If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$?

2010-11-15T17:31:26Z

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 $I$.

Let $H^*$ denote the space of continuous linear functionals on $H$. By the Riesz representation theorem, $$H^* = \{ \langle k, \cdot \rangle : k \in H \}.$$ By the assumption that $\gamma$ has covariance operator $I$, for all $k \in H$, $$\int_H |\langle k, h \rangle|^2 \, \mathrm{d}\gamma(h) = \langle k, k \rangle < \infty,$$ so $H^*$ is contained in the separable Hilbert space $L^2(H)$.

My question: Is $H^*$ dense in $L^2(H)$?

Answer by Nate Eldredge for If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$?

2010-11-15T19:08:13Z

The closure of $H^*$ in $L^2(H)$ does not contain the constants, or any other nonlinear function, since $L^2$ convergence preserves linearity.

If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$? - MathOverflow

If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$?

Answer by Nate Eldredge for If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$?