# Relationship between "Radial" Fourier transform and Fourier transform, especially at infinity

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 $\limsup_{\ell \to \infty} |\widetilde{\phi}(\ell)| > 0$.

• 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. Sep 6, 2018 at 4:09
• 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 $\limsup_{\ell \to \infty}|\widetilde{\phi}(\ell)| > 0.$ Sep 6, 2018 at 4:54
• 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. Sep 6, 2018 at 12:43
• @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. Sep 6, 2018 at 16:07

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.
• Would you mind outlining the calculation for the behaviour of $\widetilde{\phi}$ that you assert? Sep 7, 2018 at 1:02
• 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 Sep 7, 2018 at 2:29
• @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. Sep 7, 2018 at 3:01
• @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? Sep 7, 2018 at 3:06