5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.