Fourier coefficients of a trigonometric series : identical to trigonometric coefficients?

Question

In this thread :

Is $H(\theta) = \sum \limits_{k=1}^{\infty} \frac{1}{k} \cos (2\pi n_k \theta)$ for a given sequence $n_k$ equal a.e. to a continuous function?

Brad Rodgers answers the question, but I don't understand why the Fourier coefficients of the function $\sum_{k \neq 0} \frac{1}{|k|}e^{2i\pi n_k}$, are the $\frac{1}{|k|}$ !

Of course it seems obvious but it is not easy to prove I think. There is trigometric series with Fourier coefficients that are not their trigonometric coefficients.

How can we prove this ? Thanks.

Answer 1

The series that defines $H$ converges in $L^2[0,1],$ hence in $L^1[0,1].$ Therefore

$$\hat H (n) = \int_0^1 (\sum_{k \neq 0} \frac{1}{|k|}e^{2i\pi n_kt})e^{-i2\pi nt}\, dt = \sum_{k \neq 0} \frac{1}{|k|}(\int_0^1 e^{2i\pi n_kt}e^{-i2\pi nt}\, dt)$$

for all $n.$ Hence $\hat H (n_k) = 1/|k|, k \ne 0,$ and $\hat H (n) = 0$ for all other $n.$