Best Proximity Points of Cyclic -Contractions on Reflexive Banach Spaces

Abstract

We provide a positive answer to a question raised by Al-Thagafi and Shahzad (Nonlinear Analysis, 70 (2009), 3665-3671) about best proximity points of cyclic -contractions on reflexive Banach spaces.

1. Introduction

As a generalization of Banach contraction principle, Kirk et al. proved, in 2003, the following fixed point result; see [1].

Theorem 1.1.

Let and be nonempty closed subsets of a complete metric space . Suppose that is a map satisfying , and there exists such that for all and . Then, has a unique fixed point in .

Let and be nonempty closed subsets of a metric space and a strictly increasing map. We say that is a cyclic -contraction map [2] whenever , and

(1.1)

for all and , where . Also, is called a best proximity point if . As a special case, when , in which is a constant, is called cyclic contraction.

In 2005, Petruel proved some periodic point results for cyclic contraction maps [3]. Then, Eldered and Veeramani proved some results on best proximity points of cyclic contraction maps in 2006 [4]. They raised a question about the existence of a best proximity point for a cyclic contraction map in a reflexive Banach space. In 2009, Al-Thagafi and Shahzad gave a positive answer to the question [2]. More precisely, they proved some results on the existence and convergence of best proximity points of cyclic contraction maps defined on reflexive (and strictly convex) Banach spaces [2, Theorems 9, 10, 11, and 12]. They also introduced cyclic -contraction maps and raised the following question in [2].

Question 1.

It is interesting to ask whether Theorems 9 and 10 (resp., Theorems 11 and 12) hold for cyclic -contraction maps where the space is only reflexive (resp., reflexive and strictly convex) Banach space.

In this paper, we provide a positive answer to the above question. For obtaining the answer, we use some results of [2].

2. Main Results

First, we give the following extension of [4, Proposition 3.3] for cyclic -contraction maps, where is unbounded.

Theorem 2.1.

Let be a strictly increasing unbounded map. Also, let and be nonempty subsets of a metric space , a cyclic -contraction map, and for all . Then, the sequences and are bounded.

Proof.

Suppose that (the proof when is similar). By [2, Theorem 3], . Hence, it is sufficient to prove that is bounded. Since is unbounded, there exists such that

(2.1)

If is not bounded, then there exists a natural number such that

(2.2)

Then, we have

(2.3)

Since for all and , we obtain

(2.4)

Since , we have

(2.5)

Thus, we obtain . Since

(2.6)

. Hence, . This contradiction completes the proof.

Since the proof of last result was classic, we presented it separately. Here, we provide our key result via a special proof which is a general case of Theorem 2.1.

Theorem 2.2.

Let be a strictly increasing map. Also, let and be nonempty subsets of a metric space , a cyclic -contraction map, , and for all . Then, the sequences and are bounded.

Proof.

Suppose that (the proof when is similar). By [2, Theorem 3], . Hence, either and are bounded or both are unbounded. Suppose that both sequences are unbounded. Fix and define

(2.7)

for all . Since is unbounded, for all . Thus, we can choose a strictly increasing subsequence of the sequence . Since , we have

(2.8)

Again, we can choose a strictly increasing subsequence of the sequence such that . By continuing this process, for each natural number , we can choose a strictly increasing subsequence of the sequence such that . By the construction, if we consider the sequence , then , is a strictly increasing subsequence of and for all . Now, define . Also, by induction define the sequence by . Note that, the sequence is strictly increasing and . Since is a cyclic -contraction map, is a decreasing sequence. Hence by the construction of the sequence , is a decreasing sequence. Let be given. Since , we have

(2.9)

Thus,

(2.10)

for all . Hence, we have

(2.11)

for all . Since for all and , we obtain

(2.12)

for all . Consequently

(2.13)

for all . This implies that

(2.14)

for all , where

(2.15)

is a constant. But, for all . This contradiction completes the proof.

Now by using this key result, we provide our main results which give positive answer to the question. Their proofs are basically due to Al-Thagafi and Shahzad [2]. However, the crucial role is played by our key result. Weak convergence of to is denoted by .

Theorem 2.3.

Let be a strictly increasing map. Also, let and be nonempty weakly closed subsets of a reflexive Banach space and a cyclic -contraction map. Then there exists such that

(2.16)

Proof.

If , the result follows from [2, Theorem 1]. So, we assume that . For , define for all . By Theorem 2.2, the sequences and are bounded. Since is reflexive and is weakly closed, the sequence has a subsequence such that . As is bounded and is weakly closed, we can say, without loss of generality, that as . Since as , there exists a bounded linear functional such that and . For each , we have

(2.17)

Since , by using [2, Theorem 3] we obtain

(2.18)

Hence, .

Definition 2.4.

(see [2]) Let and be nonempty subsets of a normed space , , , and . We say that satisfies the proximal property if

(2.19)

Theorem 2.5.

Let be a strictly increasing map. Also, let and be nonempty subsets of a reflexive Banach space such that is weakly closed and a cyclic -contraction map. Then, there exists such that provided that one of the following conditions is satisfied

(a) is weakly continuous on .

(b) satisfies the proximal property.

Proof.

If , the result follows from [2, Theorem 1]. So, we assume that . For , define for all . By Theorem 2.2, the sequence is bounded. Since is reflexive and is weakly closed, the sequence has a subsequence such that as .

(a)Since is weakly continuous on and , we have as . So as . The rest of the proof is similar to that of Theorem 2.3.

(b)By [2, Theorem 3], we have

(2.20)

as . Since satisfies the proximal property, we have .

Theorem 2.6.

Let be a strictly increasing map. Also, let and be nonempty closed and convex subsets of a reflexive and strictly convex Banach space and a cyclic -contraction map. If , then there exists a unique such that and .

Proof.

If , the result follows from [2, Theorem 1]. So, we assume that . Since is closed and convex, it is weakly closed. By Theorem 2.3, there exists with . To show the uniqueness of , suppose that there exists another with . Since , we conclude that . As both and are convex, by the strict convexity of , we have

(2.21)

which is a contradiction. Since , we obtain, from the uniqueness of , that . Hence , and .

Theorem 2.7.

Let be a strictly increasing map. Also, let and be nonempty subsets of a reflexive and strictly convex Banach space such that is closed and convex and a cyclic -contraction map. Then, there exists a unique such that and provided that one of the following conditions is satisfied

(a) is weakly continuous on .

(b) satisfies the proximal property.

Proof.

If , the result follows from [2, Theorem 1]. So, we assume that . Since is closed and convex, it is weakly closed. By Theorem 2.5 that there exists with . Thus, . Indeed, if we assume that . Then from the convexity of and the strict convexity of , we have

(2.22)

which is a contradiction. The uniqueness of follows as in the proof of [2, Theorem  8].