Let $f$ a $2\pi$-periodic function represented by its Fourier series $\displaystyle\sum_{k=-\infty}^{+\infty}c_ke^{ikx}$. We know that $f$ is smooth if we have $\displaystyle\lim_{|n|\to +\infty}|c_n|n^k =0$ for all $k\in\mathbb{N}$; if $f$ is $C^k$ we have $\displaystyle\lim_{|n|\to +\infty}n^k|c_n|=0$ and if $c_n =o\left(\dfrac 1{|n|^{k+2}}\right)$ then $f$ is $c_k$.
The space of almost periodic functions is the closure for the uniform norm of $\mathrm{Span}\left\{e^{i\lambda x},\lambda\in\mathbb R\right\}$. We define
$$a(\lambda,f) := \lim_{T\to +\infty}\dfrac 1{2T}\int_{-T}^Tf(t)e^{-i\lambda t}dt$$ and if we put $C:=\left\{\lambda\in\mathbb{R},a(\lambda,f)\neq 0\right\}$, then $C$ is at most countable and we can associate a series $\displaystyle\sum_{\lambda\in C}a(\lambda,f)e^{i\lambda t}$. The numbers $a(\lambda,f)$ are the Fourier coefficients of $f$ and $\lambda\in C$ the Fourier exponents.
The question is: are there some properties of the Fourier coefficients and exponents which allow us to "read" the regularity of an almost periodic function?
Best Answer
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.)