1 Introduction

In 2006, Mustafa and Sims [1] introduced a generalization of metric spaces, the G-metric spaces, which assigns to each triple of elements a non-negative real number. Very recently, Jleli and Samet [2] showed that some of the fixed point theorems in G-metric spaces can be obtained from quasi-metric spaces. For some works in G-metric spaces, see [336]. In 2010, Saadati et al. [26] introduced the concept of Ω-distance and studied some nice fixed point theorems (also, see [13]). Meanwhile, Bhaskar and Lakshmikantam [37] introduced the concept of coupled fixed point and proved several fixed point theorems. Lakshmikantam and Ćirić [38] generalized the concept of coupled fixed point to the the concept of coupled coincidence point of two mappings [39]. After that, many authors established coupled fixed point results (please, see [1944]). In the present paper, we utilize the concept of Ω-distance to establish some coupled fixed point results. Also, we introduce an example to support the useability of our study.

2 Preliminaries

Definition 2.1 ([1])

Let X be an nonempty set. The mapping G:X×X×XX is called G-metric if the following axioms are fulfilled:

  1. (1)

    G(x,y,z)=0 if x=y=z (the coincidence);

  2. (2)

    G(x,x,y)>0 for all x,yX, xy;

  3. (3)

    G(x,x,z)G(x,y,z) for each triple (x,y,z) from X×X×X with zy;

  4. (4)

    G(x,y,z)=G(p{x,y,z}) for each permutation of {x,y,z} (the symmetry);

  5. (5)

    G(x,y,z)G(x,a,a)+G(a,y,z) for each x, y, z and a in X (the rectangle inequality).

Definition 2.2 ([1])

Consider X a G-metric space and ( x n ) a sequence in G.

  1. (1)

    ( x n ) is called G-Cauchy sequence if for each ϵ>0 there is a positive integer n 0 so that for all m,n,l n 0 , G( x n , x m , x l )<ϵ.

  2. (2)

    ( x n ) is said to be G-convergent to xX if for each ϵ>0 there is a positive integer n 0 such that G( x m , x n ,x)<ϵ for each m,n n 0 .

Definition 2.3 ([26])

Consider (X,G) a G-metric space and Ω:X×X×X[0,+). The mapping Ω is called an Ω-distance on X if it satisfies the three conditions in the following:

  1. (1)

    Ω(x,y,z)Ω(x,a,a)+Ω(a,y,z) for all x, y, z, a from X.

  2. (2)

    For each x, y from X, Ω(x,y,),Ω(x,,y):X[0,+) are lower semi-continuous.

  3. (3)

    for each ϵ>0 there is δ>0, so that Ω(x,a,a)δ and Ω(a,y,z)δ imply G(x,y,z)ϵ.

The following lemma [13, 26] is going to be very helpful in computing the limits of several sequences.

Lemma 2.1 Let X be a metric space, endowed with metric G, and let Ω be an Ω-distance on X. ( x n ), ( y n ) are sequences in X, ( α n ) and ( β n ) are sequences in [0,+) with lim n + α n = lim n + β n =0. If x, y, z and aX, then

  1. (1)

    If Ω(y, x n , x n ) α n and Ω( x n ,y,z) β n for nN, then G(y,y,z)<ϵ, and, by consequence, y=z.

  2. (2)

    Inequalities Ω( y n , x n , x n ) α n and Ω( x n , y m ,z) β n for m>n imply G( y n , y m ,z)0, hence y n z.

  3. (3)

    If Ω( x n , x m , x l ) α n for l,m,nN with nml, then ( x n ) is a G-Cauchy sequence.

  4. (4)

    If Ω( x n ,a,a) α n , nN, then ( x n ) is a G-Cauchy sequence.

Definition 2.4 ([37])

Consider X a nonempty set. A pair (x,y)X×X is called coupled fixed point of mapping F:X×XX if

F(x,y)=x,F(y,x)=y.

Definition 2.5 ([38])

Let X be a nonempty set. The element (x,y)X×X is a coupled coincidence point of mappings F:X×XX and g:XX if

F(x,y)=gx,F(y,x)=gy.

3 Main results

Theorem 3.1 Let (X,G) be a G-metric space and Ω an Ω-distance on X such that X is Ω-bounded. g:XX and F:X×XX are mappings. Suppose there exists k[0,1) such that for each x, y, z, x , y and z in X

Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) k max { Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) , Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) , Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) , Ω ( F ( x , y ) , g x , g z ) + Ω ( F ( y , x ) , g y , g z ) , Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) , Ω ( F ( x , y ) , F ( x , y ) , g z ) + Ω ( F ( y , x ) , F ( y , x ) , g z ) } .

Consider also that the following conditions hold true:

  1. (1)

    F(X×X)gX;

  2. (2)

    gX is a complete subspace of X with respect to the topology, induced by G;

  3. (3)

    If F(u,v)gu or F(v,u)gv, then

    inf { Ω ( g x , F ( x , y ) , g u ) + Ω ( g y , F ( y , x ) , g v ) + Ω ( g x , g u , F ( x , y ) ) + Ω ( g y , g v , F ( y , x ) ) } > 0 .

Then, F and g have a unique coupled coincidence point (u,v). Moreover, F(u,v)=gu=gv=F(v,u).

Proof Consider x 0 X and y 0 X. Because F(X×X)gX, there exist x 1 and y 1 in X such that g x 1 =F( x 0 , y 0 ) and y 1 =F( y 0 , x 0 ). By continuing the process, we obtain two sequences, ( x n ) and ( y n ), with the properties

g x n + 1 =F( x n , y n ),g y n + 1 =F( y n , x n )

Using the contraction condition, we obtain

Ω ( g x n , g x n + 1 , g x n + s ) + Ω ( g y n , g y n + 1 , g y n + s ) = Ω ( F ( x n 1 , y n 1 ) , F ( x n , y n ) , F ( x n + s 1 , y n + s 1 ) ) + Ω ( F ( y n 1 , x n 1 ) , F ( y n , x n ) , F ( y n + s 1 , x n + s 1 ) ) k max { Ω ( g x n 1 , g x n , g x n + s 1 ) + Ω ( g y n 1 , g y n , g y n + s 1 ) , Ω ( g x n , g x n 1 , g x n + s 1 ) + Ω ( g y n , g y n 1 , g y n + s 1 ) , Ω ( g x n 1 , g x n + 1 , g x n + s 1 ) + Ω ( g y n 1 , g y n + 1 , g y n + s 1 ) , Ω ( g x n , g x n , g x n + s 1 ) + Ω ( g y n , g y n , g y n + s 1 ) , Ω ( g x n , g x n + 1 , g x n + s 1 ) + Ω ( g y n , g y n + 1 , g y n + s 1 ) } .

By applying the contraction inequality repeatedly, we get that

Ω ( g x n , g x n + 1 , g x n + s ) + Ω ( g y n , g y n + 1 , g y n + s ) k n 1 max ( i , j , t ) A { Ω ( g x i , g x j , g x t ) + Ω ( g y i , g y j , g y t ) } ,
(1)

where A={(i,j,t)|1in,1jn+1,s+1tn+s1}.

Since X is Ω-bounded, there is M>0 such that Ω(x,y,z)<M for each triple (x,y,z)X×X×X. Hence, relation (1) becomes

Ω(g x n ,g x n + 1 ,g x n + s )+Ω(g y n ,g y n + 1 ,g y n + s )2 k n 1 M.

Consider now l>m>n>0, l,m,nN. The following relations hold true:

Ω ( g x n , g x m , g x l ) Ω ( g x n , g x n + 1 , g x n + 1 ) + Ω ( g x n + 1 , g x m , g x l ) Ω ( g x n , g x n + 1 , g x n + 1 ) + Ω ( g x n + 1 , g x n + 2 , g x n + 2 ) + + Ω ( g x m 1 , g x m , g x l ) ,
(2)

and, also

Ω ( g y n , g y m , g y l ) Ω ( g y n , g y n + 1 , g y n + 1 ) + Ω ( g y n + 1 , g y m , g y l ) Ω ( g y n , g y n + 1 , g y n + 1 ) + Ω ( g y n + 1 , g y n + 2 , g y n + 2 ) + + Ω ( g y m 1 , g y m , g y l ) .
(3)

Making the sum of relations (2) and (3), and using inequality (1), it follows that

Ω ( g x n , g x m , g x l ) + Ω ( g y n , g y m , g y l ) 2 M ( k n 1 + k n + + k m 2 ) 2 M k n 1 1 1 k .

Lemma 2.1, part (3), implies that (g x n ) and (g y n ) are G-Cauchy sequences. Since gX is a complete G-subspace of X, there are gu and gv in gX such that g x n gu and g y n gv.

Let ϵ>0. From the lower semi-continuity of Ω, we get

Ω(g x n ,g x m ,gu) lim inf p + Ω(g x n ,g x m ,g x p )ϵ,mn,
(4)
Ω(g y n ,g y m ,gv) lim inf p + Ω(g y n ,g y m ,g y p )ϵ,mn,
(5)
Ω(g x n ,gu,g x l ) lim inf p + Ω(g x n ,g x p ,g x l )ϵ,ln,
(6)
Ω(g y n ,gv,g y l ) lim inf p + Ω(g y n ,g y p ,g y l )ϵ,ln.
(7)

Suppose that F(u,v)gu or F(v,u)gv. Applying hypotheses (3) of the theorem, and using inequalities (4)-(7), we obtain

0 < inf { Ω ( g x n , F ( x n , y n ) , g u ) + Ω ( g y n , F ( y n , x n ) , g v ) + Ω ( g x n , g u , F ( x n , y n ) ) + Ω ( g y n , g v , F ( y n , x n ) ) } 4 ϵ

for each ϵ>0, which is a contradiction.

Therefore, F(u,v)=gu and F(v,u)=gv.

Using the contraction condition from the hypotheses, we get

Ω ( F ( u , v ) , g x n + 1 , g x n + 1 ) + Ω ( F ( v , u ) , g y n + 1 , g y n + 1 ) = Ω ( F ( u , v ) , F ( x n , y n ) , F ( x n , y n ) ) + Ω ( F ( v , u ) , F ( y n , x n ) , F ( y n , x n ) ) k max { Ω ( g u , g x n , g x n ) + Ω ( g v , g y n , g y n ) , Ω ( g x n , g u , g x n ) + Ω ( g y n , g v , g y n ) , Ω ( g u , g x n + 1 , g x n ) + Ω ( g v , g y n + 1 , g y n ) Ω ( g x n , g u , g x n ) + Ω ( g y n , g v , g y n ) } .

We apply repeatedly the contraction inequality, and we obtain

Ω ( F ( u , v ) , g x n + 1 , g x n + 1 ) + Ω ( F ( v , u ) , g y n + 1 , g y n + 1 ) k n max { Ω ( g u , g x i , g x 1 ) + Ω ( g v , g y i , g y 1 ) , Ω ( g x j , g u , g x 1 ) + Ω ( g y j , g v , g y 1 ) | 1 i n , 1 j n + 1 } .

Since X is Ω-bounded, it follows that

Ω ( F ( u , v ) , g x n + 1 , g x n + 1 ) +Ω ( F ( v , u ) , g y n + 1 , g y n + 1 ) 2M k n .
(8)

In a similar manner, it can be proved that

Ω ( g x n + 1 , F ( u , v ) , F ( v , u ) ) +Ω ( g y n + 1 , F ( v , u ) , F ( u , v ) ) 2M k n .
(9)

Taking into account (8), (9) and the first statement of Lemma 2.1, we get gu=gv.

We will prove now the uniqueness of the coupled coincidence point of F and g.

Suppose (u,v) and ( u , v ) are coupled coincidence points of F and g. Using the contraction condition, we obtain

Ω(gu,gu,gu)+Ω(gv,gv,gv)k ( Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) ) ,

hence Ω(gu,gu,gu)=Ω(gv,gv,gv)=0.

On the other hand,

Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) k max { Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) , Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) }
(10)

and

Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) k max { Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) , Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) } .
(11)

Relations (10) and (11) imply that Ω(g u ,gu,gu)=Ω(gu,g u ,gu)=0 and also Ω(g v ,gv,gv)=Ω(gv,g v ,gv)=0. Lemma 2.1 imposes that gu=g u and gv=g v , and the uniqueness is proved. □

If we take g=I d X in Theorem 3.1, we easily get the following.

Corollary 3.1 Let (X,G) be a complete G-metric space, and let Ω be an Ω-distance on X such that X is Ω-bounded. Suppose F:X×XX is a mapping for which there exists k[0,1) such that for each x, y, z, x , y and z in X

Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) k max { Ω ( x , x , z ) + Ω ( y , y , z ) , Ω ( x , x , z ) + Ω ( y , y , z ) , Ω ( x , F ( x , y ) , z ) + Ω ( y , F ( y , x ) , z ) , Ω ( F ( x , y ) , x , z ) + Ω ( F ( y , x ) , y , z ) , Ω ( x , F ( x , y ) , z ) + Ω ( y , F ( y , x ) , z ) , Ω ( F ( x , y ) , F ( x , y ) , z ) + Ω ( F ( y , x ) , F ( y , x ) , z ) } .

Consider also that if F(u,v)u or F(v,u)v, then

inf { Ω ( x , F ( x , y ) , u ) + Ω ( y , F ( y , x ) , v ) + Ω ( F ( x , y ) , u , x ) + Ω ( F ( y , x ) , v , y ) } > 0 .

Then, F has a unique coupled fixed point (u,v). Moreover, F(u,v)=u=v=F(v,u).

Corollary 3.2 Let (X,G) be a G-metric space, and let Ω be an Ω-distance on X such that X is Ω-bounded. g:XX and F:X×XX are mappings. Suppose that there exists k 1 , k 2 , k 3 , k 4 , k 5 , k 6 [0,1) with k 1 + k 2 + k 3 + k 4 + k 5 + k 6 <1 such that for each x, y, z, x , y and z in X

Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) k 1 ( Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) ) + k 2 ( Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) ) + k 3 ( Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) ) + k 4 ( Ω ( F ( x , y ) , g x , g z ) + Ω ( F ( y , x ) , g y , g z ) ) + k 5 ( Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) ) + k 6 ( Ω ( F ( x , y ) , F ( x , y ) , g z ) + Ω ( F ( y , x ) , F ( y , x ) , g z ) ) .

Consider also that the following conditions hold true:

  1. (1)

    F(X×X)gX;

  2. (2)

    gX is a complete subspace of X with respect to the topology induced by G;

  3. (3)

    If F(u,v)gu or F(v,u)gv, then

    inf { Ω ( g x , F ( x , y ) , g u ) + Ω ( g y , F ( y , x ) , g v ) + Ω ( g x , g u , F ( x , y ) ) + Ω ( g y , g v , F ( y , x ) ) } > 0 .

Then, F and g have a unique coupled coincidence point (u,v). Moreover, F(u,v)=gu=gv=F(v,u).

Proof Follows from Theorem 3.1 by noting that

k 1 Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) + k 2 Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) + k 3 Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) + k 4 Ω ( F ( x , y ) , g x , g z ) + Ω ( F ( y , x ) , g y , g z ) + k 5 Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) , + k 6 Ω ( F ( x , y ) , F ( x , y ) , g z ) + Ω ( F ( y , x ) , F ( y , x ) , g z ) k max { Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) , Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) , Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) , Ω ( F ( x , y ) , g x , g z ) + Ω ( F ( y , x ) , g y , g z ) , Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) , Ω ( F ( x , y ) , F ( x , y ) , g z ) + Ω ( F ( y , x ) , F ( y , x ) , g z ) } .

 □

If we take g=I d X in Corollary 3.2, we easily get the following.

Corollary 3.3 Let (X,G) be a complete G-metric space, and let Ω be an Ω-distance on X such that X is Ω-bounded. Suppose F:X×XX is a mapping, for which there exists k 1 , k 2 , k 3 , k 4 , k 5 , k 6 [0,1) with k 1 + k 2 + k 3 + k 4 + k 5 + k 6 <1 such that for each x, y, z, x , y and z in X

Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) k 1 ( Ω ( x , x , z ) + Ω ( y , y , z ) ) + k 2 ( Ω ( x , x , z ) + Ω ( y , y , z ) ) + k 3 ( Ω ( x , F ( x , y ) , z ) + Ω ( y , F ( y , x ) , z ) ) + k 4 ( Ω ( F ( x , y ) , x , z ) + Ω ( F ( y , x ) , y , z ) ) + k 5 ( Ω ( x , F ( x , y ) , z ) + Ω ( y , F ( y , x ) , z ) ) + k 6 ( Ω ( F ( x , y ) , F ( x , y ) , z ) + Ω ( F ( y , x ) , F ( y , x ) , z ) ) .

Consider also that if F(u,v)u or F(v,u)v, then

inf { Ω ( x , F ( x , y ) , u ) + Ω ( y , F ( y , x ) , v ) + Ω ( F ( x , y ) , u , x ) + Ω ( F ( y , x ) , v , y ) } > 0 .

Then, F has a unique coupled fixed point (u,v). Moreover, F(u,v)=u=v=F(v,u).

By modifying the contraction condition, we get the following theorem.

Theorem 3.2 Let (X,G) be a G-metric space, and let Ω be an Ω-distance on X such that X is Ω-bounded. g:XX and F:X×XX are mappings. Suppose that there exist k 1 , k 2 [0,1) with k 1 + k 2 <1 such that for each x, y, z, x , y and z in X

Ω ( F ( x , y ) , g x , F ( z , z ) ) + Ω ( F ( y , x ) , g y , F ( z , z ) ) k 1 max { Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) , Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) , Ω ( F ( x , y ) , g x , g z ) + Ω ( F ( y , x ) , g y , g z ) , Ω ( g x , F ( x , y ) , g z ) + Ω ( g y , F ( y , x ) , g z ) } + k 2 ( Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) ) ,

and the conditions (1)-(3) from Theorem  3.1 hold.

Then, F and g have a unique coupled coincidence point (u,v). Moreover, F(u,v)=gu=gv=F(v,u).

Proof Let x 0 and y 0 be elements of X. Since F(X×X)gX, there exist x 1 and y 1 in X such that g x 1 =F( x 0 , y 0 ) and y 1 =F( y 0 , x 0 ). Repeating this procedure, we obtain two sequences, ( x n ) and ( y n ), with the properties

g x n + 1 =F( x n , y n ),g y n + 1 =F( y n , x n ).

The contraction condition implies that

Ω ( g x n , g x n + 1 , g x n + s ) + Ω ( g y n , g y n + 1 , g y n + s ) = Ω ( F ( x n 1 , y n 1 ) , g x n + 1 , F ( x n + s 1 , y n + s 1 ) ) + Ω ( F ( y n 1 , x n 1 ) , g y n + 1 , F ( y n + s 1 , x n + s 1 ) ) k 1 max { Ω ( g x n 1 , g x n + 1 , g x n + s 1 ) + Ω ( g y n 1 , g y n + 1 , g y n + s 1 ) , Ω ( g x n + 1 , g x n 1 , g x n + s 1 ) + Ω ( g y n + 1 , g y n 1 , g y n + s 1 ) , Ω ( g x n , g x n + 1 , g x n + s 1 ) + Ω ( g y n , g y n + 1 , g y n + s 1 ) , Ω ( g x n + 1 , g x n , g x n + s 1 ) + Ω ( g y n + 1 , g y n , g y n + s 1 ) } + k 2 ( Ω ( g x n , g x n + 1 , g x n + s ) + Ω ( g y n , g y n + 1 , g y n + s ) ) ,

which leads us to

Ω ( g x n , g x n + 1 , g x n + s ) + Ω ( g y n , g y n + 1 , g y n + s ) k max { Ω ( g x n 1 , g x n + 1 , g x n + s 1 ) + Ω ( g y n 1 , g y n + 1 , g y n + s 1 ) , Ω ( g x n + 1 , g x n 1 , g x n + s 1 ) + Ω ( g y n + 1 , g y n 1 , g y n + s 1 ) , Ω ( g x n , g x n + 1 , g x n + s 1 ) + Ω ( g y n , g y n + 1 , g y n + s 1 ) , Ω ( g x n + 1 , g x n , g x n + s 1 ) + Ω ( g y n + 1 , g y n , g y n + s 1 ) } ,

where k= k 1 1 k 2 <1.

Following the same steps, as we did in Theorem 3.1, the conclusion is straightforward. □

Theorem 3.2 leads us to a coupled fixed point property, by considering g=I d X .

Corollary 3.4 Let (X,G) be a complete G-metric space, and let Ω be an Ω-distance on X such that X is Ω-bounded. Suppose that F:X×XX is a mapping, for which there exist k 1 , k 2 [0,1) with k 1 + k 2 <1 such that for each x, y, z, x , y and z in X

Ω ( F ( x , y ) , x , F ( z , z ) ) + Ω ( F ( y , x ) , y , F ( z , z ) ) k 1 max { Ω ( x , x , z ) + Ω ( y , y , z ) , Ω ( x , x , z ) + Ω ( y , y , z ) , Ω ( F ( x , y ) , x , z ) + Ω ( F ( y , x ) , y , z ) , Ω ( x , F ( x , y ) , z ) + Ω ( y , F ( y , x ) , z ) } + k 2 ( Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) ) ,

and if F(u,v)u or F(v,u)v, then

inf { Ω ( x , F ( x , y ) , u ) + Ω ( y , F ( y , x ) , v ) + Ω ( x , u , F ( x , y ) ) + Ω ( y , v , F ( y , x ) ) } > 0 .

Then, F has a coupled fixed point (u,v). Moreover, F(u,v)=u=v=F(v,u).

Theorem 3.3 Let (X,G) be a G-metric space, and let Ω be an Ω-distance on X. Consider F:X×XX, g:XX and ϕ:gX R + such that

Ω ( g x , F ( x , y ) , F ( z , z ) ) + Ω ( g y , F ( y , x ) , F ( z , z ) ) ϕ ( g x ) + ϕ ( g y ) + ϕ ( g z ) + ϕ ( g z ) ϕ ( F ( x , y ) ) ϕ ( F ( y , x ) ) ϕ ( F ( z , z ) ) ϕ ( F ( z , z ) )

for all x,y,z, z X. Suppose that the following conditions are fulfilled:

  1. (1)

    F(X×X)gX.

  2. (2)

    gX is a complete subspace of X with respect to the topology, induced by G.

  3. (3)

    There exists k>0 such that Ω(x,x,y)kΩ(x,y,y) holds for all x,yX.

  4. (4)

    If F(u,v)gu or F(v,u)gv, then

    inf { Ω ( g x , F ( x , y ) , g u ) + Ω ( g y , F ( y , x ) , g v ) + Ω ( g x , g u , F ( x , y ) ) + Ω ( g y , g v , F ( y , x ) ) } > 0 .

Then F and g have a coupled coincidence point (u,v).

Proof Consider ( x 0 , y 0 ) a pair from X×X. As F(X×X)gX, there exist ( x 1 , y 1 )X×X so that g x 1 =F( x 0 , y 0 ), g y 1 =F( y 0 , x 0 ).

We continue the process, and we obtain two sequences ( x n ), ( y n ) from X, having the properties that

g x n + 1 =F( x n , y n ),g y n + 1 =F( y n , x n ).

Using the contraction condition, we get

Ω ( g x n , g x n + 1 , g x n + 1 ) + Ω ( g y n , g y n + 1 , g y n + 1 ) = Ω ( g x n , F ( x n , y n ) , F ( x n , y n ) ) + Ω ( g y n , F ( y n , x n ) , F ( y n , x n ) 2 Ω ( g x n ) + 2 Ω ( g y n ) 2 Ω ( g x n + 1 ) 2 Ω ( g y n + 1 ) .
(12)

For m>n, the first part of the definition Ω-distance and (12) yields

Ω ( g x n , g x m , g x m ) + Ω ( g y n , g y m , g y m ) k = n m 1 [ Ω ( g x k , g x k + 1 , g x k + 1 ) + Ω ( g y k , g y k + 1 , g y k + 1 ) ] .
(13)

Let

S n = k = 0 n [ Ω ( g x k , g x k + 1 , g x k + 1 ) + Ω ( g y k , g y k + 1 , g y k + 1 ) ] .

According to (12),

S n 2ϕ(g x 0 )+2ϕ(g y 0 )2ϕ(g x n + 1 )2ϕ(g y n + 1 )2ϕ(g x 0 )+2ϕ(g y 0 ).

Thus, ( S n ) is an increasing bounded sequence, so

lim n + s n = n = 0 [ Ω ( g x k , g x k + 1 , g x k + 1 ) + Ω ( g y k , g y k + 1 , g y k + 1 ) ]

exists.

Now, we shall show that (g x n ) and (g y n ) are G-Cauchy sequences in gX. Consider ϵ>0. By part (3) of the definition of an Ω-distance, we choose δ>0 such that if Ω(x,a,a)<δ and Ω(x,y,z)<δ, then G(x,y,z)<ϵ. Let η=min{δ, δ k }.

Using the fact that

n = 0 + [ Ω ( g x k , g x k + 1 , g x k + 1 ) + Ω ( g y k , g y k + 1 , g y k + 1 ) ] <+

and letting n+ in (13), we choose n 0 N such that

Ω(g x n ,g x m ,g x m )+Ω(g y n ,g y m ,g y m )<ηδ

for all m>n n 0 .

Thus,

Ω(g x n ,g x m ,g x m )<δ

and

Ω(g y n ,g y m ,g y m )<δ

for all m>n n 0 . Also we have

Ω(g x m ,g x m ,g x l )+Ω(g y m ,g y m ,g y l )k [ Ω ( g x m , g x m , g x l ) + Ω ( g y m , g y m , g y l ) ] <kηδ

for all l>m n 0 . Thus,

Ω(g x m ,g x m ,g x l )<δ

and

Ω(g y m ,g y m ,g y l )<δ

for all m>n n 0 . Thus, by part (3) of the definition of Ω-distance, we have

G(g x n ,g x m ,g x l )<ϵ

and

G(g y n ,g y m ,g y l )<ϵ

for l>m>n n 0 .

Therefore, (g x n ) and (g y n ) are G-Cauchy sequences. As gX is G-complete, it follows that there are u, vX so that lim n + g x n =gu and lim n + g v n =gv.

Since Ω is lower semi-continuous in its second and third variable, we obtain, for ϵ>0

Ω(g x n ,g x m ,gu) lim inf p + Ω(g x n ,g x m ,g x p )ϵ,mn,
(14)
Ω(g y n ,g y m ,gv) lim inf p + Ω(g y n ,g y m ,g y p )ϵ,mn,
(15)
Ω(g x n ,gu,g x l ) lim inf p + Ω(g x n ,g x p ,g x l )ϵ,ln,
(16)
Ω(g y n ,gv,g y l ) lim inf p + Ω(g y n ,g y p ,g y l )ϵ,ln.
(17)

We make the sum of inequalities (14), (15), (16) and (17). It follows that

0 < inf { Ω ( g x n , F ( x n , y n ) , g u ) + Ω ( g y n , F ( y n , x n ) , g v ) + Ω ( g x n , g u , F ( x n , y n ) ) + Ω ( g y n , g v , F ( y n , x n ) ) } 4 ϵ ,

for each ϵ>0, which contradicts the hypothesis.

Hence, F(u,v)=gu and F(v,u)=gv, that is, (u,v) is a coupled coincidence point of F and g. □

By considering g=I d X , we get the following corrolary.

Corollary 3.5 Let (X,G) be a complete G-metric space, and let Ω be an Ω-distance on X. Consider F:X×XX and ϕ:X R + such that

Ω ( x , F ( x , y ) , F ( z , z ) ) + Ω ( y , F ( y , x ) , F ( z , z ) ) ϕ ( x ) + ϕ ( y ) + ϕ ( z ) + ϕ ( z ) ϕ ( F ( x , y ) ) ϕ ( F ( y , x ) ) ϕ ( F ( z , z ) ) ϕ ( F ( z , z ) )

for all x,y,z, z X. Suppose that the following conditions are fulfilled:

  1. (1)

    There exists k>0 such that Ω(x,x,y)kΩ(x,y,y) holds for all x,yX.

  2. (2)

    If F(u,v)u or F(v,u)v, then

    inf { Ω ( x , F ( x , y ) , u ) + Ω ( y , F ( y , x ) , v ) + Ω ( x , u , F ( x , y ) ) + Ω ( y , v , F ( y , x ) ) } > 0 .

Then F has coupled fixed point (u,v).

Now, we introduce the following example to support the useability of our result.

Example 3.1 Let X=[0,1]. Define

G:X×X×X R + ,G(x,y,z)=|xy|+|xz|+|yz|

and

Ω:X×X×X R + ,Ω(x,y,z)=|xy|+|xz|.

Also define

F:X×XX,F(x,y)= 1 2 x;g:XX,gx=x;ϕ:X R + ,ϕ(x)=4x.

Then,

  1. (1)

    (X,G) is a complete G-metric space.

  2. (2)

    Ω is Ω-distance.

  3. (3)

    Ω(x,x,y)2Ω(x,y,y) for all x,yX.

  4. (4)

    F(X×X)gX.

  5. (5)

    For x,y,z, z X we have

    Ω ( x , F ( x , y ) , F ( z , z ) ) + Ω ( y , F ( y , x ) , F ( z , z ) ) ϕ ( x ) + ϕ ( y ) + ϕ ( z ) + ϕ ( z ) ϕ ( F ( x , y ) ) ϕ ( F ( y , x ) ) ϕ ( F ( z , z ) ) ϕ ( F ( z , z ) ) .
  6. (6)

    If F(u,v)u or F(v,u)v, then

    inf { Ω ( x , F ( x , y ) , u ) + Ω ( y , F ( y , x ) , v ) + Ω ( x , u , F ( x , y ) ) + Ω ( y , v , F ( y , x ) ) } > 0 .

Proof The proof of (1), (2), (3) and (4) is clear. To prove (5) given x,y,z, z X.

Ω ( x , F ( x , y ) , F ( z , z ) ) + Ω ( y , F ( y , x ) , F ( z , z ) ) = Ω ( x , 1 2 x , 1 2 z ) + Ω ( y , 1 2 y , 1 2 z ) = 1 2 x + | x 1 2 z | + 1 2 y + | y 1 2 z | 3 2 x + 1 2 z + 3 2 y + 1 2 z 2 x + 2 y + 2 z + 2 z = ϕ ( x ) + ϕ ( y ) + ϕ ( z ) + ϕ ( z ) ϕ ( F ( x , y ) ) ϕ ( F ( y , x ) ) ϕ ( F ( z , z ) ) ϕ ( F ( z , z ) ) .

To prove (6), let F(u,v)u or F(v,u)v. Then u0 or v0. Thus,

inf { Ω ( x , F ( x , y ) , u ) + Ω ( y , F ( y , x ) , v ) + Ω ( x , u , F ( x , y ) ) + Ω ( y , v , F ( y , x ) ) : x , y X } = inf { Ω ( x , 1 2 x , u ) + Ω ( y , 1 2 y , v ) + Ω ( x , u , 1 2 x ) + Ω ( y , v , 1 2 y ) : x , y X } = inf { x + 2 | x u | + y + 2 | y v | : x , y X } = inf { x + 2 | x u | : x X } + inf { y + 2 | y v | : y X } u + v > 0 .

So, F and g satisfy all the hypotheses of Corollary 3.5. Hence the mappings F and g have a coupled coincidence point, Here (0,0) is the coupled coincidence point of F and g. □