Let \(f\in L^1(\mathbb R)\) and define \[ \widehat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi i x\xi}\,dx. \] Prove that \(\widehat f\in C_0(\mathbb R)\); that is, \(\widehat f\) is continuous and tends to \(0\) at infinity. Do not quote the Riemann–Lebesgue lemma without a proof sketch.
Solution
1. Continuity.
Let \(\xi_n\to\xi\). For every \(x\), \[ f(x)e^{-2\pi i x\xi_n}\longrightarrow f(x)e^{-2\pi i x\xi}, \] and \[ |f(x)e^{-2\pi i x\xi_n}|=|f(x)|\in L^1(\mathbb R). \] Dominated convergence gives \[ \widehat f(\xi_n)\longrightarrow\widehat f(\xi). \] Thus \(\widehat f\) is continuous.
2. Vanishing at infinity for smooth compactly supported functions.
Let \(g\in C_c^1(\mathbb R)\). Integration by parts gives, for \(\xi\ne0\), \[ \widehat g(\xi) =\frac1{2\pi i\xi}\int_{\mathbb R}g'(x)e^{-2\pi i x\xi}\,dx, \] because the boundary term vanishes. Hence \[ |\widehat g(\xi)|\le\frac{\|g'\|_1}{2\pi|\xi|}\longrightarrow0 \qquad(|\xi|\to\infty). \]
3. Approximate an arbitrary \(L^1\) function.
Fix \(\varepsilon>0\). Choose \(g\in C_c^1(\mathbb R)\) such that \[ \|f-g\|_1<\frac{\varepsilon}{2}, \] using density of \(C_c^1(\mathbb R)\) in \(L^1(\mathbb R)\). For every \(\xi\), \[ |\widehat f(\xi)-\widehat g(\xi)| \le\|f-g\|_1<\frac{\varepsilon}{2}. \] By 2, there exists \(R\) such that \(|\widehat g(\xi)|<\varepsilon/2\) whenever \(|\xi|>R\). Therefore, for \(|\xi|>R\), \[ |\widehat f(\xi)| \le|\widehat f(\xi)-\widehat g(\xi)|+|\widehat g(\xi)| <\varepsilon. \] Thus \(\widehat f(\xi)\to0\) as \(|\xi|\to\infty\). Together with continuity, this proves \[ \widehat f\in C_0(\mathbb R). \]