1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence in distribution (weak convergence)

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the function [tex] f_n(x)=n\cdot I\left[|x|<\frac{1}{2n}\right] [/tex] considered as a distribution in [itex] D'(\mathbb{R})[/itex], where [itex]I[/itex] denotes the indicator function. Recall that [itex] f_n[/itex] converges to [itex]\delta_0[/itex], the delta distribution, in [itex]D'(\mathbb{R})[/itex]. Show that [itex]f_n^2-n\delta_0[/itex] converges in [itex]D'(\mathbb{R})[/itex].

    3. The attempt at a solution

    By a few calculations, we can show that for any test function [itex]\phi[/itex] and natural [itex]n>0[/itex], that
    [tex] (f_n^2,\phi)=n(f_n-\delta_0,\phi).[/tex]
    At this point, I would be tempted to say that this converges to zero (as a sequence of real numbers) because [itex] (f_n-\delta_0,\phi)[/itex] "must" dominate the convergence of [itex] n [/itex] to infinity. However, this isn't rigorous, and is more of a hunch than anything. However, any attempts I've made to prove it have not gone anywhere. I feel as though it must have something to do with the smoothness of the test functions, but I can't see where to incorporate this.

    At this point, I would appreciate a nudge in the right direction.
    Thanks in advance!
     
  2. jcsd
  3. Mar 8, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Try using the second-order Taylor expansion of ##\phi(x)## in the evaluation of ##(f_n^2,\phi)##.
     
  4. Mar 11, 2013 #3
    Yep, that worked. Thanks! It's actually very easy once you apply the Taylor expansion :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Convergence in distribution (weak convergence)
Loading...