$L^2$ convergence from convergence in distribution and uniform integrability Is it true that if $X_n \to X$ in distribution and $\{X_n^2\}$ are uniformly integrable then $X_n \to X$ in $L^2$ 

No: let $X$ be a random variable such that $\mathbb P\{X=1\}=\mathbb P\{X=-1\}=1/2$ and $X_n=X$ for $n$ even, $X_n=-X$ for $n$ odd. Since $X$ is symmetric, the sequence $\left(X_n\right)_{n\geqslant 1}$ converges in distribution to $X$. Since $X_n^2$ equals $1$ almost surely, the sequence $\left(X_n^2\right)_{n\geqslant 1}$ is uniformly integrable. But 
$X_{2n+1}-X=-2X$, which does not converge to $0$ in $\mathbb L^2$. 
