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Why is $\sum_{k=1}^n \frac{\sin kx}{k}$ uniformly convergent on $[\delta,\pi]$ for all $0<\delta<\pi$ from this nice property of Fourier series?

Furthermore, why is the partial sum $\sum_{k=1}^n \frac{\sin kx}{k}$ uniformly bounded for all $n$ and $x \in (0,\pi)$ as discussed in this question.

RRL
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Wonder
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  • It is shown in the problem statement that uniform convergence on $[\delta, \pi]$ follows from the Dirichlet test since $1/k \to 0$ monotonically and uniformly (with no dependence on $x$) and $|\sum_{k=1}^n \sin kx| \leqslant 1/ \sin(\delta/2)$ for all $x,n$. Do you not know how the bound for $\sum_{k=1}^n \sin kx$ is obtained? – RRL May 13 '20 at 17:23
  • yes. and I do not know the Dirichlet test@RRL. I know Dirichlet test, but the version i know can only be used to test the convergence of a series, but not uniformly convergence\ – Wonder May 13 '20 at 22:23
  • Its an easy extension to uniform convergence. The series $\sum g_n(x) h_n(x)$ converges uniformly for $x \in D$ if $g_n(x) \to 0$ monotonically and uniformly for $x \in D$ and the partial sums $\sum_{k=1}^n h_k(x)$ are uniformly bounded for all $n$ and all $x \in D$. – RRL May 13 '20 at 22:31
  • ok thanks, how is the bounded obtained? – Wonder May 13 '20 at 22:34
  • I get it, thank you. – Wonder May 13 '20 at 23:00
  • Do you want the derivation of the bound on $\sum_{k=1}^n \sin kx$? – RRL May 13 '20 at 23:07
  • no, thank you, I got how to prove it. But I have a question about https://math.stackexchange.com/questions/2777250/understanding-uniform-bound-for-partial-sum/2777270#2777270. how to get the bound for the first term? – Wonder May 13 '20 at 23:14
  • thank you for your generous reply, but it seems that the bound depends on x, how to claim the uniform bound?@RRL – Wonder May 14 '20 at 00:27
  • Let us continue this discussion in chat. – Wonder May 14 '20 at 00:29
  • I showed there is a bound of the absolute value of the first sum that does not depend on $x$. It is $\pi$ as shown in the last line. $Mx < \frac{\pi}{x} x = \pi$. – RRL May 14 '20 at 00:30
  • but the second sum depends on x?@RRL – Wonder May 14 '20 at 00:32
  • I also edited your question as this is really about the uniform boundedness of the sum for $x \in (0,\pi)$. Hopefully this clears it up for you. – RRL May 14 '20 at 01:07
  • thank you for your generous help – Wonder May 14 '20 at 01:24

1 Answers1

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Note that for $x > 0$ we have $|\sin (nx)| \leqslant |nx| = nx$ and

$$\left|\sum_{k=1}^M\frac{\sin (nx)}{n}\right| \leqslant \sum_{k=1}^M\frac{|\sin (nx)|}{n} \leqslant \sum_{k=1}^M \frac{nx}{n} = Mx,$$

and since $M = \min(N, \lfloor \frac{\pi}{x}\rfloor)\leqslant \lfloor \frac{\pi}{x}\rfloor \leqslant \frac{\pi}{x}$ we get

$$\left|\sum_{k=1}^M\frac{\sin (nx)}{n}\right| \leqslant \frac{\pi}{x} x = \pi$$

For the second sum in the linked question it was shown that

$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant \frac{1}{|\sin(x/2)|}\left(\frac{1}{N} + \frac{1}{M+1} + \sum_{n=M+1}^{N-1}\left(\frac{1}{n} - \frac{1}{n+1} \right) \right) = \frac{2}{(M+1)|\sin(x/2)|}$$

For $z \in (0,\pi/2)$ we have the inequality $\frac{2z}{\pi} \leqslant \sin z < z$ and, thus, for $x \in (0,\pi)$ we have $\sin(x/2) \geqslant x /\pi$. Hence,

$$\frac{2}{(M+1)|\sin(x/2)|} < \frac{2\pi}{(M+1)x}$$

In bounding this sum we can assume that $M < N$ (otherwise there are no terms) and, hence, $M = \lfloor \frac{\pi}{x} \rfloor$ implying that $M \leqslant \frac{\pi}{x} < M+1$ and $\frac{1}{(M+1)x} < \frac{1}{\pi}$.

Therefore,

$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant\frac{2}{(M+1)|\sin(x/2)|} < \frac{2\pi}{(M+1)x}< \frac{2\pi}{\pi} = 2$$

RRL
  • 90,707