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Convergence of a sequense

  1. Apr 24, 2010 #1
    consider the sequence
    fn(x)=n when 0<=x<=1/n
    =0 else

    Does fn converges:
    1. point-wise a.e.
    2. in measure
    3. in Lp
    4. weakly in Lp
    5. uniformly
     
  2. jcsd
  3. Apr 24, 2010 #2

    mathman

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    1. yes, 3 & 5, no. I have forgotten the definitions for 2 and 4.
     
  4. Apr 24, 2010 #3
    How did you get that 3 and 5 are not true?

    For 2: http://en.wikipedia.org/wiki/Convergence_in_measure

    For 4: a sequence fn in Lp converges weakly in Lp to f in Lp if:
    [tex]\int[/tex] fn g converges to \int fg, for any g in Lp', 1/p+1/p'=1.
     
  5. Apr 25, 2010 #4

    mathman

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    The limit function f is 0 (the delta function is not a legitimate function).
    For 3, you have ||fn||=||fn-f||=1. Therefore no convergence.
    For 5, the problem is what happens at 0 cannot be made uniform.

    For 4, it looks like you can have a convergence if you widen the class to include distributions, so that fn -> delta function at 0.

    For 2, it looks like it is true. The interval around 0 where fn differs from f can be made as small as you want.
     
    Last edited: Apr 25, 2010
  6. Apr 26, 2010 #5
    It says Lp not L1 so the answer may change depending on p, right?
     
  7. Apr 26, 2010 #6

    mathman

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    No, my assertions are for all p, although ||fn||=||fn-f||=n(p-1)/p for Lp.
     
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