When LCS is isomorphic to subspace of some function space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:35:59Z http://mathoverflow.net/feeds/question/95475 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95475/when-lcs-is-isomorphic-to-subspace-of-some-function-space When LCS is isomorphic to subspace of some function space? yaoliang 2012-04-29T00:43:51Z 2012-04-30T02:27:04Z <p>Updated: Following Michael's suggestion, I rephrase the question slightly. </p> <p>Given a locally convex (Hausdorff) topological vector space (LCTVS), when is it isomorphic to a subspace of some function space \$Y^X\$ (equipped with the product topology), where \$Y\$ is, say, some Banach space (if it helps simplify things, can assume \$Y=\mathbb{C}\$, the complex field), and \$X\$ is some set. We are free to choose X and Y. </p> <p>If not all LCTVS have this property, then what kind of conditions do we need?</p> <p>Any reference would be appreciated, thanks!</p> http://mathoverflow.net/questions/95475/when-lcs-is-isomorphic-to-subspace-of-some-function-space/95480#95480 Answer by Nik Weaver for When LCS is isomorphic to subspace of some function space? Nik Weaver 2012-04-29T02:37:08Z 2012-04-29T02:37:08Z <p>Yes, every LCTVS can be realized as a function space: every LCTVS is isomorphic to the dual of its dual space equipped with the weak* topology. This is a consequence of a version of the Hahn-Banach theorem which states that in a real LCTVS two disjoint closed convex sets, one of which is compact, can be strictly separated. See Section IV.3 of <em>A Course in Functional Analysis</em> by Conway.</p> http://mathoverflow.net/questions/95475/when-lcs-is-isomorphic-to-subspace-of-some-function-space/95504#95504 Answer by Alberto Abbondandolo for When LCS is isomorphic to subspace of some function space? Alberto Abbondandolo 2012-04-29T12:39:38Z 2012-04-29T12:39:38Z <p>If I interpret the question correctly, Yaoliang would like to know which LCTVS are isomorphic to \$\mathbb{C}^X\$, where \$X\$ is a set (no topology), and \$\mathbb{C}^X\$ is given the product topology. If this is so, the answer is: very few. Actually, spaces of this form are fully determined by the cardinality of \$X\$.</p> <p>Just to make an example: no infinite dimensional normed space can be isomorphic to a space of this form. Indeed, in this case the set \$X\$ would have to be infinite, but then every neighborhood of \$0\$ in \$\mathbb{C}^X\$ would contain a proper vector subspace, and this is never true for normed spaces.</p>