%0 Journal Article
%T Fixed-Point Theorems for Mean Nonexpansive Mappings in Banach Spaces
%A Zhanfei Zuo
%J Abstract and Applied Analysis
%D 2014
%I Hindawi Publishing Corporation
%R 10.1155/2014/746291
%X 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¨C5]. 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
%U http://www.hindawi.com/journals/aaa/2014/746291/