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Given any separable Banach space $B$ and a centered Gaussian measure $Q$ on it with Cameron-Martin space $H$, does there exist a Hilbert space $G$ and a Gaussian measure $W$ on it such that following hold

1) $B$ is a dense subspace of $G$ and restriction of $W$ to $B$ is $Q$ (that makes $W$ supported on $B$).

2) $(B,Q)$ and $(G,W)$ have the same Cameron-Martin space $H$.

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    $\begingroup$ It seems to me that your question is simply "can any separable Banach space be densely embedded into some Hilbert space". The answer to this is obviously yes. $\endgroup$
    – Martin Hairer
    Commented Dec 15, 2013 at 17:50
  • $\begingroup$ @Martin: How does one prove that? I was thinking in the same direction. $\endgroup$
    – Nate Eldredge
    Commented Dec 15, 2013 at 19:54
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    $\begingroup$ Take a sequence $\ell_n$ of linear functionals of norm $1$ on $B$ such that $\|x\|_B = \sup_{n} \ell_n(x)$. (This exists by separability.) Then complete $B$ under the norm $\|x\|_H^2 = \sum_n n^{-2} |\ell_n(x)|^2$. $\endgroup$
    – Martin Hairer
    Commented Dec 16, 2013 at 8:34
  • $\begingroup$ B may still have measure 0 in G wrt W measure in this construction. $\endgroup$
    – user44179
    Commented Dec 17, 2013 at 7:59
  • $\begingroup$ @Nate Supposing the assertion were true, wont it reduce the study of Gaussian measures on Banach spaces to that on Hilbert spaces? $\endgroup$
    – user44179
    Commented Dec 17, 2013 at 8:10

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