Show that a series defines a function continuous at all $x≠2k$

Question

I've been given the following statement to prove:

If $\{a_k\}_{k=1}^{\infty}$ is monotonic and $$\lim_{k\to \infty}{a_k}=0$$ Then the series $$\sum_{k=1}^{\infty}{a_k\sin(kx)}$$ Defines a function continuous for all $x\neq 2\pi k, \hspace{1mm}k\in \mathbb{Z}.$

My strategy would be to use the fact that $$\left|\sum_{k=1}^{n}{\sin(kx)}\right|\leq \frac{2}{|\sin(x/2)|}$$ and given any $x_0\neq2\pi k$, I could construct an intervall $I_{x_0}$ around $x_0$ such that $\frac{2}{|\sin(x/2)|}$ would be bounded on $I_{x_0}$. Then the series would converge uniformly on $I_{x_0}$ by Dirichlet's test and since all functions $a_k\sin(kx)$ are continuous at $x_0$, the series would also be continuous at $x_0$

My main question is if the construction of these intervalls $I_{x_0}$ can legitimately be used to prove what I'm trying to prove? Is it not the case that I need to show that the series converges uniformly on the entire set $\mathbb{R}-\{2\pi k\}_{k\in \mathbb{Z}}$ before I can claim that it is continuous on it? I sense however that it isnt uniformly convergent on that set.

Thanks in advance.

Answer 1

My main question is if the construction of these intervalls $I_{x_0}$ can legitimately be used to prove what I'm trying to prove?

Yes, and that's the natural way to prove it.

Is it not the case that I need to show that the series converges uniformly on the entire set $\mathbb{R}-\{2\pi k\}_{k\in \mathbb{Z}}$ before I can claim that it is continuous on it? I sense however that it isnt uniformly convergent on that set.

You sense rightly in general. Since the terms of the series are continuous functions, if the convergence is uniform on $\mathbb{R} \setminus 2\pi\mathbb{Z}$, then it is in fact uniform on $\mathbb{R}$. Of course this can happen (e.g. if $\sum \lvert a_k\rvert < +\infty$), but typically the sum function will have discontinuities at the points of $2\pi\mathbb{Z}$, so the convergence cannot be uniform on all of $\mathbb{R}$.

However, continuity is a local property, whether a function $f$ is continuous at a point $x_0$ depends only on the values of $f$ on an arbitrarily small neighbourhood of $x_0$. And thus, if a sequence or series of continuous functions converges uniformly on some neighbourhood of $x_0$, however small that neighbourhood is, the limit function is continuous at $x_0$.

That is precisely what your argument with the $I_{x_0}$ gives you, locally uniform convergence on $\mathbb{R}\setminus 2\pi\mathbb{Z}$, and consequently the continuity of $$\sum_{k = 1}^{\infty} a_k\sin (kx)$$ on that set.