Exercise 7 of chapter 3 of stein's book on fourier analysis

Question

I need to show that the following trigonometric series converges for every x yet it is not the fourier series of a Riemann integrable function. $\sum_{n=2}^\infty \frac{\sin(nx)}{\log n}$

So far, I wrote the series as the imaginary part of $\sum_{n=2}^\infty \frac{e^{inx}}{\log(n)}$ and we know this series has fourier coefficients $\frac{1}{log(n)}$ . But I am not sure how to show it is convergent.

To show it is not Riemann integrable, I assume I would have to show it is not bounded. But since I dont know what f is,I am unsure how to proceed.

A Riemann integrable function on $[-\pi,\pi]$ must be bounded, which forces it to be in $L^2$, and that makes the Fourier coefficients square-summable, which yours are not because $\sum_{n=1}^{\infty}\frac{1}{(\log(n))^2} = \infty$. — Disintegrating By Parts, Jan 28 '18 at 23:17
Thanks @DisintegratingByParts. Do u know why it converges for every x? — usere5225321, Jan 29 '18 at 14:15
@DisintegratingByPartsd sorry the sum starts at n=2. I edited the question — usere5225321, Jan 30 '18 at 14:00

score 1 · Answer 1 · answered Apr 09 '18 at 00:03

To show that $\sum_{n \geq 2}^{\infty} \frac{\sin(nx)}{\log(n)}$ is convergent, use the Dirichlet test for series convergence (see Wikipedia. Chapter 2, Exc. 7 of Stein & Shakarchi also states this test).

Satisfying the pre-conditions for applying Dirichlet's test: $\frac{1}{\log(n)}$ is a monotonic decreasing sequence, and one can show that $\sum_{n = 2}^{N} \sin(nx)$ is bounded above for all $N$ and any fixed $x$. One way to do it is to multiply the partial sum $\sum_{n = 2}^{N} \sin(nx)$ by $\sin(x)$ and use the identity $\cos(a - b) - \cos(a + b) = 2\sin(a)\sin(b)$. Independently of $N$, the sum telescopes into finitely many cosines, so we get the desired result.

To show the second part of the question, apply Parseval's identity / Bessel's inequality (see DisintegratingByParts' comment).

Exercise 7 of chapter 3 of stein's book on fourier analysis

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