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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

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2 Answers

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.

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What's "originally bounded"? I've never heard of that one! –  Andrew Stacey Apr 29 '10 at 14:13
    
It means "bounded in the original topology" –  Johannes Hahn Apr 29 '10 at 15:23
    
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) –  Yemon Choi Apr 29 '10 at 16:28
    
@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. –  Andrew Stacey Apr 29 '10 at 17:15
    
@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. –  Ulrich Pennig Apr 30 '10 at 7:20
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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.

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