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.



Convergence of Gaussian measures
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.


