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?

The proof for the line can be applied to the circle using inequality $(4)$.

Preliminary Inequalities for $\boldsymbol{0\lt x\lt\frac\pi2}$
Inequality $\bf{1}$:
$$
\begin{align}
\frac{\sin^2(x)}{x^2}
&=\prod_{k=1}^\infty\cos^2\left(\frac{x}{2^k}\right)\tag{1a}\\
&=\prod_{k=1}^\infty\left(1-\sin^2\left(\frac{x}{2^k}\right)\right)\tag{1b}\\
&\ge\prod_{k=1}^\infty\left(1-\frac{x^2}{4^k}\right)\tag1
\end{align}
$$
Explanation:
$\text{(1a)}$: Induction, $\frac{\sin(x)}{2\sin(x/2)}=\cos(x/2)$, and $\lim\limits_{n\to\infty}2^n\sin\left(x/2^n\right)=x$
$\text{(1b)}$: $\cos^2(x)=1-\sin^2(x)\vphantom{\lim\limits_{n\to\infty}}$
$\phantom{\text{b}}\text{(1)}$: $\sin^2(x)\le x^2$ 
Inequality $\bf{2}$:
$$
\begin{align}
\frac{\tan(x)}{x}
&=\prod_{k=1}^\infty\frac1{1-\tan^2\left(x/2^k\right)}\tag{2a}\\
&\ge\prod_{k=1}^\infty\frac1{1-\frac{x^2}{4^k}}\tag2
\end{align}
$$
Explanation:
$\text{(2a)}$: Induction, $\frac{\tan(x)}{2\tan(x/2)}=\frac1{1-\tan^2(x/2)}$, and $\lim\limits_{n\to\infty}2^n\tan\left(x/2^n\right)=x$
$\phantom{\text{a}}\text{(2)}$: $\tan^2(x)\ge x^2$ 
Inequality $\bf{3}$:
Multiplying inequalities $(1)$ and $(2)$ we get
$$
\frac{\sin^2(x)\tan(x)}{x^3}\ge1\tag3
$$
Inequality $\bf{4}$:
Inequality $(3)$ implies that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right)
&=\frac2{x^3}-\frac{2\cos(x)}{\sin^3(x)}\tag{4a}\\
&\ge0\tag{4b}
\end{align}
$$
Since $\frac1{\sin^2(x)}-\frac1{x^2}$ is increasing, it attains its maximum on $\left(0,\frac\pi2\right]$ at $\frac\pi2$. Therefore,
$$
\frac1{\sin^2(x)}-\frac1{x^2}\le1-\frac4{\pi^2}\tag4
$$

Hölder Bound on the Kernel
$$
\begin{align}
\int_{-1/2}^{1/2}\frac{\sin^2(\pi nx)}{n\sin^2(\pi x)}|x|^\alpha\,\mathrm{d}x
&=2\int_0^{1/2}\frac{\sin^2(\pi nx)}{n\sin^2(\pi x)}x^\alpha\,\mathrm{d}x\tag{5a}\\
&=2n^{-\alpha-2}\int_0^{n/2}\frac{\sin^2(\pi x)}{\sin^2(\pi x/n)}x^\alpha\,\mathrm{d}x\tag{5b}\\
&\le\frac2{\pi^2}n^{-\alpha}\int_0^{n/2}\frac{\sin^2(\pi x)}{x^{2-\alpha}}\,\mathrm{d}x+\left(1-\frac4{\pi^2}\right)\frac{2^{-\alpha}}{1+\alpha}n^{-1}\tag{5c}\\
&\le\frac2{\pi^2}n^{-\alpha}\left(\frac{\pi^2}{1+\alpha}+\frac1{1-\alpha}\right)+\left(1-\frac4{\pi^2}\right)\frac{2^{-\alpha}}{1+\alpha}n^{-1}\tag{5d}\\[6pt]
&=O\!\left(n^{-\alpha}\right)\tag5
\end{align}
$$
Explanation:
$\text{(5a)}$: apply symmetry
$\text{(5b)}$: substitute $x\mapsto x/n$
$\text{(5c)}$: apply $(4)$ and evaluate the integral of $x^\alpha$
$\text{(5d)}$: estimate the integral on $[0,1]$, where $\sin^2(\pi x)\le\pi^2x^2$,
$\phantom{\text{(5d):}}$ and on $[1,\infty]$, where $\sin^2(\pi x)\le1$
