Open Access

# Results on the Existence and Convergence of Best Proximity Points

Fixed Point Theory and Applications20102010:386037

DOI: 10.1155/2010/386037

Received: 24 February 2010

Accepted: 10 June 2010

Published: 6 July 2010

## Abstract

We first consider a cyclic -contraction map on a reflexive Banach space and provide a positive answer to a question raised by Al-Thagafi and Shahzad on the existence of best proximity points for cyclic -contraction maps in reflexive Banach spaces in one of their works (2009). In the second part of the paper, we will discuss the existence of best proximity points in the framework of more general metric spaces. We obtain some new results on the existence of best proximity points in hyperconvex metric spaces as well as in ultrametric spaces.

## 1. Introduction

Let be a metric space, and let be two subsets of . A mapping is said to be cyclic provided that and . In [1] Kirk et al. proved the following interesting extension of the Banach contraction principle:

Theorem 1.1 (see [1]).

Let and be two nonempty closed subsets of a complete metric space . Suppose that is a cyclic map such that
(1.1)

for some and for all . Then has a unique fixed point in .

Later on, Eldred and Veeramani [2] considered the class of cyclic contractions.

Definition 1.2 (see [2]).

Let and be two nonempty subsets of a metric space , and let , and . We say that is a cyclic contraction if
(1.2)
for some and for all , where
(1.3)

We recall that a point is said to be a best proximity point for provided that .

In the case that is a uniformly convex Banach space, Eldred and Veeramani established the following theorem.

Theorem 1.3 (see [2]).

Let and be two nonempty closed convex subsets of a uniformly convex Banach space , and let be a cyclic contraction map. For , define for each . Then there exists a unique such that and .

In 2009, Al-Thagafi and Shahzad introduced a new class of mappings, namely, the class of cyclic -contraction maps. This new class contains the class of cyclic contraction maps.

Definition 1.4 (see [3]).

Let and be two nonempty subsets of a metric space and let be a mapping such that and . is said to be a cyclic -contraction map if there exists a strictly increasing function such that
(1.4)

for all and .

In [3] the authors were able to establish some existence and convergence results for these mappings. Moreover, they proved the existence of a best proximity point for a cyclic contraction map in a reflexive Banach space (see [3, Theorems , ]). In this way they answered a question raised by Eldred and Veeramani in the affirmative. We recall that Theorem 1.3 above was proved in the setting of a uniformly convex Banach space. The authors of [3] then asked if the result stands true if we assume that is a reflexive Banach space, rather than being uniformly convex.

Al-Thagafi and N. Shahzad then stated it was interesting to ask whether Theorems and (resp., Theorems and ) held true for cyclic -contraction maps when the Banach space in question is only reflexive (resp., reflexive and strictly convex).

In this paper we first take up these questions. It turns out that under some conditions the answer is positive. In the last section we study the existence of best proximity points in spherically complete ultrametric spaces, as well as in hyperconvex metric spaces. More precisely, we will see that best proximity points exist for cyclic -contraction maps on hyperconvex metric spaces. We will also provide an existence theorem for a cyclic map which satisfies some contractive condition on an ultrametric space.

## 2. Cyclic -Contraction Maps

In this section we first provide a positive answer to the question raised by the authors of [3]. Then we present some consequences and applications. Among other things, is a common fixed point theorem for two maps. We will begin with the following lemma.

Lemma 2.1 (see [3, Lemma ]).

Let and be two nonempty subsets of a metric space and let be a cyclic -contraction map. For , define for each . Then one has

(a) for all and ,

(b) for all and ,

(c) for all .

Now we state and prove the following lemma which is key to the proof of the main result of this section.

Lemma 2.2.

Let and be two nonempty subsets of a metric space , and let be a cyclic -contraction map. For , define for each . Then the sequences , and are bounded if either of the following conditions holds:

(i) ,

(ii)

Proof.

We first show that the sequence is bounded. Suppose the contrary. Then for every positive integer , there exists such that
(2.1)
We note that
(2.2)
According to Lemma 2.1, is nonexpansive, so that (by the property of )
(2.3)
Therefore
(2.4)
But since is increasing, it follows that
(2.5)
Thus
(2.6)
This implies that for every positive integer we have
(2.7)

contradicting the hypothesis that .

We now assume that condition (ii) holds. It follows from (2.7) that
(2.8)
Since (2.8) holds for all , we conclude that
(2.9)
for all Letting now and using Theorem of [3] we conclude that
(2.10)

which contradicts the fact that is strictly increasing.

This arguments show that the sequence is bounded. But since
(2.11)

and that both terms on the right-hand side are bounded, we conclude that is bounded.

Similarly, by considering the sequence we can prove that the sequence is bounded.

We now come to the first main result of this paper generalizing Theorem of [3] to cyclic -contraction maps.

Theorem 2.3.

Let and be two nonempty weakly closed subsets of a reflexive Banach space and let be a cyclic -contraction map satisfying either of the following:

(i) ,

(ii) .

Then there exists such that .

Proof.

Let be arbitrarily chosen. We define . It follows from Lemma 2.2 that the sequences and are bounded in and in , respectively. Since is reflexive, every bounded sequence in has a weakly convergent subsequence. Assume that weakly. Since is weakly closed, . Similarly, we may assume that there is a such that , weakly. Therefore , weakly. But according to a well-known fact in basic functional analysis, we have
(2.12)

from which it follows that .

Remark 2.4.

If we assume that the function satisfies either of the conditions (i) or (ii) of Lemma 2.2, then all three theorems (Theorems , 11, and 12 of [3] can be generalized to cyclic -contraction maps. We omit the details.

The next theorem generalizes Theorem 1.1 to reflexive Banach spaces. Note that if and for some fixed , then will be a cyclic contraction map, because for all and all we have
(2.13)

Theorem 2.5.

Let and be two nonempty subsets of a reflexive Banach space such that is weakly closed. Let be a cyclic -contraction map which is weakly continuous on . For , define for each . If then has a unique fixed point and .

Proof.

Since is cyclic -contraction, and , it follows from Lemma 2.2 that is bounded in . Therefore we can find a weak convergent subsequence, say , to a point . On the other hand, is weakly continuous, so that weakly. It follows that
(2.14)

As in the proof of Theorem 2.3 we conclude that . The proof of uniqueness part is a verbatim repetition of the proof of Theorem in [3]. We omit the details.

As an application of Theorem 2.5, we will prove a theorem on the existence and approximation of common fixed points for two maps.

Theorem 2.6.

Let be a nonempty subset of a reflexive Banach space and be two maps such that is weakly closed in and . Let be a cyclic -contraction map that satisfies this property that if there exist such that , then commutes with in . Then have a common fixed point in . Moreover, if , and for each then the sequence converges to a common fixed point of .

Proof.

By Theorem 2.5 there exists a unique such that . Since , there exists such that so that . Also there exists such that , so that . Now we have
(2.15)

That is, is a fixed point for . Since the fixed point of is unique, we must have . Therefore is a fixed point of . Similarly we can show that is a fixed point of . Consequently is a common fixed point for . According to Theorem 2.5 the sequence converges to .

Example 2.7.

Let and . Let and define with and . Also consider by . Then is cyclic contraction and satisfies the conditions of Theorem 2.6. Therefore have a common fixed point. It is clear that this common fixed point is .

## 3. Cyclic -Contraction Maps in Metric Spaces

In this section we discuss the existence of best proximity points for cyclic -contraction maps in metric spaces. Indeed we prove two existence theorems on best proximity points in hyperconvex spaces, as well as in ultrametric spaces.

Lemma 3.1.

Let be two nonempty subsets of a metric space , and Let be a cyclic -contraction map. If there exists such that , then has a best proximity point.

Proof.

Since , then is fixed point for Therefore we have
(3.1)
Thus . Since is strictly increasing, we conclude that
(3.2)

In the following definition we will use the notation for the Kuratowski measure of noncompactness of a given set . For more information see the book written by Khamsi and Kirk [4].

Definition 3.2.

Let be a subset of a metric space . A mapping is said to be condensing if is bounded and continuous, moreover , for every bounded subset of for which .

Definition 3.3 (see [4]).

A metric space is called hyperconvex if for any indexed class of closed balls , of which satisfy
(3.3)

it is necessarily the case that

We recall that for a given set , the notation denotes the family of all admissible subsets of , that is, the family of subsets of that can be written as the intersection of a family of closed balls centered at points of . For further information on the subject we refer the reader to [4]. We now state and prove the first main result of this section.

Theorem 3.4.

Let be a hyperconvex metric space, and be two nonempty subsets of such that . Suppose is a cyclic -contraction map. Put and . If is a condensing map then has a best proximity point.

Proof.

Since is a hyperconvex metric space, and since , it follows from Proposition of [5] that is a hyperconvex metric space too. On the other hand, is a condensing map, thus by Theorem of [5], or has a fixed point. It now follows from Lemma 3.1 that has a best proximity point.

Definition 3.5.

A metric space is an ultrametric space if, in addition to the usual metric axioms, the following property holds for each :
(3.4)

For example if is a discrete metric space then is an ultrametric space. Ultrametric spaces arise in the study of non-Archimedean analysis, and in particular in the study of Banach space over non-Archimedean valuation fields (see [4]).

Remark 3.6.

It is immediate from Definition 3.5 that if and are two closed balls in an ultrametric space, with , then either or In particular if , then

Definition 3.7.

An ultrametric space is said to be spherically complete if every chain of closed balls in has nonempty intersection.

As a consequence of Remark 3.6, the admissible sets of coincide with the closed balls of . Here we state and prove the second main result of this section.

Theorem 3.8.

Suppose is a spherically complete ultrametric space and are two nonempty subsets of such that . Let be a cyclic map which satisfies the following condition:
(3.5)

for each and for some Then has a best proximity point.

Proof.

Let and define for . Put By Theorem of [6], Now if there exits such that , then
(3.6)

Therefore . This argument shows that has a best proximity point.

Now let for all , we have Thus
(3.7)
Then (all balls are assumed to be closed). Now by Remark 3.6 we have
(3.8)
This shows that is a descending chain of closed balls in ; in particular, each two members of this chain intersect. It is rather obvious that each member of this chain also intersects (because ). Since and is a spherically complete ultrametric space, then itself is a closed ball (see [4, page 114]). Now each two elements of the family consisting of and intersects. Therefore if we set , according to [4, page 115], there exists a point which belongs to as well. Therefore
(3.9)
But for the second term we have
(3.10)
because
(3.11)
It now follows that
(3.12)
Since the above relation holds for all then we have
(3.13)

Therefore , which means that has a best proximity point.

In the following example we will see that the condition that is spherically complete is necessary.

Example 3.9.

Let and define a metric on by
(3.14)
It is clear that is a complete ultrametric space (see [5]). Set
(3.15)
and define the mapping by It is easy to see that is cyclic and It is not difficult to see that satisfies the relation (3.5) of the previous theorem for but has no best proximity point. To see this, assume that
(3.16)

for some . Thus which is impossible. We claim that the ultrametric space is not spherically complete.

Consider the family of closed balls in Since
(3.17)
it follows from Remark 3.6 that
(3.18)
Therefore this family is a chain of closed balls in Now let
(3.19)
for some This implies that for all we have
(3.20)

which is a contradiction.

## Declarations

### Acknowledgment

After the appearance of this paper on the current journal home page, the authors have been informed by Nasser Shahzad and Shahram Rezapour that they already published paper [7], answering a question raised by the authors of [3]. The current authors would like to thank them for this piece of information.

## Authors’ Affiliations

(1)
Department of Mathematics, Imam Khomeini International University

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