Suppose you have Banach spaces $\mathcal B\_\alpha$ where $\alpha$ is in some index set $I$. Let $\mu\_\alpha$ be Gaussian measures on $\mathcal B\_\alpha$ with Cameron-Martin spaces $\mathcal H\_{\mu\_\alpha}$.


Is it then true that the product space of all the Banach spaces with the product measure (independent components), call it $\mathcal B$, has Cameron-Martin space $\mathcal H\_\mu=\{\mathbf v=\{v\_\alpha\}\_{\alpha\in I}:\sum\_{\alpha\in I}\|v\_\alpha\|\_{\mu\_\alpha}^2<\infty\}$?


I believe I read this somewhere but I can't find a reference.



Cameron-Martin space of product space
At least if $I$ is countable this should be true, but I do not have a reference:


Let $H$ be the CM space of the Frechet space $\mathcal{B} := \prod\_{\alpha \in I} \mathcal{B}\_{\alpha}$ with the product topology and the product measure. Recall that the Hilbert space reproducing kernel of $(\mathcal{B}, \mu)$ is isometrically isomorphic ($\Vert \cdot \Vert\_{L^2(\mathcal{B}^{\ast}, \mu)} = \Vert \cdot \Vert\_H$) to the CM space $H$ and is defined as the $L^2$ closure of $j(\mathcal{B}^{\ast})$ where $j$ is the canonical inclusion $\mathcal{B}^{\ast} \rightarrow L^2(\mathcal{B}, \mu)$.


In our case, the dual to $\mathcal{B}$ can be identified with the sequences in $\prod\_{\alpha \in I} \mathcal{B}\_{\alpha}^{\ast}$ with finite support and the pairing $\mathcal{B}^{\ast} \times \mathcal{B} \rightarrow \mathbb{R}$ given by


\begin{equation}
v(x) = \sum\_{\alpha \in I} v\_{\alpha} (x\_{\alpha}), ~~ v\_{\alpha} \in \mathcal{B}\_{\alpha}^{\ast}, x\_{\alpha} \in \mathcal{B}\_{\alpha} .
\end{equation}


where the sum is, of course, finite.


Restricted to $j(\mathcal{B}^{\ast})$, the $L^2$-norm is


\begin{equation}
\Vert v \Vert\_H = \Vert j(v) \Vert\_{L^2} = \int \sum\_{\alpha, \beta \in I} v\_{\alpha} (x\_{\alpha}) v\_{\beta} (x\_{\beta}) d \mu(x) = \sum\_{\alpha \in I} \int v\_{\alpha} (x\_{\alpha})^2 d \mu(x) = \sum\_{\alpha \in I} \Vert v\_{\alpha} \Vert\_{H\_{\alpha}}^2
\end{equation}


where we used the product structure of $\mu$. Hence indeed


\begin{equation}
H = \{\mathbf v=\{v\_\alpha\}\_{\alpha\in I}:\sum\_{\alpha\in I}\|v\_\alpha\|\_{\mu\_\alpha}^2<\infty\} .
\end{equation}


