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?

  • $\begingroup$ This is kind of confusing because you appear to have two, a priori potentially conflicting, definitions of $\|h\|_\mu$: via the inner product $\langle \cdot, \cdot \rangle_{\mu}$ which is only defined on $\mathcal{H}_\mu$ in the first place, and via the "characterization" on the last line. Is it the latter that you want to use here? $\endgroup$ Jan 8, 2021 at 20:47
  • $\begingroup$ In the first display, $\mathscr{U}(\ell)$ should be $C_\mu(\ell)$, right? $\endgroup$ Jan 8, 2021 at 20:49
  • $\begingroup$ Yes to the error in the first display. Regarding the other comment, perhaps I should have been more clear. The first definition via the inner product is only defined on the Cameron-Martin Hilbert space. By the "characterization" I mean that the formula holds for any element of the Cameron-Martin space. The goal is then to show that if the second definition (which as you point out makes sense for any element of the Banach space) is finite, then the element h indeed belongs to the Cameron-Martin space. Equivalently, if an element is not in the Cameron-Martin space, then that supremum is infinite $\endgroup$
    – Kyle Liss
    Jan 8, 2021 at 21:13

1 Answer 1


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 $h \in \mathcal{H}_\mu$.


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