Mean square convergence implies mean convergence

Question

I'm trying to prove that mean square convergence implies mean convergence

$E{(Xn-X)^2}$ $=>$ $E{(Xn-X)}$ in $L^2$

This is my attempt.
By cauchy-schwarz
$|E{(Xn-X)}|$ $<=$ $(E{(Xn-X)^2})^.5$

as $lim E{(Xn-X)^2}=0$ as n goes to infinity
then $lim (E{(Xn-X)^2})^.5=0$ as n goes to infinity, therefore, $lim E{(Xn-X)}$ as n goes to infinity must equal zero

thus $E{(Xn-X)^2}$ $=>$ $E{(Xn-X)}$ in $L^2$

Thanks for any help and I apologize for the terrible formatting

Answer 1

You are correct (your notation could be more precise and clearer). By the Cauchy-Schwarz inequality, we have that $$ \operatorname E|X_n-X|\le\sqrt{\operatorname E|X_n-X|^2}\to0 $$ as $n\to\infty$, which shows that convergence in mean square implies convergence in mean, i.e. $\operatorname E|X_n-X|\to0$ as $n\to\infty$.