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


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*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.
