Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
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
maybe you could have a look at the related question?… – Katja Apr 4 '11 at 9:22

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 .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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