# Small ball Gaussian probabilities with moving center

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?

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.

• 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! Sep 12 '21 at 16:05
• 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. Sep 12 '21 at 17:22