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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 ?

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up vote 7 down vote accepted

This seems to be a typical case where one can apply Prokhorov's theorem.

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.

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great: that was exactly what I was looking for. Thank you! – Alekk May 10 '10 at 20:04

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