# Cameron-Martin space of product space

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.

• Is $I$ countable? And what is the topology on $\prod_{\alpha \in I} \mathcal{B}_{\alpha}$? Commented Sep 9, 2022 at 11:59
• @G.Chiusole Not necessarily (though the finite sum requires all but countably many zero). Product topology. Commented Sep 9, 2022 at 12:44

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

$$$$v(x) = \sum_{\alpha \in I} v_{\alpha} (x_{\alpha}), ~~ v_{\alpha} \in \mathcal{B}_{\alpha}^{\ast}, x_{\alpha} \in \mathcal{B}_{\alpha} .$$$$

where the sum is, of course, finite.

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

$$$$\Vert v \Vert_H = \Vert j(v) \Vert_{L^2} = \int \sum_{\alpha, \beta \in I} v_{\alpha} (x_{\alpha}) v_{\beta} (x_{\beta}) d \mu(x) = \sum_{\alpha \in I} \int v_{\alpha} (x_{\alpha})^2 d \mu(x) = \sum_{\alpha \in I} \Vert v_{\alpha} \Vert_{H_{\alpha}}^2$$$$

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

$$$$H = \{\mathbf v=\{v_\alpha\}_{\alpha\in I}:\sum_{\alpha\in I}\|v_\alpha\|_{\mu_\alpha}^2<\infty\} .$$$$