Fixed point theory and nonexpansive mappings

Arabian Journal of Mathematics, Dec 2012

Recall that a Banach space X has the weak fixed point property if for any nonempty weakly compact subset C of X and any nonexpansive mapping T : C→ C, T has at least one fixed point. In this article, we present three recent results using the ultraproduct technique. We also provide some open problems in this area.

A PDF file should load here. If you do not see its contents the file may be temporarily unavailable at the journal website or you do not have a PDF plug-in installed and enabled in your browser.

Alternatively, you can download the file locally and open with any standalone PDF reader:

https://link.springer.com/content/pdf/10.1007%2Fs40065-012-0042-1.pdf

Fixed point theory and nonexpansive mappings

Arab J Math Fixed point theory and nonexpansive mappings Pei-Kee Lin Mathematics Subject Classification 46B08 · 46B20 · 47H10 For any closed bounded and convex subset C of a Banach space X, let rC be the function from C to R defined by for all x, y ∈ C. C is said to have the fixed point property (for nonexpansive maps) if for any nonexpansive map T : C → C, there is x ∈ C such that T x = x. X is said to have the (weak) fixed point property if every (weakly compact) closed, bounded, convex subset C of X has the fixed point property. A nonempty, closed, convex, T -invariant subset K of C is said to be minimal if K1 is a nonempty, closed, convex, T -invariant subset of K , then K1 = K . By Zorn's lemma, for any weakly compact convex subset C and any nonexpansive map T : C → C, C has a minimal (T -invariant) subset. Suppose that the closed convex hull, co(T (C)) = C. Then for any x ∈ C, - rC (x) = sup{ y − x : y ∈ C}. Then rC is a continuous convex function. The radius r (C) and diameter diam(C) of C are defined by A map T : C → C is said to be nonexpansive if T x − T y ≤ x − y rC (x) = sup{ x − y : y ∈ C} ≥ sup{ T x − T y : y ∈ C} = sup{ T x − z : z ∈ co(C)} = sup{ T x − z : z ∈ C} = rC (T x). Therefore, if β > r (C), then the set {x ∈ C : rC (x) ≤ β} is a nonempty, closed, convex, T -invariant subset of C. If C is a minimal T -invariant set, then co(T (C)) = C. We have proved the following lemma. Lemma 1.1 Let C be a closed bounded and convex subset of a Banach space X and T a nonexpansive map from C into itself. (1) Suppose that co(T (C)) = C and β > r (C). Then the set is a nonempty, closed, convex, T -invariant subset of C. (2) (Kirk) Suppose that C is minimal. Then rC is a constant function and r (C) = diam(C). Let C be a bounded, closed, convex subset of a Banach space and T : C → C a nonexpansive mapping. A sequence (xn) in C is said to be an approximate fixed point sequence for T if lim T xn − xn = 0. n→∞ By Banach’s contraction principle, if T is a nonexpansive map from a bounded closed and convex subset C of a Banach space X into itself, then T has an approximate fixed point sequence. The following lemma is due to Karlovitz [11] and Goebel [8]. Lemma 1.2 (Karlovitz–Goebel Lemma) Let C be a nonempty, closed, bounded, and convex subset of a Banach space, and assume that T : C → C is nonexpansive. Let (xn) be an approximate fixed point sequence for T and let ψ : C → R+ be the function defined by ψ(x) = lim supn→∞ x − xn . Let α = inf{ψ(x) : c ∈ C}. is a nonempty, closed, convex, T -invariant subset of C. (2) Suppose that C is minimal and weakly compact. Then ψ(x) = diam(C) for all x ∈ C. So if β > α, then the set K = {x : ψ(x) ≤ β} is a nonempty, closed, convex, T -invariant subset of C. We have proved (1). It is clear that for any x ∈ C, ψ(x) ≤ diam(C). Assume that C is minimal. By (1), ψ is constant. By passing to a subsequence of (xn), we may assume that (xn) converges to y weakly. By Lemma 1.1 (2), rC (y) = diam(C). For any > 0, there is z ∈ C such that z − y ≥ diam(C) − . Then for any x ∈ C, ψ(x) = ψ(z) = lim sup xn − z ≥ y − z ≥ diam(C) − . n→∞ Remark 1.3 Let C be a weakly compact, convex subset of a Banach space and T : C → C is nonexpansive. Suppose that C is minimal and diam(C) = 1. Lemma 1.2 (2) is equivalent to the following statement: For any x ∈ C and It has been conjectured that every Banach space has the weak fixed point property. But Alspach [2] gave the following counterexample. Example 1.4 Let C be the set Let T be the map from C to C defined by T f (t) = It is easy see that T is an (into) isometry and T has no fixed point. Maurey introduced the ultraproduct technique in the study of this problem and proved the following results [6,17]: c0 has the weakly fixed point property. Every reflexive subspace of L1 has the fixed point property. Every superreflexive Banach space with a 1-unconditional basis has the weak fixed point property. Let C be a nonempty, bounded, closed, convex subset of a superreflexive Banach space X and T : C → C an isometry (i.e., x − y = T x − T y for all x, y ∈ C). Then T has a fixed point. In this article, we consider the following three problems: Problem 1 Let (P) be a geometric property of Banach spaces. Does X have the weak fixed property if X has property (P)? Problem 2 Let (X, · ) be a Banach space. Is there an equivalent norm | · | of X such that (X, | · |) has the weak fixed point property? Problem 3 Let (X, · ) be a Banach space with the weak fixed point property. Find a constant C > 1 such that if Y is a Banach space with the Banach–Mazur distance d(X, Y ) < C, then Y has the fixed point property. It is known that every uniformly convex Banach space (or Banach space with normal structure) has the weak fixed point property. On the other hand, we do not know whether every reflexive (or superreflexive) Banach space has the fixed point property or not. Let X be a Banach space. A subset A of X is symmetrically -separated if the distance b (...truncated)


This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2Fs40065-012-0042-1.pdf

Pei-Kee Lin. Fixed point theory and nonexpansive mappings, Arabian Journal of Mathematics, 2012, pp. 495-509, Volume 1, Issue 4, DOI: 10.1007/s40065-012-0042-1