TITLE: 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 $f$
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.