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Common fixed point results for contractive mappings in complex valued metric spaces

Abstract

In this paper, we obtain some common fixed point results for the mappings satisfying rational expressions on a closed ball in complex valued metric spaces. Our results improve several well-known conventional results.

MSC:47H10, 54H25.

1 Introduction and preliminaries

Azam et al. [1] introduced new spaces called complex valued metric spaces and established the existence of fixed point theorems under the contraction condition. Subsequently, Rouzkard and Imdad [2] established some common fixed point theorems satisfying certain rational expressions in complex valued metric spaces which generalize, unify and complement the results of Azam et al. [1]. Sintunavarat and Kumam [3] obtained common fixed point results by replacing constant of contractive condition to control functions. Recently, Klin-eam and Suanoom [4] extend the concept of complex valued metric spaces and generalized the results of Azam et al. [1] and Rouzkard and Imdad [2]. For more on fixed point theory we refer the reader to [426].

The aim of this article is to extend and improve the conditions of contraction from the whole space to closed ball and establish the common fixed point theorems which are more general than the results of Klin-eam and Suanoom [4], Rouzkard and Imdad [2], and Azam et al. [1] on complex valued metric spaces.

Let be the set of complex numbers and z 1 , z 2 C. Define a partial order on as follows:

z 1 z 2 if and only ifRe( z 1 )Re( z 2 )andIm( z 1 )Im( z 2 ).

It follows that z 1 z 2 if and only if one of the following conditions is satisfied:

  1. (i)

    Re( z 1 )=Re( z 2 ) and Im( z 1 )<Im( z 2 ),

  2. (ii)

    Re( z 1 )<Re( z 2 ) and Im( z 1 )=Im( z 2 ),

  3. (iii)

    Re( z 1 )<Re( z 2 ) and Im( z 1 )<Im( z 2 ),

  4. (iv)

    Re( z 1 )=Re( z 2 ) and Im( z 1 )=Im( z 2 ).

In particular, we will write z 1 z 2 if z 1 z 2 and one of (i), (ii), and (iii) is satisfied and we will write z 1 z 2 if only (iii) is satisfied. Note that

0 z 1 z 2 | z 1 | < | z 2 | , z 1 z 2 , z 2 z 3 z 1 z 3 .

Definition 1 Let X be a nonempty set. Suppose that the mapping d:X×XC satisfies:

  1. (1)

    0d(x,y) for all x,yX; and d(x,y)=0 if and only if x=y;

  2. (2)

    d(x,y)=d(y,x) for all x,yX;

  3. (3)

    d(x,y)d(x,z)+d(z,y) for all x,y,zX.

Then d is called a complex valued metric on X and (X,d) is called a complex valued metric space.

A point xX is called an interior point of a set AX whenever there exists 0rC such that

B(x,r):= { y X : d ( x , y ) r } A,

where B(x,r) is an open ball. Then B ( x , r ) ¯ ={yX:d(x,y)r} is a closed ball.

A point xX is called a limit point of A whenever for every 0rC, we have

B(x,r) ( A { x } ) .

A subset AX is called open whenever each element of A is an interior point of A. A subset BX is called closed whenever each limit point of B belongs to B. The family

F:= { B ( x , r ) : x X , 0 r }

is a sub-basis for a Hausdorff topology τ on X.

Let { x n } be a sequence in X and xX. If for every cC with 0c there is n 0 N such that d( x n ,x)c, for all n> n 0 , then { x n } is said to be convergent and { x n } converges to x. We denote this by lim n x n =x or x n x. If for every cC with 0c there is n 0 N such that d( x n , x m )c, for all n,m> n 0 , then { x n } is called a Cauchy sequence. If every Cauchy sequence is convergent in (X,d), then (X,d) is called a complete complex valued metric space.

Example 2 Let X= X 1 X 2 where

X 1 = { z C : Re ( z ) 0  and  Im ( z ) = 0 }

and

X 2 = { z C : Re ( z ) = 0  and  Im ( z ) 0 } .

Define d:X×XC as follows:

d( z 1 , z 2 )={ 2 3 | x 1 x 2 | + i 2 | x 1 x 2 | if  z 1 , z 2 X 1 ; 1 2 | y 1 y 2 | + i 3 | y 1 y 2 | if  z 1 , z 2 X 2 ; ( 2 3 x 1 + 1 2 y 2 ) + i ( 1 2 x 1 + 1 3 y 2 ) if  z 1 X 1 , z 2 X 2 ; ( 1 2 y 1 + 2 3 x 2 ) + i ( 1 3 y 1 + 1 2 x 2 ) if  z 1 X 2 , z 2 X 1 ,

where z 1 = x 1 +i y 1 , z 2 = x 2 +i y 2 X. Then (X,d) is a complete complex valued metric space.

Lemma 3 [1]

Let (X,d) be a complex valued metric space and let { x n } be a sequence in X. Then { x n } converges to x if and only if |d( x n ,x)|0 as n.

Lemma 4 [1]

Let (X,d) be a complex valued metric space and let { x n } be a sequence in X. Then { x n } is a Cauchy sequence if and only if |d( x n , x n + m )|0 as n.

Definition 5 [27]

Two families of self-mappings { T i } 1 m and { S i } 1 n are said to be pairwise commuting if:

  1. (1)

    T i T j = T j T i for all i,j{1,2,,m};

  2. (2)

    S k S l = S l S k for all k,l{1,2,,n};

  3. (3)

    T i S k = S k T i for all i{1,2,,m}, k{1,2,,n}.

2 Main result

In our main result, we discuss the existence of the common fixed point of the mappings satisfying a contractive condition on the closed ball. This result is very useful in the sense that it requires the contractiveness of the mappings only on a closed ball instead of the whole space.

Theorem 6 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, D and E be five nonnegative reals such that A+B+C+2D+2E<1. Let S,T:XX satisfy

d ( S x , T y ) A d ( x , y ) + B d ( x , S x ) d ( y , T y ) 1 + d ( x , y ) + C d ( y , S x ) d ( x , T y ) 1 + d ( x , y ) + D d ( x , S x ) d ( x , T y ) 1 + d ( x , y ) + E d ( y , S x ) d ( y , T y ) 1 + d ( x , y )
(1)

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,S x 0 )|(1λ)|r|,
(2)

where λ=max{ A + D 1 B D , A + E 1 B E }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Su=Tu.

Proof Let x 0 be an arbitrary point in X and define

x 2 k + 1 =S x 2 k and x 2 k + 2 =T x 2 k + 1 ,where k=0,1,2,.

We will prove that x n B ( x 0 , r ) ¯ for all nN by mathematical induction.

Using inequality (2) and the fact that λ=max{ A + D 1 B D , A + E 1 B E }<1, we have

|d( x 0 ,S x 0 )||r|.

It implies that x 1 B ( x 0 , r ) ¯ . Let x 2 ,, x j B ( x 0 , r ) ¯ for some jN. If j=2k+1, where k=0,1,2,, j 1 2 or j=2k+2 where k=0,1,2,, j 2 2 , we obtain by using inequality (1)

d ( x 2 k + 1 , x 2 k + 2 ) = d ( S x 2 k , T x 2 k + 1 ) A d ( x 2 k , x 2 k + 1 ) + B d ( x 2 k + 1 , T x 2 k + 1 ) d ( x 2 k , S x 2 k ) 1 + d ( x 2 k , x 2 k + 1 ) + C d ( x 2 k , T x 2 k + 1 ) d ( x 2 k + 1 , S x 2 k ) 1 + d ( x 2 k , x 2 k + 1 ) + D d ( x 2 k , T x 2 k + 1 ) d ( x 2 k , S x 2 k ) 1 + d ( x 2 k , x 2 k + 1 ) + E d ( x 2 k + 1 , T x 2 k + 1 ) d ( x 2 k + 1 , S x 2 k ) 1 + d ( x 2 k , x 2 k + 1 ) .

Now x 2 k + 1 =S x 2 k implies that d( x 2 k + 1 ,S x 2 k )=0, so we have

d ( x 2 k + 1 , x 2 k + 2 ) A d ( x 2 k , x 2 k + 1 ) + B d ( x 2 k + 1 , x 2 k + 2 ) d ( x 2 k , x 2 k + 1 ) 1 + d ( x 2 k , x 2 k + 1 ) + D d ( x 2 k , x 2 k + 2 ) d ( x 2 k , x 2 k + 1 ) 1 + d ( x 2 k , x 2 k + 1 ) .

This implies that

| d ( x 2 k + 1 , x 2 k + 2 ) | A | d ( x 2 k , x 2 k + 1 ) | + B | d ( x 2 k + 1 , x 2 k + 2 ) | | d ( x 2 k , x 2 k + 1 ) | | 1 + d ( x 2 k , x 2 k + 1 ) | + D | d ( x 2 k , x 2 k + 2 ) | | d ( x 2 k , x 2 k + 1 ) | | 1 + d ( x 2 k , x 2 k + 1 ) | .

Since |1+d( x 2 k , x 2 k + 1 )|>|d( x 2 k , x 2 k + 1 )|, we have

|d( x 2 k + 1 , x 2 k + 2 )|A|d( x 2 k , x 2 k + 1 )|+B|d( x 2 k + 1 , x 2 k + 2 )|+D|d( x 2 k , x 2 k + 2 )|.

This implies by the triangular inequality that

|d( x 2 k + 1 , x 2 k + 2 )| A + D 1 B D |d( x 2 k , x 2 k + 1 )|.
(3)

Similarly, we get

|d( x 2 k + 2 , x 2 k + 3 )| A + E 1 B E |d( x 2 k + 2 , x 2 k + 1 )|.
(4)

Putting λ=max{ A + D 1 B D , A + E 1 B E }, we obtain

|d( x j , x j + 1 )| λ j |d( x 0 , x 1 )|for all jN.
(5)

Now

| d ( x 0 , x j + 1 ) | | d ( x 0 , x 1 ) | + + | d ( x j , x j + 1 ) | | d ( x 0 , x 1 ) | + + λ j | d ( x 0 , x 1 ) | = | d ( x 0 , x 1 ) | [ 1 + + λ j 1 + λ j ] ( 1 λ ) | r | ( 1 λ j + 1 ) 1 λ | r |

gives x j + 1 B ( x 0 , r ) ¯ . Hence x n B ( x 0 , r ) ¯ for all nN and

|d( x n , x n + 1 )| λ n |d( x 0 , x 1 )|

for all nN. Without loss of generality, we take m>n, then

| d ( x n , x m ) | | d ( x n , x n + 1 ) | + | d ( x n + 1 , x n + 2 ) | + + | d ( x m 1 , x m ) | [ λ n + λ n + 1 + + λ m 1 ] | d ( x 0 , x 1 ) | [ λ n 1 λ ] | d ( x 0 , x 1 ) | 0 as  m , n .

This implies that the sequence { x n } is a Cauchy sequence in B ( x 0 , r ) ¯ . Therefore, there exists a point u B ( x 0 , r ) ¯ with lim n x n =u.

We prove that u=Su. Let us consider

| d ( u , S u ) | | d ( u , x 2 k + 2 ) | + A | d ( x 2 k + 1 , u ) | + B | d ( x 2 k + 1 , T x 2 k + 1 ) | | d ( u , S u ) | | 1 + d ( u , x 2 k + 1 ) | + C | d ( x 2 k + 1 , S u ) | | d ( u , T x 2 k + 1 ) | | 1 + d ( u , x 2 k + 1 ) | + D | d ( u , T x 2 k + 1 ) | | d ( u , S u ) | | 1 + d ( u , x 2 k + 1 ) | + E | d ( x 2 k + 1 , T x 2 k + 1 ) | | d ( x 2 k + 1 , S u ) | | 1 + d ( u , x 2 k + 1 ) | .

Notice that lim n |d(u, x 2 k + 2 )|= lim n |d( x 2 k + 1 ,u)|=|d( x 2 k + 1 ,Su)|=0. Hence |d(u,Su)|=0, that is, u=Su. It follows similarly that u=Tu. For uniqueness, assume that u in B ( x 0 , r ) ¯ is a second common fixed point of S and T. Then

| d ( u , u ) | A | d ( u , u ) | + B | d ( u , S u ) | | d ( u , T u ) | | 1 + d ( u , u ) | + C | d ( u , S u ) | | d ( u , T u ) | | 1 + d ( u , u ) | + D | d ( u , S u ) | | d ( u , T u ) | | 1 + d ( u , u ) | + E | d ( u , S u ) | | d ( u , T u ) | | 1 + d ( u , u ) | .

Since |1+d(u, u )|>|d(u, u )|, so we have

|d ( u , u ) |(A+C)|d ( u , u ) |.

This is contradiction because A+C<1. Hence u =u. Therefore, u is a unique common fixed point of S and T. □

By setting S=T in Theorem 6, we get the following corollary.

Corollary 7 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, D and E be five nonnegative reals such that A+B+C+2D+2E<1. Let T:XX satisfy

d ( T x , T y ) A d ( x , y ) + B d ( x , T x ) d ( y , T y ) 1 + d ( x , y ) + C d ( y , T x ) d ( x , T y ) 1 + d ( x , y ) + D d ( x , T x ) d ( x , T y ) 1 + d ( x , y ) + E d ( y , T x ) d ( y , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,T x 0 )|(1λ)|r|,

where λ=max{ A + D 1 B D , A + E 1 B E }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Tu.

Remark 8 The conclusion of Theorem 6 remains true if the condition (2) is replaced by the condition |d( x 0 ,T x 0 )|(1λ)|r|.

By choosing E=0 in Theorem 6, we get the following corollary.

Corollary 9 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, D be four nonnegative reals such that A+B+C+2D<1. Let S,T:XX satisfy

d(Sx,Ty)Ad(x,y)+B d ( x , S x ) d ( y , T y ) 1 + d ( x , y ) +C d ( y , S x ) d ( x , T y ) 1 + d ( x , y ) +D d ( x , S x ) d ( x , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,S x 0 )|(1λ)|r|,

where λ=max{ A + D 1 B D , A 1 B }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Su=Tu.

By setting S=T in Corollary 9, we get the following corollary.

Corollary 10 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, D be four nonnegative reals such that A+B+C+2D<1. Let T:XX satisfy

d(Tx,Ty)Ad(x,y)+B d ( x , T x ) d ( y , T y ) 1 + d ( x , y ) +C d ( y , T x ) d ( x , T y ) 1 + d ( x , y ) +D d ( x , T x ) d ( x , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,T x 0 )|(1λ)|r|,

where λ=max{ A + D 1 B D , A 1 B }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Tu.

By choosing D=0 in Theorem 6, we get the following corollary.

Corollary 11 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C and E be five nonnegative reals such that A+B+C+2E<1. Let S,T:XX satisfy

d(Sx,Ty)Ad(x,y)+B d ( x , S x ) d ( y , T y ) 1 + d ( x , y ) +C d ( y , S x ) d ( x , T y ) 1 + d ( x , y ) +E d ( y , S x ) d ( y , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,S x 0 )|(1λ)|r|,

where λ=max{ A 1 B , A + E 1 B E }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Su=Tu.

By setting S=T in Corollary 11, we get the following corollary.

Corollary 12 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, and E be five nonnegative reals such that A+B+C+2E<1. Let T:XX satisfy

d(Tx,Ty)Ad(x,y)+B d ( x , T x ) d ( y , T y ) 1 + d ( x , y ) +C d ( y , T x ) d ( x , T y ) 1 + d ( x , y ) +E d ( y , T x ) d ( y , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,S x 0 )|(1λ)|r|,

where λ=max{ A 1 B , A + E 1 B E }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Tu.

Remark 13 By equating A, B, C, D, and E to 0 in all possible combinations, one can derive a host of corollaries which include the Banach fixed point theorem for self-mappings on the closed ball in complex valued metric spaces.

By choosing D=E=0 in Theorem 6, we get the extension of Theorem 2.1 of [2] to the closed ball as follows.

Corollary 14 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C be three nonnegative reals such that A+B+C<1. Let S,T:XX satisfy

d(Sx,Ty)Ad(x,y)+B d ( x , S x ) d ( y , T y ) 1 + d ( x , y ) +C d ( y , S x ) d ( x , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,S x 0 )|(1λ)|r|,

where λ= A 1 B , then there exists a unique point u B ( x 0 , r ) ¯ such that u=Su=Tu.

By setting S=T in Corollary 14, we get Corollary 2.3 of [16] on the closed ball as follows.

Corollary 15 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C be three nonnegative reals such that A+B+C<1. Let T:XX satisfy

d(Tx,Ty)Ad(x,y)+B d ( x , T x ) d ( y , T y ) 1 + d ( x , y ) +C d ( y , T x ) d ( x , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,T x 0 )|(1λ)|r|,

where λ= A 1 B , then there exists a unique point u B ( x 0 , r ) ¯ such that u=Tu.

By choosing C=D=E=0 in Theorem 6, we get the extension of Theorem 4 of [1] to the closed ball as follows.

Corollary 16 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B be nonnegative reals such that A+B<1. Let S,T:XX satisfy

d(Sx,Ty)Ad(x,y)+B d ( x , S x ) d ( y , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,S x 0 )|(1λ)|r|,

where λ= A 1 B , then there exists a unique point u B ( x 0 , r ) ¯ such that u=Su=Tu.

By setting S=T in Corollary 16, we get Corollary 2.3 of [1] on the closed ball as follows.

Corollary 17 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B be nonnegative reals such that A+B<1. Let T:XX satisfy

d(Tx,Ty)Ad(x,y)+B d ( x , T x ) d ( y , T y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ . If

|d( x 0 ,T x 0 )|(1λ)|r|,

where λ= A 1 B , then there exists a unique point u B ( x 0 , r ) ¯ such that u=Tu.

As an application of Theorem 6, we prove the following theorem for two finite families of mappings.

Theorem 18 If { T i } 1 m and { S i } 1 n are two finite pairwise commuting finite families of self-mapping defined on a complete complex valued metric space (X,d) such that the mappings S and T (with T= T 1 T 2 T m and S= S 1 S 2 S n ) satisfy the contractive conditions (1) and (2), then the component maps of the two families { T i } 1 m and { S i } 1 n have a unique common fixed point.

Proof From Theorem 6, we can say that the mappings T and S have a unique common fixed point u i.e. Tu=Su=u. Now our requirement is to show that u is a common fixed point of all the component mappings of both families. In view of pairwise commutativity of the families { T i } 1 m and { S i } 1 n (for every 1km), we can write T k u= T k Tu=T T k u and T k u= T k Su=S T k u which show that T k u (for every k) is also a common fixed point of T and S. By using the uniqueness of common fixed point, we can write T k u=u (for every k) which shows that u is a common fixed point of the family { T i } 1 m . Using the same argument one can also show that (for every 1kn) S k u=u. Thus the component maps of the two families { T i } 1 m and { S i } 1 n have a unique common fixed point. □

By setting T 1 = T 2 == T m =F and S 1 = S 2 == S n =G in Theorem 18, we get the following corollary.

Corollary 19 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, D and E be five nonnegative reals such that A+B+C+2D+2E<1. Let F,G:XX satisfy

d ( F m x , G n y ) A d ( x , y ) + B d ( x , F m x ) d ( y , G n y ) 1 + d ( x , y ) + C d ( y , F m x ) d ( x , G n y ) 1 + d ( x , y ) + D d ( x , F m x ) d ( x , G n y ) 1 + d ( x , y ) + E d ( y , F m x ) d ( y , G n y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ and

|d ( x 0 , G n x 0 ) |(1λ)|r|,

where λ=max{ A + D 1 B D , A + E 1 B E }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Fu=Gu.

By setting m=n and F=G=T in Corollary 19, we get the following corollary.

Corollary 20 Suppose that (X,d) is a complete complex valued metric space and x 0 X. Let 0rC and A, B, C, D and E be five nonnegative reals such that A+B+C+2D+2E<1. Let T:XX satisfy

d ( T n x , T n y ) A d ( x , y ) + B d ( x , T n x ) d ( y , T n y ) 1 + d ( x , y ) + C d ( y , T n x ) d ( x , T n y ) 1 + d ( x , y ) d ( T n x , T n y ) + D d ( x , T n x ) d ( x , T n y ) 1 + d ( x , y ) + E d ( y , T n x ) d ( y , T n y ) 1 + d ( x , y )

for all x,y B ( x 0 , r ) ¯ and

|d ( x 0 , T n x 0 ) |(1λ)|r|,

where λ=max{ A + D 1 B D , A + E 1 B E }, then there exists a unique point u B ( x 0 , r ) ¯ such that u=Tu.

Now we give an example satisfying our main result.

Example 21 Let X 1 ={zC:Re(z)0 and Im(z)=0} and X 2 ={zC:Re(z)=0 and Im(z)0} and let X= X 1 X 2 . Consider a metric d:X×XC as follows:

d( z 1 , z 2 )={ 2 3 | x 1 x 2 | + i 2 | x 1 x 2 | if  z 1 , z 2 X 1 ; 1 2 | y 1 y 2 | + i 3 | y 1 y 2 | if  z 1 , z 2 X 2 ; 2 9 ( x 1 + y 2 ) + i 6 ( x 1 + y 2 ) if  z 1 X 1 , z 2 X 2 ; i 3 ( x 2 + y 1 ) + 2 i 9 ( x 2 + y 1 ) if  z 1 X 2 , z 2 X 1 ,

where z 1 = x 1 +i y 1 , z 2 = x 2 +i y 2 X. Then (X,d) is a complex valued metric space. Take z 0 = 1 2 +0i and r= 1 3 + 1 4 i. Then

B ( z 0 , r ) ¯ = { z X 1 : 0 Re ( z ) 1 } { z X 2 : 0 Im ( z ) 1 } .

Define S,T:XX by

S z = { 0 + x 4 i if  z X 1  with  0 Re ( z ) 1 , Im ( z ) = 0 ; 5 x 6 + 0 i if  z X 1  with  Re ( z ) > 1 , Im ( z ) = 0 ; y 5 + 0 i if  z X 2  with  0 Im ( z ) 1 , Re ( z ) = 0 ; 0 + 4 y 5 i if  z X 2  with  Im ( z ) > 1 , Re ( z ) = 0 ; T z = { 0 + x 6 i if  z X 1  with  0 Re ( z ) 1 , Im ( z ) = 0 ; 4 x 5 + 0 i if  z X 1  with  Re ( z ) > 1 , Im ( z ) = 0 ; y 7 + 0 i if  z X 2  with  0 Im ( z ) 1 , Re ( z ) = 0 ; 0 + 5 y 6 i if  z X 2  with  Im ( z ) > 1 , Re ( z ) = 0 .

By a routine calculation, one can verify that the mappings S and T satisfy the conditions (1) and (2) of Theorem 6 with A= 1 6 , B= 1 24 , C= 1 2 , D= 1 25 and E= 1 26 . Hence S and T are contractions on B ( z 0 , r ) ¯ and 0+0i B ( z 0 , r ) ¯ is a unique common fixed point of mappings S and T.

It is interesting to notice that S and T are not contractions on the whole space X for z 1 = z 2 = 3 2 +0i B ( z 0 , r ) ¯ as

d ( S z 1 , T z 2 ) = 1 30 + 1 40 i 33 , 859 3 , 744 , 000 + 4 , 837 156 , 000 i = A d ( z 1 , z 2 ) + B d ( z 1 , S z 1 ) d ( z 2 , T z 2 ) 1 + d ( z 1 , z 2 ) + C d ( z 2 , S z 1 ) d ( z 1 , T z 2 ) 1 + d ( z 1 , z 2 ) + D d ( z 1 , S z 1 ) ( z 1 , T z 2 ) 1 + d ( z 1 , z 2 ) + E d ( z 2 , S z 1 ) d ( z 2 , T z 2 ) 1 + d ( z 1 , z 2 ) .

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Acknowledgements

The authors thank the editor and the referees for their valuable comments and suggestions, which improved greatly the quality of this paper. The research of the third author is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.

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Correspondence to Satit Saejung.

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All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Ahmad, J., Azam, A. & Saejung, S. Common fixed point results for contractive mappings in complex valued metric spaces. Fixed Point Theory Appl 2014, 67 (2014). https://doi.org/10.1186/1687-1812-2014-67

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Keywords

  • common fixed point
  • contractive type mapping
  • complex valued metric space