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Remarks on contractive mappings via Ω-distance

Abstract

Very recently, some authors discovered that some fixed point results in thecontext of a G-metric space can be derived from the fixed point resultsin the context of a quasi-metric space and hence the usual metric space. In thisarticle, we investigate some fixed point results in the framework of aG-metric space via Ω-distance that cannot be obtained by theusual fixed point results in the literature. We also add an application toillustrate our results.

MSC: 47H10, 54H25, 46J10, 46J15.

1 Introduction and preliminaries

Very recently, Jleli and Samet [1] and Samet et al.[2] proved that some fixed point results in the setting of G-metricspaces, introduced by Sims and Mustafa [3], are consequences of the well-known fixed point theorem in the context ofthe usual metric space. Indeed, authors in [1, 2] noticed that G(x,y,y)=q(x,y) is a quasi-metric and obtained that the results arejust a characterization of existence results in the framework of a quasi-metric. Onthe other hand, a G-metric was introduced as a generalization of the(usual) metric. Basically, G-metrics claim the geometry of three pointsinstead of two points. Consequently, Jleli and Samet [1] and Samet et al.[2] concluded that if the expression in the fixed point theorem can bereduced to two points, then it can be written as a consequence of the relatedexistence result in the literature.

Recently, Saadati et al.[4] introduced the concept of Ω-distance on a complete G-metricspace as a generalized notion of ω-distance due to Kada etal.[5]. In these papers, the authors investigate the existence/uniqueness of afixed point of certain operators in this setting. In this paper, we revise somepublished papers (see, e.g., [6, 7]) and improve the statements in a way that cannot be manipulated by thetechniques used in [1, 2] (see also [810]).

We first recall some necessary definitions and basic results on the topics in theliterature.

Definition 1 ([3])

Let X be a non-empty set. A function G:X×X×X[0,) is called a G-metric if the followingconditions are satisfied:

  1. (i)

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

  2. (ii)

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

  3. (iii)

    G(x,x,z)G(x,y,z) for all x,y,zX, with zy,

  4. (iv)

    G(x,y,z)=G(p{x,y,z}), where p is a permutation of x, y, z (symmetry),

  5. (v)

    G(x,y,z)G(x,a,a)+G(a,y,z) for all x,y,z,aX (rectangle inequality).

A G-metric is said to be symmetric if G(x,y,y)=G(y,x,x) for all x,yX.

Definition 2 ([3])

Suppose that (X,G) is a G-metric space.

  1. (1)

    A sequence { x n } in X is said to be G-Cauchy sequence if, for each ε>0, there exists a positive integer n 0 such that for all n,m,l n 0 , G( x n , x m , x l )<ε.

  2. (2)

    A sequence { x n } in X is said to be G-convergent to a point xX if, for each ε>0, there exists a positive integer n 0 such that for all m,n n 0 , G( x m , x n ,x)<ε.

Definition 3 ([4])

Let (X,G) be a G-metric space. Then a functionΩ:X×X×X[0,) is called an Ω-distance on X if thefollowing conditions are satisfied:

  1. (a)

    Ω(x,y,z)Ω(x,a,a)+Ω(a,y,z) for all x,y,z,aX,

  2. (b)

    Ω(x,y,),Ω(x,,y):X[0,) are lower semi-continuous for any x,yX,

  3. (c)

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

Example 4 ([4])

Suppose that (X,d) is a metric space. Let G: X 3 [0,) be defined as follows:

G(x,y,z)=max { d ( x , y ) , d ( y , z ) , d ( x , z ) }

for all x,y,zX. Then one can easily show thatΩ=G is an Ω-distance on X.

Example 5 ([4])

Let X=R and (X,G) be a G-metric, where

G(x,y,z)= 1 3 ( | x y | + | y z | + | x z | )

for all x,y,zX. If we define Ω: R 3 [0,) as follows:

Ω(x,y,z)= 1 3 ( | z x | + | x y | )

for all x,y,zX, then it is an Ω-distance on .

We refer, e.g., to [4, 11] for more details and examples on the topic.

Lemma 6[4]

Suppose that(X,G)is aG-metric space and Ω is an Ω-distanceonX. Let{ x n }, { y n }be sequences inXand{ α n }, { β n }be sequences in[0,)converging to zero andx,y,z,aX. Then

  1. (a)

    ifΩ(y, x n , x n ) α n andΩ( x n ,y,z) β n fornN, thenG(y,y,z)<ε, and hencey=z;

  2. (b)

    ifΩ( y n , x n , x n ) α n andΩ( x n , y m ,z) β n form>n, thenG( y n , y m ,z)0, and hence y n z;

  3. (c)

    ifΩ( x n , x m , x l ) α n for anyl,m,nNwithnml, then{ x n }is aG-Cauchy sequence;

  4. (d)

    ifΩ( x n ,a,a) α n for anynN, then{ x n }is aG-Cauchy sequence.

Definition 7 ([4])

Suppose that (X,G) is a G-metric space and Ω is anΩ-distance on X. (X,G) is called Ω-bounded if there is a constantC>0 with Ω(x,y,z)C for all x,y,zX.

Definition 8 Let (X,) be a partially ordered set. A self-mappingT:XX is said to be non-decreasing if, forx,yX,

xyT(x)T(y).

The tripled (X,G,) is called a partially ordered G-metric spaceif (X,) is a partially ordered set endowed with aG-metric on X; see also [12, 13].

2 Fixed point theorems on partially ordered G-metric spaces

We start this section with the following classes of mappings:

Φ = { ϕ | ϕ : [ 0 , ) [ 0 , )  continuous, non-decreasing } and  Ψ = { ψ | ψ : [ 0 , ) [ 0 , )  continuous, non-decreasing }

with ϕ 1 ({0})= ψ 1 ({0})={0}.

Definition 9 Let (X,) be a partially ordered space. Suppose that thereexists a G-metric on X such that (X,G) is a complete G-metric space. A self-mappingT:XX is said to be a generalized weak-contraction mappingif it satisfies the following condition:

ψ ( Ω ( T x , T 2 x , T y ) ) ψ ( Ω ( x , T x , y ) ) ϕ ( Ω ( x , T x , y ) ) for all x,yX, with xy,

where ψΨ and ϕΦ.

Theorem 10Let(X,G,)be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mappingT:XXis a generalized weak-contraction mapping, that is,

ψ ( Ω ( T x , T 2 x , T y ) ) ψ ( Ω ( x , T x , y ) ) ϕ ( Ω ( x , T x , y ) ) for allx,yX, with xTx,

withψΨandϕΦ. Suppose also thatinf{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,Tx,y):xTx}>0for everyyXwithyTy. If there exists x 0 Xwith x 0 T x 0 , thenThas a unique fixed point, sayuX. Moreover, Ω(u,u,u)=0.

Proof If x 0 =T x 0 , then the proof is finished. Suppose that x 0 T x 0 . Since x 0 T x 0 and T is non-decreasing, we obtain

x 0 T x 0 T 2 x 0 T n + 1 x 0 .

Now, if for some nN, Ω( T n x 0 , T n + 1 x 0 , T n + 1 x 0 )=0, then

ψ ( Ω ( T n + 1 x 0 , T n + 2 x 0 , T n + 2 x 0 ) ) ψ ( Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) ) ϕ ( Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) ) ,

then Ω( T n + 1 x 0 , T n + 2 x 0 , T n + 2 x 0 )=0. Due to [(a), Definition 3], we haveΩ( T n x 0 , T n + 2 x 0 , T n + 2 x 0 )=0. On the other hand, by [(c), Definition 3], weeasily derive that G( T n x 0 , T n + 2 x 0 , T n + 2 x 0 )=0, which completes the proof.

Consequently, throughout the proof, we suppose that Ω( T n x 0 , T n + 1 x 0 , T n + 1 x 0 )>0 for all nN. Hence, we have

ψ ( Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) ) ψ ( Ω ( T n 1 x 0 , T n x 0 , T n x 0 ) ) ϕ ( Ω ( T n 1 x 0 , T n x 0 , T n x 0 ) ) ,
(2.1)

which yields that

ψ ( Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) ) ψ ( Ω ( T n 1 x 0 , T n x 0 , T n x 0 ) ) .

As a result, we conclude that {Ω( T n x 0 , T n + 1 x 0 , T n + 1 x 0 )} is non-increasing. Thus, there existsr0 such that

lim n Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) =r.

We shall show that r=0. Suppose, on the contrary, thatr>0. Then we have ϕ(r)>0. Letting n on (2.1), we obtain

ψ(r)ψ(r)ϕ(r),

a contraction. Hence, we have

lim n Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) =0.
(2.2)

Recursively, we obtain that

lim n Ω ( T n x 0 , T n + 1 x 0 , T n + t x 0 ) =0
(2.3)

for every tN.

Let lmn with m=n+k and l=m+t (k,tN). By the triangle inequality, we derive that

Ω ( T n x 0 , T m x 0 , T l x 0 ) Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) + Ω ( T n + 1 x 0 , T m x 0 , T l x 0 ) Ω ( T n x 0 , T n + 1 x 0 , T n + 1 x 0 ) + Ω ( T n + 1 x 0 , T n + 2 x 0 , T n + 2 x 0 ) + + Ω ( T m 1 x 0 , T m x 0 , T l x 0 ) .

Letting n in the inequality above, by keeping the limits (2.2)and (2.3), we obtain

lim n , m , l Ω ( T n x 0 , T m x 0 , T l x 0 ) =0.

Therefore, { T n x 0 } is a G-Cauchy sequence. Since X isG-complete, { T n x 0 } converges to a point uX. Now, for ε>0 and by the lower semi-continuity of Ω,

Ω ( T n x 0 , T m x 0 , u ) lim inf p Ω ( T n x 0 , T m x 0 , T p x 0 ) ε,mn,

and

Ω ( T n x 0 , u , T l x 0 ) lim inf p Ω ( T n x 0 , T p x 0 , T l x 0 ) ε,ln.

Assume that uTu. Since T n x 0 T n + 1 x 0 ,

0<inf { Ω ( T n x 0 , u , T n x 0 ) + Ω ( T n x 0 , u , T n + 1 x 0 ) + Ω ( T n x 0 , T n + 1 x 0 , u ) : n N } 3ε,

a contraction. Hence, we have u=Tu.

We shall show that u is the unique fixed point of T. Suppose, onthe contrary, that v is another fixed point of T. So, we have

ψ ( Ω ( u , u , v ) ) = ψ ( Ω ( T u , T 2 u , T v ) ) ψ ( Ω ( u , T u , v ) ) ϕ ( Ω ( u , T u , v ) ) = ψ ( Ω ( u , u , v ) ) ϕ ( Ω ( u , u , v ) ) < ψ ( Ω ( u , u , v ) ) ,

a contraction. Thus, the fixed point u is unique. Now, sinceu=Tu, we have

ψ ( Ω ( u , u , u ) ) = ψ ( Ω ( T u , T 2 u , T u ) ) ψ ( Ω ( u , T u , u ) ) ϕ ( Ω ( u , T u , u ) ) = ψ ( Ω ( u , u , u ) ) ϕ ( Ω ( u , u , u ) ) .

So, Ω(u,u,u)=0. □

Definition 11 Let (X,) be a partially ordered space. Suppose that thereexists a G-metric on X such that (X,G) is a complete G-metric space. A self-mappingT:XX is said to be a weak-contraction mapping if itsatisfies the following condition:

Ω ( T x , T 2 x , T y ) Ω(x,Tx,y)ϕ ( Ω ( x , T x , y ) ) for all x,yX, with xy,

where ϕΦ.

Corollary 12Let(X,G,)be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mappingT:XXis a weak-contraction mapping, that is,

Ω ( T x , T 2 x , T y ) Ω(x,Tx,y)ϕ ( Ω ( x , T x , y ) ) for all x,yX, with xTx,

whereϕΦ. Suppose also thatinf{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,Tx,y):xTx}>0for everyyXwithyTy. If there exists x 0 Xwith x 0 T x 0 , thenThas a unique fixed point, sayuX. Moreover, Ω(u,u,u)=0.

If we take ϕ(t)=kt, where k[0,1), we derive Theorem 2.2 [4] as the following corollary.

Corollary 13Let(X,G,)be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that there existsk[0,1)such that

Ω ( T x , T 2 x , T y ) kΩ(x,Tx,y)for all x,yX, with xTx.

Suppose also thatinf{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,Tx,y):xTx}>0for everyyXwithyTy. If there exists x 0 Xwith x 0 T x 0 , thenThas a unique fixed point, sayuX. Moreover, Ω(u,u,u)=0.

Definition 14 Let (X,) be a partially ordered space. Suppose that thereexists a G-metric on X such that (X,G) is a complete G-metric space. A self-mappingT:XX is said to be a Ćirić-type contractionmapping if it satisfies that there exists 0k<1 such that

Ω ( T x , T 2 x , T y ) kM(x,x,y),

where

M(x,x,y)=max { Ω ( x , T x , T x ) , Ω ( y , T y , T y ) , 1 2 Ω ( x , T y , T y ) }

for all x,yX with xy.

Theorem 15Let(X,G,)be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mappingT:XXis a Ćirić-type contraction mapping.

  1. (i)

    For everyxXandyXwithyT(y), inf{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,Tx,y):xT(x)}>0,

  2. (ii)

    There exists x 0 Xsuch that x 0 T( x 0 ),

thenThas a fixed pointuinXandΩ(u,u,u)=0.

Proof By assumption (ii), there exists x 0 X such that x 0 T( x 0 ). We fix x 1 X such that x 1 =T( x 0 ). Since T is a non-decreasing mapping,T x 0 T x 1 . There exists x 2 X such that T x 1 = x 2 . Recursively, we construct the sequence{ x n } in the following way:

x n + 1 =T x n T x n + 1 = x n + 2 for all n0.

Since T is a Ćirić-type contraction mapping, by replacingx= x n and y= x n + 1 , we get that

Ω( x n + 1 , x n + 2 , x n + 2 )=Ω(T x n ,T x n + 1 ,T x n + 1 )kM( x n , x n , x n + 1 ),
(2.4)

where

M ( x n , x n , x n + 1 ) = max { Ω ( x n , T x n , T x n ) , Ω ( x n + 1 , T x n + 1 , T x n + 1 ) , 1 2 Ω ( x n , T x n + 1 , T x n + 1 ) } = max { Ω ( x n , x n + 1 , x n + 1 ) , Ω ( x n + 1 , x n + 2 , x n + 2 ) , 1 2 Ω ( x n , x n + 2 , x n + 2 ) } max { Ω ( x n , x n + 1 , x n + 1 ) , Ω ( x n + 1 , x n + 2 , x n + 2 ) , 1 2 [ Ω ( x n , x n + 1 , x n + 1 ) + Ω ( x n + 1 , x n + 2 , x n + 2 ) ] } = max { Ω ( x n , x n + 1 , x n + 1 ) , Ω ( x n + 1 , x n + 2 , x n + 2 ) } .

Notice that if M( x n , x n , x n + 1 )Ω( x n + 1 , x n + 2 , x n + 2 ), then (2.4) yields a contradiction sincek<1.

Thus, M( x n , x n , x n + 1 )Ω( x n , x n + 1 , x n + 1 ) and inequality (2.4) and k<1 turn into

Ω( x n + 1 , x n + 2 , x n + 2 )kΩ( x n , x n + 1 , x n + 1 ).
(2.5)

Upon the discussion above, we conclude that the sequence {Ω( x n , x n + 1 , x n + 1 )} is non-increasing and bounded below. Therefore, thereexists r0 such that

lim n Ω( x n , x n + 1 , x n + 1 )=r.

We shall show that r=0. By a standard calculation, using inequality (2.5)and keeping k<1 in mind, we obtain lim n Ω( x n , x n + 1 , x n + 1 )=0. We claim that the sequence { x n } is G-Cauchy. Let lmn with m=n+k and l=m+t (k,tN). By the triangle inequality, we derive that

Ω ( x n , x m , x l ) Ω ( x n , x n + 1 , x n + 1 ) + Ω ( x n + 1 , x m , x l ) Ω ( x n , x n + 1 , x n + 1 ) + Ω ( x n + 1 , x n + 2 , x n + 2 ) + + Ω ( x m 1 , x m , x l ) .
(2.6)

On the other hand, we have

Ω ( x m 1 , x m , x m + t ) k M ( x m 2 , x m 2 , x m + t 1 ) = k max { Ω ( x m 2 , x m 1 , x m 1 ) , Ω ( x m + t 1 , x m + t , x m + t ) , 1 2 Ω ( x m 2 , x m + t , x m + t ) } k max { Ω ( x m 2 , x m 1 , x m 1 ) , Ω ( x m + t 1 , x m + t , x m + t ) , 1 2 [ Ω ( x m 2 , x m 1 , x m 1 ) + Ω ( x m 1 , x m , x m ) + + Ω ( x m + t 1 , x m + t , x m + t ) ] } .
(2.7)

By combining expressions (2.6) and (2.7), we find that

Ω ( x n , x m , x l ) Ω ( x n , x n + 1 , x n + 1 ) + Ω ( x n + 1 , x n + 2 , x n + 2 ) + + Ω ( x m 2 , x m 1 , x m 1 ) + k max { Ω ( x m 2 , x m 1 , x m 1 ) , Ω ( x m + t 1 , x m + t , x m + t ) , 1 2 [ Ω ( x m 2 , x m 1 , x m 1 ) + Ω ( x m 1 , x m , x m ) + + Ω ( x m + t 1 , x m + t , x m + t ) ] } .
(2.8)

Taking n in (2.8), we conclude that

lim n , m , l Ω( x n , x m , x l )=0,

and hence { x n } is a G-Cauchy sequence due to expression (c)of Lemma 6. Since X is G-complete, { x n } converges to a point uX. Thus, for ε>0 and by the lower semi-continuity of Ω, we have

Ω( x n , x m ,u) lim inf p Ω( x n , x m , x p )ε,mn,

and

Ω( x n ,u, x l ) lim inf p Ω( x n , x p , x l )ε,ln.

Assume that uTu. Since x n + 1 x n + 2 ,

0<inf { Ω ( x n + 1 , u , x n + 1 ) + Ω ( x n + 1 , u , x n + 2 ) + Ω ( x n + 1 , x n + 2 , u ) : n N } 3ε

for every ε>0, that is a contraction. Therefore, we haveu=Tu and Ω(u,u,u)=0. □

Definition 16 Let (X,) be a partially ordered space andf,g:XX. We say that g is an f-monotonemapping if

x,yX,f(x)f(y)g(x)g(y).

Theorem 17Let(X,G,)be a partially ordered completeG-metric space, and let Ω be anΩ-distance onXsuch thatXis Ω-bounded. Letf:XXandg:f(X)Xcommute, fbe non-decreasing andgbe anf-monotone mapping such that:

  1. (a)

    gf(X) f 2 (X);

  2. (b)

    Ω(gfx,gy, g 2 x)kM(x,x,y), whereM(x,x,y)=max{Ω( f 2 x,fy,fgx),Ω(fy,fy,gy),Ω( f 2 x, f 2 x,fgx)}for allx,yXwithf(x)f(y)and0k<1;

  3. (c)

    for everyxXandzXwith f 2 zgfz,

    inf { Ω ( x , z , x ) + Ω ( x , x , z ) + Ω ( f 2 x , g x , g f x ) : f 2 x g f x } >0;
  4. (d)

    there exists x 0 f(X)such thatf( x 0 )g( x 0 );

thenfandghave a unique common fixed pointuinXandΩ(u,u,u)=0.

Proof Let x 0 f(X) such that f( x 0 )g( x 0 ). By part (a), we can choose x 1 f(X) such that f( x 1 )=g( x 0 ). Again from part (a), we can choose x 2 f(X) such that f( x 2 )=g( x 1 ). Continuing this process, we can construct sequences{ x n } in f(X) and { z n } in f 2 (X) such that

y n =g x n =f x n + 1 ,
(2.9)

and

z n =g y n 1 =gf x n =fg x n =f y n .
(2.10)

Since f( x 0 )g( x 0 ) and f( x 1 )=g( x 0 ), we have f( x 0 )f( x 1 ). Then by Definition 16, g( x 0 )g( x 1 ). Continuing, we obtain

g x n g x n + 1 ,n0.
(2.11)

So, by (2.9) and (2.11), for all t1, f x n f x n + t . Now, for all s0,

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

Then, for s=0,

Ω( z n , z n , z n + 1 )kΩ( z n 1 , z n 1 , z n ).

For s=1,

Ω( z n , z n + 1 , z n + 1 ) k 1 + 1 max { Ω ( z n 1 , z n , z n ) , Ω ( z n 1 , z n 1 , z n ) } .

For s=2,

Ω( z n , z n + 2 , z n + 1 ) k 1 + 2 max { Ω ( z n 1 , z n + 1 , z n ) , Ω ( z n 1 , z n 1 , z n ) }

and

Ω ( z n 1 , z n 1 , z n ) k max { Ω ( z n 2 , z n 2 , z n 1 ) , Ω ( z n 2 , z n 2 , z n 1 ) , Ω ( z n 2 , z n 2 , z n 1 ) } = k Ω ( z n 2 , z n 2 , z n 1 ) k n 1 Ω ( z 0 , z 0 , z 1 ) .

Therefore, for all n1 and s0,

Ω( z n , z n + s , z n + 1 ) k n + s max { Ω ( z n 1 , z n + s 1 , z n ) , Ω ( z 0 , z 0 , z 1 ) } .
(2.12)

Notice that if Ω( z n , z n + s , z n + 1 ) k n + s Ω( z 0 , z 0 , z 1 ), so for all s0, lim n Ω( z n , z n + s , z n + 1 )=0. If Ω( z n , z n + s , z n + 1 ) k n + s Ω( z n 1 , z n + s 1 , z n ), so {Ω( z n 1 , z n + s 1 , z n )} is non-increasing and bounded below. Therefore, thereexists r0 such that

lim n Ω( z n 1 , z n + s 1 , z n )=r.

We shall show that r=0. By a standard calculation, using inequality (2.12)and keeping k<1 in mind, we obtain lim n Ω( z n 1 , z n + s 1 , z n )=0. Now, for any lmn with m=n+k and l=m+t (k,tN), we have

Ω ( z n , z m , z l ) Ω ( z n , z n + 1 , z n + 1 ) + Ω ( z n + 1 , z m , z l ) Ω ( z n , z n + 1 , z n + 1 ) + Ω ( z n + 1 , z n + 2 , z n + 2 ) + + Ω ( z m 1 , z m , z l ) Ω ( z n , z n + 1 , z n + 1 ) + Ω ( z n + 1 , z n + 2 , z n + 2 ) + + Ω ( z m 1 , z m , z m ) + Ω ( z m , z m + 1 , z m + 1 ) + + Ω ( z m + t 1 , z m , z m + t ) .

So,

lim n , m , l Ω( z n , z m , z l )=0,

and consequently, by Part (3) of Lemma 6, { z n } is a G-Cauchy sequence. Since X isG-complete, { z n } converges to a point zX. Thus, for ε>0 and by the lower semi-continuity of Ω, we have

Ω( z n , z m ,z) lim inf p Ω( z n , z m , z p )ε,mn,

and

Ω( z n ,z, z l ) lim inf p Ω( z n , z p , z l )ε,ln.

Assume that f 2 zgfz. Since f is non-decreasing, we obtain

z n = f 2 x n + 1 =f(f x n + 1 )f(f x n + 2 )=gf x n + 1 = z n + 1 ,

then z n z n + 1 . Also, for all n1,

Ω ( f 2 z n , g z n , g f z n ) = Ω ( g f z n 1 , g z n , g 2 z n 1 ) k max { Ω ( f 2 z n 1 , f z n , f g z n 1 ) , Ω ( f z n , f z n , g z n ) , Ω ( f 2 z n 1 , f 2 z n 1 , f g z n 1 ) } = k max { Ω ( g f z n 2 , g z n 1 , g 2 z n 2 ) , Ω ( f z n , f z n , g z n ) , Ω ( g f z n 2 , g f z n 2 , g 2 z n 2 ) } k 3 max { Ω ( f 2 z n 2 , f z n 1 , f g z n 2 ) , Ω ( f z n 1 , f z n 1 , g z n 1 ) , Ω ( f 2 z n 2 , f 2 z n 2 , f g z n 2 ) , Ω ( f z n , f z n , g z n ) , Ω ( f 2 z n 2 , f 2 z n 2 , f g z n 2 ) , Ω ( f 2 z n 2 , f 2 z n 2 , g f z n 2 ) , Ω ( f 2 z n 2 , f 2 z n 2 , f g z n 2 ) } = k 3 max { Ω ( f 2 z n 2 , f z n 1 , f g z n 2 ) , Ω ( f z n 1 , f z n 1 , g z n 1 ) , Ω ( f 2 z n 2 , f 2 z n 2 , f g z n 2 ) , Ω ( f z n , f z n , g z n ) } k 2 n + 1 max { Ω ( f 2 z 1 , g z 1 , g f z 1 ) , Ω ( f 2 z 1 , f 2 z 1 , f g z 1 ) , Ω ( f z i , f z i , g z i ) , 0 i n } k 2 n + 1 C ,

where C=max{Ω( f 2 z 1 ,g z 1 ,gf z 1 ),Ω( f 2 z 1 , f 2 z 1 ,fg z 1 ),Ω(f z i ,f z i ,g z i ),0in}, and consequently lim n Ω( f 2 z n ,g z n ,gf z n )=0. Therefore,

0<inf { Ω ( z n , z , z n ) + Ω ( z n , z n , z ) + Ω ( f 2 z n , g z n , g f z n ) : n N } 3ε

for every ε>0, that is a contraction. So, we have f 2 z=gfz. Then, by (b),

Ω ( g f 2 z , g ( g f z ) , g 2 f z ) k max { Ω ( f 2 f z , f ( g f z ) , f g ( f z ) ) , Ω ( f ( g f z ) , f ( g f z ) , g ( g f z ) ) , Ω ( f 2 ( f z ) , f 2 ( f z ) , f g ( f z ) ) } .

So, Ω(g f 2 z,g(gfz), g 2 fz)=0. Since X is Ω-bounded,Ω(g f 2 z,g(gfz), g 2 fz)=0<M. Similarly, Ω(g f 2 z,gfz, g 2 fz)kΩ( f 2 z, f 2 z, f 2 z)<M. By part (c) of Definition 3,G(g f 2 z,gfz, g 2 fz)=0. Then g 2 fz=gfz, which implies that gfz is a fixed point for g. Now,

f(gfz)=g f 2 z= g 2 fz=gfz.

Then u=gfz is a common fixed point of f andg.

Uniqueness. Assume that there exists vX such that fv=gv=v. Hence, we have

Ω(v,v,v)kΩ(v,v,v),

and so Ω(v,v,v)=Ω(u,u,u)=0. Also, Ω(v,u,v)=0. Then, by Part (c) of Definition 3,u=v and Ω(u,u,u)=0. □

The following corollary is a generalization of Theorem 2.1 [14].

Denote by Λ the set of all functions λ:[0,+)[0,+) satisfying the following hypotheses:

  1. (i)

    λ is a Lebesgue-integrable mapping on each compact subset of [0,+),

  2. (ii)

    for every ε>0, we have 0 ε λ(s)ds>0,

  3. (iii)

    λ<1, where λ denotes the norm of λ.

Now, we have the following corollary.

Corollary 18Let(X,G,)be a partially ordered completeG-metric space, let Ω be anΩ-distance onX, and letT:XXbe a non-decreasing self-mapping. Suppose thatψΨandϕΦsuch that

0 ψ ( Ω ( T x , T 2 x , T y ) ) λ(s)ds 0 ψ ( Ω ( x , T x , y ) ) λ(s)ds 0 ϕ ( Ω ( x , T x , y ) ) λ(s)ds,
(2.13)

for allxTx, yX, whereλΛ. Also, for everyxX,

inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } >0

for everyyXwithyTy. If there exists x 0 Xwith x 0 T x 0 , thenThas a unique fixed point.

Proof Define γ:[0,+)[0,+) by γ(t)= 0 t λ(s)ds, then from inequality (2.13), we have

γ ( ψ ( Ω ( T x , T 2 x , T y ) ) ) γ ( ψ ( Ω ( x , T x , y ) ) ) γ ( ϕ ( Ω ( x , T x , y ) ) ) ,

which can be written as

ψ 1 ( Ω ( T x , T 2 x , T y ) ) ψ 1 ( Ω ( x , T x , y ) ) ϕ 1 ( Ω ( x , T x , y ) ) ,

where ψ 1 =γψ and ϕ 1 =γϕ. Since the functions ψ 1 and ϕ 1 satisfy the properties of ψ andϕ, by Theorem 10, T has a unique fixedpoint. □

Corollary 19Let(X,G,)be a partially ordered completeG-metric space, let Ω be anΩ-distance onX, and letT:XXbe a non-decreasing self-mapping. Suppose that thereexists0k<1such that

0 ψ ( Ω ( T x , T 2 x , T y ) ) kλ(s)ds 0 M ( x , x , y ) λ(s)ds
(2.14)

for allxTx, yX, where

M(x,x,y)=max { Ω ( x , T x , T x ) , Ω ( y , T y , T y ) , 1 2 Ω ( x , T y , T y ) }

andλΛ. Also, for everyxX,

inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } >0

for everyyXwithyTy. If there exists x 0 Xwith x 0 T x 0 , thenThas a unique fixed point.

3 Application

In this section, we give an existence theorem for a solution of the followingintegral equations:

x(t)= 0 1 K ( t , s , x ( s ) ) ds+g(t),t[0,1].
(3.1)

Let X=C([0,1]) be the set of all continuous functions defined on[0,1]. Define G:X×X×XR by

G(x,y,z)=xy+yz+zx,

where x=sup{|x(t)|:t[0,1]}. Then (X,G) is a complete G-metric space. LetΩ=G. Then Ω is an Ω-distance on X.Define an ordered relation ≤ on X by

xyiffx(t)y(t),t[0,1].

Then (X,) is a partially ordered set. Now, we prove thefollowing result.

Theorem 20Suppose the following hypotheses hold:

  1. (1)

    K:[0,1]×[0,1]× R + R + andg:[0,1]Rare continuous mappings,

  2. (2)

    Kis non-decreasing in its first coordinate andgis non-decreasing,

  3. (3)

    There exists a continuous function G:[0,1]×[0,1][0,+) such that

    | K ( t , s , u ) K ( t , s , v ) | G(t,s)|uv|

for every comparableu,v R + ands,t[0,1]with sup t [ 0 , 1 ] 0 1 G(t,s)ds 1 2 ,

  1. (4)

    There exist continuous, non-decreasing functionsϕ,ψ:[0,)(0,)with ψ 1 ({0})= ϕ 1 ({0})={0}andψ(r)ψ(2r)ϕ(2r)for allr[0,).

Then the integral equation has a solution inC([0,1]).

Proof Define Tx(t)= 0 1 K(t,s,x(s))ds+g(t). By hypothesis (2), we have that T isnon-decreasing.

Now, if

inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } =0

for every yX with yTy, then for each nN, there exists x n C([0,1]) with x n T x n such that

Ω( x n ,y, x n )+Ω( x n ,y,T x n )+Ω( x n ,T x n ,y) 1 n .

Then we have

Ω( x n ,y,T x n )= sup t [ 0 , 1 ] | x n y|+ sup t [ 0 , 1 ] |yT x n |+ sup t [ 0 , 1 ] |T x n x n | 1 n .

Thus,

lim n x n (t)=y(t), lim n T x n (t)=y(t).

By the continuity of K, we have

y ( t ) = lim n T x n ( t ) = 0 1 K ( t , s , lim n x n ( s ) ) d s + g ( t ) = 0 1 K ( t , s , y ( s ) ) d s + g ( t ) = T y ( t ) ,

which is a contradiction. Therefore,

inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } >0.

Now, for x,yX with xTx, we have

ψ ( Ω ( T x , T 2 x , T y ) ) = ψ ( sup t [ 0 , 1 ] | T x ( t ) T 2 x ( t ) | + sup t [ 0 , 1 ] | T 2 x ( t ) T y ( t ) | + sup t [ 0 , 1 ] | T y ( t ) T x ( t ) | ) ψ ( sup t [ 0 , 1 ] 0 1 | K ( t , s , x ( s ) ) K ( t , s , T x ( s ) ) | d s + sup t [ 0 , 1 ] 0 1 | K ( t , s , T x ( s ) ) K ( t , s , y ( s ) ) | d s + sup t [ 0 , 1 ] 0 1 | K ( t , s , y ( s ) ) K ( t , s , x ( s ) ) | d s ) ψ ( sup t [ 0 , 1 ] ( 0 1 G ( t , s ) | x ( s ) T x ( s ) | d s ) + sup t [ 0 , 1 ] ( 0 1 G ( t , s ) | T x ( s ) y ( s ) | d s ) + sup t [ 0 , 1 ] ( 0 1 G ( t , s ) | y ( s ) x ( s ) | d s ) ) ψ ( sup t [ 0 , 1 ] ( | x ( t ) T x ( t ) | ) sup t [ 0 , 1 ] 0 1 G ( t , s ) d s + sup t [ 0 , 1 ] ( | T x ( t ) y ( t ) | ) sup t [ 0 , 1 ] 0 1 G ( t , s ) d s + sup t [ 0 , 1 ] ( | y ( t ) x ( t ) | ) sup t [ 0 , 1 ] 0 1 G ( t , s ) d s ) ψ ( 1 2 sup t [ 0 , 1 ] ( | x ( t ) T x ( t ) | ) + 1 2 sup t [ 0 , 1 ] ( | T x ( t ) y ( t ) | ) + 1 2 sup t [ 0 , 1 ] ( | y ( t ) x ( t ) | ) ) ψ ( 1 2 Ω ( x , T x , y ) ) ψ ( Ω ( x , T x , y ) ) ϕ ( Ω ( x , T x , y ) ) .

Thus, by Theorem 10, there exists a solution uC[0,1] of integral equation (3.1). □

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Gholizadeh, L., Karapınar, E. Remarks on contractive mappings via Ω-distance. J Inequal Appl 2013, 457 (2013). https://doi.org/10.1186/1029-242X-2013-457

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Keywords

  • Ω-distance
  • fixed point
  • G-metric space