If $f$ satisfies a Lipschitz condition, then $S_Nf\to f$ uniformly as $N\to \infty$ where $S_N$ is the $N$-th partial sum of the Fourier series.
I learnt the 'proof' of the above theorem from this lecture note. Let me outline its main idea:
\begin{aligned} \left|S_{N} f(x)-f(x)\right| &=\left|\frac{1}{2 \pi} \int_{0}^{2 \pi}(f(x-t)-f(x)) D_{N}(t) d t \right|\\ &=\left|\frac{1}{2 \pi} \int_{-\pi}^{\pi}(f(x-t)-f(x)) D_{N}(t) d t \right|\\ &=\left|\frac{1}{2 \pi} \int_{-\pi}^{\pi}(f(x-t)-f(x)) \frac{\sin \left(N+\frac{1}{2}\right) t}{\sin (t / 2)} d t\right|\\&\le \frac{1}{2 \pi}\left|\int_{-\pi}^{\pi}(f(x-t)-f(x)) \frac{\cos (t / 2)}{\sin (t / 2)} \sin N t d t\right| + \frac{1}{2 \pi}\left|\int_{-\pi}^{\pi}(f(x-t)-f(x)) \cos N t d t\right| \end{aligned}
Let $\color{red}{h(t)=(f(x-t)-f(t)) \frac{\cos (t / 2)}{\sin (t / 2)}}$ for $t\neq 0$, $\color{red}{k(t)=f(x-t)-f(t)}$, then the Lipschitz condition says $|h(t)|$ is continuous and bounded except for the point $0$, thus $|h|^2$ is integrable. So is $|k|^2$. By the corollary to the Bessel's inequality, $\frac{1}{2 \pi} \int_{-\pi}^{\pi} h(t) \sin N t d t\to 0$. Similarly, $\frac{1}{2 \pi} \int_{-\pi}^{\pi} k(t) \cos N t d t \to0$. But these are exactly the two terms on the right hand side. QED
However, I am skeptical about whether the above proof establishes the uniform convergence of $S_Nf(x)$. As the definition of $h(t)$ and $k(t)$ depends on $x$, how can we be sure that the $N$ can be chosen independent of $x$? I have also checked out the proof of the corollary to the Bessel's inequality (i.e. the Fourier coefficients of $f$ goes to $0$ as $N\to \infty$ provided that $|f|^2$ is integrable), it is proved by noting that $\sum a_n$ converges implies $a_n\to 0$, which is a pretty roundabout way (albeit clever) to demonstrate the result. In particular, I don't see how to modify it so that choosing a single $N$ suffices for all $x$. Please help me fill in the gaps in this proof. Many thanks!