I am doing some work regarding quasiperiodic functions but I am not able to figure out the difference between almost periodic and quasiperiodic functions. Can anyone let me know about it?
[Math] Almost periodic function vs quasi periodic function
analysisclassical-mechanicsquasiperiodic-functionspecial functions
Related Solutions
What is perhaps more useful is to think about how one proves that differentiability/decay implications in the periodic or Fourier transform case. (Think about that for a bit here. Ready?)
Suppose your function is $C^k$. Then we have that $f^{(k)}$ is uniformly bounded. Which means that $a(\lambda, f^{(k)})$ is uniformly bounded. But using that $a(\lambda, f^{(k)}) = (i\lambda)^k a(\lambda,f)$, you see that you have
$$ \sup_{\lambda \in C} |\lambda^k a(\lambda,f)| < \infty $$
follows from $f$ being $C^k$. In general this is the best you can ask for, considering the case that $f = \exp ix $. In the case of the periodic functions you can do one better, in that $ \lim_{n\to \infty} |n^k c_n| = 0$, is because you have the Riemann-Lebesgue lemma. For suitable definitions of almost periodicity, you can get something similar: if you assume your derivative function $f^{(k)}$ is almost periodic in the Besicovich sense, then you will have summability of $\lambda^k a(\lambda,f)$ which will in particular imply that
$$ \lim_{n\to \infty} \sup_{\lambda \in C, |\lambda| > n} |\lambda^k a(\lambda,f)| = 0$$
For the reverse direction, you can get something similar. If you know that $f$ is given as a convergent sum of $\sum a(\lambda,f)e^{i\lambda x}$, you see that immediately, summability of
$$ \sum_{\lambda\in C} |\lambda^k a(\lambda,f)| = S_k < \infty $$
implies that the $k$th derivative of $f$ is uniformly bounded, and that $f$ is at least $C^{k-1}$. Furthermore, if $\sup_{\lambda\in C} |\lambda| < \Lambda < \infty$, you have that your function $f^{(k)}$ is Lipschitz continuous with constant $S_k\Lambda$. So this actually implies that using, in addition, the following
$$ S_{n,k} := \sum_{\lambda \in C, |\lambda| > n} |\lambda^k a(\lambda,f)| $$
with
$$ \lim_{n \to \infty} S_{n,k} = 0 $$
that $f$ is $C^k$. (For $\epsilon$, choose $N$ large enough so that $3S_{N,k}< \epsilon$, then choose $\delta$ such that $3\delta N S_k < \epsilon$. Then do a high-low frequency splitting to show that $|f(x) - f(y)| \leq |x-y| N S_k + 2 S_{N,k}$.) So just a decay condition on the frequency is not enough: you need summability. (In the case of periodic functions, since there is a minimal spacing between frequencies, decay conditions can be directly translated to summability.)
The definition quickly implies that a nonconstant almost periodic function $f$ cannot have a limit at infinity. Indeed, pick $a,b$ such that $f(a)\ne f(b)$. Fix $\epsilon>0$ such that $$|f(a)-f(b)|>3\epsilon\tag{1}$$ Let $\ell=\ell(\epsilon) $ be as in the definition. Then for every integer $n$ there exists $\tau_n\in [n, n+\ell]$ such that $$|f(a+\tau_n)-f(a)|<\epsilon,\qquad |f(b+\tau_n)-f(b)|<\epsilon\tag{2}$$ It follows from (1) and (2) that $$|f(a+\tau_n)-f(b+\tau_n)|>\epsilon \tag{3}$$ Since $a+\tau_n\to\infty$ and $b+\tau_n \to\infty$ as $n\to\infty$, it follows that $\lim_{x\to\infty }f(x)$ does not exist.
Best Answer
I am just a master student writing a thesis in that direction, but maybe you find it helpful nonetheless.
One can show that the Bohr almost-periodic functions are the closure of the trigonometric functions in the supremum norm, i.e.
$$ \mathcal{A}:= \overline{\{ \sum_{1\leq j \leq n} a_i e^{i \nu_j x} : n \in \mathbb{N}, a_i\in \mathbb{R}\}}^{(C_b(\mathbb{R}, \mathbb{C}), \Vert \cdot\Vert_{sup})} .$$
Intuitively the difference is that the "Fourier series of a quasi-periodic function contains less independent frequencies than a general almost-periodic function". On the space of Bohr almost-periodic function we have the following sesquilinear form:
$$ \langle f, g\rangle = \lim_{x\rightarrow \infty} \frac{1}{x} \int_{0}^x f(t)\overline{g(t)}dt. $$
The frequency module $M(f)$ (the $\mathbb{Z}$-module of all frequencies that may appear in the formal Fourier series of $f$) is defined the $\mathbb{Z}$-module generated by
$$ \{ \nu\in \mathbb{R} : \langle f, e^{i\nu x}\rangle\neq 0\}.$$
We call $f$ quasi-periodic if its frequency module is finitely generated over $\mathbb{Z}$. For example, a function is periodic iff its frequency module is generated by single frequency.
There is an alternative characterization of quasi-periodic functions. Namely, $f$ is quasi-periodic if there exist a continuous map $Q:\mathbb{T}^n \rightarrow \mathbb{C}$ and a "frequency vector" $\omega=(\omega_j)_{j=1}^n$ such that $f(x)=Q(x\cdot \omega)=Q(x\omega_1, \dots, x\cdot \omega_n)$. Hence, the motion of the quasi-periodic function "lives on a finite dimensional torus". E.g.
$$ f(x)= \sin\left(\frac{2}{7}2\pi x\right) + \sin(\sqrt{2}\cdot 2\pi x) $$
has frequency module $\{ \frac{7}{2}k + \frac{1}{\sqrt{2}}l : k, l \in \mathbb{Z} \}$ and "lives on a 2-dimensional torus". Where a general almost-periodic function can be thought of "living on an infinite-dimension torus".