I'm in the following situation: 


Consider a centred Gaussian measure $\mu\_0$ on a separable Hilbert space $X$ with covariance operator $Q \in \mathcal{L}(X)$ (positive definite, self-adjoint, trace class). Denote the Cameron-Martin space of $\mu\_0$ by $E := Q^\frac12(X)$ with $\|\cdot\|\_E := \|Q^{-\frac12}\cdot\|$. Let $(u\_n)\_{n\in\mathbb{N}}$ be a sequence in $X$ that is unbounded with respect to $\|\cdot\|\_E$, but converges towards $\bar u \in E$ with respect to $\|\cdot\|\_X$. Moreover, let $\varepsilon\_n \to 0$ with $\varepsilon\_n > 0$ for all $n \in \mathbb{N}$.


Now I want to show, that $\lim\sup\_{n \to \infty} \frac{\mu\_0(B\_{\varepsilon\_n}(u\_n))}{\mu\_0(B\_{\varepsilon\_n}(\bar u))} \le 1$, where $B\_\varepsilon(u) := \{x \in X: \|x - u\|\_X \le \varepsilon\}$.


Are there ideas how to prove this or can anyone recommend me according literature?


Remark: We have $\lim\_{n\to\infty} \frac{\mu\_0(B\_{\varepsilon\_n}(0))}{\mu\_0(B\_{\varepsilon\_n}(\bar u))} = e^{\frac12\|\bar u\|\_E^2}$, so we can consider $\frac{\mu\_0(B\_{\varepsilon\_n}(u\_n))}{\mu\_0(B\_{\varepsilon\_n}(0))}$ alternatively.



Probabilities of small balls with convergent center points under Gaussian measure
Masoumeh Dashti kindly provided the following proof, which I reproduce here in my words and notation:


Let $u \mapsto W\_u$ denote the unique linear extension of $E \to L^2(X,\mu\_0)$, $u \mapsto \widetilde{W}\_u = (Q^{-\frac12}u,\cdot)\_X$, to $X$.
We first note that $Z := Q(X)$ is dense in $E = Q^{\frac12}(X)$, and that for every $w \in Z$ the linear functional $W\_{Q^{-\frac12}w} = (Q^{-1}w,\cdot)\_X$ is continuous. Now by the Cameron-Martin Theorem and Anderson's inequality,


$$ \begin{align\*}
\mu\_0(B\_{\varepsilon\_n}(u\_n))
&= \int\_{B\_{\varepsilon\_n}(u\_n - w)} \exp\left(-\|w\|\_E^2 + W\_{Q^{-\frac12}w}(v)\right) \mu\_0(dv) \\
&\le e^{-\frac12\|w\|\_E^2} \sup\_{u \in B\_{\varepsilon\_n}(u\_n - w)} \left\{\exp((Q^{-1}w,v)\_X)\right\} \mu\_0(B\_{\varepsilon\_n}(u\_n - w)) \\
&\le e^{-\frac12\|w\|\_E^2} \sup\_{u \in B\_{\varepsilon\_n}(u\_n - w)} \left\{\exp((Q^{-1}w,v)\_X)\right\} \mu\_0(B\_{\varepsilon\_n}(0))
\end{align\*} $$
holds for all $w \in Z$ and $n \in \mathbb{N}$. On the other hand, the symmetry of $B\_{\varepsilon\_n}(0)$ implies


$$ \begin{align\*}
\mu\_0(B\_{\varepsilon\_n}(\bar u)) 
&= e^{-\frac12\|\bar u\|\_E^2} \int\_{B\_{\varepsilon\_n}(0)} \exp\left(W\_{Q^{-\frac12}\bar u}(v)\right) \mu\_0(dv) \\
&= e^{-\frac12\|\bar u\|\_E^2} \int\_{B\_{\varepsilon\_n}(0)} \frac12 \left(\exp\left(W\_{Q^{-\frac12}\bar u}(v)\right) + \exp\left(-W\_{Q^{-\frac12}\bar u}(v)\right)\right) \mu\_0(dv) \\
&\ge e^{-\frac12\|\bar u\|\_E^2} \mu\_0(B\_{\varepsilon\_n}(0)).
\end{align\*} $$


Using the continuity of $(Q^{-1}w,\cdot)\_X$ and the convergence $u\_n \to \bar u$ in $X$, we obtain


$$ \begin{align\*}
{\lim\sup}\_{n \to \infty} \frac{\mu\_0(B\_{\varepsilon\_n}(u\_n))}{\mu\_0(B\_{\varepsilon\_n}(\bar u))} 
&\le e^{\frac12\|\bar u\|\_E^2 - \frac12\|w\|\_E^2} \exp((Q^{-1}w, \bar u - w)\_X) \\
&= e^{\frac12\|\bar u\|\_E^2 - \frac12\|w\|\_E^2} \exp((w, \bar u - w)\_E)
\end{align\*} $$


for all $w \in Z$. In particular, if we consider a sequence $\{w\_j\}\_{j\in\mathbb{N}} \subset Z$ with $w\_j \to \bar u$ in $E$ as $j \to \infty$, the previous estimate leads to


$$ {\lim\sup}\_{n \to \infty} \frac{\mu\_0(B\_{\varepsilon\_n}(u\_n))}{\mu\_0(B\_{\varepsilon\_n}(\bar u))} \le 1. $$


