Probabilities of small balls with convergent center points under Gaussian measure 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.
 A: 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
$$ {\lim\sup}_{n \to \infty} \frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(\bar u))} \le 1. $$
