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Multi-valued version of S C C , S K C , K S C , and C S C conditions in Ptolemy metric spaces

Journal of Inequalities and Applications20142014:471

https://doi.org/10.1186/1029-242X-2014-471

Received: 16 February 2014

Accepted: 13 November 2014

Published: 26 November 2014

Abstract

In this paper, multi-valued version of S C C , S K C , K S C , and C S C conditions in Ptolemy metric space are presented. Then the existence of a fixed point for these mappings in a Ptolemy metric space are proved. Finally, some examples are presented.

MSC:47H10.

Keywords

C A T ( 0 ) spacesfixed pointC conditionPtolemy metric space

1 Introduction

The definition of a Ptolemy metric space is introduced by Schoenberg [1, 2]. In order to define it, we need to recall the definition of a Ptolemy inequality as follows.

Definition 1.1 [1]

Let ( X , d ) be a metric space, the inequality
d ( x , y ) d ( z , p ) d ( x , z ) d ( y , p ) + d ( x , p ) d ( y , z )

is called a Ptolemy inequality, where x , y , z , p X .

Now, the definition of Ptolemy metric space is as follows.

Definition 1.2 [1]

A Ptolemy metric space is a metric space where the Ptolemy inequality holds.

Schoenberg proved that every pre-Hilbert space is Ptolemaic and each linear quasinormed Ptolemaic space is a pre-Hilbert space (see [1] and [3]). Moreover, Burckley et al. [4] proved that C A T ( 0 ) spaces are Ptolemy metric spaces. They presented an example to show the converse is not true. Espinola and Nicolae in [5] proved a geodesic Ptolemy space with a uniformly continuous midpoint map is reflexive. With respect to this, they proved some fixed point theorems.

In 2008, Suzuki [6] introduced the C condition.

Definition 1.3 Let T be a mapping on a subset K of a metric space X, then T is said to satisfy C condition if
1 2 d ( x , T x ) d ( x , y ) implies d ( T x , T y ) d ( x , y ) ,

for all x , y K .

Karapınar and Taş [7] presented some new definitions which are modifications of Suzuki’s C condition as follows.

Definition 1.4 Let T be a mapping on a subset K of a metric space X.
  1. (i)
    T is said to satisfy the S C C condition if
    1 2 d ( x , T x ) d ( x , y ) implies d ( T x , T y ) M ( x , y ) ,
    where
    M ( x , y ) = max { d ( x , y ) , d ( x , T x ) , d ( T y , y ) , d ( T x , y ) , d ( x , T y ) } for all  x , y K .
     
  2. (ii)
    T is said to satisfy the S K C condition if
    1 2 d ( x , T x ) d ( x , y ) implies d ( T x , T y ) N ( x , y ) ,
    where
    N ( x , y ) = max { d ( x , y ) , 1 2 { d ( x , T x ) + d ( T y , y ) } , 1 2 { d ( T x , y ) + d ( x , T y ) } } for all  x , y K .
     
  3. (iii)
    T is said to satisfy the K S C condition if
    1 2 d ( x , T x ) d ( x , y ) implies d ( T x , T y ) 1 2 { d ( x , T x ) + d ( T y , y ) } .
     
  4. (iv)
    T is said to satisfy the C S C condition if
    1 2 d ( x , T x ) d ( x , y ) implies d ( T x , T y ) 1 2 { d ( T x , y ) + d ( x , T y ) } .
     

It is clear that every nonexpansive mapping satisfies the S K C condition [[7], Proposition 9]. There exist mappings which do not satisfy the C condition, but they satisfy the S C C condition as the following example shows.

Example 1.5 [8]

Define a mapping T on [ 0 , 3 ] with d ( x , y ) = | x y | by
T ( x ) = { 0 if  x 3 , 2 if  x = 3 .

Karapınar and Taş [7] proved some fixed point theorems as follows.

Theorem 1.6 Let T be a mapping on a closed subset K of a metric space X. Assume T satisfies the S K C , K S C , S C C or C S C condition, then F ( T ) is closed. Moreover, if X is strictly convex and K is convex, then F ( T ) is also convex.

Theorem 1.7 Let T be a mapping on a closed subset K of a metric space X which satisfying the S K C , K S C , S C C or C S C condition, then d ( x , T y ) 5 d ( T x , x ) + d ( x , y ) holds for x , y K .

Hosseini Ghoncheh and Razani [8] proved some fixed point theorems for the S C C , S K C , K S C , and C S C conditions in a single-valued version in Ptolemy metric space. In this paper, the notation of S C C , S K C , K S C , and C S C conditions are generalized for multi-valued mappings and some new fixed point theorems are obtained in Ptolemy metric spaces.

Let X be a metric space and { x n } be a bounded sequence in X. For x X , let
r ( x , { x n } ) = lim sup n d ( x , x n ) .
The asymptotic radius r ( { x n } ) of { x n } in K is given by
r ( K , { x n } ) = inf x K r ( x , { x n } ) ,
and the asymptotic center A ( { x n } ) of { x n } in K is the set
A ( K , { x n } ) = { x K : r ( x , { x n } ) } = r ( K , { x n } ) .

Definition 1.8 [9]

A sequence { x n } in a C A T ( 0 ) space X is said to be Δ-convergent to x X , if x is the unique asymptotic center of every subsequence of { x n } .

Lemma 1.9
  1. (i)

    Every bounded sequence in X has a Δ-convergent subsequence [[10], p.3690].

     
  2. (ii)

    If C is a closed convex subset of X and if { x n } is a bounded sequence in C, then the asymptotic center of { x n } is in C [[11], Proposition  2.1].

     
  3. (iii)

    If C is a closed convex subset of X and if f : C X is a nonexpansive mapping, then the conditions, { x n } Δ -converges to x and d ( x n , f ( x n ) ) 0 , imply x C and f ( x ) = x [[10], Proposition  3.7].

     

Lemma 1.10 [12]

If { x n } is a bounded sequence in X with A ( { x n } ) = { x } and { u n } is a subsequence of { x n } with A ( { u n } ) = { u } and the sequence { d ( x n , u ) } converges, then x = u .

The next lemma and theorem play main roles for obtaining a fixed point in the Ptolemy metric spaces.

Lemma 1.11 [13]

Let { z n } and { w n } be bounded sequences in metric space K and λ ( 0 , 1 ) . Suppose z n + 1 = λ w n + ( 1 λ ) z n and d ( w n + 1 , w n ) d ( z n + 1 , z n ) for all n N . Then lim sup n d ( w n , z n ) = 0 .

Theorem 1.12 [5]

Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, { x n } X a bounded sequence and K X nonempty closed and convex. Then { x n } has a unique asymptotic center in K.

2 Main results

Let X be complete geodesic Ptolemy space and P ( X ) denote the class of all subsets of X. Denote
P f ( X ) = { A X : A  has property  f } .

Thus P b d , P c l , P c v , P c p , P c l , b d , P c p , c v denote the classes of bounded, closed, convex, compact, closed bounded, and compact convex subsets of X, respectively. Also T : K P f ( X ) is called a multi-valued mapping on X. A point u X is called a fixed point of T if u T u .

Definition 2.1 [14]

Let K be a subset of a C A T ( 0 ) space X. A map T : X P ( X ) is said to satisfy the C condition if for each x K , u x T x , and y K
1 2 d ( x , u x ) d ( x , y )
there exists a u y T y such that
d ( u x , u y ) d ( x , y ) .

Espinola and Nicolae [5] used the C condition as follows.

Theorem 2.2 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K a nonempty bounded, closed, and convex subset of X. Suppose T : K P c p ( K ) is a multi-valued mapping satisfying the C condition, then F ( T ) .

Now, we extend the S C C , S K C , K S C , and C S C conditions to multi-valued versions.

Definition 2.3 Let K be a subset of a geodesic Ptolemy space X. A map T : X P ( X ) is said to satisfy conditions (i) S C C , (ii) S K C , (iii) K S C , (iv) C S C if for each x K , u x T x , and y K
1 2 d ( x , u x ) d ( x , y )
there exists a u y T y such that
  1. (i)
    d ( u x , u y ) M ( x , y ) , where
    M ( x , y ) = max { d ( x , y ) , d ( x , u x ) , d ( u y , y ) , d ( u x , y ) , d ( x , u y ) } ,
     
  2. (ii)
    d ( u x , u y ) N ( x , y ) , where
    N ( x , y ) = max { d ( x , y ) , 1 2 { d ( x , u x ) + d ( u y , y ) } , 1 2 { d ( u x , y ) + d ( x , u y ) } } ,
     
  3. (iii)

    d ( u x , u y ) 1 2 { d ( x , u x ) + d ( u y , y ) } ,

     
  4. (iv)

    d ( u x , u y ) 1 2 { d ( u x , y ) + d ( x , u y ) } .

     

Remark 2.4 Notice that any K S C or C S C map is a S K C map.

Lemma 2.5 Let X be a complete geodesic Ptolemy space, and K a nonempty closed subset of X. Suppose T : K P c p ( K ) is a multi-valued mapping satisfying the S K C condition, then for every x , y K , u x T ( x ) and u x x T ( u x ) the following hold:
  1. (i)

    d ( u x , u x x ) d ( x , u x ) ,

     
  2. (ii)

    either 1 2 d ( x , u x ) d ( x , y ) or 1 2 d ( u x , u x x ) d ( u x , y ) ,

     
  3. (iii)

    either d ( u x , u y ) N ( x , y ) or d ( u y , u x x ) N ( u x , y ) ,

     
where
N ( u x , y ) = max { d ( u x , y ) , 1 2 { d ( u x x , u x ) + d ( u y , y ) } , 1 2 { d ( u x x , y ) + d ( u y , u x ) } } .
Proof The first statement follows from the S K C condition. Indeed, we always have
1 2 d ( x , u x ) d ( x , u x ) ,
which yields
d ( u x , u x x ) N ( x , u x ) ,
(2.1)
where
N ( x , u x ) = max { d ( x , u x ) , 1 2 { d ( u x , x ) + d ( u x x , u x ) } , 1 2 { d ( u x , u x ) + d ( u x x , x ) } } = max { d ( x , u x ) , 1 2 { d ( u x , x ) + d ( u x x , u x ) } , 1 2 d ( u x x , x ) } .
If N ( x , u x ) = d ( x , u x ) we are done. If N ( x , u x ) = 1 2 { d ( u x , x ) + d ( u x x , u x ) } then (2.1) turns into
d ( u x , u x x ) N ( x , u x ) = 1 2 { d ( u x , x ) + d ( u x x , u x ) } .
(2.2)
By simplifying (2.2), one can get (i). For the case N ( x , u x ) = 1 2 d ( u x x , x ) (2.1) turns into
d ( u x , u x x ) N ( x , u x ) = 1 2 d ( u x x , x ) 1 2 { d ( u x , x ) + d ( u x x , u x ) } ,
which implies (i). It is clear that (iii) is a consequence of (ii). To prove (ii), assume the contrary, that is,
1 2 d ( u x , x ) > d ( x , y ) and 1 2 d ( u x x , u x ) > d ( u x , y )
hold for all x , y K . Thus by triangle inequality and (i), we have
d ( x , u x ) d ( x , y ) + d ( y , u x ) < 1 2 { d ( u x , x ) + d ( u x x , u x ) } 1 2 d ( u x , x ) + 1 2 d ( u x , x ) = d ( x , u x ) .

 □

Theorem 2.6 Let X be a complete geodesic Ptolemy space, K a nonempty closed subset of X. Suppose T : K P c p ( K ) is a multi-valued mapping satisfying S K C condition, then d ( x , u y ) 7 d ( u x , x ) + d ( x , y ) for all x , y K , u x T x , and u y T y .

Proof The proof is based on Lemma 2.5; it is proved that
d ( u x , u y ) N ( x , y ) or d ( u y , u x x ) N ( u x , y )
holds, where
N ( u x , y ) = max { d ( u x , y ) , 1 2 { d ( u x x , u x ) + d ( u y , y ) } , 1 2 { d ( u x x , y ) + d ( u y , u x ) } } .
Consider the first case. If N ( x , y ) = d ( x , y ) , then we have
d ( x , u y ) d ( x , u x ) + d ( u x , u y ) d ( x , u x ) + d ( x , y ) .
For N ( x , y ) = 1 2 { d ( u x , x ) + d ( u y , y ) } one can observe
d ( x , u y ) d ( x , u x ) + d ( u x , u y ) d ( x , u x ) + 1 2 { d ( u x , x ) + d ( u y , y ) } 3 2 d ( u x , x ) + 1 2 d ( u y , y ) 3 2 d ( u x , x ) + 1 2 { d ( u y , x ) + d ( x , y ) } .
Thus,
1 2 d ( x , u y ) 3 2 d ( x , u x ) + 1 2 d ( x , y ) if and only if d ( x , u y ) 3 d ( x , u x ) + d ( x , y ) .
For N ( x , y ) = 1 2 { d ( u x , y ) + d ( u y , x ) } one can obtain
d ( x , u y ) d ( x , u x ) + d ( u x , u y ) d ( x , u x ) + 1 2 { d ( u x , y ) + d ( u y , x ) } d ( u x , x ) + 1 2 { d ( u x , x ) + d ( x , y ) } + 1 2 d ( u y , x ) .
Thus
1 2 d ( x , u y ) 3 2 d ( x , u x ) + 1 2 d ( x , y ) if and only if d ( x , u y ) 3 d ( x , u x ) + d ( x , y ) .
Take the second case into account. For N ( u x , y ) = d ( u x , y )
d ( x , u y ) d ( x , u x ) + d ( u x , u x x ) + d ( u x x , u y ) d ( x , u x ) + ( u x , x ) + d ( u x , y ) = 2 d ( x , u x ) + d ( u x , y ) 2 d ( x , u x ) + d ( u x , x ) + d ( x , y ) = 3 d ( x , u x ) + d ( x , y ) .
If N ( u x , y ) = 1 2 { d ( u x x , u x ) + d ( u y , y ) } then
d ( x , u y ) d ( x , u x ) + d ( u x , u x x ) + d ( u x x , u y ) 2 d ( x , u x ) + 1 2 { d ( u x x , u x ) + d ( u y , y ) } 5 2 d ( x , u x ) + 1 2 d ( u y , y ) 5 2 d ( x , u x ) + 1 2 { d ( u y , x ) + d ( x , y ) } ;
then
1 2 d ( x , u y ) 5 2 d ( x , u x ) + 1 2 d ( x , y ) if and only if d ( x , u y ) 5 d ( x , u x ) + d ( x , y ) .
For the last case, N ( u x , y ) = 1 2 { d ( u x x , y ) + d ( u y , u x ) } and we have
d ( x , u y ) d ( x , u x ) + d ( u x , u x x ) + d ( u x x , u y ) 2 d ( x , u x ) + 1 2 { d ( u x x , y ) + d ( u y , u x ) } 2 d ( x , u x ) + 1 2 { d ( u x x , u x ) + d ( u x , x ) + d ( x , y ) } + 1 2 { d ( u y , x ) + d ( x , u x ) } 7 2 d ( x , u x ) + 1 2 d ( u y , x ) + 1 2 d ( x , y ) .
Thus
1 2 d ( x , u y ) 7 2 d ( x , u x ) + 1 2 d ( x , y ) if and only if d ( x , u y ) 7 d ( x , u x ) + d ( x , y ) .

Hence, the result follows from all the above cases. □

Corollary 2.7 Let X be a complete geodesic Ptolemy space, K a nonempty closed subset of X. Suppose T : K P c p ( K ) is a multi-valued mapping satisfying S C C condition, then d ( x , u y ) 7 d ( u x , x ) + d ( x , y ) for all x , y K , u x T x , and u y T y .

Theorem 2.8 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K a nonempty, bounded, closed, and convex subset of X. Suppose T : K P c p ( K ) is a multi-valued mapping satisfying the S K C condition and x n is a sequence in K with lim n d ( x n , u x n ) = 0 , where u x n T x n , then F ( T ) .

Proof By Theorem 1.12, x n has unique asymptotic center denoted by x. Let n N . Applying Theorem 2.6 for x n , x, and u x n , respectively, it follows that there exists u z n T x such that d ( x n , u z n ) 7 d ( x n , u x n ) + d ( x n , x ) .

Let u z n k be a subsequence of u z n that converges to some u z T x , then
d ( x n k , u z ) d ( x n k , u z n k ) + d ( u z n k , u z ) 7 d ( x n k , u x n k ) + d ( x n k , x ) + d ( u z n k , u z ) ,

taking the superior limit as k and knowing that the asymptotic center of { x n k } is precisely x. Thus we obtain x = u z T x . Hence the proof is complete. □

By the same idea of [[4], p.6] we construct a function T : X P ( X ) , which is S K C and has a fixed point.

Example 2.9 Consider the space
X = { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) , ( 1 , 2 ) }
with l metric,
d ( ( x 1 , y 1 ) , ( x 2 , y 2 ) ) = max { | x 1 x 2 | , | y 1 y 2 | } .

X is a geodesic Ptolemy space, but it is not a C A T ( 0 ) space (see [4]).

Define a mapping T on X by
T ( x , y ) = { { ( 1 , 1 ) , ( 0 , 0 ) } if  ( x , y ) ( 0 , 0 ) , { ( 0 , 1 ) } if  ( x , y ) = ( 0 , 0 ) .
T satisfies the S K C condition. Suppose x = ( 0 , 0 ) and y = ( 1 , 1 ) , thus T x = { ( 0 , 1 ) } , then u x = ( 0 , 1 ) , so
1 2 d ( x , u x ) = 1 2 d ( ( 0 , 0 ) , ( 0 , 1 ) ) = 1 2 d ( x , y ) = d ( ( 0 , 0 ) , ( 1 , 1 ) ) = 1 ,
and we can choose u y = ( 0 , 0 ) ,
N ( ( 0 , 0 ) , ( 1 , 1 ) ) = max { d ( ( 0 , 0 ) , ( 1 , 1 ) ) , 1 2 [ d ( ( 0 , 0 ) , ( 0 , 1 ) ) + d ( ( 0 , 0 ) , ( 1 , 1 ) ) ] , 1 2 [ d ( ( 0 , 1 ) , ( 1 , 1 ) ) + d ( ( 0 , 0 ) , ( 0 , 0 ) ) ] } = 1 ;
thus
d ( u x , u y ) = d ( ( 0 , 1 ) , ( 0 , 0 ) ) = 1 N ( x , y ) = N ( ( 0 , 0 ) , ( 1 , 1 ) ) = 1 .

One can check the S K C condition holds for the other points of the space X.

Note that ( 1 , 1 ) T ( 1 , 1 ) ; thus F ( T ) = { ( 1 , 1 ) } .

Corollary 2.10 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K a nonempty bounded, closed, and convex subset of X. Suppose T : K P c p ( K ) is a multi-valued mapping satisfying the condition S C C and x n is a sequence in K with lim n d ( x n , T x n ) = 0 , then F ( T ) .

One can find in [15] the multi-valued version of the E μ and C λ conditions.

Definition 2.11 Let K be a subset of a metric space ( X , d ) . A map T : K P c l , b d ( X ) is said to satisfy the E μ condition provided that
dist ( x , T y ) μ dist ( x , T x ) + d ( x , y ) , x , y K ;

we say that T satisfies the E condition whenever T satisfies E μ for some μ 1 .

One can replace the metric space with a Ptolemy space in the following definition.

Definition 2.12 Let K be a subset of a metric space ( X , d ) and λ ( 0 , 1 ) . A map T : K P ( X ) is said to satisfy the C λ condition if for each x , y K ,
λ dist ( x , T x ) d ( x , y )
implies
H ( T x , T y ) d ( x , y ) ,

where H ( , ) stands for the Hausdorff distance.

Theorem 2.13 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K be a nonempty bounded, closed, and convex subset of X. Suppose T : K P c l , b d ( K ) is a multi-valued mapping satisfying E and C λ conditions, then F ( T ) .

Proof We find an approximate fixed point for T. Take x 0 K , since T x 0 we can choose y 0 T x 0 . Define
x 1 = ( 1 λ ) x 0 λ y 0 .
Since K is convex, x 1 K . Let y 1 T x 1 be chosen such that
d ( y 0 , y 1 ) = dist ( y 0 , T x 1 ) .
Similarly, set
x 2 = ( 1 λ ) x 1 λ y 1 .
Again we choose y 2 T x 2 such that
d ( y 1 , y 2 ) = dist ( y 1 , T x 2 ) .
By the same argument, we get y 2 K . In this way we find a sequence { x n } K such that
x n + 1 = ( 1 λ ) x n λ y n ,
where y n T x n and
d ( y n 1 , y n ) = dist ( y n 1 , T x n ) .
For every n N
λ d ( x n , y n ) = d ( x n , x n + 1 ) ,
for which it follows that
λ dist ( x n , T x n ) λ d ( x n , y n ) = d ( x n , x n + 1 ) ;
since T satisfies the C λ condition,
H ( T x n , T x n + 1 ) d ( x n , x n + 1 ) ,
this implies
d ( y n + 1 , y n ) = dist ( y n , T x n + 1 ) H ( T x n , T x n + 1 ) d ( x n , x n + 1 ) .
Now, we apply Lemma 1.11 to conclude lim n d ( x n , y n ) = 0 , where y n T x n . The bounded sequence { x n } is Δ-convergent, hence by passing to a subsequence Δ - lim n x n = v K . We choose z n T v such that
d ( x n , z n ) = dist ( x n , T v ) .
Since Tv is compact, the sequence { z n } has a convergent subsequence { z n k } with lim k z n k = w T v . Moreover, z n K , and K is closed; then w K . By the E condition
dist ( x n k , T v ) μ dist ( x n k , T x n k ) + d ( x n k , v ) for some  μ 1 .
Note that
d ( x n k , w ) d ( x n k , z n k ) + d ( z n k , w ) μ dist ( x n k , T x n k ) + d ( x n k , v ) + d ( z n k , w ) ;
this implies
lim sup n d ( x n k , w ) lim sup n d ( x n k , v ) .

Thus by the Opial property, w = v T v . □

Example 2.14 [15]

Let X = R and D = [ 0 , 7 2 ] . Define a mapping T on D with d ( x , y ) = | x y | by
T ( x ) = { [ 0 , x 7 ] if  x 7 2 , { 1 } if  x = 7 2 .
First we show T satisfies the C λ condition. Let x , y [ 0 , 7 2 ) , then
H ( T x , T y ) = | x y 7 | | x y | .
Let x [ 0 , 5 2 ] and y = 7 2 , then
H ( T x , T y ) = 1 7 2 x .
Let x ( 5 2 , 7 2 ) and y = 7 2 , then dist ( x , T x ) = 6 x 7 , thus
1 2 dist ( x , T x ) = 6 x 14 > 30 28 > 1 > | x y |
and
1 2 dist ( y , T y ) = 5 4 > 1 > | x y | .
Thus T satisfies the C λ condition with λ = 1 2 . Let x , y D , then
dist ( x , T y ) 3 d ( x , T x ) + | x y | ,

this shows T satisfies the E condition. Since T ( 0 ) = { 0 } , 0 F ( T ) .

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran

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© Hosseini Ghoncheh and Razani; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

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