  lim   max{S (Txn 1 , Txn 1 , Txn ), S (Tyn 1 , Tyn 1 , Tyn )}  n 

 lim  (t )  r

(9)

t r

which is a contradiction, so we deduce that lim max{S (Txn , Txn , Txn 1 ), S (Tyn , Tyn , Tyn 1 )}  0

(10)

n 

Now, we shall prove that {Txn } and {Tyn } are Cauchy sequences. Suppose, on the contrary, {Txn } or {Tyn } is not a Cauchy sequences, that is, lim S (Txm , Txm , Txn )  0 n,m

or lim S (Tym , Tym , Tyn )  0 n,m

This means that there exist   0 for which we can find subsequences of integers (mk ) and (nk ) with

nk  mk  k such that max{S (Txmk , Txmk , Txnk ), S (Tymk , Tymk , Tynk )}  

(11)

Again correspoidng to mk we can find nk in such a way that it is the smallest integers with nk  mk and satisfying (11). Then max{S (Txmk , Txmk , Txnk 1 ), S (Tymk , Tymk , Tynk 1 )}   with the help of triangular inequality and (12), we get S (Txmk , Txmk , Txnk )  S (Txmk , Txmk , Txmk 1 )  S (Txmk , Txmk , Txmk 1 )  S (Txnk , Txnk , Txmk 1 )

 00

(12)

.

Thus by (10), we get lim S (Txmk , Txmk , Txnk )  

(13)

Similarly, we get lim S (Tymk , Tymk , Tynk )  

(14)

k 

k 

Also, S (Txmk 1 , Txmk 1 , Txnk 1 )  S (Txmk 1 , Txmk 1 , Txmk  2 )  S (Txmk 1 ,Txmk 1 ,Txmk  2 )  S (Txnk 1 ,Txnk 1 ,Txmk  2 )  2S (Txmk 1 , Txmk 1 , Txmk  2 )  

Using (10), we get lim S (Txmk 1 , Txmk 1 , Txnk  2 )  

(15)

Similarly, we get lim S (Tymk 1 , Tymk 1 , Tymk  2 )  

(16)

k 

k 

Using (11) and (13)-(16), we get lim max{S (Txmk , Txmk , Txnk ), S (Tymk , Tymk , Tynk )} n 

 lim max{S (Txmk 1 , Txmk 1 , Txnk 1 ), S (Tymk 1 , Tymk 1 , Tynk 1 )} n 

 Now, by inequality (1), we get S (Txmk , Txmk , Txnk )  S (TF ( xmk 1 , ymk 1 ), TF ( xmk 1 , ymk 1 ), TF ( xnk 1 , ynk 1 ))  [max{S (Txmk 1 , Txmk 1 , Txnk 1 ), S (Tymk 1 , Tymk 1 , Tynk 1 )}]

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(17)

(18)

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Coupled Fixed Point Theorems in S-metric Spaces and S (Tymk , Tymk , Tynk )  S (TF ( ymk 1 , xmk 1 ), TF ( ymk 1 , xmk 1 ), TF ( ynk 1 , xnk 1 ))  [max{S (Tymk 1 , Tymk 1 , Tynk 1 ), S (Txmk 1 , Txmk 1 , Txnk 1 )}]

Now, by (18) and (19), we obtain that max{S (Txmk , Txmk , Txnk ), S (Tymk , Tymk , Tynk )}  [max{S (Txmk 1 , Txmk 1 , Txnk 1 ), S (Tymk 1 , Tymk 1 , Tynk 1 )}]

(19)

(20)

Assuming k   , in (20), and using (17), we obtain 0    lim  (t )   t 

which is a contradiction. Hence {Txn } and {Tyn } are Cauchy sequences in ( X , S ) . Since X is complete Smetric space, {Txn } and {Tyn } are convergent sequences. Since T is an ICS mapping, there exist x, y  X such that lim xn  x and lim yn  y

n 

(21)

n 

Sicne T is continuous, we get lim Txn  Tx and lim Tyn  Ty n 

(22)

n 

Now, assume that the assumption (a) holds, that is F is continuous. By (3), (21) and (22) we have x  lim xn 1 n 

 lim F ( xn , yn ) n 

 F  lim xn , lim yn   n n   F ( x, y ) and y  lim yn 1 n 

 lim F ( yn , xn ) n 

 F  lim yn , lim xn   n  n   F ( y, x) Thus we have show that F has a coupled fixed point. Suppose, now the assumption (b) holds. Since {xn } is non decreasing with xn  x and also { yn } is non increasing with yn  y . Then by assumption (b), we get

xn  x and yn  y for all n . Now,

S (Tx, Tx, TF ( x, y))  S (Tx, Tx, Txn 1 )  S (Tx, Tx, Txn 1 )  S (TF ( x, y), TF ( x, y), Txn 1 )  2S (Tx, Tx, Txn 1 )  S (TF ( x, y), TF ( x, y), TF ( xn , yn ))  2S (Tx, Tx, Txn 1 )   (max{S (Tx, Tx, Txn ), S (Ty, Ty, Tyn )}) taking lim and using (22), right hand side of this inequality is equal to zero. n 

Hence S (Tx, Tx, TF ( x, y))  0 . So Tx  TF ( x, y) and sine T is injective so x  F ( x, y) . Similarly, we can show that y  F ( y, x) . Thus, we have shown F has a coupled fixed point in X . Corollary 1. Let ( X , ) be a partially ordered set and suppose those is a metric S on X such that ( X , S ) is a complete S-metric space. Suppose T : X  X is an ICS mapping and F : X  X  X is such that F has the mixed monotone property. Assume that there exists    such that

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Coupled Fixed Point Theorems in S-metric Spaces S (Tx, Tx, Tu)  S (Ty, Ty, Tv)  S (TF ( x, y), TF ( x, y), TF (u, v))     2   for any x, y, u, v  X for which x  u and v  y . Suppose either (a) F is continuous or (b) X has the following properties: (i) if non-decreasing sequence xn  x , then xn  x for all n . (ii) if non-increasing sequence yn  y , then y  yn for all n . If there exist x0 , y0  X such that

x0  F ( x0 , y0 ) and F ( y0 , x0 )  y0 then there exist x, y  X such that

F ( x, y)  x and F ( y, x)  y , that is F has a coupled fixed point. Proof. If can be easily shown as S (Tx, Tx, Tu )  S (Ty, Ty, Tv)  max{S (Tx, Tx, Tu ), S (Ty, Ty, Tv)} . 2 Then apply Theorem 1, because that  is non-decreasing. Corollary 2. Let ( X , ) be a partially ordered set and suppose that there is a metric S on X such that ( X , S ) is a complete S-metric space. Suppose T : X  X is an ICS mapping and F : X  X  X is such that F has the mixed monotone property. Assume that there exist k  [0,1) such that

S (TF ( x, y), TF ( x, y), TF (u, v))  k max{S (Tx, Tx, Tu), S (Ty, Ty, Tv)} for any x, y, u, v  X for which x  u and v  y . Suppose either (a) F is continuous or (b) X has the following properties: (i) if non-decreasing sequence xn  x , then xn  x for all n . (ii) if non-decreasing sequence yn  y , then y  yn for all n . If there exist x0 , y0  X such that

x0  F ( x0 , y0 ) and F ( y0 , x0 )  y0 then there exist x, y  X such that

F ( x, y)  x and F ( y, x)  y , that is F has a coupled fixed point. Proof. It follows by  (t )  kt in Theorem 1. Corollary 3. Let ( X , ) be a partially ordered set and suppose that there is a metric S on X such that ( X , S ) is a complete S-metric on X . Suppose T : X  X is an ICS mapping and F : X  X  X is such that F has the mixed monotone property. Assume that there exist k  [0,1) such that k ( S (Tx, Tx, Tu ), S (Ty , Ty , Tv )) 3 for any x, y, u, v  X for which x  u and v  y . Suppose either S (TF ( x, y ), TF ( x, y ), TF (u , v )) 

(a) F is continuous or (b) X has the following properties:

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Coupled Fixed Point Theorems in S-metric Spaces (i) if non-decreasing sequence xn  x , then xn  x for all n . (ii) if non-increasing sequence yn  y , then y  yn for all n . If there exist x0 , y0  X such that

x0  F ( x0 , y0 ) and F ( y0 , x0 )  y0 then there exist x, y  X such that

F ( x, y)  x and F ( y, x)  y , that is F has a coupled fixed point. Proof. It follows by  (t )  kt in Theorem 1. Theorem 2. In addition to the condition of Theorem 1, suppose that for all ( x, y),(u, v)  X  X , there exist

(a, b)  X  X such that ( F (a, b), F (b, a)) is comparable to ( F ( x, y), F ( y, x)) and ( F (u, v), F (v, u)) then F has a unique coupled fixed point ( x, y) . Proof. We know that the set of coupled fixed poit of F is not empty by Theorem 1. Suppose, now ( x, y) and

(u, v) are two coupled fixed point of F , that is F ( x, y)  x ; F ( y, x)  y and F (u, v)  u , F (v, u)  v . We shall prove that ( x, y) and (u, v) are equal. By our supositon there exist (a, b)  X  X such that ( F (a, b), F (b, a)) is comparable to ( F ( x, y), F ( y, x)) and

( F (u, v), F (v, u)) . Now construct sequences {an } and {bn } such that a0  a , b0  b and for any n  1 .

an  F (an1 , bn1 ) , bn  F (bn1 , an 1 ) for all n

(23)

Again set x0  x , y0  y and u0  u , v0  v and on the same way construct the sequence {xn } , { yn } , {un } and {vn } . Then

xn  F ( x, y), yn  F ( y, x), un  F (u, v), vn  F (v, u) for all n  1

(24)

Since ( F ( x, y), F ( y, x))  ( x1, y1 )  ( x, y) is comparable to ( F (a, b), F (b, a))  (a1, b1 ) then it is easy to show

( x, y)  (a1 , b1 ) . Similarly, we have that ( x, y)  (an , bn ) for all n

(25)

From (25) and (1), we get S (Tx, Tx, Tan1 )  S (TF ( x, y), TF ( x, y), TF (an , bn ))

  (max{S (Tx, Tx, Tan ), S (Ty, Ty, Tbn )})

(26)

and S (Ty, Ty, Tbn 1 )  S (Tbn 1 , Tbn 1 , Ty)

 S (TF (bn , an ), TF (bn , an ), TF ( y, x))   (max{S (Tbn , Tbn , Ty), S (Tan , Tan , Tx)})

(27)

It follows, from (26) and (27), that {(Tx, Tx, Tan1 ), S (Ty, Ty, Tbn1 )}  [max{S (Tx, Tx, Tan ), S (Ty, Ty, Tbn )}] So for all n  1 max{S (Tx, Tx, Tan ), S (Ty, Ty, Tbn )}   n max{S (Tx, Tx, Ta0 ), S (Ty, Ty, Tb0 )}

(28)

But we know that  (t )  t and lim  (r )  t implies r t

lim  (t )  0 for all t  0 . n

n

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Coupled Fixed Point Theorems in S-metric Spaces Hence from (28), we get lim max{S (Tx, Tx, Tan ), S (Ty , Ty , Tbn )}  0 n 

This gives lim{S (Tx, Tx, Tan )}  0

(29)

n 

and lim{S (Ty, Ty, Tbn )}  0

(30)

Similarly we can show that lim{S (Tu , Tu , Tan )}  0

(31)

n 

n 

and lim{S (Tv, Tv, Tbn )}  0

(32)

n 

Adding (29)-(32), we get (Tx, Ty) and (Tu, Tv) are equal. The fact that T is injective gives us x  u and

y  v.

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