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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!

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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, $$ \|S_{N}(f)\|_{\infty}\le \|f\|_{\infty}+\frac{2K}{N+1}\left[\frac{1}{1}+\frac{2}{2}+\cdots+\frac{N}{N}\right] = \|f\|_{\infty}+\frac{2KN}{N+1}. $$

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  • $\begingroup$ @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

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