Let $f(x)=\sum_{k=2}^{+\infty}\frac{\sin(kx)}{\log(k)}$, for $x\in [0,2\pi]$. My aim is to show that $f$ is discontinuous at $0$ and $2\pi$.
I have showed that $\sum_{k=1}^{n}\sin(kx)$ is pointwise bounded by $\frac{1}{\sin(x/2)}$ for $x\neq 0, 2\pi$ and $f(0)=f(2\pi)=0$.
Since $\frac{1}{\log(k)}$ decreases to $0$, by Dirichlet's criterion we have that the series defined by $f$ is is pointwise convergent on $[0,2\pi]$.
Moreover, I have showed that $\sum_{k=1}^{n}\sin(kx)$ is uniformly bounded by $\frac{1}{\sin(\delta/2)}$ on the interval $[\delta, 2\pi-\delta]$, where $\delta \in (0,\pi)$.
Once again, by Dirichlet's criterion we have that the series defined by $f$ is uniformly convergent on $[\delta, 2\pi-\delta]$, for any $[\delta, 2\pi-\delta]$. This implies that this series converges uniformly on compact subsets of $(0,2\pi)$, hence $f$ is continuous on $(0,2\pi)$.
I have also showed that the convergence is not uniform; if it were, by extending $f$ to a $2\pi$-periodic function which happens to be an odd function, we see that $f$ is continuous, we can calculate $\widehat{{f(k)}}$ and Parseval's identity yields the result, since then the $L^1$ norm of $f$ is then infinite, which cannot happen.
Any thoughts on this?