When is trigonometric series the Fourier series of some function? (related to the book "Fourier and wavelet analysis"-Bachman)

Question

This question is related to relationship between trigonometric and Fourier series (book "Fourier and Wavelet Analysis" from Bachman, Narici, Beckenstein, pages 218-220). It says on page 218:

"If the trigonometric series converges uniformly on $[-\pi, \pi]$, then it must be Fourier series." (***)

(and the limits is 2$\pi$ -periodic continuous function).

1. On page 219 there is an example of series which uniformly converges on $[a, 2 \pi -a]$, where $a \in (0, \pi)$ but is not Fourier series: $\sum_{n \geq 2}\frac{sin( nx)}{ln (n)}$. They proved uniform convergence using Dirichlet's test and concluded that the limit function is continuous on $[a, 2\pi -a]$. Then they used Theorem from page 214: $\sum_{n \geq 2} \frac{1}{n ln(n)} $ diverges, so $\sum_{n \geq 2}\frac{sin( nx)}{ln (n)}$ is not the Fourier series . I don't understand why (***) does not apply here.

2. In some other source I found the proof that there doesn't exist Riemann-integrable function whose Fourier series is $\sum_{n \geq 2}\frac{sin( nx)}{ln (n)}$. Parseval's identity is used, they prove that $ \frac{1}{2 \pi} \int |f(x)|^2dx = \sum_{n \geq 2} |c_n|^2 = \sum_{n \geq 2}|\frac{1}{ln (n)}|^2 = \infty $. What should be the limits of integral here? And which criterium should I use to prove that there is no Lebesgue-integrable function whose Fourier series is $\sum_{n \geq 2}\frac{sin( nx)}{ln (n)}$ ? What are the criteria for Riemann- and Lebesgue-integrable functions, are they different?

I am learning on my own, so please help. I would be very thankful.

Answer 1

$(***)$ doesn't apply because the convergence is not uniform on $[-\pi,\pi]$ with is an hypothesis of $(***)$.

if $\sum_{n \geq 2}\frac{\sin( nx)}{\ln (n)}$ was the Fourier series of a Riemann integrable map, Parseval identity would imply that $$\frac{1}{2 \pi} \int |f(x)|^2dx = \sum_{n \geq 2} |c_n|^2 = \sum_{n \geq 2}\left\vert\frac{1}{\ln (n)}\right\vert^2$$ would be finite as the square of a Riemann integrable map is Riemann integrable. This is not the case.

For a Lebesgue integrable map, the sum

$$\sum_{n=1}^\infty \frac{c_n-c_{-n}}{n}$$ has to be finite. See this article of my website for more details.