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let $ f(z) \in Hol(\mathbb{C}) $ holomorphic function such that for each $ z_0 \in \mathbb{C} $ there exists $ N(z_0) $ such that $ f^{(N(z_0))}(z_0) = 0 $
Prove: that f is a polynom

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2 Answers 2

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Modifying Davide's argument but avoiding Baire's category theorem: Let $\overline{\mathbb E}$ be the closed unit ball and $$F_n := \{ z \in \overline{\mathbb E} : f^{(n)}(z) = 0 \}.$$ Then, by assumption, $\overline{\mathbb E} = \bigcup_n F_n$. Since $\overline{\mathbb E}$ is uncountable, not all $F_n$ can be finite. Let $n \in \mathbb N$ s.t. $F_n$ is infinite. Since $\overline{\mathbb E}$ is compact, $F_n$ has a limit point in $\overline{\mathbb E}$, i.e. there is a $z_0 \in \overline{\mathbb E}$ and a sequence $(z_k)$ in $F_n \setminus \{z_0\}$ that converges to $z_0$. We have $f^{(n)}(z_k) = 0$ for all $k$, hence $f^{(n)} = 0$ by the identity theorem, so $f$ is a polynomial.

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+1 Damn, that was beautiful! –  DonAntonio Aug 2 '12 at 21:06
    
$f^n=0$ means $f$ is a polynomial? how? –  Une Femme Douce Aug 3 '12 at 4:15
    
$f^{(n)}$ means $f$ power n or what? –  Une Femme Douce Aug 3 '12 at 4:27
    
$f^{(n)}$ denotes the $n$-th derivative of $f$. I assumed that you know that since you used that notation in your question. To see that $f^{(n)} = 0$ implies that $f$ is a polynomial just look at the Taylor expansion of $f$. –  marlu Aug 3 '12 at 15:19

Let $F_n:=\{z\in\Bbb C, f^{(n)}(z)=0\}$. Since $f^{(n)}$ is continuous, $F_n$ is closed and by hypothesis, $\Bbb C=\bigcup_{n\in\Bbb N}F_n$. By Baire's theorem, there exists a $n_0$ such that $F_{n_0}$ has a non-empty interior. Hence $f^{(n_0)}$ vanishes on a ball, and it's necessarily the null function. This proves that $f$ is a polynomial.

Note that there is a similar result for $C^{\infty}$ functions from $\Bbb R$ to $\Bbb R$, but it's harder to show. It has been discussed here.

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is there a proof without baire theorem? i didnt learn that in the complex analysis course –  bar Jul 31 '12 at 11:33
    
And what did you learn? In which chapter such an exercise appear? Actually, Baire is more a theorem from topology than complex analysis. –  Davide Giraudo Jul 31 '12 at 11:38
    
i think the relevant theorems for this exercise are: uniqeness theorem: if $ f\in Hol(D) $ D is a domain and there is U a subset of D such that U has accumulation point and $ \forall z \in U f(z)=0 $ then $ \forall z\in D f(z) = 0 $ –  bar Jul 31 '12 at 11:43

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