Let $B$ be the closed unit ball of a Banach space $X$. Then $B$ is weakly compact if and only if it is weakly sequentially compact.

Forward direction:

We first assume $B$ is compact. Kakutani's Theorem tells us that $X$ is reflexive. Every bounded sequence in $X$ has a weakly convergent subsequence. Since $B$ is weakly closed, $B$ is weakly sequentially compact.

Why can we assume $B$ is weakly closed? I'm guessing this follows from $B$ being strongly closed but I'm unclear on the relationship of the two.

I think weak closure implies strong closure: $B$ is weakly closed so its compliment is weakly open. Thus $B^C$ is certainly open in a stronger topology - such as the norm topology. Thus $B$ is closed in the norm topology. I don't see how the other direction can be true though because we are "coming from a finer topology".

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    $\begingroup$ The assumption in the forward direction is that $B$ is weakly compact. The author probably should have put "weakly" in that first sentence. $\endgroup$ – Nate Eldredge Jun 29 at 21:24
  • $\begingroup$ @NateEldredge Indeed, or $X$ is finite-dimensional and everything trivialises. $\endgroup$ – Henno Brandsma Jun 29 at 21:30
  • $\begingroup$ I did not see through to the consequences as Henno did, but I was able to pick this up! The author does this at least twice in this proof. $\endgroup$ – yoshi Jun 29 at 21:33
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    $\begingroup$ In discussions where several topologies are involved, I prefer to be very specific in what topology some notion is meant. It matters greatly. $\endgroup$ – Henno Brandsma Jun 29 at 21:41
  • $\begingroup$ the relation is quite clear: weakly closed implies strongly closed, but the reverse only holds for some sets, like convex ones, see here. That's another way to see that $B$ (which is strongly closed and convex) is weakly closed too. $\endgroup$ – Henno Brandsma Jun 30 at 9:11

$B$ weakly compact implies $B$ weakly closed, merely by $X$ being a Hausdorff space in the weak topology (which actually holds because we can separate points by functionals, Hahn-Banach etc.). This is just a special case of "compact sets are closed in Hausdorff spaces"! (general topology is actually useful, see?)

From this question (and general Banach space theory) we see that a strongly closed and convex subset is in fact weakly closed too. This applies to the closed unit ball $B$; if you already know the result you could use that as well. Of course, I like the general topology approach better, as it uses less.


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