Fixed-Point Theorems for Mean Nonexpansive Mappings in Banach Spaces

Abstract

We define a mean nonexpansive mapping T on X in the sense that , . It is proved that mean nonexpansive mapping has approximate fixed-point sequence, and, under some suitable conditions, we get some existence and uniqueness theorems of fixed point.

1. Introduction

Let be a Banach space, a nonempty bounded closed convex subset of , and  :  a nonexpansive mapping; that is, We say that has the fixed-point property if every nonexpansive mapping defined on a nonempty bounded closed convex subset of has a fixed point. In 1965, Kirk [1] proved that if is a reflexive Banach space with normal structure, then has the fixed-point property.

Let be a nonempty subset of real Banach space and a mapping from to . is called mean nonexpansive if for each , In 1975, Zhang [2] introduced this definition and proved that has a fixed point in , where is a weakly compact closed convex subset and has normal structure. For more information about mean nonexpansive mapping, one can refer to [3–5].

2. Main Results

Lemma 1. Let be a mean nonexpansive mapping of the Banach space . If is continuous and , then T has a unique fixed point.

Proof. The proof is similar to the proof of the Banach contractive theorem.

If we let and as in Lemma 1, then the condition that is continuous may not be needed. Firstly, we recall the following two lemmas.

Lemma 2. Let be a nonempty subset of Banach space and a mean nonexpansive self-mapping on with and . Let be a nonempty subset of ; one defines for any ; if the set is bounded, then is also bounded.

Proof. Let and set as fixed; then for any , we have This implies that Hence, is bounded. The proof is complete.

Lemma 3. Let be a nonempty subset of Banach space and a mean nonexpansive self-mapping on . If and , then for any , one has the following inequality: where and are two positive integers such that .

Proof. By the definition of mean nonexpansive mapping, we have that this implies that where is an integer.
When , the result is obvious. Suppose that (5) is true for ; that is, By the inequality (2) and (7), we have This implies from and that which follows that By induction, this completes the proof.

Theorem 4. Let be a nonempty closed subset of Banach space and a mean nonexpansive self-mapping on . If and , then has a unique fixed point.

Proof. For any , set ; by the definition of mean nonexpansive mapping, we have that Thus, the sequence is nonincreasing and bounded below, so exists.
Suppose that ; then we have by Lemma 2 that the set is bounded, so there exists a positive number such that where and are two integers. Since and , for any , there exists an integer such that : and
Since , there exists an integer N such that for ; we have . Setting and , then and . Thus, from (14) and Lemma 3, we have that on the other hand, by condition (12), we have that , which is a contradiction, so . We next show that exists. In fact, since , we have This implies that That is, the sequence is a Cauchy sequence in . Since is complete, thus there exists such that .
Finally, we prove that . We have from and (2) that it follows as that , which implies by that and the proof is complete.

We now consider the approximate fixed-point sequence. A sequence is called an approximate fixed-point sequence for if as . It is easy to prove that if is a nonempty bounded closed convex subset of Banach space , and is a nonexpansive mapping from to , then has an approximate fixed point sequence in . For mean nonexpansive mapping, we have the same result. Firstly, we give the following lemma.

Lemma 5 (see [6]). Let a real number and be a sequence in Banach space . Then, for any positive integer , If is the real line and for all , one has the special case

Theorem 6. Let be a nonempty bounded closed subset of Banach space and a mean nonexpansive self-mapping on . Let be fixed and the sequence defined by If , then converges strongly to 0 as .

Proof. Since is a mean nonexpansive mapping, from (21) and , we get that Thus, the sequence is nonincreasing and bounded below, so exists. Suppose that . That is, for any , there exists an integer such that Since is bounded and , there exists an integer such that where .
Now setting and for all positive integers , we get from (21) and (23) that Hence, by Lemma 5, (19), (20), and (23), we get that This implies from (24) that Since for , we have Hence, we have Since is an arbitrary positive number, it follows that . This contradiction completes the proof.

Corollary 7. Let be a bounded closed convex subset of Banach space and a mean nonexpansive self-mapping on C. Then, has an approximate fixed-point sequence in .

Proof. For any , define , and let , where ; then the sequence may be written as , so the conditions of Theorem 6 are satisfied. Hence, we have that the sequence is an approximate fixed-point sequence. The proof is complete.

Next, we consider the Opial condition. Related to the problem of existence of a fixed point for mapping and its approximation, in 1967, Opial [7] introduced the following inequality.

Definition 8. Let be a Banach space; satisfies Opial’s condition if for each in and each sequence weakly convergent to holds for .

This definition is motivated by the fact that this property implies that asymptotic center of sequence coincides with its weak limit, which of course fails in for   (and more generally in Orlicz spaces ; see [8, 9]).

Opial’s condition is connected to the following fixed-point property.

Theorem 9. Let be a real reflexive Banach space which satisfies Opial’s condition, a nonempty bounded closed convex subset of , and a mean nonexpansive. Then has a fixed point.

Proof. Let be mean nonexpansive. By Corollary 7, we have that has an approximate fixed-point sequence in ; that is, there exists a sequence of such that Since is a weakly compact convex subset of , there exists a subsequence such that weakly convergent to .
Now, we show that . Suppose, by way of contradiction, that ; then Since is mean nonexpansive, we have from (2) that It follows that This is a contradiction. Therefore, ia a fixed point of and the proof is complete.

In fact, spaces which satisfy Opial’s condition not only have the fixed point property, but also satisfy the so-called demiclosedness principle for the mean nonexpansive mapping.

Corollary 10. If is a reflexive Banach space which satisfies Opial’s condition, let be a nonempty closed convex subset of and suppose that is mean nonexpansive. For any sequence in with and , then .

Proof. The proof of this corollary is the same as in Theorem 9.

In order to understand the connection between nonexpansive mapping and mean nonexpansive mapping better, we have the following remark.

Remark 11. It is easy to see that the nonexpansive mappings and contractive mappings both are uniformly continuous and mean nonexpansive; the converse does not hold. Examples will be given to support our point of view.(1)Let be the unit interval defined by and the norm is the ordinary Euclidean distance on the line. Here, is discontinuous at ; consequently, is neither nonexpansive mapping nor contractive mapping. Now, we prove that is mean nonexpansive.

Case 1 (). By the definition of , This implies that .

Case 2 ( and ). In this case, we have This implies that .

Case 3 ( and ). The proof is the same as in Case 2.

Case 4 (). The proof is the same as in Case 1.
Hence, is mean nonexpansive by taking , .(2)Mean nonexpansive mappings, however, may be continuous only at their fixed points. For example, the map defined by is a mean nonexpansive mapping on unit interval by taking , and is continuous only at its fixed point .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The paper is a Project supported by the Scientific and Technological Research Program of Chongqing Municipal Education Commission (Grant no. KJ131104) and key Laboratory for Nonlinear Science and System Structure, Chongqing Three Gorges University, Wanzhou, China.