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Fourier tranforms

  1. May 8, 2008 #1
    If a function belongs to G(R) but has points that are jump discontinuities, it's Fourier transform will not belong to G(R).
    But would it be correct to claim that if a function in G(R) is continuous than its Fourier transform also belongs to G(R)? I guess it's not true, but can someone suggest a counterexample?
  2. jcsd
  3. May 8, 2008 #2


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    Frst, what do YOU mean by "G(R)"?
  4. May 8, 2008 #3
    Sorry, I wasn't sure whether this sign is well known.
    G(R) is the space of functions that might have points of discontinuity only of first kind, and which are absolutely integrable.
  5. May 8, 2008 #4
    ideas? anyone?...:frown:
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