Timeline for Relationship between "Radial" Fourier transform and Fourier transform, especially at infinity
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2018 at 21:34 | answer | added | Mateusz Kwaśnicki | timeline score: 2 | |
| Sep 6, 2018 at 16:07 | comment | added | Willie Wong | @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 14:19 | history | edited | MichaelGaudreau | CC BY-SA 4.0 |
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| Sep 6, 2018 at 12:43 | comment | added | Mateusz Kwaśnicki | 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 4:54 | comment | added | MichaelGaudreau | 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 4:09 | comment | added | Willie Wong | 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 3:29 | history | asked | MichaelGaudreau | CC BY-SA 4.0 |