2
$\begingroup$

The elements of the closure of $\{ \sum_{j=1}^n a_j e^{i\nu_j x}: a_j\in \mathbb{C}, \nu_j\in \mathbb{R} \}$ in the supremum-norm are called almost periodic functions. An almost periodic function $f$ is called quasi-periodic iff the frequency module

$$ \mathcal{M}(f)=\left\langle \left\{ \nu\in \mathbb{R}: \lim_{t\rightarrow \infty} \frac{1}{t} \int_0^t f(x)e^{-i\nu x}dx\neq 0 \right\} \right\rangle_{\mathbb{Z}} $$

is finitely generated as $\mathbb{Z}$-module. By definition we can approximate $f$ in supremum-norm by functions of the form $\sum_{j=1}^n a_j e^{i\nu_j x}$. I'd like to know whether one can preserve the frequencies.

Question: Is it always possible to approximate $f$ in the supremum-norm by functions $\sum_{j=1}^n a_j e^{i\nu_j x}$ with $\nu_j\in \mathcal{M}(f)$?

$\endgroup$
  • 1
    $\begingroup$ Did you forget dividing by $t$ or was that intentional? $\endgroup$ – lcv Jun 5 '17 at 11:05
  • $\begingroup$ @lcv Thanks for pointing out my mistake. I fixed it. $\endgroup$ – Severin Schraven Jun 5 '17 at 11:54
  • $\begingroup$ I would say $\displaystyle\mathcal{M}(f)= \left\{ \nu\in \mathbb{R}: \exists k \in \mathbb{N}, \lim_{t\rightarrow \infty} \frac{1}{t} \int_0^t (f(x))^ke^{-i\nu x}dx\neq 0 \right\} $ which is a $\mathbb{Z}$-module, and I would look at the Fourier transform of $f$ in the sense of distributions $\endgroup$ – reuns Jun 11 '17 at 9:35
  • $\begingroup$ @reuns I'm sorry, I don't see, how I can use the Fourier transform to get an approximation in the supremum-norm. Could you tell me, how you would proceed? $\endgroup$ – Severin Schraven Jun 11 '17 at 9:56
3
$\begingroup$

Yes. This is Satz XIV, p. 160 of Harald Bohr, Zur Theorie der Fastperiodischen Funktionen: II, Acta Math. 46 (1925) 101-214. (Page 162 attributes the quasi-periodic case to Bohl.)

$\endgroup$
  • $\begingroup$ Thank you, very much. This was exactly what I was looking for. $\endgroup$ – Severin Schraven Jun 5 '17 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.