Let \(f\) be a continuously differentiable \(2\pi\)-periodic function on \(\mathbb R\), with Fourier coefficients \[ \widehat f(n)=\frac1{2\pi}\int_0^{2\pi}f(t)e^{-int}\,dt. \] Show that the Fourier series \[ \sum_{n\in\mathbb Z}\widehat f(n)e^{int} \] is absolutely convergent for every \(t\).
Solution
Because \(f\) is \(2\pi\)-periodic and \(C^1\), integration by parts gives, for every \(n\ne0\), \[ \begin{aligned} \widehat{f'}(n) &=\frac1{2\pi}\int_0^{2\pi}f'(t)e^{-int}\,dt\\ &=\frac1{2\pi}\bigl[f(t)e^{-int}\bigr]_0^{2\pi} +in\widehat f(n)\\ &=in\widehat f(n), \end{aligned} \] since \(f(2\pi)=f(0)\).
Now \(f'\in L^2([0,2\pi])\). By Bessel’s inequality, \[ \sum_{n\in\mathbb Z}|\widehat{f'}(n)|^2<\infty. \] Hence \[ \sum_{n\ne0}n^2|\widehat f(n)|^2<\infty. \] Therefore Cauchy–Schwarz yields \[ \begin{aligned} \sum_{n\ne0}|\widehat f(n)| &=\sum_{n\ne0}\frac1{|n|}\,|n\widehat f(n)|\\ &\le \left(\sum_{n\ne0}\frac1{n^2}\right)^{1/2} \left(\sum_{n\ne0}n^2|\widehat f(n)|^2\right)^{1/2} <\infty. \end{aligned} \] Adding the single term \(|\widehat f(0)|\) gives \[ \sum_{n\in\mathbb Z}|\widehat f(n)|<\infty. \] Since \(|e^{int}|=1\), for every \(t\), \[ \sum_{n\in\mathbb Z}|\widehat f(n)e^{int}| =\sum_{n\in\mathbb Z}|\widehat f(n)|<\infty. \] Thus the Fourier series converges absolutely for every \(t\).