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May 14, 2020 at 15:51 vote accept Carlos Esparza
Nov 12, 2019 at 13:12 comment added Michael Bächtold I should have written $D(I^{m+k})\subset I^m$. It's not obvious but follows from the equality $[f_1,[f_2,[\cdots[f_n,D]\cdots]]=\sum\limits_{s\subseteq \{1,2,\ldots,n\}} (-1)^{|s|}\left(\prod\limits_{i\in s}f_i\right)D\prod\limits_{k\in \{1,2,\ldots,n\}\setminus s}f_k$ which can be proven by induction.
Nov 12, 2019 at 0:51 comment added Qfwfq Is it obvious that $D(I^{m+k})=I^m$?
Nov 11, 2019 at 15:16 history edited Michael Bächtold CC BY-SA 4.0
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Nov 11, 2019 at 13:46 comment added Carlos Esparza @IgorKhavkine Yes, that is what I have in mind. I will update my question to be more precise
Nov 11, 2019 at 13:42 comment added Igor Khavkine The first paragraph of the question recalls the smooth Peetre theorem: $D$ is a morphism of sheaves of smooth functions $\iff$ $D$ is locally a differential operator. I presume that translating this statement to the complex world replaces "sheaves of smooth functions" by "sheaves of holomorphic functions" (meaning that $D$ depends only on the germ at any point and maps germs of holomorphic functions to germs of holomorphic functions). The notion of differential operator is the usual one, as you've discussed. Hence, Q1: is this translated statement an actual theorem?
Nov 11, 2019 at 12:05 comment added Igor Khavkine For Q1, I think the intended hypothesis on the operator is that its action is known only on holomorphic functions, $D\colon \mathcal{O} \to \mathcal{O}$, rather than all smooth functions. This is a strictly weaker hypothesis than is needed to apply the smooth Petree theorem in the way that you suggest.
Nov 11, 2019 at 9:57 history answered Michael Bächtold CC BY-SA 4.0