Let $f$ be a Riemann integrable function on $[-\pi,\pi]$ such that $\hat{f}(n)\leq \frac{K}{|n|}$ for some constant $K$, for all $n\neq 0$. Show that $|S_N(f)|_\infty\leq |f|_\infty+2K$.

Here$\hat{f}(n)$ denotes the n-th Fourier coefficient of $f$, $S_N(f)(x)=\sum_{|n|\leq N}\hat{f}(n) e^{inx}$ is the N-th partial sum of the Fourier series of $f$, $|\cdot|_\infty$ is the sup-norm on $[-\pi,\pi]$.

I have tried expressing $S_N(f)$ as the convolution of $f$ with the Dirichlet kernel, but I couldn't obtain any expressions that allow application of the hypothesis $\hat{f}(n)\leq \frac{K}{|n|}$. A closer attempt was made by imitating the proof of Tauber's theorem: the Fourier series of $f$ converges to $f$ in the Abel-summable sense at points of continuity of $f$, if $\hat{f}(n)=o(\frac{1}{|n|})$ Tauber's theorem says the Fourier series converges to $f$ in the usual sense; now we have $\hat{f}(n)=O(\frac{1}{|n|})$ but a modification of the proof shows that $|$Fourier series of $f|_\infty\leq|f|_\infty+2K$. But there are 2 problems with this approach: (1) $f$ has to be continuous (2) the LHS is sup-norm of Fourier series of $f$ instead of sup-norm of its partial sum.

This question came up in one of the past papers while I was revising for my Fourier analysis course, sadly the TA and the professor don't seem to have much of an idea on how to approach this. Any help is appreciated!


If $F_N(f)$ denotes the Fejer integral, then, by definition, \begin{align} F_N(f)&=\frac{1}{N+1}\sum_{n=0}^{N}S_{N}(f) \\ &=\hat{f}(0)+\frac{N}{N+1}[\hat{f}(-1)e^{-ix}+\hat{f}(1)e^{ix}] \\ &+\frac{N-1}{N+1}[\hat{f}(-2)e^{-2ix}+\hat{f}(2)e^{2ix}]+\cdots+\frac{1}{N+1}[\hat{f}(-N)e^{-iNx}+\hat{f}(N)e^{iNx}] \\ &=S_{N}(f)-\frac{1}{N+1}[\hat{f}(-1)e^{-ix}+\hat{f}(1)e^{ix}]-\frac{2}{N+1}[\hat{f}(-2)e^{-2ix}+\hat{f}(2)e^{2ix}]-\cdots. \end{align} Because of the positivity of the Fejer kernel, $$ \|F_{N}(f)\|_{\infty}\le \|f\|_{\infty}. $$ Therefore, $$ \|S_{N}(f)\|_{\infty}\le \|f\|_{\infty}+\frac{2K}{N+1}\left[\frac{1}{1}+\frac{2}{2}+\cdots+\frac{N}{N}\right] = \|f\|_{\infty}+\frac{2KN}{N+1}. $$

  • $\begingroup$ @Tsang : I just noticed that I dropped the numerators when applying inequalities. Now the inequality fits very well with the result stated in paper you were reading. Thanks for posting this problem. I had not thought to do such a thing before, but the form of the inequality suggested it. $\endgroup$ – DisintegratingByParts Oct 29 '16 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.