Hölder continuous implying the rate of convergence of the Cesàro mean Let $f$ be a periodic function which is Hölder continuous of order $0<\alpha<1$
And $\sigma_nf$ be its $n^{th}$ Cesàro mean. Let $ \|f \|=\|f\|_\infty+\sup_{x \neq y}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}$.
The I have to show that $\|\sigma_nf-f\|_{\infty} \leq C\|f\|n^{-\alpha}$ for all natural number $n$ and a constant $C$.
I tried to use the convolution and the Fejér kernel to show this inequality but failed in every attempt and it is unbearably frustrating…
Could anyone please help me?

Hint: If $K_n$ is the Fejer kernel then $\frac1{2\pi}\int K_n=1$, so for example
$$\sigma_nf(0)-f(0)=\frac1{2\pi}\int_{-\pi}^\pi(f(t)-f(0))K_n(t)\,dt.$$Now use the fact that $|f(t)-f(0)|\le c|t|^\alpha$...
Edit: So we need to show that $$\int_{-\pi}^\pi|t|^\alpha K_n(t)\,dt\le c n^{-\alpha}.$$
The OP says he got this far and doubts that $K_n$ is nice enough to satisfy this inequality. The inequality is true, and it can be proved with no big inspiration, just fairly straightforward inequalities.
But it's a slight pain. This sort of thing is often simpler on the line instead of the circle, because formulas on the circle that are sort of analogous to dilations become actual dilations on the line.
So for example the corresponding gizmo on the line would be $k_n$, where $$k_n(t)=nk(nt)$$and $k\in L^1$. In this context we have by a simple change of variable $$\int |t|^\alpha k_n(t)=\int |t|^\alpha nk(nt)=\int(|t|/n)^\alpha)k(t)=cn^{-\alpha}.$$
Now on the circle it doesn't quite work like that, since $K_n(t)\ne nK(nt)$. But $K_n(t)$ is at least analogous to $nK(nt)$, close enough to make it all work. Note first that if $t\in[-\pi,\pi]$ then $|\sin(t/2)|\ge c|t|$. So
$$\int_{-\pi}^\pi|t|^\alpha K_n(t)\le c\int_{-\pi}^\pi |t|^\alpha n\frac{\sin^2(nt/2)}{(nt)^2},$$and you can use the change of variable  on that last integral.
