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Are Fourier transforms unique?

  1. Sep 28, 2010 #1


    for all real k, can I conclude that g(x) = h(x) for all real x?

    If the answer is yes, please help me see it, if it is not too much trouble.
  2. jcsd
  3. Sep 28, 2010 #2


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    By setting k = 0 you know they differ by a constant. Show that this constant is 0.
  4. Sep 28, 2010 #3
    You have to rely on Fourier inversion theorem, that you can find in any book that talks about Fourier transforms. In general, two functions (square integrable) that have the same Fourier transform are equal "almost everywhere", that is, everywhere except on a set of points with zero Lesbegue measure.

    For k = 0 you only know they have the same average value, not that they differ by a constant.
  5. Sep 28, 2010 #4


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    My apologies, you are right of course.
  6. Sep 28, 2010 #5


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    Assuming the functions are nice, so that the inverse Fourier transform exists, then g(x)=f(x) almost everywhere. The expression for the inverse transform looks just like the expression for the transform except for a sign reversal in the exponent.

    In other words, let H(t) be the Fourier transform of f(x) or g(x), then the inverse transform of H(t) (call it h(x)) equals f(x) (and g(x)) almost everywhere.
  7. Sep 29, 2010 #6

    Thanks! I will have to read up on Lesbegue measure.
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