I do study about Fourier Analysis with Stein and Shakarchi’s text book. Recently, I get the striking result that is, if $f$ is an integrable function on the circle, then The partial sum of Fourier series of function $f$ is converging to $f$ at a point $\theta_0$ on which $f$ is differentiable.
Indeed, this statement still holds if we only assume that $f$ satisfies a Lipschitz condition at $\theta_0$.
My question is follow. What is the minimum condition of (uniformly/pointwisely) converge of partial sum of Fourier series of function $f$ ??
You can write the truncated Fourier series in terms of the Dirichlet kernel $$ D_N(\theta)=\frac{\sin((n+1/2)\theta)}{\sin(\theta/2)}. $$ The truncated Fourier series $S_N^f(\theta)$ that involves $1,\cos(n\theta),\sin(n\theta)$ for $1 \le n \le N$ becomes $$ S_N^f(\theta)-L = \frac{1}{\pi}\int_{0}^{\pi}D_N(\theta')\left[\frac{f(\theta+\theta')+f(\theta-\theta')}{2}-L\right]d\theta'. $$ As $N\rightarrow\infty$, the integral over $[\delta,\pi]$ of the right side always tends to $0$ because of the Riemann-Lebesgue lemma, assuming only that $f$ is absolutely integrable on $[0,2\pi]$. So the Fourier series at $\theta$ converges to $L$ if, for every $\epsilon > 0$ there exists $\delta > 0$ and $N_0$ such that $$ \left|\frac{1}{\pi}\int_{0}^{\delta}D_N(\theta')\left[\frac{f(\theta+\theta')+f(\theta-\theta')}{2}-L\right]d\theta'\right| < \epsilon,\;\;\; N > N_0. $$ This is about the best you can say in general. The above does hold if, for example, the Dirichlet-Dini condition holds, meaning that the term in square brackets is absolutely integrable on $[0,\delta]$ for some $\delta$. A Holder condition on $f$ at $\theta$ for any small positive Holder exponent will certainly imply this absolute integrability, but such a condition is also too strong, meaning that it is not necessary for convergence. No one has found a "best" condition. If the graph of $f$ is nearly a straight line at $\theta$, then you get convergence; the bracketed quantity measures the asymmetry of the graph of $f$ near $\theta'=\theta$. This is why conditions of differentiability easily imply convergence.