8
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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
7
$\begingroup$

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.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.