Let \[ f(x)=e^{-|x|},\qquad x\in\mathbb R. \]
Is \(\widehat f\in C^\infty(\mathbb R)\)? Prove your answer.
Prove that \[ |\widehat f(\xi)|\longrightarrow0 \qquad(|\xi|\to\infty). \]
Solution
1. The Fourier transform is smooth.
For every integer \(k\ge0\), \[ \int_{\mathbb R}|x|^k e^{-|x|}\,dx<\infty. \] Thus \(x^k f(x)\in L^1(\mathbb R)\) for every \(k\).
Using the convention \[ \widehat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi i x\xi}\,dx, \] the \(k\)th formal derivative is \[ \frac{d^k}{d\xi^k}\widehat f(\xi) =\int_{\mathbb R}(-2\pi i x)^k f(x)e^{-2\pi i x\xi}\,dx. \] For each fixed \(k\), the absolute value of the integrand is bounded by \[ (2\pi)^k|x|^k e^{-|x|}\in L^1(\mathbb R). \] Dominated convergence therefore justifies differentiation under the integral sign, and the resulting derivative is continuous by another application of dominated convergence. Hence \[ \widehat f\in C^\infty(\mathbb R). \] The same conclusion holds for any of the standard Fourier-transform normalizations, with the obvious change in constants.
2. The Fourier transform vanishes at infinity.
Since \(f=e^{-|x|}\in L^1(\mathbb R)\), the Riemann–Lebesgue lemma gives \[ \widehat f(\xi)\longrightarrow0 \qquad(|\xi|\to\infty). \] The lemma follows by approximating \(f\) in \(L^1\) by a function \(g\in C_c^1(\mathbb R)\). For such \(g\), integration by parts gives \[ |\widehat g(\xi)|\le \frac{\|g'\|_1}{2\pi|\xi|} \qquad(\xi\ne0), \] while \[ |\widehat f(\xi)-\widehat g(\xi)|\le\|f-g\|_1. \] First choose \(g\) so that the latter bound is small, then let \(|\xi|\to\infty\). This proves the claimed decay.