1 Introduction

The study of metric spaces has played a vital role in many branches of pure and applied sciences. We can find useful applications of metric spaces in mathematics, biology, medicine, physics and computer science (see [13]). Several mathematicians improved, generalized and extended the concept of metric spaces to vector-valued metric spaces of Perov [4], G-metric spaces of Mustafa and Sims [5], cone metric spaces of Huang and Zhang [6], modular metric spaces of Chistyakov [7], partial metric spaces of Matthews [8] and others. Since Banach [9] introduced his contraction principle in complete metric spaces in 1922, this field of fixed point theory has been rapidly growing. It has been very useful in many fields such as optimization problems, control theory, differential equations, economics and many others. A number of papers in this field have been dedicated to the improvement and generalization of Banach’s contraction mapping principle in many spaces and ways (see [1013]).

Recently, Azam et al. [14] introduced a new space, the so-called complex-valued metric space, and established a fixed point theorem for some type of contraction mappings as follows.

Theorem 1.1 (Azam et al. [14])

Let (X,d) be a complete complex-valued metric space and S,T:XX be two mappings. If S and T satisfy

d(Sx,Ty)λd(x,y)+ μ d ( x , S x ) d ( y , T y ) 1 + d ( x , y )
(1.1)

for all x,yX, where λ, μ are nonnegative reals with λ+μ<1, then S and T have a unique common fixed point in X.

Theorem 1.1 of Azam et al. in [14] is an essential tool in the complex-valued metric space to claim the existence of a common fixed point for some mappings. However, it is most interesting to find another new auxiliary tool to claim the existence of a common fixed point. Some other works related to the results in a complex-valued metric space are [15, 16].

In this paper, we introduce the concept of a C-Cauchy sequence and C-complete in complex-valued metric spaces and also prove some common fixed point theorems for new generalized contraction mappings in C-complete complex-valued metric spaces.

On the other hand, integral equations arise naturally from many applications in describing numerous real world problems. These equations have been studied by many authors. Existence theorems for the Urysohn integral equations can be obtained applying various fixed point principles.

As applications, we show the existence of a common solution of the following system of Urysohn integral equations by using our common fixed point results:

x(t)= a b K 1 ( t , s , x ( s ) ) ds+g(t),
(1.2)
x(t)= a b K 2 ( t , s , x ( s ) ) ds+h(t),
(1.3)

where t[a,b]R, x,g,hC([a,b], R n ) and K 1 , K 2 :[a,b]×[a,b]× R n R n .

2 Preliminaries

In this section, we discuss some background of the complex-valued metric spaces of Azam et al. in [14] and give some notions for our results. Also, some essential lemmas which are useful for our results are given.

Let ℂ be the set of complex numbers. For z 1 , z 2 C, we will define a partial order ≾ on ℂ as follows:

z 1 z 2 Re( z 1 )Re( z 2 )andIm( z 1 )Im( z 2 ).

We note that z 1 z 2 if one of the following holds:

(C1) Re( z 1 )=Re( z 2 ) and Im( z 1 )=Im( z 2 );

(C2) Re( z 1 )<Re( z 2 ) and Im( z 1 )=Im( z 2 );

(C3) Re( z 1 )=Re( z 2 ) and Im( z 1 )<Im( z 2 );

(C4) Re( z 1 )<Re( z 2 ) and Im( z 1 )<Im( z 2 ).

It obvious that if a,bR such that ab, then azbz for all zC.

In particular, we write z 1 z 2 if z 1 z 2 and one of (C2), (C3) and (C4) is satisfied, and we write z 1 z 2 if only (C4) is satisfied. The following are well known:

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

Definition 2.1 [14]

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

(d1) 0d(x,y) for all x,yX and d(x,y)=0 if and only if x=y;

(d2) d(x,y)=d(y,x) for all x,yX;

(d3) 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.

Example 2.2 Let X=C. Define the mapping d:X×XC by

d( z 1 , z 2 )=| x 1 x 2 |+| y 1 y 2 |i,

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

Example 2.3 Let X= X 1 X 2 , where

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

and

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

Define the mapping d:X×XC by

d( z 1 , z 2 )={ 4 3 | x 1 x 2 | + i | x 1 x 2 | , z 1 , z 2 X 1 , | y 1 y 2 | + 2 i 3 | y 1 y 2 | , z 1 , z 2 X 2 , 4 3 x 1 + y 2 + i ( x 1 + 2 3 y 2 ) , z 1 X 1 , z 2 X 2 , 4 3 x 2 + y 1 + i ( x 2 + 2 3 y 1 ) , z 1 X 2 , z 2 X 1 ,

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

Example 2.4 Let X= X 1 X 2 , where

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

and

X 2 = { z C : Re ( z ) = 0 , 0 Im ( z ) 1 } .

Define the mapping d:X×XC by

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

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

Definition 2.5 [14]

Let (X,d) be a complex-valued metric space.

  1. (1)

    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.
  2. (2)

    A point xX is called a limit point of A whenever, for all 0rC,

    B(x,r)(AX).
  3. (3)

    A set AX is called open whenever each element of A is an interior point of A.

  4. (4)

    A set AX is called closed whenever each limit point of A belongs to A.

  5. (5)

    A sub-basis for a Hausdorff topology τ on X is the family

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

Definition 2.6 [14]

Let (X,d) be a complex-valued metric space, { x n } be a sequence in X and let xX.

  1. (1)

    If, for any cC with 0c, there exists NN such that, for all n>N, d( x n ,x)c, then { x n } is said to be convergent to a point xX or { x n } converges to a point xX and x is the limit point of { x n }. We denote this by lim n x n =x or x n x as n.

  2. (2)

    If, for any cC with 0c, there exists NN such that, for all n>N, d( x n , x n + m )c, where mN, then { x n } is called a Cauchy sequence in X.

  3. (3)

    If every Cauchy sequence in X is convergent, then (X,d) is said to be a complete complex-valued metric space.

Next, we give some lemmas which are an essential tool in the proof of main results.

Lemma 2.7 [[14], see Definition 2.5]

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

Lemma 2.8 [14]

Let (X,d) be a complex-valued metric space and { 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, where mN.

Definition 2.9 Let S and T be self-mappings of a nonempty set X.

  1. (1)

    A point xX is called a fixed point of T if Tx=x.

  2. (2)

    A point xX is called a coincidence point of S and T if Sx=Tx and the point wX such that w=Sx=Tx is called a point of coincidence of S and T.

  3. (3)

    A point xX is called a common fixed point of S and T if x=Sx=Tx.

Lemma 2.10 [17]

Let X be a nonempty set and T:XX be a function. Then there exists a subset EX such that T(E)=T(X) and T:EX is one-to-one.

3 Common fixed points (I)

Throughout this paper, ℝ denotes a set of real numbers, C + denotes a set {cC:0c} and Γ denotes the class of all functions γ: C + [0,1) which satisfies the condition: for any sequences { x n } in C + ,

γ( x n )1| x n |0.

The following are examples of the function in Γ:

  1. (1)

    γ 1 (x)=k, where k[0,1);

  2. (2)

    γ 2 (x)= 1 1 + k | x | , where k(0,).

Now, we introduce the concepts of a C-Cauchy sequence and C-complete in complex-valued metric spaces.

Definition 3.1 Let (X,d) be a complex-valued metric space and { x n } be a sequence in X.

  1. (1)

    If, for any cC with 0c, there exists NN such that, for all m,n>N, d( x n , x m )c, then { x n } is called a C-Cauchy sequence in X.

  2. (2)

    If every C-Cauchy sequence in X is convergent, then (X,d) is said to be a C-complete complex-valued metric space.

Next, we prove our main results.

Theorem 3.2 Let (X,d) be a C-complete complex-valued metric space and S,T:XX be mappings. If there exist two mappings α,β: C + [0,1) such that, for all x,yX,

  1. (a)

    α(x)+β(x)<1;

  2. (b)

    the mapping γ: C + [0,1) defined by γ(x):= α ( x ) 1 β ( x ) belongs to Γ;

  3. (c)

    d(Sx,Ty)α(d(x,y))d(x,y)+ β ( d ( x , y ) ) d ( x , S x ) d ( y , T y ) 1 + d ( x , y ) .

Then S and T have a unique common fixed point in X.

Proof Let x 0 be an arbitrary point in X. We construct the sequence { x n } in X such that

x 2 n + 1 =S x 2 n , x 2 n + 2 =T x 2 n + 1
(3.1)

for all n0. For all n0, we get

d ( x 2 n + 1 , x 2 n + 2 ) = d ( S x 2 n , T x 2 n + 1 ) α ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , x 2 n + 1 ) + β ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , S x 2 n ) d ( x 2 n + 1 , T x 2 n + 1 ) 1 + d ( x 2 n , x 2 n + 1 ) = α ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , x 2 n + 1 ) + β ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , x 2 n + 1 ) d ( x 2 n + 1 , x 2 n + 2 ) 1 + d ( x 2 n , x 2 n + 1 ) = α ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , x 2 n + 1 ) + β ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n + 1 , x 2 n + 2 ) ( d ( x 2 n , x 2 n + 1 ) 1 + d ( x 2 n , x 2 n + 1 ) ) α ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , x 2 n + 1 ) + β ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n + 1 , x 2 n + 2 ) ,

which implies that

d ( x 2 n + 1 , x 2 n + 2 ) ( α ( d ( x 2 n , x 2 n + 1 ) ) 1 β ( d ( x 2 n , x 2 n + 1 ) ) ) d ( x 2 n , x 2 n + 1 ) = γ ( d ( x 2 n , x 2 n + 1 ) ) d ( x 2 n , x 2 n + 1 ) .
(3.2)

Similarly, for all n0, we get

d ( x 2 n + 2 , x 2 n + 3 ) = d ( x 2 n + 3 , x 2 n + 2 ) = d ( S x 2 n + 2 , T x 2 n + 1 ) α ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 1 ) + β ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , S x 2 n + 2 ) d ( x 2 n + 1 , T x 2 n + 1 ) 1 + d ( x 2 n + 2 , x 2 n + 1 ) = α ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 1 ) + β ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 3 ) d ( x 2 n + 1 , x 2 n + 2 ) 1 + d ( x 2 n + 1 , x 2 n + 2 ) = α ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 1 ) + β ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 3 ) ( d ( x 2 n + 1 , x 2 n + 2 ) 1 + d ( x 2 n + 1 , x 2 n + 2 ) ) α ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 1 ) + β ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 3 ) ,

which implies that

d ( x 2 n + 2 , x 2 n + 3 ) ( α ( d ( x 2 n + 2 , x 2 n + 1 ) ) 1 β ( d ( x 2 n + 2 , x 2 n + 1 ) ) ) d ( x 2 n + 2 , x 2 n + 1 ) = γ ( d ( x 2 n + 2 , x 2 n + 1 ) ) d ( x 2 n + 2 , x 2 n + 1 ) .
(3.3)

From (3.2) and (3.3), we have

d( x n , x n + 1 )γ ( d ( x n 1 , x n ) ) d( x n 1 , x n )

for all nN. Therefore, we get

| d ( x n , x n + 1 ) | γ ( d ( x n 1 , x n ) ) | d ( x n 1 , x n ) | | d ( x n 1 , x n ) |
(3.4)

for all nN. This implies the sequence { | d ( x n 1 , x n ) | } n N is monotone non-increasing and bounded below. Therefore, |d( x n 1 , x n )|d for some d0.

Next, we claim that d=0. Assume to the contrary that d>0. In (3.4), taking n, we have

γ ( d ( x n 1 , x n ) ) 1.

Since γΓ, we get |d( x n 1 , x n )|0, which is a contradiction. Therefore, we have d=0, that is,

| d ( x n 1 , x n ) | 0.
(3.5)

Next, we prove that { x n } is a C-Cauchy sequence. According to (3.5), it is sufficient to show that the subsequence { x 2 n } is a C-Cauchy sequence. On the contrary, assume that { x 2 n } is not a C-Cauchy sequence. By Definition 3.1(1), there is cC with 0c for which, for all kN, there exists m k > n k k such that

d( x 2 n k , x 2 m k )c.
(3.6)

Further, corresponding to n k , we can choose m k in such a way that it is the smallest integer with m k > n k k satisfying (3.6). Then we have

d( x 2 n k , x 2 m k )c
(3.7)

and

d( x 2 n k , x 2 m k 2 )c.
(3.8)

By (3.7), (3.8) and the notion of a complex-valued metric, we have

c d ( x 2 n k , x 2 m k ) d ( x 2 n k , x 2 m k 2 ) + d ( x 2 m k 2 , x 2 m k 1 ) + d ( x 2 m k 1 , x 2 m k ) c + d ( x 2 m k 2 , x 2 m k 1 ) + d ( x 2 m k 1 , x 2 m k ) .

This implies

|c| | d ( x 2 n k , x 2 m k ) | |c|+ | d ( x 2 m k 2 , x 2 m k 1 ) | + | d ( x 2 m k 1 , x 2 m k ) | .

On taking limit as k, we have

| d ( x 2 n k , x 2 m k ) | |c|.
(3.9)

Further, we have

d ( x 2 n k , x 2 m k ) d ( x 2 n k , x 2 m k + 1 ) + d ( x 2 m k + 1 , x 2 m k ) d ( x 2 n k , x 2 m k ) + d ( x 2 m k , x 2 m k + 1 ) + d ( x 2 m k + 1 , x 2 m k ) ,

and then

| d ( x 2 n k , x 2 m k ) | | d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 m k + 1 , x 2 m k ) | | d ( x 2 n k , x 2 m k ) | + | d ( x 2 m k , x 2 m k + 1 ) | + | d ( x 2 m k + 1 , x 2 m k ) | .

Passing to the limit when k and using (3.5) and (3.9), we get

| d ( x 2 n k , x 2 m k + 1 ) | |c|.
(3.10)

Now, from the triangle inequality for a complex-valued metric d, we obtain that

d ( x 2 n k , x 2 m k + 1 ) d ( x 2 n k , x 2 n k + 1 ) + d ( x 2 n k + 1 , x 2 m k + 2 ) + d ( x 2 m k + 2 , x 2 m k + 1 ) = d ( x 2 n k , x 2 n k + 1 ) + d ( S x 2 n k , T x 2 m k + 1 ) + d ( x 2 m k , x 2 m k + 1 ) d ( x 2 n k , x 2 n k + 1 ) + α ( d ( x 2 n k , x 2 m k + 1 ) ) d ( x 2 n k , x 2 m k + 1 ) + β ( d ( x 2 n k , x 2 m k + 1 ) ) d ( x 2 n k , S x 2 n k ) d ( x 2 m k + 1 , T x 2 m k + 1 ) 1 + d ( x 2 n k , x 2 m k + 1 ) + d ( x 2 m k , x 2 m k + 1 ) = d ( x 2 n k , x 2 n k + 1 ) + α ( d ( x 2 n k , x 2 m k + 1 ) ) d ( x 2 n k , x 2 m k + 1 ) + β ( d ( x 2 n k , x 2 m k + 1 ) ) d ( x 2 n k , x 2 n k + 1 ) d ( x 2 m k + 1 , x 2 m k + 2 ) 1 + d ( x 2 n k , x 2 m k + 1 ) + d ( x 2 m k , x 2 m k + 1 ) ,

which implies that

| d ( x 2 n k , x 2 m k + 1 ) | | d ( x 2 n k , x 2 n k + 1 ) | + α ( d ( x 2 n k , x 2 m k + 1 ) ) | d ( x 2 n k , x 2 m k + 1 ) | + β ( d ( x 2 n k , x 2 m k + 1 ) ) | d ( x 2 n k , x 2 n k + 1 ) d ( x 2 m k + 1 , x 2 m k + 2 ) 1 + d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 m k , x 2 m k + 1 ) | | d ( x 2 n k , x 2 n k + 1 ) | + α ( d ( x 2 n k , x 2 m k + 1 ) ) | d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 n k , x 2 n k + 1 ) d ( x 2 m k + 1 , x 2 m k + 2 ) 1 + d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 m k , x 2 m k + 1 ) | | d ( x 2 n k , x 2 n k + 1 ) | + α ( d ( x 2 n k , x 2 m k + 1 ) ) 1 β ( d ( x 2 n k , x 2 m k + 1 ) ) | d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 n k , x 2 n k + 1 ) d ( x 2 m k + 1 , x 2 m k + 2 ) 1 + d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 m k , x 2 m k + 1 ) | | d ( x 2 n k , x 2 n k + 1 ) | + γ ( d ( x 2 n k , x 2 m k + 1 ) ) | d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 n k , x 2 n k + 1 ) d ( x 2 m k + 1 , x 2 m k + 2 ) 1 + d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 m k , x 2 m k + 1 ) | | d ( x 2 n k , x 2 n k + 1 ) | + | d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 n k , x 2 n k + 1 ) d ( x 2 m k + 1 , x 2 m k + 2 ) 1 + d ( x 2 n k , x 2 m k + 1 ) | + | d ( x 2 m k , x 2 m k + 1 ) | .

Taking k, we have

|c| ( lim k γ ( d ( x 2 n k , x 2 m k + 1 ) ) ) |c||c|,

that is,

lim k γ ( d ( x 2 n k , x 2 m k + 1 ) ) =1.

Since γΓ, we get |d( x 2 n k , x 2 m k + 1 )|0, which contradicts 0c. Therefore, we can conclude that { x 2 n } is a C-Cauchy sequence and hence { x n } is a C-Cauchy sequence. By the completeness of X, there exists a point zX such that x n z as n.

Next, we claim that Sz=z. If Szz, then d(z,Sz)>0. By the notion of a complex-valued metric d, we have

d ( z , S z ) d ( z , x 2 n + 2 ) + d ( x 2 n + 2 , S z ) = d ( z , x 2 n + 2 ) + d ( T x 2 n + 1 , S z ) = d ( z , x 2 n + 2 ) + d ( S z , T x 2 n + 1 ) d ( x 2 n + 2 , z ) + α ( d ( z , x 2 n + 1 ) ) d ( z , x 2 n + 1 ) + β ( d ( z , x 2 n + 1 ) ) d ( z , S z ) d ( x 2 n + 1 , T x 2 n + 1 ) 1 + d ( z , x 2 n + 1 ) = d ( x 2 n + 2 , z ) + α ( d ( z , x 2 n + 1 ) ) d ( z , x 2 n + 1 ) + β ( d ( z , x 2 n + 1 ) ) d ( z , S z ) d ( x 2 n + 1 , x 2 n + 2 ) 1 + d ( z , x 2 n + 1 ) d ( x 2 n + 2 , z ) + d ( z , x 2 n + 1 ) + d ( z , S z ) d ( x 2 n + 1 , x 2 n + 2 ) 1 + d ( z , x 2 n + 1 ) ,
(3.11)

which implies that

| d ( z , S z ) | | d ( x 2 n + 2 , z ) | + | d ( z , x 2 n + 1 ) | + | d ( z , S z ) d ( x 2 n + 1 , x 2 n + 2 ) 1 + d ( z , x 2 n + 1 ) | .

Taking n, we have |d(z,Sz)|=0, which is a contradiction. Thus, we get Sz=z. It follows similarly that Tz=z. Therefore, z=Sz=Tz, that is, z is a common fixed point of S and T.

Finally, we show that z is a unique common fixed point of S and T. Assume that there exists another point z ˆ such that z ˆ =S z ˆ =T z ˆ . Now, we have

d ( z , z ˆ ) = d ( S z , T z ˆ ) α ( d ( z , z ˆ ) ) d ( z , z ˆ ) + β ( d ( z , z ˆ ) ) d ( z , S z ) d ( z ˆ , T z ˆ ) 1 + d ( z , z ˆ ) = α ( d ( z , z ˆ ) ) d ( z , z ˆ ) .

Hence |d(z, z ˆ )|α(d(z, z ˆ ))|d(z, z ˆ )|. Since 0α(d(z, z ˆ ))<1, we get |d(z, z ˆ )|=0 and then z= z ˆ . Therefore, z is a unique common fixed point of S and T. This completes the proof. □

Corollary 3.3 Let (X,d) be a C-complete complex-valued metric space and S,T:XX be mappings. If S and T satisfy

d(Sx,Ty)λd(x,y)+ μ d ( x , S x ) d ( y , T y ) 1 + d ( x , y )
(3.12)

for all x,yX, where λ, μ are nonnegative reals with λ+μ<1, then S and T have a unique common fixed point in X.

Proof We can prove this result by applying Theorem 3.2 by setting α(x)=λ and β(x)=μ. □

Corollary 3.4 Let (X,d) be a C-complete complex-valued metric space and T:XX be a mapping. If there exist two mappings α,β: C + [0,1) such that, for all x,yX,

  1. (a)

    α(x)+β(x)<1;

  2. (b)

    the mapping γ: C + [0,1) defined by γ(x):= α ( x ) 1 β ( x ) belongs to Γ;

  3. (c)

    d(Tx,Ty)α(d(x,y))d(x,y)+ β ( d ( x , y ) ) d ( x , T x ) d ( y , T y ) 1 + d ( x , y ) .

Then T has a unique fixed point in X.

Proof We can prove this result by applying Theorem 3.2 with S=T. □

Corollary 3.5 Let (X,d) be a C-complete complex-valued metric space and T:XX be a mapping. If T satisfies

d(Tx,Ty)λd(x,y)+ μ d ( x , T x ) d ( y , T y ) 1 + d ( x , y )
(3.13)

for all x,yX, where λ, μ are nonnegative reals with λ+μ<1, then T has a unique fixed point in X.

Proof We can prove this result by applying Corollary 3.4 with α(x)=λ and β(x)=μ. □

Theorem 3.6 Let (X,d) be a C-complete complex-valued metric space and T:XX. If there exist two mappings α,β: C + [0,1) such that, for all x,yX,

  1. (a)

    α(x)+β(x)<1;

  2. (b)

    the mapping γ: C + [0,1) defined by γ(x):= α ( x ) 1 β ( x ) belongs to Γ;

  3. (c)

    d( T n x, T n y)α(d(x,y))d(x,y)+ β ( d ( x , y ) ) d ( x , T n x ) d ( y , T n y ) 1 + d ( x , y ) for some nN.

Then T has a unique fixed point in X.

Proof From Corollary 3.4, we get T n has a unique fixed point z. Since

T n (Tz)=T ( T n z ) =Tz,

we know that Tz is a fixed point of T n . Therefore, Tz=z by the uniqueness of a fixed point of T n . Therefore, z is also a fixed point of T. Since the fixed point of T is also a fixed point of T n , the fixed point of T is also unique. □

Corollary 3.7 Let (X,d) be a C-complete complex-valued metric space and S,T:XX be mappings. If T satisfy

d ( T n x , T n y ) λd(x,y)+ μ d ( x , T n x ) d ( y , T n y ) 1 + d ( x , y )
(3.14)

for all x,yX for some nN, where λ, μ are nonnegative reals with λ+μ<1, then T has a unique fixed point in X.

Proof We can prove this result by applying Theorem 3.6 with α(x)=λ and β(x)=μ. □

Remark 3.8 It is easy to see that Corollaries 3.3, 3.5 and 3.7 hold in complete complex-valued metric spaces. Therefore, Corollaries 3.3, 3.5 and 3.7 become Theorem 4, Corollary 5 and Corollary 6 of Azam et al. [14] in complete complex-valued metric spaces.

4 Common fixed points (II)

In this section, we prove a common fixed point theorem for weakly compatible mappings in C-complete complex-valued metric spaces.

Since Banach’s fixed point theorem, many authors have improved, extended and generalized Banach’s fixed point theorem in several ways. Especially, in [18], Jungck generalized Banach’s fixed point theorem by using the concept of commuting mappings as follows.

Theorem J Let (X,d) be a complete metric space. Then a continuous mapping S:XX has a fixed point in X if and only if there exists a number α(0,1) and a mapping T:XX such that T(X)S(X), S and T are commuting (i.e., TSx=STx for all x in X),

d(Tx,Ty)αd(Sx,Sy)

for all x,yX. Further, S and T have a unique common fixed point in X (i.e., there exists a unique point z in X such that Sz=Tz=z).

Note that if we put S= I X (the identity mapping on X) in Theorem J, we have Banach’s fixed point theorem.

Since Theorem J, in 1986, Jungck [18] introduced more generalized commuting mappings in metric spaces, called compatible mappings, which also are more general than weakly commuting mappings (that is, the mappings S,T:XX are said to be weakly commuting if d(STx,TSx)d(Sx,Tx) for all xX) introduced by Sessa [19] as follows.

Definition 4.1 Let S and T be mappings from a metric space (X,d) into itself. The mappings S and T are said to be compatible if

lim n d(ST x n ,TS x n )=0

whenever { x n } is a sequence in X such that lim n S x n = lim n T x n =z for some zX.

In general, commuting mappings are weakly commuting and weakly commuting mappings are compatible, but the converse is not necessarily true; some examples can be found in [18, 2022].

Also, some authors introduced some kind of generalizations of compatible mappings in metric spaces and other spaces (see [2124]) and they proved common fixed point theorems using these kinds of compatible mappings in metric spaces and other spaces.

In [25], Jungck and Rhoades introduced the concept of weakly compatible mappings in symmetric spaces (X,d) and proved some common fixed point theorems for these mappings in symmetric spaces as follows.

Definition 4.2 Let S and T be mappings from a metric space (X,d) into itself. The mappings S and T are said to be weakly compatible if they commute at coincidence points of S and T.

In Djoudi and Nisse [26], we can find an example to show that there exist weakly compatible mappings which are not compatible mappings in metric spaces.

Now, we give the main result in this section.

Theorem 4.3 Let (X,d) be a complex-valued metric space and S,T:XX be such that T(X)S(X) and S(X) is C-complete. If there exist two mappings α,β: C + [0,1) such that, for all x,yX,

  1. (a)

    α(x)+β(x)<1;

  2. (b)

    the mapping γ: C + [0,1) defined by γ(x):= α ( x ) 1 β ( x ) belongs to Γ;

  3. (c)

    d(Tx,Ty)α(d(Sx,Sy))d(Sx,Sy)+ β ( d ( S x , S y ) ) d ( S x , T x ) d ( S y , T y ) 1 + d ( S x , S y ) .

Then S and T have a unique point of coincidence in X. Moreover, S and T have a unique common fixed point in X if S and T are weakly compatible.

Proof Consider the mapping S:XX. By Lemma 2.10, there exists EX such that S(E)=S(X) and S:EX is one-to-one.

Next, we define a mapping W:S(E)S(E) by W(Sx)=Tx for all SxS(E). Therefore, is well defined since S is one-to-one on E. Since WS=T, using (c), we get

d ( W ( S x ) , W ( S y ) ) α ( d ( S x , S y ) ) d ( S x , S y ) + β ( d ( S x , S y ) ) d ( S x , W ( S x ) ) d ( S y , W ( S y ) ) 1 + d ( S x , S y )
(4.1)

for all Sx,SyS(E). Since S(E)=S(X) is C-complete and (4.1) holds, we can apply Corollary 3.4 with a mapping . Therefore, there exists a unique fixed point zS(X) such that Wz=z. It follows from zS(X) that z=S z for some z X. So, W(S z )=S z , that is, T z =S z . Therefore, T and S have a unique point of coincidence.

Next, we show that S and T have a common fixed point. Now, we have z=T z =S z . Since S and T are weakly compatible, we get

Sz=ST z =TS z =Tz.

This implies Sz=Tz is a point of coincidence of S and T. But z is a unique point of coincidence of S and T. Therefore, we conclude that z=Sz=Tz, which implies that z is a common fixed point of S and T.

Finally, we prove the uniqueness of a common fixed point of S and T. Assume that z ¯ is another common fixed point of S and T. So, z ¯ =S z ¯ =T z ¯ , and then z ¯ is also a point of coincidence of S and T. However, we know that z is a unique point of coincidence of S and T. Therefore, we get z ¯ =z, that is, z is a unique common fixed point of S and T. This completes the proof. □

5 Urysohn integral equations

In this section, we show that Theorem 3.2 can be applied to the existence of a common solution of the system of the Urysohn integral equations.

Theorem 5.1 Let X=C([a,b], R n ), a>0, and d:X×XC be defined by

d(x,y)= max t [ a , b ] x ( t ) y ( t ) 1 + a 2 e i tan 1 a .

Consider the Urysohn integral equations

x(t)= a b K 1 ( t , s , x ( s ) ) ds+g(t),
(5.1)
x(t)= a b K 2 ( t , s , x ( s ) ) ds+h(t),
(5.2)

where t[a,b]R, x,g,hX and K 1 , K 2 :[a,b]×[a,b]× R n R n .

Suppose that K 1 , K 2 are such that F x , G x X for all xX, where

F x ( t ) = a b K 1 ( t , s , x ( s ) ) d s , G x ( t ) = a b K 2 ( t , s , x ( s ) ) d s

for all t[a,b].

If there exist two mappings α,β: C + [0,1) such that for all x,yX the following hold:

  1. (a)

    α(x)+β(x)<1;

  2. (b)

    the mapping γ: C + [0,1) defined by γ(x):= α ( x ) 1 β ( x ) belongs to Γ;

  3. (c)

    F x ( t ) G y ( t ) + g ( t ) h ( t ) 1 + a 2 e i tan 1 a α( max t [ a , b ] A(x,y)(t))A(x,y)(t)+β( max t [ a , b ] A(x,y)(t))B(x,y)(t), where

    A ( x , y ) ( t ) = x ( t ) y ( t ) 1 + a 2 e i tan 1 a , B ( x , y ) ( t ) = F x ( t ) + g ( t ) x ( t ) G y ( t ) + h ( t ) y ( t ) 1 + d ( x , y ) 1 + a 2 e i tan 1 a ,

then the system of integral equations (5.1) and (5.2) has a unique common solution.

Proof Define two mappings S,T:XX by Sx= F x +g and Tx= G x +h. Then we have

d ( S x , T y ) = max t [ a , b ] F x ( t ) G y ( t ) + g ( t ) h ( t ) 1 + a 2 e i tan 1 a , d ( x , S x ) = max t [ a , b ] F x ( t ) + g ( t ) x ( t ) 1 + a 2 e i tan 1 a

and

d(y,Ty)= max t [ a , b ] G y ( t ) + h ( t ) y ( t ) 1 + a 2 e i tan 1 a .

We can show easily that for all x,yX,

d(Sx,Ty)α ( d ( x , y ) ) d(x,y)+ β ( d ( x , y ) ) d ( x , S x ) d ( y , T y ) 1 + d ( x , y ) .

Now, we can apply Theorem 3.2. Therefore, we get the Urysohn integral equations (5.1) and (5.2) have a unique common solution. □