Proof of the Dirichlet–Dini Criterion for Pointwise convergence of Fourier series I have tried and failed to prove the Dirichlet–Dini Criterion  for Pointwise convergence of Fourier series which is as follows (source: Wikipedia)

if $f$ is $2\pi$–periodic, locally integrable and satisfies
$$
\int_{0}^{\pi}\left|\frac{f\left(x_{0}+t\right)+f\left(x_{0}-t\right)}{2}-\ell\right| \frac{\mathrm{d} t}{t}<\infty,
$$
Then $S_nf(x_0)\to \ell$.

I will appreciate a proof for this theorem or a reference to one - I couldn't do either.
 A: I'm gonna simplify the problem by assuming that $\ell=x_0=0$: notice that
$$
\dfrac{f(x_0+t)+f(x_0-t)}{2}-\ell = \dfrac{(f(x_0+t)-\ell)+(f(x_0-t)-\ell)}{2}.
$$
Hence Dini hypothesis is the same as saying that the function $f_1(t) = f(x_0+t)-\ell$ satisfies
$$
\int^{2\pi}_0\left|\dfrac{f_1(t)+f_1(-t)}{2}\right|\dfrac{dt}{t}<\infty.
$$
As it's proved later in the lemma, translating functions by a complex number $\ell$ and translating the axis $[0,2\pi]$ by $x_0$ acts nicely on the corresponding Fourier series so, if I can find the limit of $S_N(f_1;0)$, I can find the limit of $S_N(f;x_0)$ as well. Thus I'll proof the case $\ell= x_0 =0$ and then use this "nice action" for the general case. Consider 
$$
g(t)=\dfrac{f(t)+f(-t)}{1-e^{i t}} 
$$
Let'us see that $g$ is integrable.
Notice that $h(t) = \dfrac{t}{1-e^{it}}$ is continuous at $[0,2\pi]$ (using Hôpital for instance) and take 
$$
K=\max_{[0,2\pi]} h(t)<\infty.
$$
Now
$$
\int_0^{2\pi}\left|g(t)\right|\,dt= \int^{2\pi}_0\left|\dfrac{f(t)+f(-t)}{t}\right|\left |h(t)\right| \,dt\leq 2K\int^{2\pi}_0\left|\dfrac{f(t)+f(-t)}{2}\right| \,\dfrac{dt}{t}
$$
which is finite by hypothesis, thus it's integrable. 
Compute now $\hat f(n)+\hat f(-n)$ for every $n$. Notice that
\begin{align*}
\hat f(n)+\hat f(-n)&=\int ^{2\pi}_0 f(t)e^{-itn}\,dt + \int ^{2\pi}_0 f(t)e^{itn}\,dt\\
 &\overset{(*)}{=}\int ^{2\pi}_0 f(t)e^{-itn}\,dt+\int ^{2\pi}_0 f(-t)e^{-itn}\,dt\\
 &=\int ^{2\pi}_0 (f(t)+ f(-t))e^{-itn}\,dt\\
 &=\int ^{2\pi}_0 g(t)(1-e^{it})e^{-itn}\,dt\\
 &=\hat g(n)-\hat g(n-1).
\end{align*}
I used in $(*)$ the change of coordinates $t\rightarrow t-2\pi$ for the second integral (recall that $e^{it}$ and $f_1(t)$ are $2\pi$- periodic).
In particular 
$$
2S_N(f;0)= 2 \sum_{|n|\leq N} \hat f (n)=\sum_{|n|\leq N}\hat f(n)+\hat f(-n)=\hat g(N)-\hat g(-N-1).
$$
since we got a telescoping sum. Riemann-Lebesgue lemma applies to $g(t)$, hence
$$
\lim_{|N|\rightarrow \infty}\hat g(N) = 0
$$
and 
$$
\lim_{N\rightarrow \infty} 2S_N(f;0) = 0=\ell.
$$
For the general case reduce to the first one by considering $f_1(t)=f(t+x_0)-\ell$. Indeed
$$
\int^{2\pi}_0\left|\dfrac{f_1(t)+f_1(-t)}{2}\right| \,\dfrac{dt}{t}=\int^{2\pi}_0\left|\dfrac{f(t+x_0)+f(-t+x_0)-2\ell}{2}\right| \,\dfrac{dt}{t}
=\int^{2\pi}_0\left|\dfrac{f(t+x_0)+f(-t+x_0)}{2}-\ell\right| \,\dfrac{dt}{t}
$$
which is finite by hypothesis.
By the discussion above
$$
\lim_{N\rightarrow \infty}S_N(f_1;0) = 0.
$$
It remains to point out that 
Lemma
$S_N(f_1;0) = S_N(f;x_0) - \ell$ for every $N\geq 0$.
Proof
\begin{align*}
\hat f_1(n)&=\dfrac{1}{2\pi}\int^{2\pi}_0 f_1(t) e^{-itn}\,dt\\
&=\dfrac{1}{2\pi}\int^{2\pi}_0 (f(t+x_0)-\ell) e^{-itn}\,dt\\
&\overset{(*)}{=}\dfrac{1}{2\pi}\int^{2\pi}_0 (f(t)-\ell) e^{-itn+ix_0n}\,dt\\
&=\dfrac{e^{ix_0n}}{2\pi}\int^{2\pi}_0 (f(t)-\ell) e^{-itn}\,dt\\
&=e^{ix_0n} \hat f(n) - \dfrac{e^{ix_0n}\ell}{2\pi}\int^{2\pi}_0e^{-int}\,dt
\end{align*}
I use at (*) the change of variables $\,t\rightarrow t-x_0$ and that $f$ is $2\pi$-periodic.
Thus
$$
\hat f_1(n)=
\begin{cases}
e^{ix_0n}\hat f(n),\quad \text{if $n\neq0$}\\
\hat f(0) -\ell,\quad \text{if $n= 0$}
\end{cases}
$$
In particular
\begin{align*}
S_N(f_1;0)&=\sum_{|n|\leq N} \hat f_1(n) e^{in 0}\\
&=\sum_{|n|\leq N} \hat f_1(n)\\   
&=f(0)-\ell+\sum_{0<|n|N} \hat f(n)e^{inx_0}\\
&=S_N(f; x_0)-\ell
\end{align*}
QED
Hence, 
$$
\lim_{N\rightarrow \infty}S_N(f;x_0) -\ell=\lim_{N\rightarrow \infty}S_N(f_1;0)=0.
$$
A: See A. Zygmund, Trigonometric Series, Third edition, Volumes I & II combined, Cambridge Mathematical Library, Cambridge University Press, 2002 on page 52.
A: I think this article might help you. Pointwise Convergence of Fourier Series, Charles Fefferman, Annals of Mathematics, Second Series, Vol. 98, No. 3 (Nov., 1973), pp. 551-571:
http://www.jstor.org/discover/10.2307/1970917?sid=21105651264483&uid=4&uid=2&uid=3737760
A: Start with convergence criterion:
A necessary and sufficient condition for the Fourier series T(x) of  f  to converge pointwise to c(x) on E  is that there exists a fixed $\delta$ such that $ 0 < \delta < \pi$ and
$\int_0^\delta  {{g_{c(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $  pointwise on E.
Here ${g_{c(x)}}(u) = \frac{1}{2}\left( {f(x + u) + f(x - u) - 2c(x)} \right)$.
If $\frac{{{g_c}(u)}}{u}$ is integrable,which is your given condition then by the  the Riemann Lebesgue Theorem,$\int_0^\delta  {{g_{c(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $. This proves pointwise convergence. Here you can take any  $\delta > 0$.
 For details see convergence criterion and Theorem 25 inConvergence of Fourier Series https://037598a680dc5e00a4d1feafd699642badaa7a8c.googledrive.com/host/0B4HffVs7117IbmZ2OTdKSVBZLVk/Fourier%20Series/Convergence%20of%20Fourier%20Series.pdf
