Let \(f\in L^1(\mathbb R)\). Show directly, without citing general properties of the Fourier transform, that \[ \widehat f(\xi)=\int_{\mathbb R}e^{-ix\xi}f(x)\,dx \] is uniformly continuous and satisfies \[ \widehat f(\xi)\longrightarrow0 \qquad(|\xi|\to\infty). \]
Solution
1. \(\widehat f\) is uniformly continuous.
For \(h,\xi\in\mathbb R\), \[ \widehat f(\xi+h)-\widehat f(\xi) =\int_{\mathbb R}e^{-ix\xi}(e^{-ixh}-1)f(x)\,dx. \] Therefore \[ |\widehat f(\xi+h)-\widehat f(\xi)| \le \int_{\mathbb R}|e^{-ixh}-1|\,|f(x)|\,dx. \] The right-hand side is independent of \(\xi\). For each fixed \(x\), \[ |e^{-ixh}-1|\to0 \qquad(h\to0), \] and \[ |e^{-ixh}-1|\,|f(x)|\le2|f(x)|\in L^1(\mathbb R). \] By dominated convergence, \[ \int_{\mathbb R}|e^{-ixh}-1|\,|f(x)|\,dx\longrightarrow0. \] Hence \[ \sup_{\xi\in\mathbb R}|\widehat f(\xi+h)-\widehat f(\xi)|\longrightarrow0, \] which is uniform continuity.
2. Interval indicators have transforms tending to zero.
If \(a<b\), then for \(\xi\ne0\), \[ \widehat{\mathbf1_{[a,b]}}(\xi) =\int_a^b e^{-ix\xi}\,dx =\frac{e^{-ia\xi}-e^{-ib\xi}}{i\xi}. \] Thus \[ \left|\widehat{\mathbf1_{[a,b]}}(\xi)\right| \le\frac{2}{|\xi|}\longrightarrow0 \qquad(|\xi|\to\infty). \] The same conclusion holds for every finite linear combination of indicators of bounded intervals.
3. Approximate an arbitrary \(L^1\) function.
Finite linear combinations of indicators of bounded intervals are dense in \(L^1(\mathbb R)\). Given \(\varepsilon>0\), choose such a function \(s\) with \[ \|f-s\|_1<\varepsilon. \] For every \(\xi\), \[ |\widehat f(\xi)-\widehat s(\xi)| \le\int_{\mathbb R}|f-s| =\|f-s\|_1<\varepsilon. \] By 2, there is \(R>0\) such that \[ |\widehat s(\xi)|<\varepsilon \qquad(|\xi|>R). \] Hence, for \(|\xi|>R\), \[ |\widehat f(\xi)| \le|\widehat f(\xi)-\widehat s(\xi)|+|\widehat s(\xi)| <2\varepsilon. \] Therefore \(\widehat f(\xi)\to0\) as \(|\xi|\to\infty\).