convergence in mean square implies convergence of variance

I need some hints for the following question:

Suppose $X,X_1,X_2, \cdots \in L^2(\Omega)$ are random variables that converge in mean square. Show that $Var[X_n] \rightarrow Var[X]$.

Convergence in mean square implies that as $n \rightarrow \infty$ we have that $\mathbb{E}[(X_n-X)^2] \rightarrow 0$.

I tried to use the definition of variance $Var[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ and trying to prove that $|Var[X_n]-Var[X]|\leq \mathbb{E}[(X_n-X)^2]$ but I don't get any result.

Solution 1:

Using the Cauchy-Schwarz inequality, we have that $$ |\operatorname E(X_n-X)|\le\bigl(\operatorname E|X_n-X|^2\bigr)^{1/2}\to0 $$ and $\operatorname EX_n\to\operatorname EX$. We also have that $\operatorname EX_n^2\to\operatorname EX^2$ using the continuity of the norm $(\operatorname E|X|^2)^{1/2}$ of $L_2(\Omega)$. Finally, $$ \operatorname{Var}X_n=\operatorname EX_n^2-(\operatorname EX_n)^2\to\operatorname EX^2-(\operatorname EX)^2=\operatorname{Var}X. $$