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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 $\nu_j\in \mathcal{M}(f)$?

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  • $\begingroup$ 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
    Commented Jun 11, 2017 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
    Commented Jun 11, 2017 at 9:56

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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.)

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  • $\begingroup$ Thank you, very much. This was exactly what I was looking for. $\endgroup$
    – Severin Schraven
    Commented Jun 5, 2017 at 12:53

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