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}}


\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)


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