How to show for $\alpha\in (0,1)$, any $f\in C^\alpha([0,1]/{\sim})$ has a Fourier series $S_nf$ uniformly converging to $f$ Technically homework(a midterm) but its over and I'm itching to know the solution. I know how to show it for $\alpha>1/2$ (the Fourier series will converge absolutely), but apparently its true for any $\alpha$; the question guided me as follows:


*

*Show that if a equicontinuous sequence of functions ($f_n$) converges pointwise to $f$, then $f_n$ converges uniformly to $f$.

*Show for $f∈ C^\alpha([0,1]/{\sim})$ that $S_nf → f$ pointwise.

*Show that the sequence $(S_nf)$ is equicontinuous and conclude.


1 and 2 posed no problems to me but I could not do 3. Any help? In addition, I would not mind other ways to prove the result.
 A: Suppose that $|f(x)|\le C$ and $|f(x)-f(y)|\le C|x-y|^\alpha$.

Express the Difference Using the Dirichlet Kernel
Using the Dirichlet Kernel, we get
$$
\begin{align}
|S_nf(x)-f(x)|
&=\left|\,\int_{-1/2}^{1/2}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|\\
&=\left|\,\sum_{k=-n}^n\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|\tag{1}
\end{align}
$$

Estimate each Integral Using the smoothness of $\boldsymbol{f}$
Since $\left|\,\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}\,\right|\le\frac{2n+1}{\big|2|k|-1\big|}$ and each interval is $\frac1{2n+1}$ wide, we can bound
$$
\begin{align}
\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|
&\le\frac{C}{\big|2|k|-1\big|}\left(\frac{2|k|+1}{4n+2}\right)^\alpha\tag{2}
\end{align}
$$

Estimate each Integral Using Cancellation from $\boldsymbol{\sin((2n+1)\pi x)}$
For $|y|\le\frac12$, we have $|2y|\le|\sin(\pi y)|\le|\pi y|$, and because
$$
\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\,\mathrm{d}y=0\tag{3}
$$
and
$$
\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}|\sin((2n+1)\pi y)|\,\mathrm{d}y=\frac2{(2n+1)\pi}\tag{4}
$$
if we let $m_k$ be the middle of the range of $\frac{f(x-y)-f(x)}{\sin(\pi y)}$ on $\left[\frac{2k-1}{4n+2},\frac{2k+1}{4n+2}\right]$, for $k\ne0$, we can bound
$$
\begin{align}
&\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\frac{f(x-y)-f(x)}{\sin(\pi y)}\,\mathrm{d}y\,\right|\\
&=\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\left[\frac{f(x-y)-f(x)}{\sin(\pi y)}-m_k\right]\,\mathrm{d}y\,\right|\\
&\le\frac1{(2n+1)\pi}\frac{\overbrace{\pi\frac{2|k|+1}{4n+2}}^{\sin(\pi y)}\overbrace{C(2n+1)^{-\alpha}\vphantom{\frac{|}2}}^{\Delta (f(x-y)-f(x))}+\overbrace{2C\vphantom{()^1}}^{f(x-y)-f(x)}\overbrace{\pi(2n+1)^{-1}}^{\Delta\sin(\pi y)}}{\underbrace{\frac{4k^2-1}{(2n+1)^2}}_{\sin^2(\pi y)}}\\
&=\frac{C(2n+1)^{-\alpha}}{4|k|-2}+\frac{2C}{4k^2-1}\tag{5}
\end{align}
$$

Use each Estimate in its Proper Place
If we use estimate $(2)$ for $k\le m=n^{\frac{\alpha}{\alpha+1}}$ and estimate $(5)$ for $k\gt m$, then we get
$$
\begin{align}
\sum_{|k|\le m}\frac{C}{\big|2|k|-1\big|}\left(\frac{2|k|+1}{4n+2}\right)^\alpha
&\le\frac{C}{(4n+2)^\alpha}\left[1+6\sum_{k=1}^m(2k+1)^{\alpha-1}\right]\\
&\le\frac{C}{(4n+2)^\alpha}\frac3\alpha(2m+1)^\alpha\\
&\sim\frac{3C}{\alpha2^\alpha}n^{-\frac\alpha{\alpha+1}}\tag{6}
\end{align}
$$
and
$$
\begin{align}
\sum_{m\lt|k|\le n}\frac{C(2n+1)^{-\alpha}}{4|k|-2}
&\le\frac{C}{2^{\alpha+1}}\frac{H_n}{n^\alpha}\\
&\sim\frac{C}{2^{\alpha+1}}\frac{\log(n)}{n^\alpha}\\
&=o\left(n^{-\frac{\alpha}{\alpha+1}}\right)\tag{7}
\end{align}
$$
and
$$
\begin{align}
\sum_{m\lt|k|\le n}\frac{2C}{4k^2-1}
&\le C\sum_{k=m}^\infty\frac1{k^2-1}\\
&=\frac{C}{2}\sum_{k=m}^\infty\left(\frac1{k-1}-\frac1{k+1}\right)\\
&=\frac{C}{2}\left(\frac1{m-1}+\frac1m\right)\\
&\sim Cn^{-\frac{\alpha}{\alpha+1}}\tag{8}
\end{align}
$$

Put Everything Together
Therefore, we have uniform convergence:
$$
|S_nf(x)-f(x)|\le\left(1+\frac3{\alpha2^\alpha}\right)Cn^{-\frac{\alpha}{\alpha+1}}\tag{9}
$$
A: While I accepted the answer above, this is how my lecturer (and later my friend) explained it to me (the exam is tomorrow). We first define $$g_n(x):=f(x) - S_n f(x)$$ just to remind ourselves that we need to be careful of cancellations. Then uniform convergence of $S_nf$ to $f$ is equivalent to showing $g_n→ 0$ uniformly; since we know (part 2) that $g_n(x) → 0$ pointwise, it suffices to show $g_n$ is uniformly continuous (by part 1). 
Since $\newcommand{\d}{\text{d}}\newcommand{\intT}{∫_{-1/2}^{1/2}}g_n(x) = f(x)\times 1 - \intT f(z-x) D_n(z) \ \d z = \intT [f(x) -f(z-x)] D_n(z)\ \d z$ , 
\begin{align}
|g_n(x) - g_n(y)| ≤ \intT |D_n(z)|\underbrace{|f(x)  - f(z-x) - f(y) + f(z-y)|}_{(\star)}\ \d z
\end{align}
We now need to find bounds independent of $n$ . We use a simple bound for the Dirichlet kernel $D_n$: as there is $C_0$ such that $|\sin(2π z)|>C_0|z|$ on $[-1/2,1/2]$,
$$|D_n(z)| < \frac{C_1}{|z|} $$
Since we don't gain too much form a simple bound we need to bound $(\star)$. The trick is to use two different bounds, each good on different sets:
\begin{align} |\color{red}{f(x)  - f(z-x)} - \color{blue}{f(y) + f(z-y)}| &\leq C_3|z|^\alpha \\
|\color{red}{f(x)}  - \color{blue}{f(z-x)} - \color{red}{f(y)} + \color{blue}{f(z-y)}| 
&\leq C_3|x-y|^\alpha \\
\end{align}
Thus $$|g_n(x) - g_n(y)| \leq ∫_{|z|\leq|x-y|} C_4|z|^{\alpha-1} \ \d z + |x-y|^\alpha ∫_{|x-y|<|z|<1/2}\frac{C_5}{z} \ \d z = I_1 + I_2 $$
Now $I_1$ is $\mathcal{O}(|x-y|)$ because $|z|^{\alpha-1}$ is $L^1([-1/2,1/2])$. The second we compute,
$$I_2 = C_5 |x-y|^\alpha\left(\log\frac{1}{2} + log\frac{1}{|x-y|}\right) $$
And we win because polynomials beat logarithms.
