If $H$ is a separable Hilbert space, is $L^2(H)$ separable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:38:29Z http://mathoverflow.net/feeds/question/46094 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46094/if-h-is-a-separable-hilbert-space-is-l2h-separable If $H$ is a separable Hilbert space, is $L^2(H)$ separable? Tom LaGatta 2010-11-15T01:02:09Z 2010-11-15T01:37:52Z <p>Let $H$ be a separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$.</p> <p>Is the Hilbert space $L^2(H,\gamma)$ separable?</p> http://mathoverflow.net/questions/46094/if-h-is-a-separable-hilbert-space-is-l2h-separable/46097#46097 Answer by Byron Schmuland for If $H$ is a separable Hilbert space, is $L^2(H)$ separable? Byron Schmuland 2010-11-15T01:25:19Z 2010-11-15T01:37:52Z <p>By Example 7.14.13 in Volume 2 of Bogachev's <em>Measure Theory</em>, every Radon measure on $H$ is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable. </p>