Fourier series of almost periodic functions and regularity 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? 
 A: 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.)
