Timeline for Do Degree Zero Pseudo-Differential Operators on a Manifold Send Smooth Functions to Smooth Functions?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 13, 2018 at 5:22 | answer | added | Bombyx mori | timeline score: 3 | |
| Mar 12, 2018 at 16:11 | comment | added | Liviu Nicolaescu | For the specific operators you mentioned, they do map smooth sections to smooth sections. | |
| Mar 12, 2018 at 4:12 | comment | added | YYY | @Kashayar: This is not trivial. See math.mit.edu/~rbm/18.157-F05-Chapter5.ps, or math.mit.edu/~rbm/iml90c2.ps, for example. Your manifold is compact, so the estimates are not difficult. | |
| Mar 12, 2018 at 4:02 | comment | added | Kashayar | @YYY I'm sorry, but I don't see how that follows. Could you please explain some more? | |
| Mar 12, 2018 at 3:29 | comment | added | YYY | One way to see it is that smooth functions is within the intersection of Sobolev spaces. So Deane Yang's comment just follow. | |
| Mar 12, 2018 at 3:14 | comment | added | Kashayar | Thanks for your comment. Do you have a reference for why smoothness of symbol results in the p.d.operator to send smooth functions to themselves? Also, if the symbol is say, analytic, can I say the same thing? So it seems in case of the Dolbeault operator, if I understand correctly, the symbol at $(x,\zeta)$ is $\frac{1}{ i \Vert \zeta \Vert }$, which is smooth when $\zeta \neq 0$. | |
| Mar 12, 2018 at 0:36 | comment | added | Deane Yang | If the symbol of the pseudodifferential operator is smooth away from the zero section of the cotangent bundle, then it always maps smooth function to smooth functions. The order tells you how it maps between $L^2$ Sobolev spaces. An operator of order $K$ is a bounded map from $W^{2,j+k}$ to $W^{2,j}$ for any $j$. | |
| Mar 11, 2018 at 23:59 | history | asked | Kashayar | CC BY-SA 3.0 |