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Sufficient condition for bounded Fourier transform

  1. Sep 26, 2014 #1
    Hello,

    Let's suppose we are given a function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex], and we assume its Fourier transform [itex]F=\mathcal{F}(f)[/itex] exists and has compact support.

    What sufficient condition could we impose on f, in order to be sure that F is also bounded?
     
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  3. Sep 26, 2014 #2

    mathman

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    f is L1 is certainly sufficient, even if F does not have compact support.
     
  4. Sep 30, 2014 #3
    Thanks a lot!
    I think I figured out a way to prove it, but it relies on the Fourier shift theorem and the compact support assumption.
    The proof I sketched out is more or less like this:

    - Let's suppose F is not bounded and has compact support
    - Then there must be at least one point in the support of F where the Fourier transform goes to infinity.
    - Let u be that point, and translate the Fourier transform so that u coincides with the origin.
    - This means that the inverse FT of the translated F must have integral=infinity.
    - But from the Fourier shift-theorem we know that such a function is simply f itself multplied by the complex sinusoid e-iux
    -
    Note that [itex]|\int_{-\infty}^{+\infty}f(x)e^{-iux}dx| \leq \int_{-\infty}^{+\infty}|f(x)|dx[/itex] which means that f is not absolute integrable.
    - Now if we assumed f was in L1 we would reach a contradiction

    However I don't see how to generalize this argument to the case when F does not have compact support, e.g. F(u) might go to infinity only for u-->infinity which would render the "translation trick" useless.
     
  5. Sep 30, 2014 #4

    mathman

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    I don't understand your argument. If f is in L1 then, as you noted, the Fourier transform is bounded at every point by the integral of |f(x)|.
     
  6. Sep 30, 2014 #5
    Yes. I have just realized that I in my proof I don't need the statement "there must be at least one point in the support of F where the Fourier transform goes to infinity". The inequality that I wrote simply says that the magnitude of the F at each point is bounded by the absolute integral of f.
    Thanks again for clarifying this.
     
  7. Oct 1, 2014 #6

    WWGD

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    Cant we use a continuity argument together with compactness to argue boundedness of the transform?
     
  8. Oct 1, 2014 #7

    mathman

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    If you look at the reverse question. F bounded with compact support implies f is L1.
     
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