Relation between Fixed Point and Asymptotical Center of Nonexpansive Maps

Abstract

We introduce the concept of asymptotic center of maps and consider relation between asymptotic center and fixed point of nonexpansive maps in a Banach space.

1. Introduction

Many topics and techniques regarding asymptotic centers and asymptotic radius were studied by Edelstein [1], Bose and Laskar [2], Downing and Kirk [3], Goebel and Kirk [4], and Lan and Webb [5]. Now, We recall that definitions of asymptotic center and asymptotic radius.

Let be a nonempty subset of a Banach space and a bounded sequence in . Consider the functional defined by

(1.1)

The infimum of over is said to be the asymptotic radius of with respect to and is denoted by . A point is said to be an asymptotic center of the sequence with respect to if

(1.2)

The set of all asymptotic centers of with respect to is denoted by .

We present new definitions of asymptotic center and asymptotic radius that is for a mapping and obtain new results.

Definition 1.1.

Let be a bounded closed convex subset of . A sequence is said to be an asymptotic center for a mapping if, for each ,

(1.3)

Definition 1.2.

Let be a nonempty subset of . We say that has the fixed-point property for continuous mappings of with asymptotic center if every continuous mapping admitting an asymptotic center has a fixed point.

Definition 1.3.

Let be a nonempty subset of . We say that has Property if for every bounded sequence , the set is a nonempty and compact subset of .

Example 1.4.

Let be a normed space and a nonempty subset of . It is clear that

(i)if is a compact set, then in nonempty compact set and so has Property ;

(ii)if is a open set, since , therefore is empty and so fail to have Property .

2. Main Results

Our new results are presented in this section.

Proposition 2.1.

Let be a Banach space and let be a nonempty closed bounded and convex subset of . If satisfies Property , then every continuous mapping asymptotically admitting a center in has a fixed point.

Proof.

Assume that is a continuous mapping and is a asymptotic center. Let has set of asymptotic center . Since has Property , is nonempty and compact and it is easy to see that it is also convex. In order to obtain the result, it will be enough to show that is -invariant since in this case we may apply Schauder's Fixed-Point Theorem [4, Theorem  18.10]. Indeed, let . Since is a asymptotic center for , we have

(2.1)

Therefore .

Theorem 2.2.

Let be a Banach space and let be a nonempty closed bounded and convex subset of . If has the fixed-point property for continuous mappings admitting an asymptotic center, then has Property .

Proof.

Suppose that fails to have Property . There exists such that either or is noncompact. In the second case, by Klee's theorem  in [6] there exists a continuous function without fixed points (). Since a closed convex subset of a normed space is a retract of the space, there exists a continuous mapping such that for all . Define by . Clearly is a continuous mapping. Moreover,

(2.2)

that is, is an asymptotic center for . Therefore, by Proposition 2.1, has a fixed point in , . Hence sets a contradiction.

Concerning the first case we proceed as follows.

Let . We take such that . For each positive integer , we consider the following nonempty sets:

(2.3)

where

(2.4)

Since , we have that

(2.5)

Fix an arbitrary and define, by induction, a sequence such that and the segment does not meet . Given , there exists a unique positive integer such that . In this case we define

(2.6)

It is a routine to check that is a continuous mapping from to . Furthermore, for every .

Let be a continuous retraction from into the closed convex subset . We can define by . It is clear that is a asymptotic center for and that is fixed-point free.

Proposition 2.1 (Theorem 2.2) is a generalizations of Theorem  3.1 (Theorem  3.3) in [1]. It can be verified that definition of space is not necessary here.

As an easy consequence of both Proposition 2.1 and Theorem 2.2, we deduce the following result.

Corollary 2.3.

Let be a nonempty closed bounded and convex subset of a Banach space . The following conditions are equivalent.

(1) has the fixed-point property for continuous mappings admitting asymptotic center in .

(2) has Property .

Let be a nonempty closed convex bounded subset of a Banach space . By we denote the family of all nonempty compact convex subsets of . On we consider the well-known Hausdorff metric . Recall that a mapping is said to be nonexpansive whenever

(2.7)

Theorem 2.4.

Let be a Banach space and let be a nonempty closed convex and bounded subset of satisfying Property . If is a nonexpansive mapping, then has a fixed point.

Proof.

Let be a nonexpansive mapping. The multivalued analog of Banach's Contraction Principle allows us to find a sequence in such that .

For each , the compactness of guarantees that there exists satisfying .

Now we are going to show that for every ,

(2.8)

Taking any , from the compactness of we can find such that

(2.9)

By compactness again we can assume that converges to a point . From above it follows that

(2.10)

Therefore .

Now we define the mapping by . Since the mapping is upper semicontinuous and for every is a compact convex set we can apply the Kakutani-Bohnenblust-Karlin Theorem in [5] to obtain a fixed point for and hence for .

Let be a metric space and a mapping. Then a sequence in is said to be an approximating fixed-point sequence of if .

Let be a bounded closed and convex subset of a Banach space , a nonexpansive mapping and . Then a mappings define by is always asymptotically regular, that is, for every , .

Proposition 2.5.

Let be a Banach space and a closed bounded convex subset of , and . If is a nonexpansive mapping, then the sequence is an asymptotic center for .

Proof.

The above comments guarantee that is an approximated fixed-point sequence for . Let us see that the sequence an asymptotic center for . Given we have

(2.11)

Therefore is asymptotic center for .

Theorem 2.6.

Let be a normed space, a nonexpansive mapping with an approximating fixed point sequence and be a nonempty subset of such that is a nonempty star-shaped subset of . Then has an approximating fixed-point sequence in .

Proof.

Suppose . Therefore

(2.12)

and so .

Now, let be the star center of . For every define by

(2.13)

For every , is a contraction, so there exists exactly one fixed point of . Now

(2.14)

Therefore is the approximating fixed-point sequence in of .

Corollary 2.7.

Let be a normed space, a nonexpansive mapping with an approximating fixed-point sequence and be a nonempty subset of such that . Suppose is a nonempty weakly compact star-shaped subset of . If is demiclosed, then has a fixed point in .

Proof.

By the last theorem, has an approximating fixed-point sequence . Because is weakly compact, there exists a subsequence of such that . Since is demiclosed on and , it follows that . Therefore, .