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There is a classical theorem of Jackson stating that the $N$-th partial sum $S_N f$ of the Fourier series of a Lipschitz continuous function $f$ (which is periodic with period 1) satisfies $$ |f(x) - S_N f(x)| \leq c \frac{K \log N}{N} $$ uniformly for all $x \in [0,1]$, where $c$ is an absolute constant and $K$ is the constant in the Lipschitz condition. (This is stated in a more general case in the Wikipedia article here: http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Uniform_convergence )

Question: does anybody know where I can find a numeric value for the constant $c$? (It is particularly important for me that $c$ is independent of the function $f$ - this fact cannot be seen from the statement in Zygmund's book, for example).

Be careful: Jackson's theorem is often stated in the form of the error between $f$ and the "best approximation" of $f$ - this is not the same as the error in the approximation by the partial sum of the Fourier series!

If you know where I can find the requested inequality, please let me know.

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The estimate in the wikipedia article you linked does what you want because the modulus of continuity satisfies $\omega(2\pi/N)\le 2\pi L/N$ ($L$ = Lipschitz constant), but here's a derivation from scratch:

For convenience, let's focus on $x=0$, and assume that $f(0)=0$. We want to bound $$ (S_Nf)(0) = \int_{-\pi}^{\pi} f(x) D_N(x)\, dx , $$ where $D_N(x)=\sin(N+1/2)x/\sin x/2$ is the Dirichlet kernel.

The part of the integral over $|x|\le 1/N$ is $\lesssim L/N$, with an absolute ($f$ independent) constant because $|f(x)/\sin x/2|\lesssim L$. To estimate the contributions coming from $|x|>1/N$, we integrate by parts, integrating $\sin(N+1/2)x$ and differentiating the rest. The boundary terms are $O(L/N)$ and $$ \frac{1}{N+1/2}\left| \int_{1/N\le|x|\le \pi} \left( \frac{f(x)}{\sin x/2}\right)' \cos(N+1/2)x\, dx \right| \lesssim \frac{L\ln N}{N} , $$ because $|f'|\le L$ by assumption. (Also, when you differentiate $\sin x/2$ in the denominator, use that $|f(x)|\le L|x|$ in the estimate.)

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    $\begingroup$ Yes, of course I saw that the article in Wikipedia contains what I want, but I can't cite a statement from Wikipedia in a scientific paper. Anyway, your explanation is very fine, thank you very much for that. $\endgroup$ – Kurisuto Asutora Nov 13 '14 at 10:41

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