About Rudin's outline to proving that Lipschitz functions have converging Fourier Series I'm trying to do the following exercise from Rudin's Real and Complex Analysis:

Suppose $f\in C(T)$ and $f\in \text{Lip }\alpha$ for some $\alpha>0$. Prove that the Fourier series of $f$ converges to $f(x)$, by completing the following outline: It is enough to consider the case $x=0$, $f(0)=0$. The difference between the partial sums $s_n(f;0)$ and the integrals
  $$\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\frac{\sin nt}{t}\:dt$$
  tends to $0$ as $n\to\infty$. The function $f(t)/t$ is in $L^1(T)$. Apply the Riemann-Lebesgue lemma. More careful reasoning shows that the convergence is actually uniform on $T$.

I already know how to prove the result as in Lipschitz Continuity and Hölder Continuity helps Fourier series to converge.
However, I don't understand Rudin's outline.


*

*When he says that it is enough to consider the case $x=0$, $f(0)=0$, is it simply because it changes almost nothing to the proof?

*Since we can prove that $s_n(f;0)$ tends to $0$ by proving that $f(t)\cot(t/2)$ is in $L^1$, it seems to me that Rudin's outline is only better if it is easier to prove that $f(t)/t\in L^1$ and that the difference between $s_n(f;0)$ and the integrals tends to $0$. But I don't see any easy way to do the latter.

*Why is the convergence uniform and why does it matter?


I would appreciate if anyone could answer those 3 questions.
 A: *

*The aim is to show that $s_n(f,a) \rightarrow f(a) $ for all $a$, however it is easy to show that if you let $g(x)=g(x+a)-g(a)$,
$$s_n(f,a)-s_n(g,0) = \sum_{-n}^n \frac1{2\pi}\int_{-\pi}^\pi (f(t)e^{ik(a-t) } - g(t)e^{-ikt})dt = f(a)$$
so if you showed that $s_n(g,0)\rightarrow 0$ you may deduce that $s_n(f,a)\rightarrow f(a)$.

*I think I agree with what you say. As 
$$s_n(f,0)=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{-n}^n e^{-ikt}\right)dt = \frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sin(nt)\cot(t/2)+\cos(nt)\right)dt.$$
There is no additional complexity in showing that $t\mapsto f(t)\cot(t/2)$ is $L^1$ compared to $t\mapsto f(t)/t$, and then the result comes directly from Riemann-Lebesgue lemma. My only guess is that Rudin might be implicitly referring to the approximation of the Dirichlet kernel by the sinus cardinal function:
$$D_n(x)=\frac{2\sin (nx)}x+\left(\sin (nx) \left(\operatorname{cotan} \frac x 2-\frac2x\right)+\cos(nx)\right)$$
where $D_n(t)=\sum_{-n}^n e^{-ikt}$.

*Uniform convergence is much stronger than pointwise convergence, it definitely makes a difference!
A: Since it may be helpful to someone, here is my complete answer to this question (done, of course, with a lot of help from @FXV).

Firstly, if $g(t)=f(t+x)-f(x)$, then by linearity we have that
  $$s_n(f;x)-s_n(g;0)=f(x).$$
  In other words, if $s_n(g;0)\to 0$, then $s_n(f;x)\to f(x)$ so it suffices to consider the case $x=0$, $f(0)=0$.
Now we use the usual formula for $\sin(x+y)$ to write
  $$D_n(t)=\frac{\sin (n+\frac{1}{2})t}{\sin(t/2)}=2\cdot\frac{\sin nt}{t}+\left[\sin nt\left(\cot (t/2)-\frac{2}{t}\right)+\cos nt\right].$$
  Since $D_n$ is even, this implies that the difference in the exercise's statement is equal to
  $$\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\left(\cot (t/2)-\frac{2}{t}\right)\sin nt\:dt+\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\cos nt\:dt.$$
  But $\cot(t/2)-2/t\to 0$ as $t\to 0$, which implies that this function has only a removable singularity. Thus, by defining it to be $0$ for $t=0$ we see that both
  $$f(t)\left(\cot (t/2)-\frac{2}{t}\right) \quad\text{and}\quad f(t)$$
  are continuous functions and belong to $L^1(T)$. The Riemann-Lebesgue lemma then implies that both integrals tend to $0$ as $n\to\infty$.
Since $f\in \text{Lip }\alpha$,
  $$\left|\frac{f(t)}{t}\right|=\frac{|f(t)-f(0)|}{|t-0|}\leq M_f |t-0|^{\alpha-1}=M_f|t|^{\alpha-1}$$
  for all $t\neq 0$. This implies that $|f(t)/t|$ is integrable and so $f(t)/t\in L^1(T)$. Finally, as we saw,
  $$s_n(f;0)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\frac{\sin nt}{t}\:dt+x_n,$$
  where $x_n\to 0$ as $n\to \infty$. A third application of the Riemann-Lebesgue lemma implies that $s_n(f;0)\to 0$.
In view of Arzelà-Ascoli's theorem (theorem 7.25 in Principles of Mathematical Analysis), it suffices to show that $\{s_n(f;x)-f(x)\}$ is equicontinuous to have uniform convergence. For that, we have to bound
  $$|[s_n(f;x)-f(x)]-[s_n(f;y)-f(y)]|.$$
  Since $\int_{-\pi}^\pi D_n(t)\:dt=2\pi$, this is equal to
  $$\left|\int_{-\pi}^\pi \left\{[f(x-t)-f(x)]-[f(y-t)-f(y)]\right\} D_n(t)\:dt\right|.$$
  Using the triangular inequality twice we bound this integral by
  \begin{multline*}
    \int_A \left\{ |f(x-t)-f(x)|+|f(y-t)-f(y)| \right\} |D_n(t)|\:dt+\\
    \int_B \left\{ |f(x-t)-f(y-t)|+|f(x)-f(y)| \right\} |D_n(t)|\:dt,
\end{multline*}
  where $A=\{t\in [-\pi,\pi]\::\: |t|<|x-y|\}$ and $B=\{t\in [-\pi,\pi]\::\: |t|>|x-y|\}$. As $f\in \text{Lip }\alpha$, it follows that this is bounded by
  $$\int_A \left\{2M_f |t|^\alpha\right\} |D_n(t)|\:dt+\int_B \left\{2M_f |x-y|^\alpha\right\}|D_n(t)|\:dt.$$
  Now, since $(t/2)/\sin(t/2)\to 1$ as $t\to 0$, there exists $\delta>0$ such that
  $$|D_n(t)|<4|t|^{-1}$$
  for all $0<|t|<\delta$. We conclude that, if $|x-y|<\min(\delta,\pi)$, the integral over $A$ is bounded by
  $$8M_f\int_A|t|^{\alpha-1}\:dt$$
  and that the integral over $B$ is bounded by
  $$2M_f|x-y|^{\alpha} \underbrace{\int_B |D_n(t)|\:dt}_{\text{bounded}}.$$
  Finally, since $-1<\alpha-1\leq 0$, we can estimate $\int_A |t|^{\alpha-1}\:dt$ in the following way: let $|x-y|=a$ and observe that
  \begin{align*}
    \int_A |t|^{\alpha-1}\:dt &= 2 \int_0^{a} t^{\alpha-1}\:dt\\
    &= 2\left(\int_{a/2}^{a} t^{\alpha-1}\:dt+\int_{a/4}^{a/2} t^{\alpha-1}\:dt+\dotsc \right)\\
    &\leq 2\left(\frac{a}{2}\left(\frac{a}{2}\right)^{\alpha-1}+\frac{a}{4}\left(\frac{a}{4}\right)^{\alpha-1}+\dotsc\right)\\
    &=K|x-y|^{\alpha},
\end{align*}
  for some constant $K>0$. Putting it all together, we get that if $|x-y|<\min(\delta,\pi)$, then
  $$|[s_n(f;x)-f(x)]-[s_n(f;y)-f(y)]|<K'|x-y|^\alpha,$$
  for another constant $K'>0$. This implies that $\{s_n(f;x)-f(x)\}$ is equicontinuous and the result follows.

