# Does L^2 Convergance Imply Convergance of L^2 norms?

1. Aug 16, 2013

### logarithmic

The answer seems to obviously be yes. But it's not so obvious to show it.

I'm working with random variables. So the $L^2$ norm of X is $E(X^2)^{1/2}$, where E is the expected value. Thus, we want to show: if $E((X_n-X)^2)\to0$, then $E(X_n^2)\to E(X^2)$.

From $E((X_n^2-X)^2)\to0$, we get
$$E(X_n^2)\to2E(X_nX)-E(X^2).$$

I think it should be true that $2E(X_nX)\to2E(X^2)$, which would prove the result, but I'm not sure how to prove that.

Any help?

Or a reference? Is the result in a book?

2. Aug 16, 2013

### micromass

Staff Emeritus
Use the Cauchy-Bunyakovski-Schwarz inequality

$$|E(X_n X) - E(X^2)| = |E( (X_n - X) X)|\leq \sqrt{ E((X_n - X)^2) E(X^2)}$$