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Timeline for Gaussian measure on Banach spaces

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Dec 17, 2013 at 8:49 comment added Nate Eldredge @user44179: I think the idea is to define $W$ as the extension of $Q$ to $G$ (so that $W(G \setminus B) = 0$). In other words, $W$ is the pushforward of $Q$ under the inclusion map. For your second comment, note that many of the interesting aspects of $Q$ have to do with its interaction with the topology of $B$; the interaction of $W$ with the topology of $H$ may not be as interesting. For example, if $Q$ is Wiener measure, asking about the uniform norm and the $L^2$ norm of a Brownian motion gives us very different information.
Dec 17, 2013 at 8:10 comment added user44179 @Nate Supposing the assertion were true, wont it reduce the study of Gaussian measures on Banach spaces to that on Hilbert spaces?
Dec 17, 2013 at 7:59 comment added user44179 B may still have measure 0 in G wrt W measure in this construction.
Dec 16, 2013 at 8:34 comment added Martin Hairer 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 $\|x\|_H^2 = \sum_n n^{-2} |\ell_n(x)|^2$.
Dec 15, 2013 at 19:54 comment added Nate Eldredge @Martin: How does one prove that? I was thinking in the same direction.
Dec 15, 2013 at 17:50 comment added Martin Hairer 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.
S Dec 15, 2013 at 11:15 history suggested Davide Giraudo CC BY-SA 3.0
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Dec 14, 2013 at 22:43 history edited user44179 CC BY-SA 3.0
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Dec 14, 2013 at 19:41 review First posts
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Dec 14, 2013 at 19:21 history asked user44179 CC BY-SA 3.0