Assume we have a Gaussian measure $\mu$ supported on a Banach space $X$. Can we always find a Hilbert space $H$ embedded in $X$ sch that $\mu$ is also supported on $H$?
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2 Answers
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No, not necessarily. It is shown in Examples 3.6.6 and 3.6.7 of Bogachev's Gaussian Measures that if $X = C([0,1])$ and $\mu$ is classical Wiener measure, then for any Hilbert space $H$ embedded in $X$, we have $\mu(H) =0$, so that $\mu$ is not supported on $H$.
answered Jan 6, 2017 at 5:05
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2$\begingroup$ There's one thing that confuses me about that argument, though. If I am reading it right, it reaches a contradiction by showing that the embedding of the Cameron-Martin space for $X$, which is $H^1_0((0,1])$, into $L^2([0,1])$ would have to be trace class, and it isn't. However, this question seems to say that the embedding is trace class. What's going on? $\endgroup$ Commented Jan 6, 2017 at 6:28
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$\begingroup$ I believe that the embedding of H^1 into L^2 is NOT trace class (it is just on the boundary, ie is in the p-Schaten classes for $p>1$). $\endgroup$ Commented Jan 7, 2017 at 9:02
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For the always existence of Hilbert space $H$ need some conditions about Banach space and structure of Gaussian measure to be satisfied as montioned in the below theorem where the uniqness is the strong condition for existence but in your case it's not always Hilbert space $H$ exist :
Theorem: Suppose that $E$ is a separable, real Banach space and that is a centered Gaussian measure which is non-degenerate.Then there exists a unique Hilbert space H such that $(H;E;W)$ is an abstract Wiener.
For more informations look this paper about "Gaussian Measures on a Banach Space" theorem 8.2.5 with a complet proof.
Edit: I edited my answer because I don't meant the existence of Hilbert space $H$ related to the given question but i meant the existence will be satisfied if and only if the conditions cited in theorem 8.2.5 hold
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1$\begingroup$ This is almost identical to an answer posted yesterday by T. Amdeberhan and then deleted. It doesn't answer the question because in this case you get $\mu(H)=0$, so the measure is not supported on $H$. Also, Theorem 8.2.1 of that paper is a completely different theorem; you meant 8.2.5. $\endgroup$ Commented Jan 6, 2017 at 16:35
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1$\begingroup$ The updated version still does not seem to address the original question, nor does it fully convince me that the material being linked to is understood $\endgroup$ Commented Jan 6, 2017 at 19:08
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$\begingroup$ I'm still confused. You talk about "existence of Hilbert space"; presumably you mean the existence of a Hilbert space satisfying certain properties, but I don't understand what you mean those properties to be. Can you try to give a clear and precise statement of what you are claiming is true, and explain how it relates to the question at hand? $\endgroup$ Commented Jan 6, 2017 at 19:26