0

Let $\mathcal{M}$ be a closed convex subset of a Banach space $\mathcal{B}$. Assume that $\mathcal{M}$ is weakly closed. Let

$$F:\mathcal{M}\to\mathcal{M}$$

be weakly continuous and such that $F(\mathcal{M})$ is included in a weakly compact subset of $\mathcal{B}$ (if needed, suppose it is weakly compact). According to this book and a document I am reading, it seems to be sufficient to conclude that $F$ admits a fixed point, thanks to Schauder's fixed point theorem. A similar question has been asked here.

Schauder's fixed point theorem talks about a Banach space. A priori, a compact space in a Banach space is assumed to be compact with respect to the strong topology, i.e. the topology induced by the norm. Therefore, why can we apply Schauder's fixed point theorem when we only have weakly compact subset?

EDIT: changed "weak(ly)-$\star$" to "weak(ly)".

MoebiusCorzer
  • 3,354
  • Weak-* is a topology for the dual of a Banach space. Are you assuming $\mathcal B$ is the dual of some other Banach space, or do you just mean the weak topology? – Robert Israel May 15 '17 at 18:04
  • @RobertIsrael My terminology is confusing and you are right, I meant weak topology. I shall edit my question. – MoebiusCorzer May 15 '17 at 18:21

1 Answers1

1

Actually, the Schauder fixed point theorem works in any Hausdorff topological vector space. This could be a Banach space with its weak topology.

Robert Israel
  • 470,583
  • Thank you for your prompt answer. Would you have a reference for the general Schauder's fixed point theorem (by "general", I mean a proof that it works for any Hausdorff topological vector space). – MoebiusCorzer May 15 '17 at 18:23