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In these lecture notes, exercise 3.39 is posed: Let $\tilde B, B$ be Banach spaces, and $\mu$ be a Gaussian measure on $B$. If $\tilde B$ is continuously embedded into $B$ and $\mu(\tilde B)=1$, then the "restricted" measure $\tilde \mu$ on $\tilde B$ defined by $$\tilde\mu(A)=\mu(A), \quad A\in\mathcal{B}(\tilde B)$$ is Gaussian.

Questions: How can we show $\tilde\mu$ is Gaussian? Is the Cameron-Martin spaces and corresponding Cameron-Martin norms the same?

Thoughts: One first has to show that the embedding $$j:B'\rightarrow \mathcal{R}_\mu,$$ where $B'$ is the dual space and $\mathcal{R}_\mu$ is the $L^2(B,\mu)$-closure of $B'$, extends to an embedding $$\tilde j: \tilde B'\rightarrow \mathcal{R}_\mu$$

  1. Then, to me it looks like $\|f\|_{L^2(B,\mu)}=\|f\|_{L^2(\tilde B,\tilde\mu)}$ for $f\in L^2(B,\mu)$. Does this mean that $\mathcal{R}_{\tilde \mu}$, the $L^2(\tilde B,\tilde \mu)$-closure of $\tilde B'$ equals $\mathcal{R}_\mu$?
  2. We know elements in $\mathcal{R}_\mu$ are $L^2(B,\mu)$ Gaussian random variables (by an $L^2$-limit argument of approximating sequence in $\mathcal{R}_\mu \cap B'$). To me this means $\tilde l \in \tilde B'$ satisfy $\tilde l(X)$ is Gaussian for $X\sim \mu$, so we are not there yet.

I am a little confused.

SoupMath
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1 Answers1

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It is not necessary to mess around with Cameron-Martin spaces to see that that measure is Gaussian. Let $\tilde{l} \in \tilde{B}^*$ and let $l \in B^*$ be a Hahn-Banach extension of $\tilde{l}$.

Then $$\tilde{l}^* \tilde{\mu}(A) = \tilde{\mu}(\tilde{l}^{-1}(A)) = \mu(\tilde{l}^{-1}(A)) = \mu(l^{-1}(A))$$ where the last equality follows since $B \setminus \tilde{B}$ is a $\mu$-null set. Therefore $\tilde{l}^* \tilde{\mu} = l^* \mu$ and so $\tilde{l}^* \tilde{\mu}$ is Gaussian which shows that $\tilde{\mu}$ is Gaussian since $\tilde{l} \in \tilde{B}^*$ was arbitrary.

It is then the case that the two measures have the same Cameron-Martin space, up to the natural isometric isomorphism from $L^2(B, \mu)$ to $L^2(\tilde{B}, \tilde{\mu})$. This follows since this isometric isomorphism restricts to an isometric isomorphism of $B^*$ with $\tilde{B}^*$ with the $L^2$-norms.

Rhys Steele
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  • Thanks a lot! Can you expand a little bit on the last part? I can see there is a natural isometry from $B$ to $\tilde B$, but does this extend to the closure? And why does this imply the same Cameron-Martin space? – SoupMath Nov 03 '22 at 19:48
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    You have an isometry $i: B^* \to \tilde{B}^$ with the relevant $L^2$-norms. This means you have an isometric embedding $i: B^ \to R_{\tilde{\mu}}$. Since $B^*$ is dense in $R_\mu$ this isometric embedding extends to an isometric isomorphism between $R_\mu$ and $R_{\tilde{\mu}}$. – Rhys Steele Nov 03 '22 at 19:55