2
$\begingroup$

My notes say that the theorem of Banach-Alaoglu states the following: If $X$ is a normed separable space, then every bounded sequence in $X'$ has a weak-* convergent subsequence.

How is this equivalent to the usual formulation from Wikipedia, etc - i.e. the closed unit ball being weak-*-compact, for a (not necessarily separable) normed space $X$? Or is it a special case?

$\endgroup$
1
$\begingroup$

If $X$ is separable, then the closed unit ball in $X'$ is metrizable with respect to the weak-$\ast$ topology. Hence compactness, given by Banach-Alaoglu, implies sequential compactness.

$\endgroup$
1
$\begingroup$

It's a special case. If $X$ is separable then it's not hard to show that the closed unit ball of $X'$ is metrizable in the weak-* topology (note $X'$ itself is not weak-* metrizable), so compactness is equivalent to sequential compactness in that case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.