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Convergence of a sequence of functions to zero in the L1 norm?

  1. Nov 27, 2009 #1
    I just want to make sure I'm straight on the definition.

    Am I correct in assuming that, if I want to show that a sequence [itex]\langle f_n \rangle[/itex] of functions converges to 0 in the [itex]L^1[/itex] norm, I have to show that, for every [itex]\epsilon > 0[/itex], there exists [itex]N \in \mathbb N[/itex] such that

    [tex]
    \int |f_n| < \epsilon
    [/tex]

    whenever [itex]n > N[/itex]?

    Also, is it possible for a sequence of functions to converge uniformly to 0 and yet *not* converge to 0 in the [itex]L^1[/itex] norm? (I'm pretty sure I have an example of this if the above definition is correct.)

    Thanks!
     
    Last edited: Nov 27, 2009
  2. jcsd
  3. Nov 28, 2009 #2
    Yes, if by [tex]L^1[/tex] you mean [tex]L^1(-\infty;+\infty)[/tex].
     
  4. Nov 28, 2009 #3
    Thanks! Yes, I'm only asking about [itex]L^1(\mathbb R)[/itex].

    And how about my second question concerning uniform convergence? Is your answer to that "yes" as well?
     
  5. Nov 29, 2009 #4
    Anyone? Anyone?
     
  6. Nov 29, 2009 #5

    Landau

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    Science Advisor

    I don't think so. We have [tex]\|...\|_1\leq \|...\|_\infty[/tex], so every sequence of functions converging in de supremum-norm converges in the 1-norm, with the same limit. [The converse is not true.]

    So, what's your example? :smile:
     
  7. Nov 29, 2009 #6
    Well, just consider a function that's a triangle with base [itex]2n[/itex] (centered at the origin) and height [itex]1/n[/itex]. Then the integral of all functions in the sequence is always 1, so it can't possibly converge to zero in the [itex]L^1[/itex] norm, but it converges to 0 uniformly (right?) on [itex]\mathbb R[/itex] since

    [tex]
    \sup_{x\in \mathbb R} |f_n| = 1/n \to 0 \text{ as } n\to \infty.
    [/tex]

    I may have gotten a definition wrong somewhere here, so please feel free to correct me if I have. To make sure we're on the same page, note that by the [itex]L^1[/itex] norm of [itex]f[/itex], I mean

    [tex]
    ||f||_1 = \int |f|.
    [/tex]

    Also, I'm not sure about the veracity of your claim that [tex]\|...\|_1\leq \|...\|_\infty[/tex]. What about the constant function [tex]f = 1[/tex]? We have [tex]\| f \|_\infty = 1[/tex], but [tex]\| f \|_1 = \infty[/tex].
     
    Last edited: Nov 29, 2009
  8. Nov 29, 2009 #7
    I think the reason I'm confused about this is that I'm thinking about the fact from second-semester real analysis that if a sequence of functions converges uniformly to a function, the integrals converge to the integral. So I keep thinking the sequence of functions in my example should converge to 0 in the [tex]L^1[/tex] norm. But this is only on a bounded interval [tex][a,b][/tex], correct? My example is crap if we restrict ourselves to any bounded interval, but I think it holds water if we're allowed to consider the whole real line.
     
  9. Nov 29, 2009 #8

    Landau

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    Science Advisor

    Hah, I was indeed working with [tex]\|f\|_1:=\int_0^1|f(t)dt[/tex]. I should have read the first reply, I apologize.
    So it looks like you understand it perfectly, you last statement about compact intervals is also correct.
     
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