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Uniform convergence, Lp convergence

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A stereotypical problem in measure theory.
    fn(x)=1 when 0<=x<=1/n
    =0 when 1/n<x<=1
    f(x)=0 0<=x<=1

    Under the lebesgue measure
    Does fn converge to f?
    a. for all x
    b. almost everywhere
    c. uniformly on [0,1]
    d. uniformly almost everywhere on [0,1]
    e. almost uniformly
    f. in measure
    g. in Lp

    2. Relevant equations

    3. The attempt at a solution
    My guess of the answer would be:
    a. for all x: No, doesn't converge at x=0
    b. almost everywhere: Yes, x=0 has lesbesgue measure 0
    c. uniformly on [0,1]: No
    d. uniformly almost everywhere on [0,1]: No
    e. almost uniformly: Yes
    f. in measure: Yes
    g. in Lp: shouln't it depend on p? no idea.
  2. jcsd
  3. Dec 1, 2008 #2
    My answers:
    a) No
    b) Yes
    c) No
    d) What do you mean by uniformly almost everywhere? How does this differ from almost uniformly? I've never seen this concept before.
    e) Yes
    f) Yes
    g) Yes

    To see this last one, note that for any p > 0, [tex]|f_n|^p \to 0[/tex] a.e., and [tex]|f_n|^p \leq 1[/tex], so the DCT implies that [tex]\int |f_n|^p dm \to 0[/tex].
  4. Dec 1, 2008 #3
    Sorry how does dominated convergence theorem implies that [tex]\int |f_n|^p dm \to 0[/tex].?

    All DCT says is that it is less than the integral of 1 right?
  5. Dec 1, 2008 #4


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    I think so, but you don't need it anyway. ||f_n||_p=(1/n)^(1/p). Doesn't that go to zero for all p<infinity?
  6. Dec 1, 2008 #5
    The Dominated Convergence Theorem allows you to pull the limit inside the integral, so, since [tex]|f_n|^p[/tex] is dominated by an integrable function ([tex]f\equiv1[/tex]),
    \lim \int |f_n|^p dm = \int ( \lim |f_n|^p ) dm = \int 0 dm = 0.

    Dick is of course correct in his evaluation of [tex]||f_n||_p[/tex] both methods work.
  7. Dec 1, 2008 #6


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    Right. Sorry, I didn't bother to look it up the DCT because you don't need it. But rochfor1 is correct.
  8. Dec 1, 2008 #7
    No worries! Your way is actually easier than mine...I'm just so used to doing these types of problems without being able to explicitly compute the norms, so DCT and MCT are the first things I go to.
    Last edited: Dec 1, 2008
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