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This question already has an answer here:

Let $f:\mathbb{C}\to \mathbb{C}$ be an entire function, and suppose that for every $z\in \mathbb{C}$ there exists $n_z\in \mathbb{N}$ such that $f^{(n_z)}(z)=0$.

Is $f$ necessarily a polynomial?

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marked as duplicate by Martin R, mrf, copper.hat, Hans Lundmark, Najib Idrissi May 22 '15 at 7:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ So it seems that there is $N$ and an uncountable set $A$ in $\mathbb C$ so that $f^{(N)}(z) = 0$ for all $z\in A$. $\endgroup$ – user99914 May 22 '15 at 5:12
  • $\begingroup$ Hint: Think about how to apply Baire Category theorem. $\endgroup$ – M. L. Nguyen May 22 '15 at 5:12
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Suppose $f^{(n)}$ does not vanish for all $n\in\mathbb{N}$. Then $\mathbb{C}=\bigcup_{n=1}^\infty (f^{(n)})^{-1}(0)$. But the zeros of $f^{(n)}$ are isolated, hence countable. So the RHS is a countable union of countable sets, which is still countable, while $\mathbb{C}$ is uncountable, contradiction.

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  • $\begingroup$ Why are the zeroes of $f^{(n)} $ isolated? $\endgroup$ – Marco Flores May 22 '15 at 13:59
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If an entire function $f$ satisfies such property, then $\mathbb{C}$ can be covered by countable union of closed sets. To see this, for each natural number $n$, define $B_n:=\{z\in \mathbb{C}:f^{(n)}(z)=0\}$. By the hypothesis, $B_n$ is non-empty. $f$ being entire implies that each $f^{(n)}$ is continuous, so one can easily see that each $B_n$ is a closed subset of $\mathbb{C}$. Now note that by construction, $\mathbb{C}=\cup B_n$. So by Baire Category Theorem, there exists an $n$ such that $B_n$ has a non-empty interior. This means $f^{(n)}$ vanishes on a set that has limit points. Then $f^{(n)}$ being also entire, by the Isolated Zeroes Principles, vanishes on all $\mathbb{C}$. So we can see now that $f$ is a polynomial of degree at most $n$.

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