# Dominated convergence for quasi-periodic functions

A function $f:\mathbb{R} \rightarrow \mathbb{C}$ is called almost periodic if it is the uniform limit of trigonometric polynomials. One can show that for almost periodic $f$ the following pointwise limit (called the mean value of $f$) exists $$M(f) = \lim_{x\rightarrow \infty} \frac{1}{x} \int_0^x f(t)dt.$$ We call a almost periodic function $f$ quasi-periodic if the $\mathbb{Z}$-module $$\mathcal{M}(f)=\langle \nu\in \mathbb{R} : M(f\cdot e^{-i \nu x}) \neq 0 \rangle_{\mathbb{Z}}$$ (called the frequency module of $f$) is finitely generated. Further we set $$\mathcal{A}(f)=\{ \text{ g quasi-periodic and }\mathcal{M}(g) \subseteq \mathcal{M}(f)\}.$$

My question is whether there are known results relating pointwise limit of quasi-periodic functions and the convergence of the associated mean values.

For example is there a version of Lebesgue's dominated convergence theorem for $M(\cdot)$? I think the minimal assumptions should look something like this

Proposition: Let $f$ be a quasi-periodic function and let $(f_n)_{n\in \mathbb{N}}\subseteq \mathcal{A}(f)$ be a sequence bounded in the supremum norm and $f_n \rightarrow f$ pointwise then $M(f_n)\rightarrow M(f)$.

Any reference would be greatly appreciated.

Edit: For me a trigonometric polynomial is a function of the form

$$\sum_{j=1}^n a_n e^{i \nu_j x},$$

where $n\in \mathbb{N}, a_j\in \mathbb{C}$ and $\nu_j\in \mathbb{R}$.

• Typo? ${\mathcal M}(f)$ is a subset of $\mathbb R$, so I don't think you want $f_n \in {\mathcal M}(f)$. – Robert Israel Oct 19 '16 at 19:08
• @RobertIsrael Thanks for pointing out. It's indeed a mistake. I corrected it. – Severin Schraven Oct 19 '16 at 19:42
• @user1952009 E.g. $\exp(ix) + \exp(itx)$ where $t$ is irrational. – Robert Israel Oct 19 '16 at 22:56
• It is a trigonometric polynomial. – Robert Israel Oct 19 '16 at 22:59
• @RobertIsrael ok then he has to say that a "trigonometric polynomial" is $\sum_{n=1}^N a_n e^{i \omega_n t}$ not $\sum_{n=-N}^N a_n e^{i \omega n t}$ as one might think – reuns Oct 19 '16 at 23:01

The proposition is indeed true. To see this one uses an alternative characterization of quasi-periodic functions. Let $\omega_1, \dots, \omega_m\in \mathbb{R}$ a $\mathbb{Z}$-basis of the frequency module of $f$ and set $\omega= (\omega_1, \dots, \omega_m)$. By the work of Harald Bohr, we have that for every $f_n\in \mathcal{M}(f)$ exists a unique $F_n\in C^0(\mathbb{T}^m, \mathbb{C})$ such that
$$f_n(x) = F_n(x\omega).$$
$$M(f_n) = \frac{1}{(2\pi)^m} \int_{\mathbb{T}^m} F_n(\theta)d\theta.$$
As $\{ x\omega \in \mathbb{T}^m: x\in \mathbb{R}\}$ lies dense in $\mathbb{T}^m$, pointwise convergence $f_n \rightarrow f$ implies pointwise convergence of $F_n \rightarrow F$ on the entire torus. Furthermore, the uniform boundedness of the $f_n$ carries over to the uniform boundedness of the $F_n$. Hence, the proposition follows by dominated convergence.