Entire Function Bounded by Polynomial is Polynomial

Theorem

Let $f : \C \to \C$ be an entire function such that:

$\cmod {\map f z} \le M {\cmod z}^k$

for all $z \in \C$, for some $k \in \N$ and real $M > 0$.

Then $f$ is a polynomial of degree at most $k$.

Proof

Let $r > 0$ be a real number.

Let $D = \map {B_r} 0$ be the open ball with centre $0$ of radius $r$.

By Holomorphic Function is Analytic, we have that:

$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$

for all $z \in \C$, where:

$\ds a_n = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f t} {t^{n + 1} } \rd t$

Note that for $t \in \partial D$, we have:

$\cmod {\map f t} \le M {\cmod t}^k = M r^k$

so for all $t \in \partial D$:

$\cmod {\dfrac {\map f t} {t^{n + 1} } } \le \dfrac {M r^k} {r^{n + 1} }$

From the Estimation Lemma for Contour Integrals::

\(\ds \cmod {a_n}\)

\(=\)

\(\ds \cmod {\frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f t} {t^{n + 1} } \rd t}\)

\(\ds \)

\(\le\)

\(\ds \frac 1 {2 \pi} \times 2 \pi r \times \frac {M r^k} {r^{n + 1} }\)

Estimation Lemma for Contour Integrals

\(\ds \)

\(=\)

\(\ds \frac M {r^{n - k} }\)

Then for $n > k$, we have:

$\dfrac M {r^{n - k} } \to 0$

as $r \to \infty$.

Since $r$ was arbitrary, we have:

$a_n = 0$

for all $n > k$.

That is:

$\ds \map f z = \sum_{n \mathop = 0}^k a_n z^n$

for all $z \in \C$.

That is, $f$ is a polynomial of degree at most $k$.

$\blacksquare$