In books on Malliavin calculus and stochastic PDE, I found the following result is frequently used. I state it here in the simplest form.
Given a separable Hilbert space $\left(H, \langle \cdot, \cdot \rangle\right)$ and a probability measure $\mu$ on $H$. Suppose that $H_0$ is a dense subset of $H$. Consider the following class of functionals on $H$, $$\mathcal C_0 = \left\{h \mapsto \cos(\langle h, h_0 \rangle), h \mapsto \sin(\langle h, h_0 \rangle)|h_0 \in H_0\right\}. $$ Let $\mathcal C$ be the linear span of $\mathcal C_0$, then $\mathcal C$ is dense in $L^2(\mu)$, where $L^2(\mu)$ is the Hilbert space of all real-valued functionals $F: H \rightarrow \mathbb R$ with the norm $E_\mu |F|^2$ being finite.
Of course, from the fact that $H$ is a Radon space, we can restrict the problem to a compact subspace of $H$, and apply Stone-Weierstrass Theorem to prove this. However since this result usually appears without any proof, I guess there is a very direct and easy proof. Could anyone teach me?
Thank you in advance.
We may assume that $H_0=H$ since $\lVert \langle \cdot,h_n\rangle-\langle \cdot,h_0\rangle\rVert \to 0$ if $h_n\to h_0$, by dominated convergence.
Take a function $f\in L^2(\mu)$ such that $\int fg \mathrm d\mu=0$ for each $g\in \mathcal C_0$. Decomposing $f$ into positive and negative parts ($f^+$ and $f^-$ respectively), we have for each $h_0\in H$, $$\tag{*}\int_H f^+(h)\exp\left(\langle h,h_0\rangle\right)\mathrm d\mu=\int_H f^-(h)\exp\left(\langle h,h_0\rangle\right)\mathrm d\mu.$$ Taking $h_0=0$, we can see that $f^+\mu$ and $f^-\mu$ are probability measures. By (*), these probability measures share the same characteristic functional. This proves that $f^+=f^-$ almost everywhere hence $f=0$.