Fourier tranforms

[maria clara]

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.

 

[HallsofIvy]

Frst, what do YOU mean by "G(R)"?

 

[maria clara]

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.

 

[maria clara]

ideas? anyone?...