most recent 30 from http://mathoverflow.net 2013-05-25T22:08:51Z http://mathoverflow.net/feeds/question/22965 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22965/bounded-and-weakly-bounded-sets-in-top-vector-spaces Ralf 2010-04-29T11:55:15Z 2010-09-07T09:29:00Z

<p>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.</p> <p>Thanks for your help.</p> <p>Cheers,</p> <p>Ralf</p>

http://mathoverflow.net/questions/22965/bounded-and-weakly-bounded-sets-in-top-vector-spaces/22969#22969 Ulrich Pennig 2010-04-29T12:12:36Z 2010-04-29T12:12:36Z

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

http://mathoverflow.net/questions/22965/bounded-and-weakly-bounded-sets-in-top-vector-spaces/37961#37961 VoBo 2010-09-07T09:29:00Z 2010-09-07T09:29:00Z

<p>This is direct consequence of the <strong>Mackey Theorem</strong>: 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'.</p> <p>As the original and the weak topology give the same dual, the bounded sets are identical.</p>