If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:30:51Z http://mathoverflow.net/feeds/question/46124 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46124/if-h-is-a-separable-hilbert-space-is-its-dual-dense-in-l2h If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$? Tom LaGatta 2010-11-15T17:31:26Z 2010-11-15T19:08:13Z <p>Let $H$ be an infinite-dimensional, separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$.</p> <p>Let <code>$H^*$</code> denote the space of continuous linear functionals on $H$. By the Riesz representation theorem, <code>$$H^* = \{ \langle k, \cdot \rangle : k \in H \}.$$</code> By the assumption that $\gamma$ has covariance operator $I$, for all $k \in H$, <code>$$\int_H |\langle k, h \rangle|^2 \, \mathrm{d}\gamma(h) = \langle k, k \rangle &lt; \infty,$$</code> so <code>$H^*$</code> is contained in the <a href="http://mathoverflow.net/questions/46094/if-h-is-a-separable-hilbert-space-is-l2h-separable" rel="nofollow">separable</a> Hilbert space $L^2(H)$.</p> <p><b>My question:</b> Is <code>$H^*$</code> dense in $L^2(H)$?</p> http://mathoverflow.net/questions/46124/if-h-is-a-separable-hilbert-space-is-its-dual-dense-in-l2h/46135#46135 Answer by Nate Eldredge for If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$? Nate Eldredge 2010-11-15T19:08:13Z 2010-11-15T19:08:13Z <p>The closure of $H^*$ in $L^2(H)$ does not contain the constants, or any other nonlinear function, since $L^2$ convergence preserves linearity.</p>