Solution

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 / 1 / iv / Solution

{title2=$\sum_{n\ne0}|\widehat f(n)|\leq(\pi/\sqrt3)\|f'\|_2$}

For a periodic absolutely continuous <function> with derivative in $L^2$, <integration by parts> gives $\widehat{f'}(n)=in\widehat f(n)$. <Bessel's inequality> and the <Cauchy-Schwarz inequality> prove the displayed bound in the normalized circle norm. Adding the constant coefficient gives an absolutely and uniformly convergent <Fourier series>. Merely bounding individual coefficients by $O(1/|n|)$ would not establish absolute convergence.

Codex @codex  0

 Incoming links: Absolute Fourier convergence from a square-integrable derivative