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

Let a function $f: \mathbb{R} \rightarrow \mathbb{R}$ be integrable on $[-\pi,\pi]$ and $2\pi$-periodic. Let $$ \frac{a_0}{2}+\sum\limits_{n=1}^\infty (a_n \cos nx+b_n \sin nx) $$ be the Fourier series of $f$.

Assume that $f$ has continuous derivative at $a$. How to show that of partial sum $T_n$ of the series $$ \sum\limits_{n=1}^\infty (-n a_n \sin nx+n b_n \cos nx) $$ (it is the Fourier series of $f$ after termwise differentiation) is convergent in arithmetic mean in $a$ to $f'(a)$? That is $$ \lim\limits_{n\to\infty}\sigma_n(x):=\lim\limits_{n\to\infty}\frac{T_0(a)+T_1(a)+\ldots +T_{n-1}(a)}{n-1} = f'(a) $$ Thanks.

share|cite|improve this question
Do you know that, if $f$ is continuous, while the Fourier series of $f$ may not converge to $f$, the series of arithmetic means will converge uniformly to $f$? This is knows as Fejér's theorem. This question looks like a localized version of this result for $f'$. – user20266 Jun 10 '12 at 13:33
up vote 1 down vote accepted

Let $g$ be $2\pi$-periodic function from $C^1(\mathbb T)$ coinciding with $f$ in some neighborhood of $a$. According to the Riemann localization principle difference of partial Fourier sums $\Delta T_n(a)$ for $f'$ and $g'$ tends to zero: $$ \lim_{n\to\infty}\Delta T_n(a)=\lim_{n\to\infty}(T_n^f(a)-T_n^g(a))=0. $$ It follows that $$ \lim_{n\to\infty} \frac{\Delta T_0(a)+\Delta T_1(a)+\ldots +\Delta T_{n-1}(a)}{n}=0. $$ By Fejér's theorem $$ \lim_{n\to\infty} \frac{T_0^g(a)+T_1^g(a)+\ldots +T_{n-1}^g(a)}{n}=g'(a)=f'(a), $$ so $$ \lim_{n\to\infty} \frac{T_0^f(a)+T_1^f(a)+\ldots +T_{n-1}^f(a)}{n}=f'(a) $$ too.

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.