I have read this:

We have a map $S:W_0 \to W_0$. Moreover $W_0$ is not empty, convex, and weakly compact in $W$. Thus we can apply Schauder's fixed point theorem:

Schauder's fixed point theorem: If $E$ is convex compact subset of a Banach space and if $S:E \to E$ is continuous then there is a fixed point of $S$.

So we just need to prove that $S$ is weakly continuous (from $W_0 \to W_0$).

Why the author checks for weakly continuous instead of the stronger continuity? Can someone give me a good definition of weakly compact (in terms of sequences and boundedness)? Thanks.

  • 1
    $\begingroup$ The reason is that you need to check continuity with respect to the topology that you are taking into account. Here, the author examines (in the context of the problem) the weak topology (because he says that he has a "...weakly compact set W"). So the theorem can be applied to functions that are continuous with respect to this topology (the weak topology). $\endgroup$
    – Jimmy R.
    Commented Mar 13, 2014 at 0:15

2 Answers 2


I will answer your last question ("Can someone give me a good definition of weakly compact (in terms of sequences and boundedness)?"), and I will try to edit the post later to see if I can answer the rest.

Let us denote by $X^*$ the space of all linear continuous functional on $X$, a normed vector space. Then, the weak topology $\sigma(X,X^*)$ on $X$ is the final topology on $X$ with respect to $X^*$. Then, a set $A$ is weakly compact if it is compact with respect to the weak topology $\sigma(X,X^*)$. Thus, $A$ is weak sequentially compact, by the Eberlien-Smulian theorem (see Proof of Eberlein–Smulian Theorem for a reflexive Banach spaces). So,

"Theorem": $A$ is weakly compact if every sequence in $A$ has a convergent subsequence whose limit is in $A$.

  • $\begingroup$ Are you sure about your theorem? If $A \subset B$ is a weakly compact then I would expect some reference to the superset $B$ $\endgroup$
    – maximumtag
    Commented Mar 10, 2014 at 13:45
  • $\begingroup$ @maximumtag Yes, there is a relation - $A\subset B$ is weakly compact iff $A$ is compact in the final topology $\sigma(B,B^{*})$. $\endgroup$
    – user122283
    Commented Mar 10, 2014 at 22:28

I think what is used here is the following version of the Schauder theorem:

Let $V$ be a separable reflexive Banach space. Let $K \subset V$ be nonempty, convex, bounded, and compact for the weak topology. Let $S : K \to K$ be weak-to-weak continuous. Then $S$ has a fixed point.

I couldn't find a reference for this. However, it seems to me that the usual proof of the Schauder theorem carries over using metrizability of the weak topology (on bounded sets of a reflexive Banach space).

Edit: Cf. Tikhonov fixed point theorem.


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