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Is it possible to construct an abstract Wiener space $(W,H,\mu)$ such that $C^{0,\frac{1}{2}}(\Omega)\subset H$ and $W$ is a normed function space such that the convergence in norm implies convergence in (Lebesgue) measure? The set $\Omega$ is a bounded subset of $\mathbb R^d$.

I know there exist Wiener spaces $(C[0,1],H^\alpha[0,1],\mu)$, with a fractional Sobolev space $H^\alpha$ with index $\alpha>1/2$, and the space $(H^{-\frac{d+1}{2}}(\Omega),L_2(\Omega),\mu)$. Is there anything in between? I would be very grateful for help.

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  • $\begingroup$ A candidate (in 1 dim) would be $H=H^{1/2}$, and the question seems to be whether the Gaussian Fourier series partial sums $\sum_1^N n^{-1/2}Z_n e^{inx}$ (with $Z_n$ i.i.d. complex Gaussian variables) converge in measure on $(0,2\pi)$. I would be surprised if it did! And the condition that $W$ be normed adds to the doubtfulness... $\endgroup$
    – Jean Duchon
    Commented Oct 10, 2015 at 15:53
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    $\begingroup$ What you are asking is equivalent to squeezing Hilbert spaces $H$ in between $C^{0,\frac{1}{2}}(\Omega)$ and $C(\Omega^\prime)$ for subsets $\Omega^\prime$ of arbitrarily large measure. Indeed, whenever a Gaussian measure lives in $L^0$ its Cameron-Martin space has to consist of functions that are continuous on sets of large measure, and conversely, for any such Cameron-Martin space, the canonical Gaussian measure lives in $W := L^1(\mu^\prime)$ for some equivalent measure $\mu^\prime$. $\endgroup$
    – Alexander Shamov
    Commented Mar 9, 2016 at 4:21
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    $\begingroup$ My not-so-educated guess is that this is impossible because the embedding $C^\alpha \to C$ should only be $2$-summing for $\alpha > d/2$. $\endgroup$
    – Alexander Shamov
    Commented Mar 9, 2016 at 4:26
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    $\begingroup$ I have just had to roll back an edit from @AmirSagiv which changed the sense/intent of the original by implicitly reversing quantifiers: the question starts with $\Omega$ and asks to find $(W,H,\mu)$, while Amir's edit starts with $(W,H,\mu)$ and asked "does there exist $\Omega$". Amir, please stop doing these kinds of edit if you are not absolutely certain about the original intent and the English idiom! $\endgroup$
    – Yemon Choi
    Commented May 8, 2016 at 13:10

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Not an answer, but a few (perhaps trivial) ideas. (Please don't upvote this, because this question should stay "unanswered".)

  • It's known that for any abstract Wiener space, the inclusion $H \subset W$ must be compact. So we could ask the purely functional-analytic question: do there even exist a Hilbert space $H$ and a Banach space $W$ such that $C^{0,\frac{1}{2}}(\Omega) \subset H \subset W \subset L^0(\Omega)$, where $L^0(\Omega)$ is the space of measurable functions on $\Omega$ with the topology of convergence in measure, all inclusions are continuous, and the inclusion $H \subset W$ is compact? If not, this resolves the present question negatively.

  • It's also known that any Hilbert–Schmidt operator on $H$ leads to a measurable norm, under which the completion $W$ admits a Gaussian measure $\mu$ whose Cameron-Martin space is $H$. So we could ask: do there exist Hilbert spaces $H,W$ such that $C^{0,\frac{1}{2}}(\Omega) \subset H \subset W \subset L^0(\Omega)$, where all inclusions are continuous and the inclusion $H \subset W$ is Hilbert–Schmidt? If yes, this resolves the present question positively.

  • I thought at first about trying to take $H = L^2(\Omega)$, but the compactness of $H \subset W$ rules this out, since one can find an $L^2$-bounded sequence for which no subsequence converges in measure (e.g. $f_n(x) = e^{inx}$). Maybe this would rule out some other candidates for $H$ as well.

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  • $\begingroup$ Your first idea, for a negative answer (which is my bet nevertheless) doesn't work, as the inclusion $H^s\subset L^2$ is compact ($s>0$, $\Omega$ bounded) and $C^{0,1/2}\subset H^s$ for $s<\frac12$. (Besides, only for $s>\frac12$ -- if $\Omega\subset\mathbb R$ -- is the inclusion $H^s\subset L^2$ Hilbert-Schmidt, so that a positive answer cannot be that simple). $\endgroup$
    – Jean Duchon
    Commented Oct 18, 2015 at 9:17
  • $\begingroup$ You could always make this answer CW... $\endgroup$
    – David Roberts ♦
    Commented Jul 3, 2017 at 5:53

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