Characteristic functions of infinite dimensional random elements I am trying to understand if it is possible to prove the convergence in distribution of a sequence of infinite dimensional random elements using characteristic functions.

Suppose that $\{X_n:n\ge1\}$ is a sequence of random elements with values in $L_2[0,1]$ (the Hilbert space of square integrable real functions defined on $[0,1]$). Is it possible to define a characteristic function of $X_n$ for each $n\ge1$ and to prove the convergence in distribution of $X_n$ by showing that the sequence of characteristic functions converges pointwise?

There is a somewhat related example on Wikipedia of a characteristic function of a stochastic process. If $X(s)$ is a stochastic process, then
$$
\varphi_X(t)
=\operatorname E\exp\biggl\{i\int_\mathbb R t(s)X(s)ds\biggr\}
$$
for all functions $t(s)$ such that the integral  $\int_\mathbb R t(s)X(s)\mathrm ds$ converges for almost all realizations of $X$. If the characteristic functions of a sequence of stochastic processes converge pointwise, does it mean that the finite-dimensional distributions of stochastic processes converge?
Any help is much appreciated!
 A: It is possible to define characteristic functions for random variables with values in Hilbert/Banach spaces. If $\mathcal H$ is a separable Hilbert space and $X$ a random variable with values in $\mathcal H$, then we can define for $f\in \mathcal H$ the quantity 
$$\varphi_X(f)=\mathbb E\left[\exp\left(i\langle f,X\rangle\right)\right].$$
The terminology "characteristic function" still makes sense since two random variables $X$ and $Y$ such that $\varphi_X(f)=\varphi_Y(f)$ for each $f\in\mathcal H$ have the same distribution. 
However, this is not so useful to prove convergence in distribution. For example, if we take $(\xi_j)_{j\geqslant 1}$ a sequence of independent identically distributed standard Gaussian random variables and define $X_n:=\sum_{j=1}^n\xi_je_j$ (where $\mathcal H=\ell^2$ and $e_j=(0,\dots,0,1,0,\dots)$ and $1$ is at the $j$th position). We can show that for each $f\in \mathcal H$, the sequence $\left(\varphi_{X_n}(f)\right)_{n\geqslant 1}$ converges to $\exp\left(-\lVert f\rVert^2\right)$, hence we have the pointwise convergence. However, the sequence $\left(X_n\right)_{n\geqslant 1}$ is not tight in $\mathcal H$.
