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?

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    $\begingroup$ Also, why wouldn't the same decay statements give you sufficient (but maybe not necessary) conditions for smoothness? By requiring rate of decay of $a(\lambda,f)$ in $\lambda$, you can get uniform boundedness of the derivatives of $f$, no? $\endgroup$ – Willie Wong Jul 4 '11 at 0:26
  • $\begingroup$ @Willie Wong: I've corrected the definition of Fourier transform. By "rate of decay", do you mean something like $\lim_{n\to\infty}a(\lambda_n,f)\lambda_n^k =0$ for all $k\in\mathbb{N}$? $\endgroup$ – Davide Giraudo Jul 4 '11 at 9:31

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

$$ S_{n,k} := \sum_{\lambda \in C, |\lambda| > n} |\lambda^k a(\lambda,f)| $$


$$ \lim_{n \to \infty} S_{n,k} = 0 $$

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.)

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  • $\begingroup$ Nice answer. I think there is a typo: it should be $a(\lambda,f^{(k)})=(i\lambda)^ka(\lambda, f)$ instead of $a(\lambda,f^{(k)})=\lambda^ka(\lambda, f)$ (but of course it doesn't change anything). $\endgroup$ – Davide Giraudo Jul 16 '11 at 11:29
  • $\begingroup$ right you are. Fixed. $\endgroup$ – Willie Wong Jul 16 '11 at 13:03

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