Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 16, 2013 #1
    The answer seems to obviously be yes. But it's not so obvious to show it.

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

    From [itex]E((X_n^2-X)^2)\to0[/itex], we get

    I think it should be true that [itex]2E(X_nX)\to2E(X^2)[/itex], 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. jcsd
  3. Aug 16, 2013 #2
    Use the Cauchy-Bunyakovski-Schwarz inequality

    [tex]|E(X_n X) - E(X^2)| = |E( (X_n - X) X)|\leq \sqrt{ E((X_n - X)^2) E(X^2)}[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook