Let $X$ be a separable Banach space with its Borel $\sigma$-algebra $\mathcal F$. Let $x_n \to x$ in $X$. Fix a Gaussian covariance operator $K$, and let $\mathbb P_n$ and $\mathbb P$ be Gaussian measures on $X$ with covariance $K$ and means $x_n$ and $x$, respectively.
Question: How do I show that $\mathbb P_n \to \mathbb P$ weakly?
Surely this is a theorem or an exercise somewhere; e.g. in Talagrand and Ledoux's Probability in Banach Spaces or Vakhania, Tarieladze and Chobanyan's Probability Distributions in Banach Spaces.
The characteristic functions of $\mathbb P_n$ converge to those of $\mathbb P$ (simple exercise). By de Acosta's theorem, this implies that $\mathbb P_n \to \mathbb P$, provided that the family $\{\mathbb P_n\}$ is flatly concentrated. I'm not so familiar with the concept (hence this question), but I'm guessing this is related to the concentration of measure property of Gaussian measures.
Best Answer
Somehow I didn't register how strong the assumptions Tom was making were, hence the fact that my other answer missed the point.
Unless I'm still missing something, this is very easy. Say $Z$ is a Gaussian random vector in $X$ with covariance $K$ and mean $0$. You want to show that $Z+x_n \to Z+x$ weakly, i.e. $\mathbb{E} f(Z+x_n) \to \mathbb{E} f(Z+x)$ for every bounded continuous $f:X\to \mathbb{R}$. Since $f$ is both bounded and continuous, this follows immediately from dominated convergence.