Timeline for Gaussian measure on Banach spaces
Current License: CC BY-SA 3.0
11 events
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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 |
improved formatting, added tags.
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Dec 15, 2013 at 10:43 | review | Suggested edits | |||
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Dec 14, 2013 at 22:43 | history | edited | user44179 | CC BY-SA 3.0 |
added 10 characters in body
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Dec 14, 2013 at 19:41 | review | First posts | |||
Dec 14, 2013 at 19:44 | |||||
Dec 14, 2013 at 19:21 | history | asked | user44179 | CC BY-SA 3.0 |