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Example: for function $$f(x)=x^{3}(1-x)^{3}=\sum f_{s}\exp(2\pi isx)$$ Fourier series of its fourth derivative are different from derivative of its Fourier series $$f^{(4)}(x)=-360x^{2}+360x-72=\sum g_{s}\exp(2\pi i s x)$$ with $g_{s}\neq(2\pi s)^{4}f_{s}$

Related question: When $$\sum\left|f_{s}\right|^{2}j^{2p}<\infty$$ is equivalent to $f\in C^{(p)}[0,1]$? (for periodic f)

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notice that the 3rd derivative of your function (when extended from $[0,2\pi)$ periodically to $\mathbb{R}$) is not continuous. Fourier series will correctly give its distributional derivative, which includes $\delta$-functions coming from jumps. You would need to subtract (the right multiple of) the Fourier series of $\delta$, which is just $\sum_k e^{2\pi i k x}$. For your second question, the inequality means that $f^{(p)}$ is in $L^2$, so at least $f\in C^{(p-1)}(S^1)$. –  user8268 Mar 31 '11 at 15:42
    
Thank you very much! I didn't notice it! –  Katja Mar 31 '11 at 15:53
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maybe you could have a look at the related question? math.stackexchange.com/questions/30744/… –  Katja Apr 4 '11 at 9:22

1 Answer 1

An advertisement for the utility and aptness of Sobolev theory is the perfect connection between $L^2$ "growth conditions" on Fourier coefficients, and $L^2$ notions of differentiability, mediated by Sobolev's lemma that says ${1\over 2}+k+\epsilon$ $L^2$ differentiability of a function on the circle implies $C^k$-ness. Yes, there is a "loss". However, the basis of this computation is very robust, and generalizes to many other interesting situations.

That is, rather than asking directly for a comparison of $C^k$ properties and convergence of Fourier series (etc.), I'd recommend seizing $L^2$ convergence, extending this to Sobolev theory for both differentiable and not-so-differentiable functions, and to many distributions (at least compactly-supported), and only returning to the "classical" notions of differentiability when strictly necessary.

I know this is a bit avante-garde, but all my experience recommends it. A supposedly readable account of the issue in the simplest possible case, the circle, is at functions on circles .

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