Order of growth of the Cesaro sum of Fourier series Let $C(\mathbb{T})$ denote the space of continuous $1-$periodic functions. Let $g\in C(\mathbb{T})$ be differentiable at some point $x_0$. Show that
$$\limsup_{N\to\infty}\frac{|(\sigma_Ng)'(x_0)|}{\log N}<\infty$$
where
$$(\sigma_N g)(x)=g*F_N(x)$$
and $F_N(x)=\sum_{k=1-N}^{N-1}\left(1-\frac{|k|}{N}\right)e^{2\pi i kx}$ is the Fejer kernel.
It seems that there are a lot of equivalent forms of Fejer kernel, but I have no idea which one to use for this particular question. I'll be really appreciate if anybody can help.
 A: Let $F_n$ denote the Fejer kernel, then
$$F_n'(t)=\sum_{1<|j|\leq n-1}\left(1-\frac{|j|}{n}\right)e^{2\pi ij t}\quad\Longrightarrow\quad\int_{-\frac{1}{2}}^{\frac{1}{2}}F_n'(t)d t=0$$
Let $g\in C(\mathbb{T})$ be differentiable at $x_0$, then for $|y|\leq\frac{1}{2}$, there is $M>0$ s.t. 
$$|g(x_0-y)-g(x_0)|\leq M|y|$$
Note that
$$\begin{aligned}
F_n'(t)&=\left(\frac{1}{n}\frac{\sin^2(n\pi t)}{\sin^2(\pi t)}\right)'\\
&=\frac{1}{n}\frac{2n\pi\sin(n\pi t)\cos(n\pi t)}{\sin^2(\pi t)}-\frac{1}{n}\frac{2\pi\sin^2(n\pi t)}{\sin^3(\pi t)}\\
&=\frac{\pi\sin(2n\pi t)}{\sin^2\pi t}-\frac{2\pi\sin(\pi t)\sin^2(n\pi t)}{n\sin^4(\pi t)}
\end{aligned}$$
Recall that for $|t|\leq\frac{1}{2}$ we have
$$2|t|\leq|\sin (\pi t)|\leq\pi|t|$$
therefore
$$\begin{aligned}
|F_n'(t)|&\leq\frac{|\sin(2n\pi t)|}{4t^2}+\frac{2\pi^2|t||n\pi t||\sin(n\pi t)|}{16nt^4}\\
&=\frac{|\sin(2n\pi t)|}{4t^2}+\frac{\pi^3|\sin(n\pi t)|}{8t^2}
\end{aligned}$$
It follows that
$$\begin{aligned}
|(\sigma_ng)'(x_0)|=&\left|\int_{-\frac{1}{2}}^{\frac{1}{2}}g(x_0-y)F'_n(y)dy\right|=\left|\int_{-\frac{1}{2}}^{\frac{1}{2}}[g(x_0-y)-g(x_0)]F'_n(y)dy\right|\\
\leq&\int_{-\frac{1}{2}}^{\frac{1}{2}}M|y|\left(\frac{|\sin(2n\pi y)|}{4y^2}+\frac{\pi^3|\sin(n\pi y)|}{8y^2}\right)dy\\
=&M\int_{|y|\leq\frac{1}{n}}\frac{2|\sin(2n\pi y)|+\pi^3|\sin(n\pi y)|}{8y}dy\\
&+M\int_{\frac{1}{n}\leq|y|\leq\frac{1}{2}}\frac{2|\sin(2n\pi y)|+\pi^3|\sin(n\pi y)|}{8y}dy\\
\leq &M\int_{|y|\leq\frac{1}{n}}\frac{2|2n\pi y|+\pi^3|n\pi y|}{8y}d y+M\int_{\frac{1}{n}\leq|y|\leq\frac{1}{2}}\frac{2+\pi^3}{8y}dy\\
=&\frac{M}{8}\int_{|y|\leq\frac{1}{n}}4\pi n+\pi^3 ndy+\frac{M(2+\pi^3)}{4}(\log n-\log 2)\\
=&\frac{M(4\pi+\pi^3)}{8}-\frac{M(2+\pi^3)}{4}\log 2+\frac{M(2+\pi^3)}{4}\log n\\
\leq &C\log n
\end{aligned}$$
for some $C>0$ sufficiently large.
