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Convergence in the sense of distributions

  1. May 30, 2012 #1
    I have the following problem: prove that the sequence [itex]e^{inx}[/itex] tends to [itex]0[/itex], in the sense of distributions, when [itex]n\to \infty[/itex]. Here it is how I approached the problem. I have to prove this:

    [tex]\lim \int e^{inx}\phi(x)\,dx=0[/tex]

    , where [itex]\phi[/itex] is a test-function. I changed variable: [itex]nx=x'[/itex] and got:

    [tex]\lim \frac{1}{n}\int e^{ix'}\phi(x'/n)dx'[/tex]

    Now, can I exchange limit and integral? I would say yes, because of dominated convergence: the absolute value of the integrand is less than, say, [itex]c|\phi|[/itex], which is summable and the limit of the integrand exists, because [itex]\phi[/itex] is continuos. So,

    [tex]\phi(0)\lim \frac{1}{n}\int e^{ix'}dx'[/tex]

    But can I say that this last limit is zero? I mean, shouldn't the limit function be summable, again by dominated convergence? I suspect that I have a wrong understanding either of convergence in the sense of distribution or of dominated convergence. Can you clear up my doubts?
  2. jcsd
  3. May 30, 2012 #2
    The idea is right. But I don't quite understand why you use things like "summable". Can you tell us what you mean with dominated convergence exactly?
  4. May 30, 2012 #3


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    What are the limits of integration? ∫eixdx = 0 when integrating over an interval of length 2π.
  5. May 30, 2012 #4
    I think summable is an older term for integrable.
  6. May 30, 2012 #5
    Oh, that would make sense actually.
  7. May 30, 2012 #6
    Ok, by summable I mean that the integral over all space exists and it is finite. I don't know if the same terminology is used in english. Also, I didn't write the limits of integration, but they are over all space (let's say over all R or R^N). The dominated converge theorem I have studied says that if a sequence of measurable functions converges point-wise and the modulus of these functions (indexed by n) is dominated by a nonnegative summable function I can exchange limit and integral and the limit function is automatically summable:

    [tex]\lim\int f_n dx=\int f dx[/tex]

    if [itex]|f_n|<[/itex] some summable function (at least almost everywhere etc). Is this correct? In my part of the world it is late in the night, so I will check for answers tomorrow morning. Thanks again for your help!
  8. May 30, 2012 #7
    In that case, everything looks ok.
  9. May 31, 2012 #8
    What worries me is that the last integral I get (the integral of the complex exponential) does not exists, does it? Shouldn't the limit function be summable?
  10. May 31, 2012 #9
    Note that the test function [itex]\varphi[/itex] is nonzero on a compact domain K. So what you're actually doing is calculating

    [tex]\lim_{n\rightarrow +\infty} \int_K e^{inx}\varphi(x)dx[/tex]

    If you work from this then you end up with

    [tex]\int_K e^{ix}[/tex]

    which does exist.
  11. May 31, 2012 #10
    I see, but my definition of a test-function is an infinitely differentiable function that goes to zero faster than any inverse power. Is this equivalent to saying that it has a compact support?
  12. May 31, 2012 #11
    No. But it goes faster than zero than what?? Inverse powers of polynomials??

    In that case, applying dominated convergence does not seem correct here. You say that it is smaller than [itex]c|\varphi|[/itex]. But I don't see how that can be true. I get

    [tex]\left| \int e^{ix}\varphi(x/n)dx\right| \leq \int |\varphi(x/n)dx|=n\int |\varphi(u)du|=n|\varphi|[/tex]

    But this is dependent of n and is not bounded by a summable function.
  13. May 31, 2012 #12
    They must be [itex]o(||x||^{-m})[/itex] but not only that, also all their derivatives must satisfy the same condition (I was in a hurry this afternoon). And yes, you are right, my estimate is clearly wrong. My new attempted solution is this: just notice that it is a limit as [itex]n\to \infty [/itex] of the Fourier transform [itex]\hat \phi(n) [/itex]. By a lemma of Riemann and Lebesgue, the Fourier transform of a summable function vanishes at infinity.
  14. May 31, 2012 #13
    The only problem is that the fourier coefficients are for ##L1## functions on the circle. For the infinite domain case you need to appeal to the monotone convergence theorem, approximating from compact intervals.
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