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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.


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 $n.$

  • $\begingroup$ How can I have missed that ! Simple and fast, thank you so much ! $\endgroup$ – J.Mayol Nov 13 '16 at 18:11
  • $\begingroup$ You're welcome. $\endgroup$ – zhw. Nov 13 '16 at 21:41
  • $\begingroup$ Just one question ! Why is there convergence in $L^2$ (or $L^1$) ? The norm of $t \mapsto e^{2i \pi n_k t}/k$ is $1/|k|$ both $L_1$ or $L^2$ thus not summable ! $\endgroup$ – J.Mayol Nov 14 '16 at 16:03
  • $\begingroup$ No, $\int_0^1 |\sum a_n e^{2\pi i n t}|^2\, dt = \sum |a_n|^2.$ $\endgroup$ – zhw. Nov 14 '16 at 16:07
  • 1
    $\begingroup$ L^2 convergence implies L^1 convergence in any finite measure space. $\endgroup$ – zhw. Nov 14 '16 at 16:18

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