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?
On
Perhaps here is another doable proof: Let's denote $$[f]_{C^{\alpha}}:=\sup_{x\neq y}\dfrac{|f(x)-f(y)|}{|x-y|^{\alpha}},$$ and $$\|f\|_{C^{\alpha}}:=\|f\|_{\infty}+[f]_{C^{\alpha}}.$$
The Fejer Kernel has the form $$K_{n}(x):=\dfrac{1}{2\pi}\dfrac{1}{n+1}\Big(\dfrac{\sin\frac{n+1}{2}x}{\sin\frac{x}{2}}\Big)^{2}.$$
Firstly, $\sin n\alpha\leq n\sin\alpha$, so $$(1)\ \ |K_{n}(x)|=\dfrac{1}{2\pi(n+1)}\dfrac{|\sin\frac{n+1}{2}x|^{2}}{|\sin\frac{x}{2}|^{2}}\leq\dfrac{1}{2\pi(n+1)}(n+1)^{2}\dfrac{|\sin\frac{x}{2}|^{2}}{|\sin\frac{x}{2}|^{2}}=\dfrac{n+1}{2\pi}.$$
On the other hand, note that $$|\sin\frac{x}{2}|>\dfrac{|x|}{\pi}>\dfrac{|x|}{2\pi}\ \text{for}\ 0<|x|<\pi,$$ which implies that $$(2)\ \ |K_{n}(x)|=\dfrac{1}{2\pi(n+1)}\dfrac{|\sin\frac{n+1}{2}x|^{2}}{|\sin\frac{x}{2}|^{2}}\leq\dfrac{1}{2\pi(n+1)}\dfrac{1}{|\sin\frac{x}{2}|^{2}}\leq\dfrac{2\pi}{(n+1)x^{2}}.$$
Now, recall that $\int_{-\pi}^{\pi}K_{n}(y)dy=1$, then we compute as follows: \begin{align*} |\sigma_{n}(f)(x)-f(x)|&=|(K_{n}*f)(x)-f(x)|\\ &=|(f*K_{n})(x)-f(x)|\ \text{by the commutative of convolution}\\ &=\Big|\int_{-\pi}^{\pi}f(x-y)K_{n}(y)dy-\int_{-\pi}^{\pi}f(x)K_{n}(y)dy\Big|\ \text{by what we just recalled}\\ &\leq\int_{-\pi}^{\pi}|f(x-y)-f(x)|\cdot |K_{n}(y)|dy\\ &\leq\int_{-\pi}^{\pi}[f]_{\alpha}|y|^{\alpha}\cdot|K_{n}(y)|dy\\ &=[f]_{\alpha}\int_{|y|<\frac{\pi}{n+1}}|y|^{\alpha}|K_{n}(y)|dy+[f]_{\alpha}\int_{\frac{\pi}{n+1}\leq|y|\leq\pi} |y|^{\alpha}|K_{n}(y)|dy\\ &\leq\dfrac{[f]_{\alpha}(n+1)}{2\pi}\int_{|y|<\frac{\pi}{n+1}}|y|^{\alpha}dy+\dfrac{2\pi[f]_{\alpha}}{(n+1)}\int_{\frac{\pi}{n+1}\leq|y|\leq\pi}|y|^{\alpha}\cdot y^{-2}dy, \end{align*} where the last inequality was obtained by imposing the inequality $(1)$ to the first term and the inequality $(2)$ to the second term.
Now, do the normal integral, and then we will have \begin{align*} |\sigma_{n}(f)(x)-f(x)|&\leq\dfrac{[f]_{\alpha}(n+1)}{2\pi}\cdot\dfrac{2}{\alpha+1}\cdot\dfrac{\pi^{\alpha+1}}{(n+1)^{\alpha+1}}+\dfrac{2\pi[f]_{\alpha}}{(n+1)}\cdot\dfrac{2}{\alpha-1}\Big(\pi^{\alpha-1}-\Big(\dfrac{\pi}{n+1}\Big)^{\alpha-1}\Big)\\ &=\dfrac{[f]_{\alpha}\pi^{\alpha}}{(\alpha+1)(n+1)^{\alpha}}+\dfrac{4\pi^{\alpha}[f]_{\alpha}}{(n+1)(\alpha-1)}-\dfrac{4\pi^{\alpha}[f]_{\alpha}}{(n+1)^{\alpha}(\alpha-1)}\ \ (*). \end{align*}
Note that the second term is $\leq 0$ since $\alpha<1\implies (\alpha-1)<0$ and all other terms are $\geq 0$.
Therefore, \begin{align*} (*)&\leq \dfrac{[f]_{\alpha}\pi^{\alpha}}{(\alpha+1)(n+1)^{\alpha}}-\dfrac{4\pi^{\alpha}[f]_{\alpha}}{(n+1)^{\alpha}(\alpha-1)}\\ &=\dfrac{[f]_{\alpha}}{(n+1)^{\alpha}}\Big(\dfrac{\pi^{\alpha}}{\alpha+1}-\dfrac{4\pi^{\alpha}}{\alpha-1}\Big). \end{align*}
Thus, define $C_{\alpha}:=\Big(\dfrac{\pi^{\alpha}}{\alpha+1}-\dfrac{4\pi^{\alpha}}{\alpha-1}\Big)$ and then we are able to conclude that $$|\sigma_{n}(f)(x)-f(x)|\leq \dfrac{C_{\alpha}[f]_{\alpha}}{(n+1)^{\alpha}}\leq \dfrac{C_{\alpha}[f]_{\alpha}}{n^{\alpha}}.$$
In particular, notice that $C_{\alpha}\geq 0$, since $(\alpha-1)<0$ so $-\dfrac{4\pi^{\alpha}}{(\alpha-1)}>0$.
Hence, we can further write: $$|\sigma_{n}(f)(x)-f(x)|\leq \dfrac{C_{\alpha}[f]_{\alpha}}{n^{\alpha}}\leq\dfrac{C_{\alpha}([f]_{\alpha}+|f(x)|)}{n^{\alpha}},$$ since $|f(x)|\geq 0$.
Now, taking $\sup$ on the both side of the inequality over all $x$, we obtain the desired inequality that $$\|\sigma_{n}(f)-f\|_{\infty}\leq\dfrac{C_{\alpha}([f]_{\alpha}+\|f\|_{\infty})}{n^{\alpha}}=\dfrac{C_{\alpha}\|f\|_{\alpha}}{n^{\alpha}}.$$
On
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 $\sin^2(\pi x)\le1$
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.