Timeline for Generalized Friedrichs Lemma
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 25, 2012 at 15:25 | vote | accept | Boaz Haberman | ||
| Jul 24, 2012 at 20:13 | comment | added | Bazin | It is true in the bornology sense defined in my answer. | |
| Jul 23, 2012 at 22:28 | comment | added | Deane Yang | Is it possible to do it as follows: Write $J_\epsilon = I + S_\epsilon$, where $S_\epsilon \rightarrow 0$. Then $[A,J_\epsilon] = [A,S_\epsilon]$. Now show the latter remains bounded in the appropriate norm. | |
| Jul 23, 2012 at 20:41 | answer | added | Bazin | timeline score: 2 | |
| Jul 23, 2012 at 3:14 | comment | added | Boaz Haberman | There is no problem constructing these things; my favorite example is heat flow for time $\epsilon$. It is true that $[A,J_\epsilon] \in \Psi^{m-1}$ for each $\epsilon$. Actually, $[A,J_\epsilon] \in \Psi^{-\infty}$ since $J_\epsilon \in \Psi^{-\infty}$. The difficulty is in showing that $[A,J_\epsilon]$ is bounded uniformly as an operator from $H^{m-1}$ to $L^2$. To my mind there is no way a priori to bound the commutator in $\Psi^{m-1}$ in terms of the bounds for $J_\epsilon$ and $A$, despite the fact that $J_\epsilon \in \Psi^0$ uniformly. | |
| Jul 23, 2012 at 3:07 | history | edited | Boaz Haberman | CC BY-SA 3.0 |
added 2 characters in body
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| Jul 23, 2012 at 1:11 | comment | added | Deane Yang | I must be missing something. Isn't $[A,B]$ bounded in $\Psi^{m-1}$ for any pseudodifferential operator $B$ of order $-1$ or less, because both $AB$ and $BA$ are? | |
| Jul 22, 2012 at 23:51 | comment | added | Piero D'Ancona | After localization with a (finite) partition of unity, you can work in one single chart, where the mollifiers can be defined by convolution. Which step of this construction does not convince you? | |
| Jul 22, 2012 at 21:30 | history | asked | Boaz Haberman | CC BY-SA 3.0 |