Bounded and weakly bounded sets in top. vector spaces

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.

Cheers,

Ralf

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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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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.
@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