Determine whether there is a nonzero smooth compactly supported function on \(\mathbb{R}\) whose Fourier transform is also compactly supported?
Solution
No. Let \(f\) be smooth with \(\operatorname{supp}f\subseteq[-S,S]\) and put \(F(\zeta)\da\int_{-S}^Sf(x)e^{-2\pi ix\zeta}\,dx\) for \(\zeta\in\CC\), so that \(F=\widehat f\) on \(\RR\).
1. \(F\) is entire.
On \([-S,S]\times K\), for \(K\subset\CC\) compact, the integrand and its \(\zeta\)-derivative are bounded by \(\abs{f(x)}\) times \(e^{2\pi S\sup_K\abs{\Im\zeta}}\) and \(2\pi Se^{2\pi S\sup_K\abs{\Im\zeta}}\). Dominated convergence permits differentiation under the integral [@Fol13].
2. If \(\widehat f\) has compact support, then \(F\equiv0\).
\(F\) vanishes on a real interval outside the support of \(\widehat f\), which has accumulation points, so \(F\equiv0\) by the identity theorem and step 1 [@SS03].
3. Q.E.D.
By step 2, \(\widehat f\equiv0\). A smooth compactly supported function is a Schwartz function, so Fourier inversion gives \(f(x)=\int\widehat f(\xi)e^{2\pi ix\xi}\,d\xi=0\) for every \(x\) [@SS03a].