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This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this.
Given $a_n$ a decreasing sequence of positive numbers and suppose that $$\sum_{n=0}^{\infty}{a_n \sin{(nx)}}$$ Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that $n a_n \to 0$.
Clearly $\frac{1}{n}$ fits the description of $a_n$, and $n \frac{1}{n} \to 1 \neq 0$, so this would prove that there is a mistake in the problem if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$.
So my question is if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for every $x$.

(I know that the series converge uniformly for every x in $[\delta, 2\pi - \delta]$, for $0 < \delta <2\pi$ by using the Dirichlet criterion.)

alejopelaez
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  • Please let us know if you come up with something for the original exercise. I'm curious! Have you tried something? – Giuseppe Negro Mar 24 '11 at 19:49
  • @Douglas: Thanks, I changed that, $a_n$ can be any decreasing sequence of positive reals. – alejopelaez Mar 24 '11 at 22:36
  • @dissonance: I was thinking of proving that $\sum{n a_n}$ converge, and that would imply $n a_n \to 0$. For that I was thinking of using the Cauchy criterion to bound the partial sums of $\sum{n a_n}$. But I'm thinking that $\sum{n a_n}$ doesn't necessarily converge. Another approach was using the fact that $a_n \sin{(nx)} \to 0$ uniformly, so I tried to pick an x that would allow me to relate it to $n a_n$. But so far I haven't been able to solve it. – alejopelaez Mar 24 '11 at 22:54
  • I think that $\sum na_n$ needs not converge. This would imply that $\sum na_n \cos(nx)$ converges uniformly and so, once put $f(x)=\sum a_n sin(nx)$, that $f\in C^1(\mathbb{R})$ and $f'(x)=\sum n a_n cos(nx)$. That $f \in C^1$ looks like something too strong. The second avenue looks more promising: maybe you could use a result from Katznelson's harmonic analysis book: if $x$ is an irrational multiple of $2\pi$ then $(nx\ \mod\ 2\pi)$ is dense on the unit circle and so $\sin(nx)$ is a dense subset of $[-1, 1]$. This could help someway? – Giuseppe Negro Mar 25 '11 at 00:03

3 Answers3

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The sum of the series is non-continuous (you can view this as the Fourier series for a saw-tooth function; or just check the behavior around x=0), so the convergence cannot be uniform. Each summand is obviously a continuous function, and a uniformly convergent series of continuous functions is continuous.

Alon Amit
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  • What behavior around x=0 are you referring to? – Mykie Mar 27 '11 at 04:24
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    The sum is positive and bounded away from 0 for small positive x, and negative for small negative x. I think this can be shown "by hand" without resorting to Fourier analysis, in case the OP is not familiar with it. – Alon Amit Mar 27 '11 at 05:05
  • Actually, what we want to look at is near $2\pi$, not $0$, right? – Pedro May 22 '13 at 18:07
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    What's the difference? – Alon Amit May 23 '13 at 19:04
  • @AlonAmit, I am curious how you would show the behaviour around $0$ "by hand" without using Fourier Analysis? – Philipp Oct 17 '21 at 21:38
  • @Philipp look at a truncated sum through n=N, and check the value at x=pi/N and x=-pi/N. – Alon Amit Oct 17 '21 at 21:56
  • @AlonAmit, can you elaborate a bit. I don't see how it helps if I look at $\sum\limits_{n=1}^N\frac{1}{n}\sin(n\frac{\pi}{N})$ and $\sum\limits_{n=1}^N\frac{1}{n}\sin(n\frac{-\pi}{N})$. How do I see that both don't converge to $0$? – Philipp Oct 17 '21 at 23:11
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Hint:

Use Cauchy Criterion to prove that your infinite series isn't uniformly converges for all $x\in[0,2\pi]$.

Salech Alhasov
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We will use Cauchy Criterion to prove that the series $$\sum_{n=1}^{\infty}\frac{\sin{(nx)}}{n}$$ is not umiformly convergent on $[0,2\pi]$.

Actually, for any $n\geq1$, take $x_n=\frac{\pi}{4n}$, then we have $$S_{2n}(x_n)-S_n(x_n)=\frac{\sin\frac{(n+1)\pi}{4n}}{n+1}+\cdots+\frac{\sin\frac{(2n)\pi}{4n}}{2n} >\frac{\sqrt2}{2}\left(\frac{1}{n+1}+\cdots+\frac1{2n}\right) >\frac{\sqrt2}{4}.$$

Here is a general result: Let $\{a_n\}_{n=1}^\infty \subset(0,\infty)$ be non-increasing. Then the series $\sum\limits_{n=1}^\infty a_n \sin(nx)$ is uniformly convergent on $\mathbb R$ if and only if $\lim\limits_{n \to \infty} n a_n =0$.

See uniformly convergent and Section 7.2 of Edwards' book Fourier Series: A Modern Introductionfor detail!

Riemann
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