Uniform Convergence and Weierstrass M-test

Question

Find $r$ such that $K(x)=-\sum\limits_{n=1}^{\infty} \frac{1}{n^{r-1}}\sin(nx)$ does not converge uniformly for $x \in R$.

I tried tackling this problem using the Weistrass M-test to find for which $r$ the above series converges uniformly. I found that it converges uniformly for $r>2$. Therefore, I think that the correct anwser is to say that it does not converge uniformly for $r \in (-\infty, 2]$.

I would be grateful if someone could confirm this result, or could warn me if I am completely wrong about this.

Answer 1

I'll look at the $r=2$ case. The $r<2$ case will follow similarly.

Exercise: If $F_1, F_2, \dots$ are continuous functions on $\mathbb R$ and $F_N\to F$ uniformly on $\mathbb R,$ then $F_N(x_N) \to F(x)$ whenever $x_N\to x.$

Let $F_N(x) = \sum_{n=1}^{N}(\sin (nx))/n.$ Suppose $F_N \to F$ uniformly on $\mathbb R.$ Then from the above,$ F_N(x_N) \to F(0)=0$ whenever $x_N\to 0.$

Now there is a constant $c>0$ such that $\sin x \ge cx$ for $x\in [0,1].$ Let $x_N=1/N.$ Then

$$\tag 1 F_N(x_N) = \sum_{n=1}^{N}\frac{\sin(n/N)}{n} \ge \sum_{n=1}^{N}\frac{c(n/N)}{n} = c.$$

That's a contradiction and we're done.

Answer 2

You need to show that $$ -\sum\limits_{n=1}^{\infty} \frac{1}{n^{r-1}}\sin(nx) $$ does not converge uniformly for $r=2.$ Then your result will be confirmed.

It is well known that $$ \sum\limits_{n=1}^{\infty} \frac{1}{n}\sin(nx) $$ converges for all $x\in\mathbb{R}$ but does not converge uniformly.

Let $$ S_N(x)=\sum_{n=1}^N \frac{1}{n}\sin(nx). $$ We can show that \begin{align} S_N\left(\frac{2\pi}{2N+1} \right)&=\int_0^\pi\frac{\sin t}{t}dt\approx \frac{\pi}{2}\times 1.17,\\ S_N\left(-\frac{2\pi}{2N+1} \right)&=\int_0^{-\pi}\frac{\sin t}{t}dt\approx -\frac{\pi}{2}\times 1.17 . \end{align} This shows that the convergence of $\sum_{n=1}^\infty \frac{1}{n}\sin(nx)$ is not uniform.
Search "Gibbs phenomenon".