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Homework Help: 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

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    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 :)
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