Let \(f\in L^1(\mathbb R)\) and define \[ \widehat f(z)=\int_{\mathbb R}e^{-ixz}f(x)\,dx. \] Show that, unless \(f=0\) almost everywhere, the functions \(f\) and \(\widehat f\) cannot both have compact support.
Solution
Assume that \(f\) has compact support, say \[ \operatorname{supp}f\subset[-R,R]. \] Then for every complex number \(z\) the integral \[ F(z)=\int_{-R}^R e^{-ixz}f(x)\,dx \] is absolutely convergent. Moreover, for every integer \(k\ge0\), \[ F^{(k)}(z)=\int_{-R}^R(-ix)^k e^{-ixz}f(x)\,dx, \] with differentiation justified by dominated convergence on compact subsets of \(\mathbb C\). Hence \(F\) is entire, and on the real axis it agrees with the Fourier transform \(\widehat f\).
Now suppose also that \(\widehat f\) has compact support. Then there exists \(M>0\) such that \[ F(\xi)=\widehat f(\xi)=0 \qquad\text{for all real }|\xi|>M. \] Thus the entire function \(F\) vanishes on the nonempty open interval \((M,\infty)\) of the real axis. By the identity theorem for holomorphic functions, \[ F\equiv0\quad\text{on }\mathbb C. \] Therefore \(\widehat f=0\) on \(\mathbb R\). By uniqueness of the Fourier transform on \(L^1(\mathbb R)\), \[ f=0\quad\text{almost everywhere}. \] Consequently, a nonzero \(L^1\) function and its Fourier transform cannot both have compact support.