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Fourier tranforms

  • #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?
Thanks.
 

Answers and Replies

  • #2
HallsofIvy
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Frst, what do YOU mean by "G(R)"?
 
  • #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.
 
  • #4
ideas? anyone?...:frown:
 

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