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.


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

$$ {\lim\sup}_{n \to \infty} \frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(\bar u))} \le 1. $$


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