-2
$\begingroup$

Let $\phi:\mathbb{R}^n \to \mathbb{R}$ be a $C^{\infty}$ function with compact support.

What is the relationship between

$$ \widehat{\phi}(k) = \int e^{-2\pi i x \cdot k} \phi(x) dx, \quad k \in \mathbb{R}^n $$

and

$$ \widetilde{\phi}(\ell) = \int e^{-2\pi i |x| \ell} \phi(x) dx, \quad \ell \in \mathbb{R} $$

In particular, I would like to know what I can say about $\widehat{\phi}(k)$ if I know that $\limsup_{\ell \to \infty} |\widetilde{\phi}(\ell)| > 0$.

Edit:

In my attempt to simplify a problem, I have simplified it too much. As Willie Wong has pointed out, $\limsup_{\ell \to \infty} |\widetilde{\phi}(\ell)| > 0$ is never satisfied for $\phi$ smooth and compactly supported.

I'm actually interested in the situation where $\phi(x)dx$ is replaced by $d\mu(x)$, where it is indeed possible to have $\limsup_{\ell \to \infty} |\widetilde{\phi}(\ell)| > 0$.

$\endgroup$
4
  • 1
    $\begingroup$ Write $\psi(r) = \int_{\mathbb{S}^{n-1}} \phi(r\omega) d\omega$, you have that $\tilde{\phi}(\ell) = \int_0^\infty e^{-2\pi i r \ell} \psi(r) r^{n-1} dr$. Extend $\psi(r)$ to be zero for $r \leq 0$. Then the function $\psi(r) r^{n-1}$ is by definition $L^1(\mathbb{R})$. By Riemann-Lebesgue you must have $\lim_{\ell\to \infty} \tilde{\phi}(\ell) = 0$. So your assumption that the limsup is positive is vacuous. $\endgroup$ Sep 6, 2018 at 4:09
  • $\begingroup$ You're right. In my attempt to simplify the problem, I have simplified it too much. I'm actually interested in the situation where $\phi(x) dx$ is replaced by $d\mu(x)$, where it is indeed possible to have $\limsup_{\ell \to \infty}|\widetilde{\phi}(\ell)| > 0.$ $\endgroup$ Sep 6, 2018 at 4:54
  • 1
    $\begingroup$ I do not think the two have much in common: if $\mu$ is the uniform measure on the sphere, then $\tilde\phi$ oscillates, but $\hat\phi$ converges to zero. If $\mu$ is the uniform measure on the boundary of a hyper-cube, then $\tilde\phi$ converges to zero, but $\hat\phi$ oscillates in cardinal directions. $\endgroup$ Sep 6, 2018 at 12:43
  • $\begingroup$ @MateuszKwaśnicki: I think your comment should be the answer. Given that $\tilde{\phi}$ is only sensitive to the spherical symmetric part of $\mu$, what you wrote pretty succinctly captures the difference. $\endgroup$ Sep 6, 2018 at 16:07

1 Answer 1

2
$\begingroup$

(A comment turned into an answer, following Willie Wong's suggestion).

I do not think the two have much in common. If $\mu$ is the uniform measure on the sphere, then $\tilde\phi$ oscillates, but $\hat\phi$ converges to zero. On the other hand, if $\mu$ is the uniform measure on the boundary of a hyper-cube, then $\tilde\phi$ converges to zero, but $\hat\phi$ oscillates in cardinal directions.

$\endgroup$
15
  • $\begingroup$ Would you mind outlining the calculation for the behaviour of $\widetilde{\phi}$ that you assert? $\endgroup$ Sep 7, 2018 at 1:02
  • $\begingroup$ For the hyper-cube case, after integrating in the angular directions, $\tilde{\phi}$ is essentially the Fourier transform of an $L^1$ function (since the cube is not spherically symmetric the measure smears out on integration in angular directions), and the convergence follow from Riemann-Lebesgue. In the sphere case, $\tilde{\phi}$ is essentially the Fourier transform of the point mass. @MichaelGaudreau $\endgroup$ Sep 7, 2018 at 2:29
  • $\begingroup$ @WillieWong I have this intuition that oscillation of $\widetilde{\phi}$ should slow down the decay of $\widehat{\phi}$. For example, the Fourier transform of Lebesgue measure on a solid ball, $\widehat{\lambda}$, has very rapid Fourier decay. And $\widetilde{\lambda}$ also decays. But the Fourier transform of the spherical measure, $\widehat{\sigma}$, only decays like $|x|^{-(n+1)/2}$, and $\widetilde{\, \sigma }$ doesn't decay at all. So it would seem like non-decay of $\widetilde{\mu}$ is an obstacle to the decay of $\widehat{\mu}$ somehow. Continued in next. $\endgroup$ Sep 7, 2018 at 3:01
  • $\begingroup$ @WillieWong One could imagine an intermediate scenario where $\mu$ is a measure on a union of concentric spheres whose radii form a Cantor set. If it's the middle thirds Cantor set, then $\widetilde{\mu}$ will not decay at all. Then, based on this intuition that non-decay of $\widetilde{\mu}$ is an obstacle to the decay of $\widehat{\mu}$, I would expect $\widehat{\mu}$ to decay no faster than $\widehat{\sigma}$ does. That's the kinda of relationship I was hoping for. Do you think I should edit the original question? $\endgroup$ Sep 7, 2018 at 3:06
  • $\begingroup$ @MichaelGaudreau: Your question seems to be related to Fourier (and Hausdorff) dimension of measures. I know next to nothing about this; these lecture notes (and the corresponding book) seem to give a good entry point to the literature. $\endgroup$ Sep 7, 2018 at 7:51

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.