Why must the Fourier transform of a compactly support function not have compact support?

Question

I've heard this stated several times, most recently as a motivation for using the Schwartz space as test functions.

I think I can just about prove it using Heisenberg's uncertainty principle, but was hoping someone could show me a direct proof? Attempts so far have led pretty much nowhere.

Answer 1

The Fourier transform of a function with compact support is an entire function. The set of zeroes of an entire function is discrete, unless it is identically zero. Google Paley-Wiener theorem.