I stumbled upon te following result:
Let $f:\mathbb{R}\rightarrow \mathbb{C}$ be a continuous $2\pi$-periodic function and $S_n$ denote the $n^{th}$ partial sum of its Fourier series. Then $\|S_n\|_{\infty}=o(\ln(n))$.
Trying to prove it, I wrote $S_n=f\ast D_n$, where $D_n$ is the $n^{th}$ Dirichlet kernel. The upper bound $\|S_n\|_{\infty}\leq \|f\|_{\infty}\|D_n\|_1$ unfortunately only gives $\|S_n\|_{\infty}=O(\ln(n))$.
To get a tighter bound, I rewrote $$D_n(x)=\frac{2\sin (nx)}{x} + \left(\sin (nx) \left(\operatorname{cotan} \frac x 2-\frac2x\right)+\cos (nx)\right).$$ Noting that the second term of the sum is a bounded function, the result to prove is that $$\sup_{t\in [-\pi,\pi]}\left|\int_{-\pi}^{\pi}\frac{f(t-x)\sin(nx)}{x}dx \right|=o(\ln(n)).$$ I don't see how to proceed from here. I see an analogy with the the result stating that if $(\varphi_n)$ is a sequence of functions converging to the Dirac, then for any uniformly continuous function $f$, $\varphi_n\ast f\rightarrow f$ uniformly. But I don't manage to exploit this result essentially because I can't prove that for any $\epsilon>0$, $\int_{\pi >|x|>\epsilon}[\sin(nx)/x] dx\rightarrow 0$. Any ideas ? (As an additional note, I would like to avoid using distribution theory to prove the result)
I would use the density of smooth functions. Write $f=g+h$ where $g$ is smooth ($C^1$ is enough) and $\|h\|_\infty$ is small. Then $g*D_n = O(1)$, because the Fourier series of a $C^1$ function converges uniformly. And the estimate $\|h*D_n\|_\infty \le \|h\|_\infty \|D_n\|_1$ yields the desired result, since $\|h\|_\infty$ can be arbitrarily small.