Timeline for Compactness modulo symmetries of critical NLS solution
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2017 at 21:05 | answer | added | Matt Rosenzweig | timeline score: 2 | |
| Apr 28, 2017 at 16:12 | comment | added | Terry Tao | [Correction: it is $g_{n+1}(t) g_n(t)^{-1}$ that is constrained, not $g_{n+1}(t)^{-1} g_n(t)$.] | |
| Apr 28, 2017 at 16:07 | comment | added | Terry Tao | (increasing the number of balls needed to cover at the $2^{-n-1}$ stage if necessary). Iterating this, the $g_n(t)$ now converge to a usable limit $g(t)$. | |
| Apr 28, 2017 at 16:06 | comment | added | Terry Tao | Given any two vectors $f_1, f_2$ in $\dot H^1$, $\|g(t) f_1 - f_2\|_{\dot H^1}$ will converge to $(\|f_1\|_{\dot H^1}^2 + \|f_2\|_{\dot H^1}^2)^{1/2}$ as $t \to \infty$. As a consequence, if $g_n(t) u(t)$ is within $2^{-n}$ of some fixed $f_n$, and $g_{n+1}(t) u(t)$ is within $2^{-n-1}$ of some fixed $f_{n+1}$, and $u(t)$ has norm larger than, say, $10 \times 2^{-n}$, then $g_{n+1}(t)^{-1} g_n(t)$ is constrained to a compact set that depends only on $f_n$ and $f_{n+1}$. One can use this to adjust the $g_{n+1}$ and $f_{n+1}$ so that $g_{n+1}(t)$ is within, say, $2^{-n}$ of $g_n(t)$ ... | |
| Apr 28, 2017 at 6:43 | comment | added | Matt Rosenzweig | @TerryTao Dear Professor Tao, thanks for your comment. My difficulty with your suggestion, which I think is just a reformulation of my difficulty as stated in my original question, is that I do not know how to show the proposition contained in your first sentence. It's clear to me that for each, say $2^{-n}$, I can find a function $g_{n}(t)\in G$ such that $\{g_{n}(t)u(t):t\in I\}$ is contained in finitely many balls of radius $2^{-n}$ in $\dot{H}^{1}$. But I do not see why I can extract a limiting $g$ such that that $\{g(t)u(t) :t \in I\}$ is totally bounded in $\dot{H}^{1}$. | |
| Apr 28, 2017 at 5:59 | comment | added | Terry Tao | If $\{u(t): t \in I\}$ were totally bounded in $G \backslash \dot H^1$, then there would exist $g(t) \in G$ such that $\{ g(t) u(t): t \in I \}$ was totally bounded in $\dot H^1$. Now take contrapositives (and use Heine-Borel). | |
| Apr 27, 2017 at 23:04 | history | edited | Matt Rosenzweig | CC BY-SA 3.0 |
added 243 characters in body
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| Apr 27, 2017 at 16:32 | history | asked | Matt Rosenzweig | CC BY-SA 3.0 |