Abstract
In this paper, we derive a fixed-point theorem for self-mappings. That is, it is shown that every isometric self-mapping on a weakly compact convex subset of a strictly convex Banach space has a fixed point.
1. Introduction
Let be a Banach space and be a closed convex subset . Let be a self-mapping of . Recall that is said to be nonexpansive iffor all . The fixed-point set of is . We say that the subset of is said to have an approximate fixed-point sequence for a nonexpansive self-map iffor any sequence . When the closed convex subset is bounded, then such a sequence always exists; indeed, by letting , for all , be a null sequence and defining the maps where arbitrarily , one can see that , implying that is a contraction mapping with contraction constant . By the Banach contraction mapping theorem, it follows that there exists a unique such that , which implies thatfrom which we get . Given that and , it follows that is an approximate fixed-point sequence.
Another way of constructing an approximate fixed-point sequence is to require or assume that is nonempty. Now due to the assumption that , the sequence is bounded (indeed, for all and taken arbitrarily in ). Hence,and is an approximating fixed point for .
A mapping of a set in a Banach space is called isometric iffor all . Note that an isometric mapping is just a nonexpansive mapping in which the inequality is always an equality. A well-known result of Brodskii and Milman [1] asserts that if is a weakly compact convex subset of and has normal structure, then has the fixed-point property for isometric mappings. In particular, any compact convex subset of has the fixed-point property (see [2]).
References [3–10] can be consulted for fixed-point problems on isometric mappings.
Definition 1 (strictly convex Banach space). A strictly convex Banach space is a Banach space such that whenever , then if and only if for some .
An example of a strictly convex Banach space is a Hilbert space.
Definition 2 (convex linear). Let be a Banach space and be a closed convex subset of . Then, the map is said to be a convex linear iffor all and .
An example of a convex linear is a linear map.
2. Preliminaries
We introduce the following useful theorems that will be used in the proof of our main result.
Theorem 1 (Mazur’s theorem). Every nonempty convex subset of a Banach space is strongly closed if and only if it is weakly closed.
Theorem 2 (Cantor’s intersection theorem). Let be a topological space. A decreasing nested sequence of nonempty compact closed subset of has nonempty intersection. In other words, suppose that is a sequence of nonempty compact closed subset of satisfying , and it follows that
3. Main Result
We give the proof of the main result of this paper, which is accomplished in Theorem 3 below. The following lemma, corollary, and proposition shall aid us in arriving at the conclusion of the main result.
Lemma 1. Let be a strictly convex Banach space and be a closed convex subset and be an isometric mapping. Then, is convex linear on . That is, for all and .
Proof. Let and . Without loss of generality, assume . Then, implies thatSimilarly, which also implies thatFirst, we show that . To see this, we observe that if , then from (9), we haveleading to the contradiction that .
Similarly, if , then from (8), we haveleading to the contradiction that .
Now since is an isometry, it follows thatwhich implies that . Since is strictly convex and implies that there exist such thatwe obtainwhere and . We finally show that .
From (14), we have which implies thatAlso, implies thatFrom (16), which implies that . Hence, from (14), we have shown that which establishes the convex linearity of .
Corollary 1. Let be a strictly convex Banach space, be a closed convex subset, and be an isometric mapping. Then, the function is a continuous convex function.
Proof. From Lemma 1, is a convex linear. Hence, subtracting the term from both sides of for all and , we haveWe have the following evaluation:Thus, is a convex function and continuous (because is continuous).
Proposition 1. Let be continuous convex function on a weakly compact convex subset of any Banach space . Then, attains its minimum on . That is, there exist such that
Proof. Let . We show that and that for some .
Suppose that , and for each , define . For each , the set is closed (and weakly closed by Theorem 1), convex, and nonempty (since ). Therefore, forms a nested decreasing sequence of weakly compact nonempty sets. By Theorem 2, . But this implies that there is some such that for all , an impossibility. Consequently, .
So define a sequence of sets for all n belonging to N (natural numbers). As before, is a nested sequence of weakly compact nonempty sets and so . If , then and as required.
Theorem 3 (main result). Every isometric self-mapping on a weakly compact convex subset of a strictly convex Banach Space has a fixed point.
Proof. We know from Corollary 1 that is a continuous convex function. So, by Proposition 1, attains its minimum on , say . By the approximate fixed point of , it is always possible to choose a sequence such that for all . Hence, and so .
Example 1. Let and for all . Then, is an isometry, and by Theorem 3, has a fixed point.
As an application of Theorem 3, it is desired to solve the linear problemfor some nonzero scalar and constant vector , where is linear and is closed and convex. The following theorem gives a solution.
Theorem 4. Let be a closed convex subset of a Banach space and let be linear and continuous. Then, equation (20) has a solution if the following holds:(1)(2) is weakly compact(3) is strictly convex
Proof. By defining the auxiliary mappingit is clearly seen that is well defined and invariant on since is convex and . Since , then by taking norms gives usSince is bounded, thenSince is an isometry, then by Theorem 3, one can find a solution .
Data Availability
No data were used to support this study.
Conflicts of Interest
The author declares that there are no conflicts of interest.