# holomorphic function on the complex plane

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

-

## 2 Answers

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.

-

+1 Damn, that was beautiful! – DonAntonio Aug 2 '12 at 21:06
$f^n=0$ means $f$ is a polynomial? how? – Taxi Driver Aug 3 '12 at 4:15
$f^{(n)}$ means $f$ power n or what? – Taxi Driver 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.

-

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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9520666003227234, "perplexity_flag": "head"}