<s>[INST] 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?
 [/INST] 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 $\sin^2(\pi x)\le1$
</s>