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Lp weak convergence

  1. Feb 10, 2012 #1

    I am struggling to understand why L1 is not weakly compact, while Lp, p>1, is.

    The example I have seen put forward is the function u_n (x) = n if x belongs to (0, 1/n), 0 otherwise, the function being defined on (0,1).

    It is shown this u_n converging to the Dirac measure, and this shows L1 not being weakly compact (as integrating u_n times the charactersitic function of the interval yields 1 as a result, while the result is zero for a function which is zero at the boundary.

    I can not see why this should not happen for Lp, p > 1 too. In the attached short notes there is an example after which (pag. 6) it is stated that this function converges weakly to zero in Lp.

    I can not understand this.

    Many thanks


    Attached Files:

  2. jcsd
  3. Feb 10, 2012 #2


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    Sorry, not sure my response makes sense, edited it out. Will think about it some more.
    Last edited: Feb 10, 2012
  4. Feb 10, 2012 #3


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    Actually maybe what I said did make sense. So to be weakly compact any bounded sequence has to have a convergent sub-sequence from what I remember right? The sequence you defined is not bounded in Lp for p>1.
  5. Feb 10, 2012 #4
    Thanks for replying, I am not sure I understand your reply. The sequence i mention is bounded in the L1 norm, and the example I mentioned, u_n = x ^(1/2) if x belongs to (0,1/n), 0 otherwise, is bounded in L2. But L2, on the contrary to L1, is weakly compact.
  6. Feb 10, 2012 #5


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    I thought your example was

    "u_n (x) = n if x belongs to (0, 1/n), 0 otherwise"

    So my point is that this is bounded in L1, and therefore it can be used as a counterexample to show L1 is not weakly compact.

    However since it is not bounded in Lp for p>1, it can not be used as a counter example in those cases because the definition of weak compactness only puts restrictions on the behavior of bounded sequences.
  7. Feb 10, 2012 #6
    Sorry, I was very unclear. I showed the example in L1, giving for granted that when considering the L2 case its analogue, mentioned in the attachment, would have considered.
    The function u_n = x^(1/2) if x belongs to (0,1/n), 0 otherwise.
    This is bounded in L2 and must converges to an element of L2, given that L2 is weakly compact.

    But thank you anyhow, I think I discovered my silly mistake!

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