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I need help understanding the last part of a proof in Fourier analysis. Claim is that partial sums $\sum_{n=1}^N \frac{\sin (nx)}{x}$ are uniformly bounded for all $n$ and all $x \in \mathbb{R}$.

Starting with $x \in (0,\pi)$ and $M = \min(N, \lfloor \frac{\pi}{x}\rfloor) $ bounds are found for

$$\sum_{k=1}^M\frac{\sin (nx)}{n} ,\sum_{k=M+1}^N\frac{\sin (nx)}{n}$$

I understand how the first one is done but for the second one it is said without details:

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

The second inequality comes from $\sin(x/2) \geq (2/\pi)(x/2)$, but I don't see how to get the first one $(\leq 1/[(M+1)\sin(x/2)].$

I think the bound for $\sum \sin (nx)$ is used:

$$\sum_{n=1}^N \sin(nx) = \frac{\sin(Nx/2)\sin((N+1)x/2)}{\sin(x/2)}$$

but because $\sin(nx)$ changes sign I can't say

$$\sum_{n=M+1}^N\frac{\sin (nx)}{n} \leq \frac{1}{M+1}\sum_{n=M+1}^N\sin (nx)$$

I want to understand the proof and see if it is correct, not find another way to bound the partial sums.

SAS
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1 Answers1

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You are correct that the sign change prevents you from concluding that

$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant \frac{1}{M+1}\left|\sum_{n=M+1}^N \sin nx\right| $$

However, you can use summation by parts with $S_n(x) = \sum_{k=1}^n \sin kx$ to get

$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| = \left|\frac{S_N(x)}{N} - \frac{S_M(x)}{M+1}+ \sum_{n=M+1}^{N-1} S_n(x) \left(\frac{1}{n} - \frac{1}{n+1} \right) \right|$$

Since $|S_n(x)| \leqslant 1/|\sin(x/2)|$ it follows that

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

and $|\sin(x/2)| = \sin(x/2)$ for $x$ in this range.

The proof is valid, although it is unclear if the author meant to include the extra factor of $2$ found here.

RRL
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