2
$\begingroup$

Given a square-integrable function $f(t)\in \mathbb{R}$ defined on $t \in \mathbb R$ which is both

  1. Aperiodic in the sense that $f(t)=f(t+T)$ if and only if $T=0$
  2. Discretely valued in the sense that $f(t) \in \{f_1, f_2 ... \}$

where the $f_i$ are either (a) a finite number or (b) a countably infinite number of distinct real values.

My expectation is that, in both cases (a) and (b), the Fourier transform $\hat{f}(\omega)$ will be decaying as $\hat{f}(\omega)\sim O(\omega^{-1})$ for large $|\omega|$, arguing simply that $f(t)$ must have at least one discontinuity, and the Fourier transform of the step function decays this way.

Is this intuitive picture sufficient to argue this? Or are there cases where it is incorrect.

$\endgroup$
  • $\begingroup$ $L^2$, I have amended the question. $\endgroup$ – ComptonScattering Aug 5 '17 at 22:38
  • $\begingroup$ The Fourier transform is linear; for a finite number of terms this is just a finite sum of step functions. So that is OK provided the values at infinity are zero. In the countable case you should be a bit more careful. $\endgroup$ – Ian Aug 5 '17 at 22:45
  • $\begingroup$ You can drop "aperiodic" now, since a periodic function would not be square-integrable. $\endgroup$ – user357151 Aug 5 '17 at 23:13
  • $\begingroup$ And you can drop the "almost-periodic" tag, because AP functions are not square-integrable. $\endgroup$ – kimchi lover Aug 5 '17 at 23:21
  • $\begingroup$ Hm I feel I may have thrown the baby out with the bathwater here. I am a physicist a generally we are perfectly happy treating a peridoic function $\sin(x)$ as square integrable by treating $\mathrm{e}^{- \eta |x|} \sin (x)$ and taking the limit $\eta \rightarrow 0$ at the appropriate stage of the calculation. I am not familiar with the more rigorous ways in which these are resolved, but for me the distinction between not-decaying, and decaying with an infinitesimal rate are immaterial. Maybe I do not mean to restrict to $L^2$? $\endgroup$ – ComptonScattering Aug 5 '17 at 23:23
2
$\begingroup$

(b) If countably many values are allowed, the tail of the Fourier transform can be heavier than $|\omega|^{-1}$. Indeed, if $f(x) = |x|^{-\alpha}$ with $0<\alpha<1/2$ near zero (and is truncated at infinity), the Fourier transform will behave like $|\omega|^{\alpha-1}$ at infinity. (Why? The decay of Fourier transform is controlled by $\|\tau_{1/\omega} f - f\|_1$ where $\tau$ is the translation operator. Translating $|x|^{-\alpha}$ by $1/\omega$ and subtracting cancels off some of it, but it leaves a neighborhood of size $1/|\omega|$ around $0$ without much cancellation, and this neighborhood contributes $|1/\omega|^{-\alpha+1}$ to the $L^1$ norm.)

A countably-valued function can have this type of singularity at $0$ too, just discretize $|x|^{-\alpha}$ at the scales $2^{-k} < |x| < 2^{1-k}$.

(a) If a function takes on finitely many values on interval, that is, $f(x) = \sum a_k \chi_{I_k}$ where each $I_k$ is an interval, then the explicit form of the Fourier transform can be found, and it indeed decays like $|\omega|^{-1}$.

But if you allow general measurable sets, i.e, $f(x) = \sum a_k \chi_{E_k}$ for some disjoint measurable sets $E_k$, then again there's a problem because the decay of $\|\tau_{1/\omega}\chi_{E_k} - \chi_{E_k}\|_1$ can be slower than $1/|\omega|$; I think it can actually be arbitrarily slow.

$\endgroup$
  • $\begingroup$ Thanks, particularly for the example of how the singularity appears in a discretised function. $\endgroup$ – ComptonScattering Aug 6 '17 at 4:16

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.