Suppose that \(f: \RR\to\RR\) is a continuous function that vanishes outside of some finite interval. For each \(z\in \CC\), define \[ g(z) = \int_{-\infty}^\infty f(t) e^{-izt} \,dt .\]
Show that \(g\) is entire.
Solution
Choose \(R>0\) such that \(f(t)=0\) for \(|t|>R\). Then \[ g(z)=\int_{-R}^R f(t)e^{-izt}\,dt. \] Fix a compact set \(K\subset\mathbb C\). If \(M=\sup_{z\in K}|\Im z|\), then for \(z\in K\) and \(|t|\le R\), \[ |f(t)e^{-izt}| \le |f(t)|e^{MR}, \] and similarly \[ |-it\,f(t)e^{-izt}| \le R|f(t)|e^{MR}. \] These are integrable majorants on \([-R,R]\). Hence differentiation under the integral sign is valid, giving \[ g'(z)=\int_{-R}^R (-it)f(t)e^{-izt}\,dt. \] Thus \(g\) is holomorphic on all of \(\mathbb C\), i.e. \(g\) is entire.
Iterating the same argument gives, for every \(n\ge0\), \[ g^{(n)}(z)=\int_{-R}^R(-it)^n f(t)e^{-izt}\,dt. \]