Timeline for Fourier support condition in the paper 'A study guide for the $l^2$ decoupling theorem'
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22, 2019 at 10:57 | comment | added | msaBU | This sounds fine to me. Thank you very much! | |
| Feb 19, 2019 at 20:54 | comment | added | Zane Li | Break this up into two integrals $\int_{[0, 1/K^{1/2}] \cup [1 - 1/K^{1/2}, 1]}$ and the remainder. The first piece we can control like $g_1$ as above and the second piece we can multiply by $\phi(x, z)$ as you defined and is then the Fourier transform of $\phi E_{S_L}g$ is supported in a $O(K^{-1})$ neighborhood of the curve $(\xi, \xi^2 + 1)$, that is a $O(K^{-1})$ neighborhood of the curve $\eta = \xi^2+ 1$. | |
| Feb 19, 2019 at 20:54 | comment | added | Zane Li | How about this argument: I will take all notation from the study guide (including their definition of $S_L$ and $L$ is the line $\eta = 1$). Let $y$ be a fixed value and let $G_y(x, z) := \int_{1 - 1/K}^{1 + 1/K}g(x, y, z)e(\eta y)e((\eta^2 - 1)z)\, d\eta$. Since $(E_{S_L}g)(x, y, z) = \int_{0}^{1}\int_{1 - 1/K}^{1 + 1/K}g(x, y, z)e(\xi x + \eta y + (\xi^2 + \eta^2)z)\, d\eta\, d\xi$, using our definition of $G_y$ shows that $(E_{S_L}g)(x, y, z) = \int_{0}^{1}G_{y}(x, z)e(\xi x + (\xi^2 + 1)z)\, d\xi$. | |
| Feb 18, 2019 at 19:41 | comment | added | msaBU | For $\alpha_2=0$ for example, it is still possible that $\xi$ lies under the parabola $\eta=\alpha_1^2$ and therefore not in $N_{C/K}([0,1])$. Or am I missing something? | |
| Feb 18, 2019 at 19:41 | comment | added | msaBU | Sorry for answering only now. Your argument for $g_1$ seems fine to me. However, I still have a question about your argument concerning $g_2$. If I compute the Fourier support of $\phi E_{S_L,y}g_2$, I find that $\mathcal{F}(\phi E_{S_L,y}g_2)(\xi_1,\xi_2)$ vanishes unless $(\xi_1,\xi_2)\in B_{1/K}(\alpha_1,\alpha_1^2+\alpha_2^2)$ for $(\alpha_1,\alpha_2)$ in the support of $g_2$. But I don't see why this implies that $(\xi_1,\xi_2)\in N_{C/K}([0,1])$. | |
| May 6, 2018 at 3:46 | review | Late answers | |||
| May 6, 2018 at 4:24 | |||||
| May 6, 2018 at 3:31 | review | First posts | |||
| May 6, 2018 at 4:22 | |||||
| May 6, 2018 at 3:30 | history | answered | Zane Li | CC BY-SA 4.0 |