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Let $E\to M$ be a smooth vector bundle over a smooth manifold $M$.

Let $D$ be a differential operator defined on the space $\Gamma(E)$ of smooth sections of $E$. We fix a section $s\in \Gamma(E)$.

Assume that for every $x\in M$ there exist a natural number $n$ such that $D^n(s)(x)=0$.

Does this imply that $D^k(s)=0$, for some $k\in \mathbb{N}$?

If the answer is "No", what about if we assume that $D$ is an elliptic operator and $M$ is a compact manifold?

The motivation for this question is the following fact which I learned from the book of "R.P.Boas:A primer of real functions".

https://www.jstor.org/stable/10.4169/j.ctt5hh8x5

Fact: Assume that $f:\mathbb{R} \to \mathbb{R}$ is a smooth function with the property that for every $x\in \mathbb{R}$ there exist a natural number $n$ with $f^{(n)}(x)=0$.Then $f$ is a polynomial.

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  • $\begingroup$ (A version of) the real analytic result you mention is discussed in this MO question: mathoverflow.net/questions/34059/… $\endgroup$
    – Sam Hopkins
    Commented Apr 24, 2019 at 2:59
  • $\begingroup$ @SamHopkins Thanks for your comment.That is about smooth finction. Among answer one can find a reference to Boas book. I learned this result from the book of Boas when i was a master student any way the link you mentioned is about that classical theorem not about sections of vector bundles. $\endgroup$
    – Ali Taghavi
    Commented Apr 24, 2019 at 5:23
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    $\begingroup$ right, I just mention to link that in case other people were interested in learning more about the classic result you mentioned. $\endgroup$
    – Sam Hopkins
    Commented Apr 24, 2019 at 13:22
  • $\begingroup$ @SamHopkins yes I see thanks for the link. $\endgroup$
    – Ali Taghavi
    Commented Apr 25, 2019 at 6:03

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