# $\Psi$ in finite Wiener–Itô Chaos implies existence of continuous representative on neighborhood of Cameron–Martin space?

Let $$(\Theta, H, \mu)$$ be an abstract Wiener space, i.e. let $$(\Theta, \lVert \cdot \rVert_{\Theta})$$ be a separable Banach space, let $$(H, \langle \cdot, \cdot \rangle_{H})$$ be a separable Hilbert space densely embedded in $$\Theta$$, and let $$\mu$$ be a Gaussian measure on $$\Theta$$ with characteristic functional $$\exp( - \frac{1}{2} \Vert f \Vert_H^2 ), f \in \Theta^{\ast}$$. Let $$(e_i)_{i = 1}^{\infty} \subseteq \Theta^{\ast}$$ be an orthonormal basis of $$H$$ contained in $$\Theta^{\ast}$$.

Let $$\Psi: \Theta \rightarrow \mathbb{R}$$ be a measurable function contained in the $$d$$-th inhomogeneous Wiener–Itô Chaos $$\mathcal{H}^{(\leq d)}(\Theta, \mu)$$, i.e. the $$L^2(\Theta, \mu)$$ closure of the linear span of

$$\Bigl\{ \prod_{\alpha_i \in \alpha} h_{\alpha_i}(e_i) : \alpha = (\alpha_1, \alpha_2, \dotsc) \in \mathbb{N}^{\mathbb{N}}, \lvert \alpha \rvert \leq d \Bigr\}$$

where $$h_1, h_2, \dotsc$$ are the usual Hermite polynomials.

Question: In the situation above, does there necessarily exist a measurable $$\Phi: \Theta \rightarrow \mathbb{R}$$ which equals $$\Psi$$ for $$\mu$$-almost every $$\theta \in \Theta$$ and is continuous on a neighbourhood of $$H$$?

Remark: I'm actually interested in the case where $$\mathbb{R}$$ is replaced by another separable Banach space $$E$$, but I'm more than happy to understand the special case first, and I don't think the difference is too drastic for this purpose anyway.

Special Cases: In the case where $$\dim \Theta < \infty$$ (and $$\mu$$ is non-degenerate), this is of course true. Since in that case the ONB $$(e_i)_{i = 1}^{\infty}$$ is finite, the (inhomogeneous) Wiener–Itô Chaos is finite-dimensional and spanned by polynomials, $$\Psi$$ itself is continuous on all of $$\Theta$$, which constitutes a neighbourhood of $$H$$.

(Edit) Remark: Since on $$\mathbb{R}$$ with the standard normal distribution $$\gamma_1$$, an indicator function like $$1_{[0, \infty)}$$ does not have a continuous version, I tried finding a counterexample to my question by looking for a Bernoulli random variable defined on $$\Theta$$, and then modifying it to something like $$1_{\{e_1(\cdot) > 0\}}$$. However, one can show that random variables in finite Wiener–Itô Chaos are Malliavin differentiable, but a (non-trivial) Bernoulli random variable is not. So maybe the answer to my question is positive after all.

Motivation: The motivation for this comes from trying to show a large deviation principle (LDP) for the measures $$(\Psi^{\ast} \mu_{\varepsilon})_{\varepsilon>0}$$ on $$\mathbb{R}$$ (but generically $$E$$), where $$(\mu_{\varepsilon})_{\varepsilon>0}$$ is the family of measures on $$\Theta$$ defined by $$\mu_{\varepsilon}(\cdot) = \mu(\varepsilon^{-1/2}( \cdot))$$. Those of course satisfy an LDP with rate

$$I(x) = \frac{1}{2} \lVert x \rVert_H^2 , ~~x \in H$$

and $$I(x) = \infty$$ for $$x \notin H$$. Now, if $$\Psi: \Theta \rightarrow \mathbb{R}$$ were continuous, then the contraction principle would assert that the rate function of $$(\Psi^{\ast} \mu_{\varepsilon})_{\varepsilon>0}$$ is

$$I(s) = \inf \Bigl\{ \frac{1}{2} \lVert h \rVert_H^2 : \Psi(h) = s \Bigr\}, ~~ s \in \mathbb{R}.$$

However, since $$\mu(H) = 0$$, the image of $$H$$ under $$\Psi$$, which controls the rate function, could be just about anything. However, if there was a continuous $$\Phi$$ which equals $$\Psi$$ $$\mu$$-almost surely and is continuous on a neighbourhood of $$H$$, then this gives a canonical candidate for the rate function.

Alternatively, one might use an extended version of the contraction principle to get an LDP here. But that version requires that there exist measurable functions $$\Psi_N: \Theta \rightarrow \mathbb{R}$$ which are continuous on neighborhoods of $$H$$ and converge uniformly to $$\Psi$$ on that neighborhood, implying that $$\Psi$$ had to be continuous that neighborhood.

• One thing that might be helpful. Ito integral can always be converted to Stratonovich integral. However you do pick up extra $\varepsilon$ terms under the scaled measure $\mu_\varepsilon$. Sep 17 at 18:29

The answer to the question is no, even in the case $$d=1$$. Take for example $$H = \ell^2$$, $$\Theta = \{\xi\,:\, \|\xi\| = \sup_{n\ge 1} |\xi_n|/(1+\log n) < \infty\}$$, $$\mu$$ the law of i.i.d. normals, and $$\Psi(\xi) = \sum_n \xi_n/n^{3/4}$$. Assume by contradiction that a version of $$\Psi$$ exists that is continuous at $$0$$, so that there exist $$K < \infty$$ and $$\epsilon > 0$$ such that $$|\Psi(\xi)| \le K$$ for almost every $$\|\xi\| \le 2\epsilon$$. In particular, it is the case that, for every $$h$$ with $$\|h\| \le \epsilon$$, $$|\Psi(\xi + h)| \le K$$ on the ball of radius $$\epsilon$$. On the other hand, if $$h \in H$$, then the Cameron-Martin theorem tells us that the law of $$\Psi(\cdot + h)1_{\|\cdot\| \le \epsilon}$$ is equivalent to that of $$\Psi(\cdot)1_{\|\cdot\| \le \epsilon}$$ and that, writing $$f_n = n^{-3/4}$$, we have $$\Psi(\cdot+h) = \Psi(\cdot) + \langle{h,f}\rangle$$ almost surely. Since we can easily find an element $$h \in H$$ such that $$\|h\|\le \epsilon$$ but $$\langle{h,f}\rangle \ge 2K$$, we have a contradiction.