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I would like to prove (if possible, otherwise find a counterexample for) the following lemma:

Let $(X,\|\cdot \|_X)$ be a separable Banach space. Additionally, we have a centred Gaussian measure $\mu$ on $X$, with Cameron--Martin space $(E, |\cdot |_E)$.

We consider a sequence $x_n\in X$ with the following properties: $|x_n|_E\to \infty$ (wlog nondecreasing) and there exists a $C > 0$ such that $\inf_n \|x_n\|_X > C$.

Let $\delta_n$ be any nonincreasing sequence $\delta_n \to 0$ for $n\to \infty$. We define the open ball wrt $\|\cdot\|_X$: $$B_\delta (z) = \{x\in X: \|x-z\|_X < \delta\}.$$

Then (this is what I want to prove) $$ \lim_{n\to \infty} \frac{\mu(B_{\delta_n}(x_n))}{\mu(B_{\delta_n}(0))} = 0. \quad (*)$$

My thoughts so far:

For fixed $x\in X$, it is known (for example Bogachev, this is because the squared CM norm is the Onsager--Machlup functional, and it also follows from this thread: Probabilities of small balls with convergent center points under Gaussian measure) that $$ \lim_{m\to \infty} \frac{\mu(B_{\delta_m}(x))}{\mu(B_{\delta_m}(0))} = e^{-|x|_E^2/2}. $$

Plugging a sequence $x_n$ as above in here:

$$\lim_{n\to\infty} \lim_{m\to \infty} \frac{\mu(B_{\delta_m}(x_n))}{\mu(B_{\delta_m}(0))} = \lim_{n\to \infty} e^{-|x_n|_E^2/2} = 0.$$

This means that (*) amounts to "applying both limits at once" instead of "sequentially". In my opinion, the missing piece here would be uniformity (with respect to $n$) of the limit

$$ \lim_{m\to \infty} \frac{\mu(B_{\delta_m}(x_n))}{\mu(B_{\delta_m}(0))}.$$

In other words, this expression should converge uniformly over $n$ by using $\inf_n \|x_n\|_X > C$ (which means that the balls $B_{\delta_m}(x_n)$ are uniformly bounded away from the origin). But I am unable to prove this. Maybe I am wrong and the statement is wrong after all?

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This is not a full answer but maybe of interest. Given a sequence of $x_n$ with $\|x_n\|_E\to \infty$ we can always find a sequence of $\delta_n\to 0$ so that

$$\lim_{n\to\infty}\frac{\mu(B_{\delta_n}(x_n))}{\mu(B_{\delta_n}(0))}=0.$$

First note that by Cameron-Martin we have that

$$\mu(B_{\delta_n}(x_n))=\int_{B_{\delta_n}(0)}e^{x_n^\ast(\omega)-\frac12\|x_n\|^2_E}\mu(d\omega)=e^{-\frac12\|x_n\|^2_E}\int_{B_{\delta_n}(0)}e^{x_n^\ast(\omega)}\mu(d\omega),$$

where $x_n^\ast\in X^\ast$ is a continuous linear functional. As it is continuous we have that for each $x_n^\ast$ there is some $L_n>0$ so that

$$|x_n^\ast(\omega)|\leq L_n\|\omega\|_X$$

for all $\omega\in X$. Therefore on the set $B_{\delta_n}(0)$ we have that

$$|x_n^\ast(\omega)|\leq L_n\delta_n$$

and thus

$$\mu(B_{\delta_n}(x_n))\leq e^{-\frac12\|x_n\|^2_E+L_n\delta_n}\int_{B_{\delta_n}(0)}\mu(d\omega).$$

If $\delta_n$ is so that $L_n\delta_n$ doesn't diverge faster that $\frac12\|x_n\|_E^2$ then we have that

$$\lim_{n\to\infty}\frac{\mu(B_{\delta_n}(x_n))}{\mu(B_{\delta_n}(0))}\leq \lim_{n\to\infty} e^{-\frac12\|x_n\|^2_E+L_n\delta_n}=0.$$

In particular we can choose $\delta_n=1/(nL_n)$, eg.

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  • $\begingroup$ Yes, that is something along the lines of what I also thought about. Problem is that so far I have found no way of controlling the norm $L_n$ unfortunately. Also, an element $x_n\in E$ (i believe) does not necessarily correspond to a dual element $x_n^\star$, only something in the closure of $X^\star$ with respect to $L^2(X,\mu)$. Thank you anyway! $\endgroup$ Sep 12 '21 at 16:05
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    $\begingroup$ To clarify further: In this setting I am stuck with a specific combination of $x_n$ and $\delta_n$ and I cannot "accelerate" $\delta_n$ further. $\endgroup$ Sep 12 '21 at 17:22

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