6
$\begingroup$

Consider a locally convex topological vector space V over the complex numbers. Is it true that every weakly bounded subset of V is indeed bounded? If not, what additional requirements are needed for this to hold? Perhaps someone has a reference, I was not able to find something in the literature.

Thanks for your help.

Cheers,

Ralf

$\endgroup$

2 Answers 2

8
$\begingroup$

Theorem 3.18 in the excellent book by Rudin "Functional Analysis" says: In a locally convex space $X$, every weakly bounded set is originally bounded, and vice versa. The proof is based on the Banach-Alaoglu theorem (well, no surprise) and Baire's category theorem.

$\endgroup$
10
  • $\begingroup$ What's "originally bounded"? I've never heard of that one! $\endgroup$ Apr 29, 2010 at 14:13
  • $\begingroup$ It means "bounded in the original topology" $\endgroup$ Apr 29, 2010 at 15:23
  • 1
    $\begingroup$ Of course in the case where X is not just a LCTVS but a Banach space then this is the Uniform Boundedness theorem (a.k.a. Banach-Steinhaus, more or less) $\endgroup$
    – Yemon Choi
    Apr 29, 2010 at 16:28
  • $\begingroup$ @Yemon: that's what worries me a little about this answer. Banach-Steinhaus is a Big Theorem and I don't think that it holds for all LCTVS (does it even hold for incomplete nvs?). Unfortunately, I'm not in my office so can't check my sources, but I want to say something like "X needs to be barrelled", but that might only be to do with the space of functions on X, not X itself. $\endgroup$ Apr 29, 2010 at 17:15
  • $\begingroup$ @Andrew: a subset $E$ of a topological vector space $X$ is (originally) bounded if to every neighborhood $V$ of $0$ in $X$ corresponds a number $s > 0$ such that $E \subset tV$ for every $t > s$. I'm sorry, I should have stated the definition that I use in my answer, but I think, this is the common definition if you just have topological vector spaces. $\endgroup$ Apr 30, 2010 at 7:20
3
$\begingroup$

This is direct consequence of the Mackey Theorem: Having a dual pair (V,V') with V' as the dual of the locally convex space V, the bounded sets on V under any dual topology are identical. A dual topology on V is a locally convex topology $\tau$ such that (V,$\tau$)' = V'.

As the original and the weak topology give the same dual, the bounded sets are identical.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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