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 Triangle Inequality for Contour Integrals, we have:

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 $k$.