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.

• You need to get the hypotheses straight. If $g$ is differentiable only at one point then there's no such thing as the convolution $g'*F_N$. – David C. Ullrich Sep 29 '15 at 15:18
• @DavidC.Ullrich Thanks for pointing out the mistake, I've fixed it. – Kato yu Sep 29 '15 at 15:24
• Ok, this version makes sense. What form for $F_N$ should you use? Earlier in the book you can find a proof that if $f$ is continuous then $f*F_n\to f$ uniformly. Use the form for $F_N$ used in that proof. – David C. Ullrich Sep 29 '15 at 15:27
• @DavidC.Ullrich Sure, I'll try working with that. – Kato yu Sep 29 '15 at 15:31
• I think... actually this doesn't seem quite as obvious as it did a minute ago. – David C. Ullrich Sep 29 '15 at 15:47

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.