MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $H$.

share|cite|improve this question
    
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. – Martin Hairer Dec 15 '13 at 17:50
    
@Martin: How does one prove that? I was thinking in the same direction. – Nate Eldredge Dec 15 '13 at 19:54
1  
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 $\|x\|_H^2 = \sum_n n^{-2} |\ell_n(x)|^2$. – Martin Hairer Dec 16 '13 at 8:34
    
B may still have measure 0 in G wrt W measure in this construction. – user44179 Dec 17 '13 at 7:59
    
@Nate Supposing the assertion were true, wont it reduce the study of Gaussian measures on Banach spaces to that on Hilbert spaces? – user44179 Dec 17 '13 at 8:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.