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Fourier Trasform of Delta functions

  1. Dec 15, 2011 #1
    Hi All,

    I am trying to understand more rigorously why the Fourier transform of a constant functions equals the Dirac delta distribution.

    I found somewhere this is justified by imposing the self-adjointness, so that under a duality pairing <..,..> and indicating with F(f) the transform of a function f, it is required that

    <F(δ), g> = < δ , g >

    If g equals the constant unitary function my source, http://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_transform, quotes,

    <1, F (g)> = g (0)= < δ , g >

    I understand the second equality, but not sure about the first...

    Many thanks for your help
  2. jcsd
  3. Dec 19, 2011 #2
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