# Ngô Quốc Anh

## April 4, 2010

### Growth of an entire function

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247) — Ngô Quốc Anh @ 16:04

The classical Liouville theorem says that

Theorem (Liouville). Every bounded entire function must be constant. That is, every holomorphic function $f$ for which there exists a positive number $M$ such that $|f(z)| \leqslant M$  for all $z$ in $\mathbb C$ is constant.

The aim of this entry to is generalize the constant $M$, precisely, what happen if we replace $M$ by a polynomial?

Followed by this topic we can prove the following theorem.

Theorem (Generalized Liouville). Assume $f$ is an entire function. If $|f(z)| \leqslant A+B|z|^p$ for all $z \in \mathbb C$ with some positive numbers $A,B,p$, then $f$ is a polynomial of degree bounded by $p$.

Proof. Denote the Laurent expansion of $f$ by

$\displaystyle f\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {{a_n}{z^n}}$

where

$\displaystyle {a_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f\left( z \right)dz}} {{{z^{n + 1}}}}}$.

Since $f$ is entire, all coefficients $a_n$ with $n<0$ are zero, i.e.

$\displaystyle f\left( z \right) = \sum\limits_{n =0 }^{ + \infty } {{a_n}{z^n}}$.

For any integers $n>p$, one has

$\displaystyle\left| {{a_n}} \right| \leqslant \frac{1}{{2\pi }}\int\limits_{\left| z \right| = r < 1} {\left| {\frac{{f\left( z \right)}}{{{z^{n + 1}}}}} \right|dz} = \frac{1}{{2\pi {r^n}}}\int_0^{2\pi } {f(r{e^{i\theta }})d\theta } \leqslant \frac{{A + B{r^p}}}{{2\pi {r^n}}}$.

Letting $r\to \infty$ we obtain that $a_n=0$ which implies that $f$ is a polynomial of degree at most $p$.

Corollary. For a given entire function $f$, if the following limit

$\displaystyle\mathop {\lim }\limits_{|z| \to \infty } \frac{{f(z)}}{{|z{|^p}}}$

exists then $f$ is a polynomial of degree at most $p$.

Question. What happen if

$\displaystyle\mathop {\lim }\limits_{|z| \to \infty } |z{|^p}f(z)$

exists?