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Let $X$ be a Banach Space. We say that

  • A mapping $T: C\subset X \to C$ is nonexpansive if $\|Tx - Ty\| \leq \|x-y\|$, for all $x,y \in C$.

  • $X$ has the Fixed Point Property (FPP) if every nonexpansive endomorphism of a nonempty, bounded, closed and convex subset $A \subset X$ has a fixed point.

  • $X$ has the Weak Fixed Point Property (w-FPP) if every nonexpansive endomorphism of a nonempty, weakly compact, convex subset $A \subset X$ has a fixed point.

From what I have read, some authors claim that, in a Reflexive Space, both the w-FPP and the FPP coincide but I don't understand why.

I can see how the Eberlein–Šmulian theorem tell us that the FPP $\implies$ w-FPP and that I need to prove that in a Reflexive Space every nonempty, weakly compact and convex set is closed and bounded.

I was thinking that perhaps using the fact that a Banach Space $X$ is reflexive iff the closed unit ball is weakly compact would be helpful, however I do not know how to proceed.

Angel Peñaflor
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  • Do you miss a compactness assumption for FPP? Also what do you mean by endomorphism? (I am not an expert on these properties thats why I am asking) – daw May 27 '22 at 07:12
  • I don't think so, I have seen this being claimed without the assumption of compactness. By endomorphism I just mean a mapping that goes from the set $A$ to itself. – Angel Peñaflor May 27 '22 at 08:06
  • Then $X=\mathbb R$ seems to be a counterexample: $f(x):=x+1$ is nonexpansive and has no fixed point on $A=\mathbb R$, but $\mathbb R$ has the weak FPP by Brouwers fixed point theorem. – daw May 27 '22 at 08:13
  • Oh, you're absolutely right. I committed a mistake, I forgot to add bounded to the definition of the FPP. – Angel Peñaflor May 27 '22 at 08:23

1 Answers1

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Use that a convex set is norm closed iff it is weakly closed.

Using additionally that weakly compact sets are bounded, you immediately see that a weakly compact, convex subset $A\subset X$ is norm closed, bounded, and convex.

On the other hand, if a subset $A\subset X$ is convex, norm closed, and bounded, your own characterization of reflexiveness implies that $A$ is a weakly closed subset of a weakly compact set. Hence $A$ is weakly compact and convex.

This should prove both directions.

Johannes Agerskov
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