Fourier transform on the test function space.

Hi All,

I am reading many books on the Fourier transform. They all state that the Fourier transform is not a mapping from test function space ( functions of compact support) to the test function space. In one of the books I read, they explain that if is a test function, and F( ) is a test function, then is 0. Does anyone know how to start on proving this? Or can anyone give me a test function (by above, any non zero one will do!) that I can compute the FT of? I have tried, and end up with having to integrate something that even Mathematica can't do.

Thanks.

Does the Fourier transform of a function with compact support itself have compact support?

I suspect not, but I can't think of an example of this.

Don't think of examples; look at the definition. Suppose f is compactly supported, and suppose the FT of f is compactly supported, and see if you can derive a contradiction.

I suppose I am trying to deduce that supp [tex]\phi[\MATH] is the empty set, and so phi is 0, but I can't see how to proceed on your hint.

By one of the Paley–Wiener theorems, if a square-integrable function is compactly supported then its Fourier transform is the restriction to the real axis of an entire function. But if such a function has compact support then it must be identically zero, because a nonzero entire function can only have isolated zeros.