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Convergence in the L^2 norm

  1. Apr 8, 2012 #1
    Suppose there exists a sequence [itex]f_n[/itex] of square-integrable functions on [itex]\mathbb R[/itex] such that [itex]f_n(x) \to f(x)[/itex] in the L^2-norm with [itex]x \ f_n(x) \to g(x)[/itex], also in the L^2-norm. We know from basic measure theory that there's a subsequence [itex]f_{n_k}[/itex] with [itex]f_{n_k}(x) \to f(x)[/itex] for a.e. x. But my professor seems to be claiming that this somehow implies [itex]x \ f_{n_k}(x) \to g(x)[/itex] for a.e. x. I don't see why this is. Obviously, we know that [itex]x \ f_{m_k} \to g[/itex] a.e. for SOME subsequence of [itex]f_n[/itex]...but how do we know it works for the SAME subsequence? Can someone offer some guidance? Thanks!
     
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  3. Apr 8, 2012 #2

    Fredrik

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    If I understand you correctly, there's a set E such that ##\mu(E)=0## and ##xf_n(x)\to g(x)## for all x in ##E^c##. This means that for all ##x\in E^c##, ##\langle xf_n(x)\rangle_{n=1}^\infty## is a convergent sequence in ##\mathbb R##, and as you know (or can easily prove), every subsequence of a convergent sequence in ##\mathbb R## converges to the limit of the sequence.
     
  4. Apr 8, 2012 #3
    Well, if we have [itex]f_{n_k}(x) \to f(x)[/itex] off of a set [itex]E\subset \mathbb R[/itex] with [itex]\mu(E) = 0[/itex], then [itex]|f_{n_k}(x) - f(x)| \to 0[/itex] as [itex]k\to \infty[/itex] whenever [itex]x\notin E[/itex]. But I want to somehow show that this implies the existence of a [itex]E'\subset \mathbb R[/itex] (which may or may not be the same as [itex]E[/itex]) with [itex]\mu(E') = 0[/itex] such that [itex]|x \ f_{n_k}(x) - g(x)| \to 0[/itex] as [itex]k\to \infty[/itex] whenever [itex]x\notin E'[/itex]. How in the world is one supposed to get from the first statement to the one we want to prove, if the only other thing you know is that [itex]\| x \ f_n(x) - g(x) \|_{L^2(\mathbb R)} \to 0[/itex] as [itex]n\to \infty[/itex]?

    I hope that clarifies my question a bit!
     
  5. Apr 8, 2012 #4

    Fredrik

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    OK, I wasn't paying enough attention to when you were using the L^2-norm and when you were just talking about convergence almost everywhere. I will think about it.
     
  6. Apr 8, 2012 #5

    micromass

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    Isn't it obvious that if [itex]f_n[/itex] converges both to f and g in [itex]L^2[/itex] that then f=g a.e.?? That would imply it.
     
  7. Apr 8, 2012 #6

    Fredrik

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    Hehe, after my last post yesterday, I realized that I was much too tired to do any math. I decided to give it another shot "tomorrow", i.e. today, in the unlikely event that micromass wouldn't already have posted the solution. I should have realized that there was no chance that he wouldn't already have done that. :smile:
     
  8. Apr 8, 2012 #7

    micromass

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    Haha. Next time I'll let you finish it up!!
     
  9. Apr 9, 2012 #8

    morphism

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    But it's xf_n(x) that's converging to g, and not f_n.

    Something does seem fishy about the argument in the OP. Maybe we need more context?
     
  10. Apr 9, 2012 #9

    micromass

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    Oh, I see. I misread there.

    Anyway. If [itex]xf_n(x)\rightarrow g[/itex] in [itex]L^2[/itex], then the [itex]xf_{n_k}\rightarrow g[/itex] in [itex]L^2[/itex]. Thus there is a subsequence [itex]xf_{n_{k_l}}[/itex] that converges to g a.e.

    Evidently, the sequence [itex]xf_{n_k}(x)[/itex] converges a.e. (to xf(x)). And since a subsequence converges to g, it means that the sequence [itex]xf_{n_k}(x)[/itex] converges to g a.e.

    Did I do something stupid?
     
  11. Apr 9, 2012 #10

    morphism

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    That works!
     
  12. Apr 10, 2012 #11

    Fredrik

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    Oh, don't worry about that. I don't mind at all. I would probably need 7 hours to do what you can do in 7 minutes anyway. :smile:
     
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