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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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Let $$B_n=\sum_{k=0}^nb_k,\:\:F_n(x)=\sum_{k=0}^na_kb_k,\quad a_k=k^{-1},b_k=\sin(kx),\:\:(k,n)\in\mathbb N^2$$

We have $$F_n(x)=a_0b_0+\sum_{k=1}^na_k(B_k-B_{k-1})=a_nB_n+\sum_{k=0}^{n-1}a_kB_k-\sum_{k=0}^{n-1}a_{k+1}B_k\\=a_nB_n+\sum_{k=0}^{n-1}(a_k-a_{k+1})B_k,\quad \forall k\le n-1.$$

Now, $$B_k=\mathcal Im\left(\sum_{j=0}^ke^{ij\phi}\right)=\mathcal Im\left(\frac{1-e^{i\phi(k+1)}}{1-e^{i\phi}}\right),\quad \phi\in\mathbb R,\:i\in\mathbb C.\\1-e^{i\phi}=e^{i\phi/2}\left(e^{-i\phi/2}-e^{i\phi/2}\right)=-2ie^{i\phi/2}\sin\left(\frac{\phi}{2}\right).$$

Thus, whenever $\:\phi\in\{0,2\pi\mathbb Z\},\:\:F_n\:$ becomes unbounded.

User 123732
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