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I have a problem that I already received hints but still hard for me.

Let $f$ be an analytic function in a region $D$. Suppose that, for each $z \in D$ exist $n=n(z) \in \mathbb{N}$ such that $f^{(n(z))}(z)=0$. Prove that $f$ is polynomial.

What I already have is that I can define $D_n = \{ z \in D | f^{(n)}(z)=0 \}$ and I need to use exhaustion by compact sets.

I have no idea how to do that and I will apreciate if someone give me the most elementary awnser possible. (Or a really good hint)

Duda Duarte
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  • See here for an answer to a similar question: https://math.stackexchange.com/questions/1959597/show-that-fn0-0-for-infinitely-many-n-ge-0/1959609#1959609 – carmichael561 Jun 01 '17 at 02:13
  • Well an analytic function is either identically $0$ or $0$ at at most countably many points. It easily follows that a function that satisfies this condition must have an iterated derivative that is identically zero. I'm guessing you can't use the characterization of zero sets though. – Matt Samuel Jun 01 '17 at 02:21

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By "region" $D$ I assume you mean that $D$ is a non-empty connected open set. Let $S(n)=\{z\in D| f^{(n)}(z)=0\}.$ Then $D=\cup_{n\in \mathbb N}S(n)$ is uncountable, so $S(n)$ must be uncountable for some $n.$ Any uncountable subset of $\mathbb C$ has an uncountable bounded subset. An analytic function, such as $f^{(n)}$, which is $0$ on an infinite bounded subset of $D$, is $0$ on all of $D.$

DanielWainfleet
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