Does $\sum \frac{\sin^2(nx)}{n^2x}$ converges uniformly on $(0,\delta]$?

Question

I'm already aware, by Parsevsal's theorem, that $$\sum_{n=1}^{\infty} \frac{\sin^2(nx)}{n^2x}=\frac{\pi-x }{2}$$ is a pointwise convergence on $(0,\delta]$ for all $\delta\in(0,\pi)$. Now I'm wondering, is this convergence also uniform?

The first idea that popped into my mind was using Dini's theorem. However, since Dini's theorem requires all the functions to be defined on a compact set, it is not immediately available.

But is it possible to extend the definition of $\sin^2(nx)/n^2x$ to $x=0$ so as to make the redefined series fulfill all the conditions that Dini's theorem required? That's to say, can we define a series $\sum f_n(x)$ such that $f_n$ is defined on $[0,\delta]$ and that:
1). $f_n$ is continuous.
2). $f_n(x)=\sin^2(nx)/n^2x$ except at $x=0$.
3). $\sum_{1}^{\infty}f_n(0)=\pi/2$
4). $f_n\le f_{n+1}$ on $[0,\delta]$.

After doing some work I found that the answer is unfortunately no. Because by requiring (1) and (2), we must let $f_n(0)=0$ for all $n$, which contradicts (3).

So now I begin to doubt, is it the case that the convergence just can't be uniform?

Answer 1

Put $\displaystyle R_n(x)=\sum_{k\geq n}\frac{\sin(kx))^2}{k^2x}$ for $x>0$. If your series is uniformly convergent on $I=]0,\delta]$ for some $\delta >0$, then for every $\varepsilon>0$ there exists an $N$ such that for $n\geq N$ and $x\in I$, we have $R_n(x)<\varepsilon$.

Now we have $$R_n(x)\geq S_n(x)=\sum_{k=n}^{2n}\frac{\sin(kx))^2}{k^2x}$$

Put $x=1/n$, it is in $I$ for $n$ large.We have $$S_n(1/n)=\frac{1}{n}\sum_{j=0}^n\frac{(\sin(1+j/n))^2}{1+2 j/n+j^2/n^2}$$

Hence $S_n(1/n)$ is a Riemann-sum for $[0,1]$ and the function $\displaystyle f(x)=\frac{(\sin(1+x))^2}{1+2x+x^2}$. So $\displaystyle S_n(1/n)\to \int_0^1f(t)dt=c>0$, and this contradict the uniform convergence of the series.

Answer 2

Note that for $x$ small, the sum is close to $\pi / 2$ and yet as $x \to 0$, for fixed $n$ you have $\sin^2 (nx) / n^2 x \simeq (nx)^2/n^2x = x$. Thus as $x \to 0$ you need more and more terms in your series to get up to a particular fixed constant $c < \pi / 2$, so the convergence is not uniform.

Answer 3

Assume the series converges uniformly on some $(0,\delta).$ Then by the standard result,

$$\lim_{x\to 0^+} \sum_{n=1}^{\infty} \frac{\sin^2(nx)}{n^2x}= \sum_{n=1}^{\infty} \lim_{x\to 0^+}\frac{\sin^2(nx)}{n^2x} = \sum_n 0 = 0.$$

But the series equals $(\pi-x)/2,$ which has limit $\pi/2$ as $x\to 0^+.$ That's a contradiction, proving that the original assumption is wrong.