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Apr 21, 2021 at 3:39 vote accept geometricK
Apr 21, 2021 at 3:39
Apr 17, 2021 at 12:09 answer added Bazin timeline score: 2
Jun 9, 2018 at 3:55 comment added Deane Yang For an arbitrary power, It might be a nontrivial theorem but proving that the square of a pseudodifferential and operator and the square root of a positive pseudodifferential operator are pseudodifferential is straightforward using the symbol calculus.
Jun 9, 2018 at 3:24 comment added geometricK But, for instance, it is a non-trivial result of Seeley in the setting of a compact manifold that powers of elliptic operators are still pseudodifferential. I guess I'm asking a version of the question "does Seeley's result work in the non-compact setting?"
Jun 8, 2018 at 18:50 comment added Deane Yang Yes. All of the operations used to define $A$ in terms of $D$ are well defined in the space of pseudodifferential operators.
S Jun 8, 2018 at 17:27 history suggested Ali Taghavi
I add a tag.
Jun 8, 2018 at 16:51 review Suggested edits
S Jun 8, 2018 at 17:27
Jun 8, 2018 at 16:22 history asked geometricK CC BY-SA 4.0