Let $\mu$ be a centered Gaussian measure on a separable Banach space $\mathscr{U}$. The construction of the Cameron-Martin Hilbert space $H_\mu \subset \mathscr{U}$ typically goes as follows: Let $C_\mu: \mathscr{U}^* \to \mathscr{U}$ be defined by $$ C_\mu(\ell) = \int u \ell(u) \mu(du).$$ This is well-defined for $\ell \in \mathcal{U}^*$ since $\int_\mathscr{U} \|u\|^2 \mu(du) < \infty$ by the Fernique theorem. Moreover, there is a natural embedding $\mathscr{U}^* \subseteq L^2(\mu)$. Let $R_\mu$ denote the closure of $\mathscr{U}^*$ in $L^2(\mu)$. Then $R_\mu$ is a Hilbert space with respect to the usual inner product. The Cameron-Martin Hilbert space $\mathcal{H}_\mu$ is then defined as $C_\mu(R_\mu)$ and endowed with the inner product $$ \langle h_1, h_2 \rangle_\mu = \langle C_\mu \ell_1, C_\mu \ell_2 \rangle_\mu := \langle \ell_1, \ell_2 \rangle_{L^2(\mu)}. $$
Question: One can show that for any $h \in \mathcal{H}_\mu$ there holds $$\|h\|_{\mu} = \text{sup}_{\|\ell\|_{L^2(\mu)}\le 1} |\ell(h)|.$$ How does one show that $h \in \mathscr{U}$ is in the Cameron-Martin space if and only if the previous supremum is finite?
First, let's note that for all $\ell, k \in \mathscr{U}^*$, we have the identity $$\ell(C_\mu(k)) = \ell\left(\int u k(u)\,\mu(du)\right) = \int \ell(u) k(u) \,\mu(du) = \langle \ell, k \rangle_{L^2(\mu)}$$ and so by continuity and density the same identity holds for all $\ell \in \mathscr{U}^*$ and $k \in R_\mu$.
Suppose the supremum is finite. This means the map $\ell \mapsto \ell(h)$ is a bounded linear functional on $\mathscr{U}^*$ equipped with the $L^2(\mu)$ inner product. It thus extends to a bounded linear functional on the Hilbert space $R_\mu$, which by Riesz representation can be identified with some $k \in R_\mu$. This means in particular that $\langle \ell, k \rangle_{L^2(\mu)} = \ell(h)$ for all $\ell \in \mathscr{U}^*$. By the note above, we also have $\ell(C_\mu(k)) = \langle \ell, k \rangle_{L^2(\mu)}$ for all $\ell \in \mathscr{U}^*$. By the Hahn-Banach theorem, we conclude that $h = C_\mu(k)$ and so $h \in \mathcal{H}_\mu$.