Show that if $f$ has compact support, then its Fourier transform cannot have compact support unless $f=0$. [closed]
If $f$ has compact support, its Fourier transform $\hat{f}$ can be extended to a holomorphic function on a certain domain of $\mathbb{C}$ (you just have to set $g(z) = \int f(t)e^{-itz}dt$ and apply the holomorphy under integral sign theorem).
Therefore, if $\hat{f}$ has also compact support, it is zero on a set with an accumulation point, therefore it is zero everywhere by the isolated zeros theorem : $\hat{f}=0$. But then, as the fourier transform is injective, this implies $f=0$.