Timeline for A Fourier-analytic inequality used by Jean Bourgain
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2022 at 3:42 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
a minor typo
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| Jul 10, 2012 at 21:05 | vote | accept | Ian Morris | ||
| Jul 10, 2012 at 17:37 | answer | added | Terry Tao | timeline score: 37 | |
| Jul 10, 2012 at 16:18 | comment | added | Ian Morris | ^^ By "uniform" I mean "uniform with respect to $f$ in the category being considered", and by "2|\hat{f}(\xi)^2|" I mean "4|\hat{f}(\xi)|^2". | |
| Jul 10, 2012 at 16:17 | comment | added | Ian Morris | Stopple: I've tried that without achieving anything conclusive. The square of the $L^2$ norm of the difference is $\int_{\mathbb{R}^2} |\hat{f}(\xi)|^2 |e^{-\delta t \|\xi\|}-e^{-\delta^{-1} t\|\xi\|}|^2 d\xi$. As $t \to 0$ the integrand converges to zero pointwise while being bounded by the integrable function $2|\hat{f}(\xi)|^2$ so this converges to zero for fixed $f$ by the Dominated Convergence Theorem, but it is not clear to me why this would be uniform. The supremum over $\xi$ of $|e^{-\delta t \|\xi\|}-e^{-\delta^{-1} t\|\xi\|}|^2$ doesn't depend on $t$ so bounding it also doesn't help. | |
| Jul 10, 2012 at 15:53 | comment | added | Stopple | Just a quick guess: Convolution is linear, Fourier transform is an isometry which converts convolution into product, and then explicit calculation? | |
| Jul 10, 2012 at 15:09 | history | asked | Ian Morris | CC BY-SA 3.0 |