Prove or disprove uniform convergence

Question

$g_n(y)=\sum_{k=1}^{n} \frac{\sin ky}{k},x \in [0,\pi]$. So i try to find the pointwise limit function first, following this approach in this post Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$. I find it to be $\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=g(x) ,\, x\in(0,2\pi).$ I tried to prove it by Dirichlet test:
take $h_n(x)=\frac{1}{n}\quad \text{and} \quad u_n(x)=\sin(nx) , \text{then} |\sum_{k=1}^n\sin(kx)| \leq \frac{1}{\sin(\frac{x}{2})} \leq \frac{1}{\sin(\frac{\delta}{2})}$.
I want to prove or disprove that $(g_n)$ converge uniformly on $[0,\pi]$.

Answer 1

You have shown correctly with the Dirichlet test that $(g_n)$ converges uniformly on the interval $[\delta, \pi]$ for any $\delta > 0$. This relies on the fact that $\left|\sum_{k=1}^n \sin kx \right| \leqslant 1 / \sin (\delta/2)$ is uniformly bounded for all $n \in \mathbb{N}$ and $x \in [\delta,\pi]$.

However, if $\delta = 0$ the partial sums are not bounded and the Dirichlet test is not applicable. This is a clue that convergence is not uniform on $[0,\pi]$.

If the convergence were uniform, then by the Cauchy criterion for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $m > n \geqslant N$ and all $x \in [0,\pi]$ we would have

$$\tag{1}\left|\sum_{k=n+1}^{m}\frac{\sin kx }{k} \right| < \epsilon$$

We can show that this criterion is violated and the convergence is not uniform. Take $\epsilon = 1/(4\sqrt{2})$ and, for any integer $N$, no matter how large, choose $n = N$, $m = 2n$ and $x_n = \pi/(4n) \in [0,\pi]$. For any $k > n$, we have $kx_n > \pi/4$ and $\sin kx_n > \sin (\pi/4) = 1/\sqrt{2}$.

Hence,

$$\tag{2}\left|\sum_{k=n+1}^{2n}\frac{\sin kx_n }{k} \right| > \frac{1}{\sqrt{2}}\sum_{k=n+1}^{2n}\frac{1 }{k} > \frac{1}{\sqrt{2}}\cdot n \cdot \frac{1}{2n} = \frac{1}{2\sqrt{2}} > \epsilon,$$

and we have a contradiction to condition (1).

Answer 2

$\sum_{n \geq 1} \frac{\sin nx}{n}$ is exactly the Fourier series of $f(x)=\frac{\pi-x}{2}$ ($x \in [0, 2\pi)$) extended $2\pi$ periodically to $\mathbb{R}$. Were the convergence of the Fourier series uniform, then the Fourier series would in fact converge pointwise to $f$ (see this) but notice that at $x=0$ $f$ fails to coincide with its Fourier series. Hence we deduce that convergence is not uniform.