# weak convergence in infinite dimensional spaces

Weak convergence can be tricky when dealing with infinite dimensional spaces. For example, the usual Levy's continuity theorem does not extend readily to separable Banach spaces.

Consider a (separable) Hilbert space $H$: we know that the sequence of $H$-valued random variables $Z^N$ converges in laws towards the random variable $Z$. We also know that the sequence of random processes $W^N \in C([0,T],H)$ (continuous functions from $[0,T]$ to $H$) converges in laws to a Brownian motion $W$.

Question: If we can show that for any $k,h_1, \ldots, h_n \in H$ and time $t_j$ $$E[e^{i(k,Z^N)}e^{\sum_{j=1}^N i(h_j,W_{t_j})}] \to E[e^{i(k,Z^N)}] \ E[e^{\sum_{j=1}^N i(h_j,W_{t_j})}],$$ is it enough to conclude that the sequence of couples $(Z^N,W^N)$ also converge in laws to $(Z,W)$, with $Z$ and $W$ independent ?

Since both sequences $(Z^N)$ and $(W^N)$ converge in distribution, both families of distributions are tight due to Prokhorov theorem. It easily follows that the sequence of couples $(Z^N,W^N)$ is tight, and again due to Prokhorov theorem, it is relatively compact, and we have only to see that there is a unique limiting point in distribution for any subsequence of $(Z^N,W^N)$. But each limiting point has to have the characteristic functional that you give in the r.h.s., and this characterizes the limiting distribution uniquely.