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Fixed and coupled fixed point theorems of omega-distance for nonlinear contraction

Abstract

In this paper we utilize the notion of Ω-distance in the sense of Saadati et al. (Math. Comput. Model. 52:797-801, 2010) to construct and prove some fixed and coupled fixed point theorems in a complete G-metric space for a nonlinear contraction. Also, we provide an example to support our results.

MSC:47H10, 54H25.

1 Introduction

The concept of G-metric space was introduced by Mustafa and Sims [1]. After that, many authors constructed fixed point theorems in G-metric spaces. In [2] and [3], common fixed points results for mappings which satisfy the generalized (φ,ψ)-weak contraction are obtained. In [4], the author proves a common fixed point theorem for two self-mappings verifying a contractive condition of integral type in G-metric spaces. In [5, 6] and [7], tripled coincidence point results for a mixed monotone mapping in G-metric spaces are established; also see [8]. Some common fixed point results for two self-mappings, one of them being a generalized weakly G-contraction of type A and B with respect to the other mapping, are stated in [9]. Fixed point theorems for mappings with a contractive iterate at a point are formulated in [10] and in [11]. Papers [12] and [13] refer to common fixed point theorems for single-valued and multi-valued mappings which satisfy contractive conditions on G-metric spaces. In [14] and [15], theorems from G-metric spaces are used to obtain several results on complete D-metric spaces. Various contractive conditions on G-metric spaces which lead to fixed point results are stated in [16]. Paper [17] deals with the existence of fixed point results in G-metric spaces. In [18], common fixed point theorems with ϕ-maps on G-cone metric spaces are established. In [19], a general fixed point theorem for mappings satisfying an ϕ-implicit relation is proved. Paper [20] states fixed point theorems for mappings satisfying ϕ-maps in G-metric spaces. Mohamed Jleli and Bessem Samet [21] in their nice paper pointed out that the quasi-metric spaces play a major role to construct some known fixed point theorems in a G-metric space. For other recent results in G-metric spaces, please see [2224].

The coupled fixed point is one of the most interesting subjects in metric spaces. The notion of coupled fixed point was introduced by Bhaskar and Lakshmikantham [25], and the notion of coincidence coupled fixed point was introduced by Lakshmikantham and Ćirić [26]. In recent years many authors established many nice coupled and coincidence coupled fixed point theorems in metric spaces, partial metric spaces and G-metric spaces. For some works on this subject, we refer the reader to [2738].

2 Preliminaries

It is fundamental to recall the definition of G-metric spaces.

Definition 2.1 ([1])

Let X be a nonempty set. 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 to be a G-metric space and ( x n ) to be a sequence in G.

  1. (1)

    ( x n ) is called a 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 .

Now, we recall the definitions of coupled and coincidence coupled fixed points.

Definition 2.3 ([25])

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

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

Definition 2.4 ([26])

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.

In 2010, Saadati et al. [39] utilized the notion of G-metric spaces to introduce the concept of Ω-distance. Moreover, Saadati et al. [40] constructed some fixed point theorem in G-metric spaces by using the notion of Ω-distance.

Definition 2.5 ([39])

Consider (X,G) to be a G-metric space and Ω:X×X×X[0,+). Ω is called an Ω-distance on X if it satisfies the three conditions as follows:

  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 is very useful in this paper.

Lemma 2.1 ([39, 40])

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.

The following two sets are very useful to build our nonlinear contraction in this paper:

Φ = { φ : [ 0 , + ) [ 0 , + ) | φ  is continuous, increasing , φ ( t ) = 0  if and only if  t = 0 } , Ψ = { ψ : [ 0 , + ) [ 0 , + ) | ψ  is lower semi-continuous , Ψ = ψ ( t ) = 0  if and only if  t = 0 } .

For some works on fixed point theorems based on the above sets, see, for example, [1, 7, 1420, 23, 3339, 4144].

In the present paper, we utilize the concept of Ω-distance and the sets Φ, Ψ to establish some fixed and coupled fixed point theorems. Also, we introduce an example as an application of our results.

3 Main results

In the first part of the section, we introduce and prove the following fixed point theorem.

Theorem 3.1 Let (X,G) be a G-metric space and Ω be an Ω-distance on X. Consider φΦ, ψΨ and T:XX such that

φΩ(Tx,Ty,Tz)φΩ(x,y,z)ψΩ(x,y,z)
(1)

holds for each (x,y,z)X×X×X.

Suppose that if uTu, then

inf { Ω ( x , T x , u ) : x X } >0.

Then T has a unique fixed point.

Proof Let x 0 X and x n + 1 =T x n for each nN. If there is nN for which x n + 1 = x n , then x n is a fixed point of T.

In the following, we assume x n + 1 x n for each nN.

First we shall prove that lim n + Ω( x n , x n + 1 , x n + 1 )=0.

For nN, n1, we have

φ Ω ( x n , x n + 1 , x n + 1 ) = φ Ω ( T x n 1 , T x n , T x n ) φ Ω ( x n 1 , x n , x n ) ψ Ω ( x n 1 , x n , x n ) φ Ω ( x n 1 , x n , x n ) .
(2)

φ is a nondecreasing function, hence Ω( x n , x n + 1 , x n + 1 )Ω( x n 1 , x n , x n ), n1. It follows that (Ω( x n , x n + 1 , x n + 1 )) is a nondecreasing sequence, therefore there exists lim n + Ω( x n , x n + 1 , x n + 1 )=r0.

Taking n+ in inequality (2) and using the continuity of φ and the lower semi-continuity of ψ, we get

φrφr lim inf n + ψΩ( x n 1 , x n , x n )φrψr,

imposing ψr=0, that is, r=0.

Analogously, it can be proved that lim n + Ω( x n + 1 , x n , x n )=0 and also that

lim n + Ω( x n , x n , x n + 1 )=0.

The next step is to prove that lim m , n + Ω( x n , x m , x m )=0, m>n.

By reductio ad absurdum, suppose the contrary. Hence, there exist ε>0 and two sequences ( n k ) and ( m k ) such that

Ω( x n k , x m k , x m k )ε,Ω( x n k , x m k 1 , x m k 1 )<ε, m k > n k .

As lim n + Ω( x n , x n + 1 , x n + 1 )=0, it follows

ε Ω ( x n k , x m k , x m k ) Ω ( x n k , x m k 1 , x m k 1 ) + Ω ( x m k 1 , x m k , x m k ) < ε + Ω ( x m k 1 , x m k , x m k ) ε as  k + .

Therefore, lim k + Ω( x n k , x m k , x m k )=ε.

On the other hand,

ε Ω ( x n k , x m k , x m k ) Ω ( x n k , x n k + 1 , x n k + 1 ) + Ω ( x n k + 1 , x m k , x m k ) Ω ( x n k , x n k + 1 , x n k + 1 ) + Ω ( x n k + 1 , x m k + 1 , x m k + 1 ) + Ω ( x m k + 1 , x m k , x m k ) .
(3)

The contraction condition (1) yields

φ Ω ( x n k + 1 , x m k + 1 , x m k + 1 ) φ Ω ( x n k , x m k , x m k ) ψ Ω ( x n k , x m k , x m k ) φ Ω ( x n k , x m k , x m k ) ,

so Ω( x n k + 1 , x m k + 1 , x m k + 1 )Ω( x n k , x m k , x m k ), and relation (3) becomes

ε Ω ( x n k , x n k + 1 , x n k + 1 ) + Ω ( x n k + 1 , x m k + 1 , x m k + 1 ) + Ω ( x m k + 1 , x m k , x m k ) Ω ( x n k , x n k + 1 , x n k + 1 ) + Ω ( x n k , x m k , x m k ) + Ω ( x m k + 1 , x m k , x m k ) .

Letting k+, we get lim k + Ω( x n k + 1 , x m k + 1 , x m k + 1 )=ε.

Having in mind the continuity of φ and the lower semi-continuity of ψ, we obtain

φεφε lim inf k + Ω( x n k , x m k , x m k )φεψε,

which is impossible, since ε>0.

It follows that lim m , n + Ω( x n , x m , x m )=0, m>n.

In a similar manner, it can be proved that lim m , n + Ω( x n , x n , x m )=0, m>n.

Consider now l>m>n, l,m,nN. Since

Ω( x n , x m , x l )Ω( x n , x m , x m )+Ω( x m , x m , x l )0

as l,m,n+, we conclude that lim l , m , n + Ω( x n , x m , x l )=0. By Lemma 2.1, ( x n ) is a G-Cauchy sequence in the G-complete space (X,G), so it converges to uX.

Suppose uTu. Consider ε>0. As ( x n ) is a Cauchy sequence, there is n 0 N such that

Ω( x n , x m , x l )<ε,n,m,l n 0 .

Thus

lim inf l + Ω( x n , x m , x l ) lim inf l + ε=ε,n,m n 0 .

From the lower semi-continuity of Ω in its third variables, we have

Ω( x n , x m ,u) lim inf l + Ω( x n , x m , x l )ε,n,m n 0 .
(4)

Considering m=n+1 in inequality (4), we get

Ω( x n , x n + 1 ,u)ε.

On the other hand, we have

0 < inf { Ω ( x , T x , u ) : x X } inf { Ω ( x n , x n + 1 , u ) : n n 0 } < ε ,

which contradicts the hypotheses.

Therefore, u=Tu and hence u is a fixed point of T.

We shall deal now with the uniqueness of the fixed point of T.

Suppose that there are u and v in X fixed points of the mapping T.

It follows that

φΩ(v,u,u)=φΩ(Tv,Tu,Tu)φΩ(v,u,u)ψΩ(v,u,u),

which is possible only for ψΩ(v,u,u)=0, that is, Ω(v,u,u)=0.

Similarly, it can be proved that Ω(u,v,u)=0.

According to the definition of an Ω-distance, Ω(v,u,u)=0 and Ω(u,v,u)=0 imply u=v. Hence, T has a unique fixed point. □

Haghi et al. [45] in their interesting paper showed that some common fixed point theorems can be obtained from the known fixed point theorems; for other interesting article by Haghi et al., please see [46]. By using the same method of Haghi et al. [45], we get the following result.

Theorem 3.2 Let (X,G) be a G-metric space and Ω be an Ω-distance on X. Consider φΦ, ψΨ and T,S:XX such that

φΩ(Tx,Ty,Tz)φΩ(Sx,Sy,Sz)ψΩ(Sx,Sy,Sz)

holds for each (x,y,z)X×X×X.

Suppose the following hypotheses:

  1. (1)

    TXSX.

  2. (2)

    If SuTu, then

    inf { Ω ( S x , T x , S u ) : x X } >0.

Then T and S have a unique common fixed point.

As consequent results of Theorem 3.1 and Theorem 3.2, we have the following.

Corollary 3.1 Let (X,G) be a G-metric space and Ω be an Ω-distance on X. Consider ψΨ and T:XX such that

Ω(Tx,Ty,Tz)Ω(x,y,z)ψΩ(x,y,z)

holds for each (x,y,z)X×X×X.

Suppose that if uTu, then

inf { Ω ( x , T x , u ) : x X } >0.

Then T has a unique fixed point.

Corollary 3.2 Let (X,G) be a G-metric space and Ω be an Ω-distance on X. Consider ψΨ and T,S:XX such that

Ω(Tx,Ty,Tz)Ω(Sx,Sy,Sz)ψΩ(Sx,Sy,Sz)

holds for each (x,y,z)X×X×X.

Suppose the following hypotheses:

  1. (1)

    TXSX.

  2. (2)

    If SuTu, then

    inf { Ω ( S x , T x , S u ) : x X } >0.

Then T and S have a unique common fixed point.

In the second part of the section, we introduce and prove the following coincidence coupled fixed point theorem.

Theorem 3.3 Consider (X,G) to be a G-metric space endowed with an Ω-distance called  Ω. Let F:X×XX and g:XX be two mappings with the properties F(X×X)gX, and gX is a complete subspace of X with respect to the topology induced by G.

Suppose that there exist φΦ and ψΨ such that

φ ( Ω ( F ( x , y ) , F ( x , y ) , F ( z , z ) ) + Ω ( F ( y , x ) , F ( y , x ) , F ( z , z ) ) ) φ ( Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) ) ψ ( Ω ( g x , g x , g z ) + Ω ( g y , g y , g z ) )
(5)

for each (x,y),( x , y ),(z, z )X×X.

Additionally, suppose that 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 ) : x , y X } >0.

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

Proof Let ( x 0 , y 0 )X×X. Having in mind that F(X×X)gX, for each nN, there is a pair ( x n + 1 , y n + 1 )X×X such that

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

First, we prove that

lim n + Ω(g x n ,g x n + 1 ,g x n + 1 )=0and lim n + Ω(g y n ,g y n + 1 ,g y n + 1 )=0.

Using inequality (5), we get

φ ( Ω ( g x n , g x n + 1 , g x n + 1 ) + Ω ( g y n , g y n + 1 , g y n + 1 ) ) = φ ( Ω ( F ( x n 1 , y n 1 ) , F ( x n , y n ) , F ( x n , y n ) ) + Ω ( F ( y n 1 , x n 1 ) , F ( y n , x n ) , F ( y n , x n ) ) ) φ ( Ω ( g x n 1 , g x n , g x n ) + Ω ( g y n 1 , g y n , g y n ) ) ψ ( Ω ( g x n 1 , g x n , g x n ) + Ω ( g y n 1 , g y n , g y n ) ) φ ( Ω ( g x n 1 , g x n , g x n ) + Ω ( g y n 1 , g y n , g y n ) ) .
(6)

Since φ is a nondecreasing function, we obtain

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 1 , g x n , g x n ) + Ω ( g y n 1 , g y n , g y n ) , n N , n 1 ,

that is, (Ω(g x n ,g x n + 1 ,g x n + 1 )+Ω(g y n ,g y n + 1 ,g y n + 1 )) is a nondecreasing sequence. Denote by r0 its limit.

Letting n+ in relation (6), the continuity of φ and the lower semi-continuity of ψ imply

φrφr lim inf n + ψ ( Ω ( g x n , g x n + 1 , g x n + 1 ) + Ω ( g y n , g y n + 1 , g y n + 1 ) ) φrψr,

which forces φr=0, that is, r=0.

Since Ω takes nonnegative values,

lim n + Ω(g x n ,g x n + 1 ,g x n + 1 )=0and lim n + Ω(g y n ,g y n + 1 ,g y n + 1 )=0.

A similar procedure leads us to

lim n + Ω ( g x n + 1 , g x n , g x n ) = 0 , lim n + Ω ( g y n + 1 , g y n , g y n ) = 0 ; lim n + Ω ( g x n , g x n , g x n + 1 ) = 0 , lim n + Ω ( g y n , g y n , g y n + 1 ) = 0 .

Now, our purpose is to show that

lim m , n + Ω(g x n ,g x m ,g x m )=0and lim m , n + Ω(g x n ,g x m ,g x m )=0,m>n.

Supposing the contrary, there exist ε>0 and two subsequences ( n k ) and ( m k ) for which

Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ε , Ω ( g x n k , g x m k 1 , g x m k 1 ) + Ω ( g y n k , g y m k 1 , g y m k 1 ) < ε , m k > n k .

We obtain

ε Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) Ω ( g x n k , g x m k 1 , g x m k 1 ) + Ω ( g y n k , g y m k 1 , g y m k 1 ) + Ω ( g x m k 1 , g x m k , g x m k ) + Ω ( g y m k 1 , g y m k , g y m k ) < ε + Ω ( g x m k 1 , g x m k , g x m k ) + Ω ( g y m k 1 , g y m k , g y m k ) .

As k+ and lim n + (Ω(g x n ,g x n + 1 ,g x n + 1 )+Ω(g y n ,g y n + 1 ,g y n + 1 ))=0, we get

lim k + ( Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ) =0.

Also, using the properties of Ω, we have

ε Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) Ω ( g x n k , g x n k + 1 , g x n k + 1 ) + Ω ( g x n k + 1 , g x m k , g x m k ) + Ω ( g y n k , g y n k + 1 , g y n k + 1 ) + Ω ( g y n k + 1 , g y m k , g y m k ) Ω ( g x n k , g x n k + 1 , g x n k + 1 ) + Ω ( g x n k + 1 , g x m k + 1 , g x m k + 1 ) + Ω ( g x m k + 1 , g x m k , g x m k ) + Ω ( g y n k , g y n k + 1 , g y n k + 1 ) + Ω ( g y n k + 1 , g y m k + 1 , g y m k + 1 ) + Ω ( g y m k + 1 , g y m k , g y m k ) .
(7)

Taking advantage of the contraction condition, it follows

φ ( Ω ( g x n k + 1 , g x m k + 1 , g x m k + 1 ) + Ω ( g y n k + 1 , g y m k + 1 , g y m k + 1 ) ) φ ( Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ) ψ ( Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ) φ ( Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ) .

Hence

Ω ( g x n k + 1 , g x m k + 1 , g x m k + 1 ) + Ω ( g y n k + 1 , g y m k + 1 , g y m k + 1 ) Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ,

and relation (7) becomes

ε Ω ( g x n k , g x n k + 1 , g x n k + 1 ) + Ω ( g x n k + 1 , g x m k + 1 , g x m k + 1 ) + Ω ( g x m k + 1 , g x m k , g x m k ) + Ω ( g y n k , g y n k + 1 , g y n k + 1 ) + Ω ( g y n k + 1 , g y m k + 1 , g y m k + 1 ) + Ω ( g y m k + 1 , g y m k , g y m k ) Ω ( g x n k , g x n k + 1 , g x n k + 1 ) + Ω ( g x n k , g x m k , g x m k ) + Ω ( g x m k + 1 , g x m k , g x m k ) + Ω ( g y n k , g y n k + 1 , g y n k + 1 ) + Ω ( g y n k , g y m k , g y m k ) + Ω ( g y m k + 1 , g y m k , g y m k ) .

For k+, lim k + (Ω(g x n k + 1 ,g x m k + 1 ,g x m k + 1 )+Ω(g y n k + 1 ,g y m k + 1 ,g y m k + 1 ))=ε.

The properties of φ, ψ lead us to

φ ε = lim k + φ ( Ω ( g x n k + 1 , g x m k + 1 , g x m k + 1 ) + Ω ( g y n k + 1 , g y m k + 1 , g y m k + 1 ) ) φ ε lim inf k + ψ ( Ω ( g x n k , g x m k , g x m k ) + Ω ( g y n k , g y m k , g y m k ) ) φ ε ψ ε .

Since ε>0, we obtain a contradiction. Therefore, lim m , n + Ω(g x n ,g x m ,g x m )=0 and lim m , n + Ω(g y n ,g y m ,g y m )=0, m>n.

Analogously, it can be proved that lim m , n + Ω(g x n ,g x n ,g x m )=0 and also

lim m , n + Ω(g y n ,g y n ,g y m )=0,m>n.

Consider l>m>n. Then

Ω(g x n ,g x m ,g x l )Ω(g x n ,g x m ,g x m )+Ω(g x m ,g x m ,g x l )0as n,m,l+.

By Lemma 2.1, we get lim n , m , l + Ω(g x n ,g x m ,g x l )=0, l>m>n. Hence, (g x n ) is a G-Cauchy sequence in gX, which is complete. Similarly, (g y n ) converges in gX. Let gu= lim n + g x n and gv= lim n + g y n , u,vX.

Let us show now that (u,v) is a coupled coincidence point of F and g. In that respect, consider ε>0. Since (g x n ) is a Cauchy sequence, then there exists n 0 N such that for each n,m,l n 0 , Ω(g x n ,g x m ,g x l )<ε. The properties of lower semi-continuity of Ω imply

Ω(g x n ,g x m ,gu) lim inf p + Ω(g x n ,g x m ,g x p )ε,
(8)
Ω(g y n ,g y m ,gv) lim inf p + Ω(g y n ,g y m ,g y p )ε.
(9)

Considering m=n+1 in (8) and (9), we obtain

Ω ( g x n , F ( x n , y n ) , g u ) +Ω ( g y n , F ( y n , x n ) , g v ) 2ε.

On the other hand, we get

0 < inf { Ω ( g x , F ( x , y ) , g u ) + Ω ( g y , F ( y , x ) , g v ) : x , y X } inf { Ω ( g x n , F ( x n , y n ) , g u ) + Ω ( g y n , F ( y n , x n ) , g v ) : n n 0 } 2 ε ,

which is a contradiction.

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

In the following, we refer to the uniqueness of the coupled coincidence point of F and g.

Consider (u,v) and ( u , v ) to be two coupled coincidence points of F and g.

By using the contraction condition, we obtain

φ ( Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) ) = φ ( Ω ( F ( u , v ) , F ( u , v ) , F ( u , v ) ) + Ω ( F ( v , u ) , F ( v , u ) , F ( v , u ) ) ) φ ( Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) ) ψ ( Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) ) φ ( Ω ( g u , g u , g u ) + Ω ( g v , g v , g v ) ) ,

which leads us to ψ(Ω(g u ,gu,gu)+Ω(g v ,gv,gv))=0, or Ω(g u ,gu,gu)=Ω(g v ,gv,gv)=0.

In a similar manner, we prove that Ω(gu,g u ,gu)=Ω(gv,g v ,gv)=0.

Lemma 2.1 implies that gu=g u and gv=g v .

Having in mind that gu=F(u,v) and gv=F(v,u), we get

φ ( Ω ( g u , g v , g v ) + Ω ( g v , g u , g v ) ) = φ ( Ω ( F ( u , v ) , F ( v , u ) , F ( v , u ) ) + Ω ( F ( v , u ) , F ( u , v ) , F ( u , v ) ) ) φ ( Ω ( g u , g v , g v ) + Ω ( g v , g u , g v ) ) ψ ( Ω ( g u , g v , g v ) + Ω ( g v , g u , g v ) ) ,

hence ψ(Ω(gu,gv,gv)+Ω(gv,gu,gv))=0, or Ω(gu,gv,gv)=0 and Ω(gv,gu,gv)=0. Applying Lemma 2.1, it follows that gu=gv. □

Taking g=I d X , the identity mapping, in Theorem 3.3 we obtain a theorem of coupled fixed points.

Corollary 3.3 Consider (X,G) to be a complete G-metric space endowed with an Ω-distance called Ω. Let F:X×XX be a mapping.

Suppose that there exist φΦ and ψΨ such that

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

holds for each (x,y),( x , y ),(z, z )X×X.

Additionally, suppose 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 ) : x , y X } >0.

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

Taking φ= i [ 0 , + ) , the identity function, in Theorem 3.3 and Corollary 3.3, we get the following results.

Corollary 3.4 Consider (X,G) to be a G-metric space endowed with an Ω-distance called  Ω. Let F:X×XX and g:XX be two mappings with the properties F(X×X)gX, and gX is a complete subspace of X with respect to the topology induced by G.

Suppose that there exists ψΨ such that

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

holds for each (x,y),( x , y ),(z, z )X×X.

Additionally, suppose that 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 ) : x , y X } >0.

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

Corollary 3.5 Consider (X,G) to be a complete G-metric space endowed with an Ω-distance called Ω. Let F:X×XX be a mapping.

Suppose that there exist φΦ and ψΨ such that

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

holds for each (x,y),( x , y ),(z, z )X×X.

Additionally, suppose 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 ) : x , y X } >0.

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

Corollary 3.6 Consider (X,G) to be a G-metric space endowed with an Ω-distance called  Ω. Let F:X×XX and g:XX be two mappings with the properties F(X×X)gX, and gX is a complete subspace of X with respect to the topology induced by G.

Suppose that there exists k[0,1) such that

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

holds for each (x,y),( x , y ),(z, z )X×X.

Additionally, suppose that 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 ) : x , y X } >0.

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

Proof The proof follows from Corollary 3.4 by defining ψ:[0,+)[0,+) via ψ(t)=(1k)t. □

Corollary 3.7 Consider (X,G) to be a complete G-metric space endowed with an Ω-distance called Ω. Let F:X×XX be a mapping.

Suppose that there exists k[0,1) such that

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

holds for each (x,y),( x , y ),(z, z )X×X.

Additionally, suppose 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 ) : x , y X } >0.

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

Proof The proof follows from Corollary 3.5 by defining ψ:[0,+)[0,+) via ψ(t)=(1k)t. □

The following example supports our results.

Example 3.1 Take X={0,1,2,3,}. Define G:X×X×X[0,+) by the formula

G(x,y,z)={ 0 if  x = y = z ; x + y + z if  x y ,  or  x z ,  or  y z .

Define

Ω:X×X×XX,Ω(x,y,z)=x+2max{y,z}

and

T:XX,Tx={ 0 if  x = 0 , 1 ; x 1 if  x 2 .

Also, define φ:[0,+)[0,+) via φ(t)= t 2 and ψ:[0,+)[0,+) via ψ(t)=t. Then:

  1. (1)

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

  2. (2)

    φΦ and ψΨ.

  3. (3)

    Ω is an Ω-distance function.

  4. (4)

    If uTu, then

    inf { Ω ( x , T x , u ) : x X } >0.
  5. (5)

    The following inequality:

    φΩ(Tx,Ty,Tz)φΩ(x,y,z)ψΩ(x,y,z)

holds for all x,y,zX.

Proof The proofs of (1) and (2) are clear. To prove part (3), consider x,y,z,aX. Since

x+2max{y,z}x+2a+a+2max{y,z},

we get

Ω(x,y,z)Ω(x,a,a)+Ω(a,y,z).

This finishes the proof of the first item of the definition of Ω-distance.

To prove the second item of the definition of Ω-distance, let x,yX and ( z n ) be any sequence in X converging to z with respect to the topology induced by G in X. Thus z n =z for all nN except finitely many terms. Therefore

x+2max{y, z n }x+2max{y,x}as n+.

So, Ω(x,y, z n )Ω(x,y,z) and hence Ω(x,y,):X[0,+) is lower semi-continuous.

Similarly, we can show that Ω(x,,z):X[0,+) is lower semi-continuous.

To prove the last item of the definition of Ω-distance, consider ε>0. Take δ= ε 2 . Given x,y,zX such that Ω(x,a,a)δ and Ω(a,y,z)δ, by the definition of a G-metric space, we have

G ( x , y , z ) G ( x , a , a ) + G ( a , y , z ) x + 2 a + a + y + z x + 2 a + a + 2 max { y , z } = Ω ( x , a , a ) + Ω ( a , y , z ) ε .

This completes the proof of an Ω-distance.

To prove part (4), given uX such that uTu, then u0. Note that

inf { Ω ( x , T x , u ) : x X } inf { x + 2 u + : x X } 2 u > 0 .

To prove part (5), given x,y,zX, we divide the proof into the following four cases.

Case 1: x=y=z=0. Here, Ω(x,y,z)=0 and Ω(Tx,Ty,Tz)=0. Thus

φΩ(Tx,Ty,Tz)φΩ(x,y,z)ψΩ(x,y,z).

Case 2: x>0 and y=z=0. Here, Ω(x,y,z)=x and Ω(Tx,Ty,Tz)=x1. Since ( x 1 ) 2 x 2 x, we have

φΩ(Tx,Ty,Tz)φΩ(x,y,z)ψΩ(x,y,z).

Case 3: x=0 and y or z are not equal to 0. Without loss of generality, we may assume that yz. Thus y0. Here, Ω(x,y,z)=2y and Ω(Tx,Ty,Tz)=2(y1). Since 4 ( y 1 ) 2 4 y 2 2y, we have

φΩ(Tx,Ty,Tz)φΩ(x,y,z)ψΩ(x,y,z).

Case 4: x, y, z are all different from 0. Without loss of generality, we assume that yz. Then Ω(x,y,z)=x+2y and Ω(Tx,Ty,Tz)=x1+2(y1). Since ( x 1 ) 2 x 2 x and 4 ( y 1 ) 2 4 y 2 2y, we have

φ Ω ( T x , T y , T z ) = [ x 1 + 2 ( y 1 ) ] 2 = ( x 1 ) 2 + 4 ( x 1 ) ( y 1 ) + 4 ( y 1 ) 2 x 2 x + 4 x y + 4 y 2 2 y = ( x + 2 y ) 2 ( x + 2 y ) = φ Ω ( x , y , z ) ψ Ω ( x , y , z ) .

Note that Example 3.1 satisfies all the hypotheses of Theorem 3.1. Thus T has a unique fixed point. Here, 0 is the unique fixed point of T. □

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Shatanawi, W., Pitea, A. Fixed and coupled fixed point theorems of omega-distance for nonlinear contraction. Fixed Point Theory Appl 2013, 275 (2013). https://doi.org/10.1186/1687-1812-2013-275

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