# Convergence in mean of differentiated Fourier series

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.

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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

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.

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