Open Access

Best Proximity Points of Cyclic -Contractions on Reflexive Banach Spaces

Fixed Point Theory and Applications20102010:946178

DOI: 10.1155/2010/946178

Received: 8 July 2009

Accepted: 12 January 2010

Published: 20 January 2010

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, Petru el 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].

Declarations

Acknowledgment

The authors express their gratitude to the referees for their helpful suggestions concerning the final version of this paper.

Authors’ Affiliations

(1)
Department of Mathematics, Azarbaidjan University of Tarbiat Moallem
(2)
Department of Mathematics, King AbdulAziz University

References

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Copyright

© The Author(s). 2010

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