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Ω-Distance and coupled fixed point in G-metric spaces

Abstract

Saadati et al. (Math. Comput. Model. 52:797-801, 2010) introduced the concept of Ω-distance in generalized metric spaces and studied some nice fixed point theorems. Very recently, Jleli and Samet (Fixed Point Theory Appl. 2012:210, 2012) showed that some of the fixed point theorems in G-metric spaces can be obtained from quasi-metric space. In this paper, we utilize the concept of Ω-distance in the sense of Saadati et al. to establish some common coupled fixed point results. Also, we introduce an example to support the useability of our results. Note that the method of Jleli and Samet cannot be used in our results.

MSC:47H10, 54H25.

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 $G:X×X×X→X$ 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 $x,y∈X$, $x≠y$;

3. (3)

$G(x,x,z)≤G(x,y,z)$ for each triple $(x,y,z)$ from $X×X×X$ with $z≠y$;

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 $x∈X$ 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 $a∈X$, then

1. (1)

If $Ω(y, x n , x n )≤ α n$ and $Ω( x n ,y,z)≤ β n$ for $n∈N$, 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 $l,m,n∈N$ with $n≤m≤l$, then $( x n )$ is a G-Cauchy sequence.

4. (4)

If $Ω( x n ,a,a)≤ α n$, $n∈N$, 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 $F:X×X→X$ 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 $F:X×X→X$ and $g:X→X$ 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. $g:X→X$ and $F:X×X→X$ 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 $A={(i,j,t)|1≤i≤n,1≤j≤n+1,s+1≤t≤n+s−1}$.

Since X is Ω-bounded, there is $M>0$ such that $Ω(x,y,z) 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 $l>m>n>0$, $l,m,n∈N$. 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

$Ω(g x n ,g x m ,gu)≤ lim inf p → + ∞ Ω(g x n ,g x m ,g x p )≤ϵ,m≥n,$
(4)
$Ω(g y n ,g y m ,gv)≤ lim inf p → + ∞ Ω(g y n ,g y m ,g y p )≤ϵ,m≥n,$
(5)
$Ω(g x n ,gu,g x l )≤ lim inf p → + ∞ Ω(g x n ,g x p ,g x l )≤ϵ,l≥n,$
(6)
$Ω(g y n ,gv,g y l )≤ lim inf p → + ∞ Ω(g y n ,g y p ,g y l )≤ϵ,l≥n.$
(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 $F:X×X→X$ 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. $g:X→X$ and $F:X×X→X$ 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 $F:X×X→X$ 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. $g:X→X$ and $F:X×X→X$ 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

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

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

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 $F:X×X→X$ 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 $F:X×X→X$, $g:X→X$ 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 $x,y∈X$.

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

$Ω(g x n ,g x m ,g x m )+Ω(g y n ,g y m ,g y m )<η≤δ$

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 ) ]

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, $v∈X$ 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 $ϵ>0$

$Ω(g x n ,g x m ,gu)≤ lim inf p → + ∞ Ω(g x n ,g x m ,g x p )≤ϵ,m≥n,$
(14)
$Ω(g y n ,g y m ,gv)≤ lim inf p → + ∞ Ω(g y n ,g y m ,g y p )≤ϵ,m≥n,$
(15)
$Ω(g x n ,gu,g x l )≤ lim inf p → + ∞ Ω(g x n ,g x p ,g x l )≤ϵ,l≥n,$
(16)
$Ω(g y n ,gv,g y l )≤ lim inf p → + ∞ Ω(g y n ,g y p ,g y l )≤ϵ,l≥n.$
(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 $F:X×X→X$ 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 $x,y∈X$.

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

$G:X×X×X→ R + ,G(x,y,z)=|x−y|+|x−z|+|y−z|$

and

$Ω:X×X×X→ R + ,Ω(x,y,z)=|x−y|+|x−z|.$

Also define

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

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 $x,y∈X$.

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 $u≠0$ or $v≠0$. 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. □

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Correspondence to Ariana Pitea.

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Both authors contributed equally and significantly in writing this article. Both authors read and approved the final manuscript.

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