Weak convergence in Hilbert space L2 implies convergence in distribution? Does weak convergence in $L^2$ (for $X_n, X \in L^2$ we say that $X_n$ converges weakly to $X$ ($X_n \rightarrow^w X$) if for every $Y\in L^2$ we have $\mathbb{E}X_nY \rightarrow \mathbb{E}XY$) implies convergence in distribution ($X_n \rightarrow^D X$ if $F_{X_n} \rightarrow F_{X}$ in every point of continuity of $F_X$)? Maybe we need some extra assumptions? If it is so, how to prove that fact?
Thanks a lot

Weak convergence is not enough in order to give convergence in distribution. For example, take $\Omega:=[0,1]$, and $X_n\left(\omega\right):=\sin\left(\pi n\omega\right)$. Then $X_n\to 0$ weakly but not in distribution. Indeed, if a sequence $ X_n\to 0$ in distribution, then $X_n\to 0$ in probability and since $\left\lvert X_n\right\rvert\leqslant 1$, this implies that $\mathbb E\left\lvert X_n\right\rvert\to 0$, which is not the case (or more directly, in the definition of convergence in distribution given by $\mathbb Ef(X_n)\to\mathbb E f(X)$ for any continuous and bounded $f$, choose $f\colon x\mapsto \min\left\{1,\left\lvert x\right\rvert    \right\}$.               
But strong convergence in $L^2$ implies convergence in probability which implies convergence in distribution. 
