Open Access

Remarks on contractive mappings via Ω-distance

Journal of Inequalities and Applications20132013:457

https://doi.org/10.1186/1029-242X-2013-457

Received: 19 July 2013

Accepted: 7 October 2013

Published: 7 November 2013

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.

Keywords

Ω-distancefixed pointG-metric space

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 , y X , where x y ,

     
  3. (iii)

    G ( x , x , z ) G ( x , y , z ) for all x , y , z X , with z y ,

     
  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 , a X (rectangle inequality).

     

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

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 x X 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 function Ω : 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 , a X ,

     
  2. (b)

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

     
  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 , z X . Then one can easily show that Ω = 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 , z X . If we define Ω : R 3 [ 0 , ) as follows:
Ω ( x , y , z ) = 1 3 ( | z x | + | x y | )

for all x , y , z X , 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 and x , y , z , a X . Then
  1. (a)

    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 hence y = z ;

     
  2. (b)

    if Ω ( y n , x n , x n ) α n and Ω ( x n , y m , z ) β n for m > n , then G ( y n , y m , z ) 0 , and hence y n z ;

     
  3. (c)

    if Ω ( x n , x m , x l ) α n for any l , m , n N with n m l , then { x n } is aG-Cauchy sequence;

     
  4. (d)

    if Ω ( x n , a , a ) α n for any n N , 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 constant C > 0 with Ω ( x , y , z ) C for all x , y , z X .

Definition 8 Let ( X , ) be a partially ordered set. A self-mapping T : X X is said to be non-decreasing if, for x , y X ,
x y T ( 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-mapping T : X X 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 , y X ,  with  x y ,

where ψ Ψ and ϕ Φ .

Theorem 10Let ( X , G , ) be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mapping T : X X is 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 all x , y X ,  with  x T x ,

with ψ Ψ and ϕ Φ . Suppose also that inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } > 0 for every y X with y T y . If there exists x 0 X with x 0 T x 0 , thenThas a unique fixed point, say u X . 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 n N , 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 have Ω ( 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 n N . 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 exists r 0 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, that r > 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 t N .

Let l m n with m = n + k and l = m + t ( k , t N ). 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 u X . 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 , m n ,
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 , l n .
Assume that u T u . 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 } 3 ε ,

a contraction. Hence, we have u = T u .

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, since u = T u , 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-mapping T : X X is said to be a weak-contraction mapping if itsatisfies the following condition:
Ω ( T x , T 2 x , T y ) Ω ( x , T x , y ) ϕ ( Ω ( x , T x , y ) ) for all  x , y X ,  with  x y ,

where ϕ Φ .

Corollary 12Let ( X , G , ) be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mapping T : X X is a weak-contraction mapping, that is,
Ω ( T x , T 2 x , T y ) Ω ( x , T x , y ) ϕ ( Ω ( x , T x , y ) ) for all  x , y X ,  with  x T x ,

where ϕ Φ . Suppose also that inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } > 0 for every y X with y T y . If there exists x 0 X with x 0 T x 0 , thenThas a unique fixed point, say u X . Moreover, Ω ( u , u , u ) = 0 .

If we take ϕ ( t ) = k t , 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 exists k [ 0 , 1 ) such that
Ω ( T x , T 2 x , T y ) k Ω ( x , T x , y ) for all  x , y X ,  with  x T x .

Suppose also that inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } > 0 for every y X with y T y . If there exists x 0 X with x 0 T x 0 , thenThas a unique fixed point, say u X . 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-mapping T : X X is said to be a Ćirić-type contractionmapping if it satisfies that there exists 0 k < 1 such that
Ω ( T x , T 2 x , T y ) k M ( 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 , y X with x y .

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

    For every x X and y X with y T ( y ) , inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T ( x ) } > 0 ,

     
  2. (ii)

    There exists x 0 X such that x 0 T ( x 0 ) ,

     

thenThas a fixed pointuinXand Ω ( 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  n 0 .
Since T is a Ćirić-type contraction mapping, by replacing x = 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 ) k M ( 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 since k < 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 ) k Ω ( 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 r 0 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 l m n with m = n + k and l = m + t ( k , t N ). 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 u X . 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 , m n ,
and
Ω ( x n , u , x l ) lim inf p Ω ( x n , x p , x l ) ε , l n .
Assume that u T u . 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 } 3 ε

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

Definition 16 Let ( X , ) be a partially ordered space and f , g : X X . We say that g is an f-monotonemapping if
x , y X , 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. Let f : X X and g : f ( X ) X commute, fbe non-decreasing andgbe anf-monotone mapping such that:
  1. (a)

    g f ( X ) f 2 ( X ) ;

     
  2. (b)

    Ω ( g f x , g y , g 2 x ) k M ( x , x , y ) , where M ( x , x , y ) = max { Ω ( f 2 x , f y , f g x ) , Ω ( f y , f y , g y ) , Ω ( f 2 x , f 2 x , f g x ) } for all x , y X with f ( x ) f ( y ) and 0 k < 1 ;

     
  3. (c)
    for every x X and z X with f 2 z g f z ,
    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 that f ( x 0 ) g ( x 0 ) ;

     

thenfandghave a unique common fixed pointuinXand Ω ( 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 = g f x n = f g 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 , n 0 .
(2.11)
So, by (2.9) and (2.11), for all t 1 , f x n f x n + t . Now, for all s 0 ,
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 ) k Ω ( 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 n 1 and s 0 ,
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 s 0 , 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 r 0 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 l m n with m = n + k and l = m + t ( k , t N ), 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 z X . 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 , m n ,
and
Ω ( z n , z , z l ) lim inf p Ω ( z n , z p , z l ) ε , l n .
Assume that f 2 z g f z . Since f is non-decreasing, we obtain
z n = f 2 x n + 1 = f ( f x n + 1 ) f ( f x n + 2 ) = g f x n + 1 = z n + 1 ,
then z n z n + 1 . Also, for all n 1 ,
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 , 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 } , and consequently lim n Ω ( f 2 z n , g z n , g f 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 } 3 ε
for every ε > 0 , that is a contraction. So, we have f 2 z = g f z . 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 ( g f z ) , g 2 f z ) = 0 . Since X is Ω-bounded, Ω ( g f 2 z , g ( g f z ) , g 2 f z ) = 0 < M . Similarly, Ω ( g f 2 z , g f z , g 2 f z ) k Ω ( f 2 z , f 2 z , f 2 z ) < M . By part (c) of Definition 3, G ( g f 2 z , g f z , g 2 f z ) = 0 . Then g 2 f z = g f z , which implies that g f z is a fixed point for g. Now,
f ( g f z ) = g f 2 z = g 2 f z = g f z .

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

Uniqueness. Assume that there exists v X such that f v = g v = v . Hence, we have
Ω ( v , v , v ) k Ω ( 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 ) d s > 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 let T : X X be a non-decreasing self-mapping. Suppose that ψ Ψ and ϕ Φ such that
0 ψ ( Ω ( T x , T 2 x , T y ) ) λ ( s ) d s 0 ψ ( Ω ( x , T x , y ) ) λ ( s ) d s 0 ϕ ( Ω ( x , T x , y ) ) λ ( s ) d s ,
(2.13)
for all x T x , y X , where λ Λ . Also, for every x X ,
inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } > 0

for every y X with y T y . If there exists x 0 X with x 0 T x 0 , thenThas a unique fixed point.

Proof Define γ : [ 0 , + ) [ 0 , + ) by γ ( t ) = 0 t λ ( s ) d s , 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 = γ ψ and ϕ 1 = γ ϕ . 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 let T : X X be a non-decreasing self-mapping. Suppose that thereexists 0 k < 1 such that
0 ψ ( Ω ( T x , T 2 x , T y ) ) k λ ( s ) d s 0 M ( x , x , y ) λ ( s ) d s
(2.14)
for all x T x , y X , 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 ) }
and λ Λ . Also, for every x X ,
inf { Ω ( x , y , x ) + Ω ( x , y , T x ) + Ω ( x , T x , y ) : x T x } > 0

for every y X with y T y . If there exists x 0 X with 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 ) ) d s + 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 × X R by
G ( x , y , z ) = x y + y z + z x ,
where x = sup { | x ( t ) | : t [ 0 , 1 ] } . Then ( X , G ) is a complete G-metric space. Let Ω = G . Then Ω is an Ω-distance on X.Define an ordered relation ≤ on X by
x y iff x ( 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 + and g : [ 0 , 1 ] R are 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 ) | u v |
     
for every comparable u , v R + and s , t [ 0 , 1 ] with sup t [ 0 , 1 ] 0 1 G ( t , s ) d s 1 2 ,
  1. (4)

    There exist continuous, non-decreasing functions ϕ , ψ : [ 0 , ) ( 0 , ) with ψ 1 ( { 0 } ) = ϕ 1 ( { 0 } ) = { 0 } and ψ ( r ) ψ ( 2 r ) ϕ ( 2 r ) for all r [ 0 , ) .

     

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

Proof Define T x ( t ) = 0 1 K ( t , s , x ( s ) ) d s + 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 y X with y T y , then for each n N , 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 ] | y T 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 , y X with x T x , 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 u C [ 0 , 1 ] of integral equation (3.1). □

Declarations

Acknowledgements

The authors thank anonymous reviewers for their remarkable comments, suggestionsand ideas that helped to improve this paper.

Authors’ Affiliations

(1)
Department of Mathematics, Sari Branch, Islamic Azad University
(2)
Department of Mathematics, Atilim University

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© Gholizadeh and Karapınar; licensee Springer. 2013

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