Entire function bounded by a polynomial is a polynomial Suppose that an entire function $f(z)$ satisfies $\left|f(z)\right|\leq k\left|z\right|^n$ for sufficiently large $\left|z\right|$, where $n\in\mathbb{Z^+}$ and $k>0$ is constant. Show that $f$ is a polynomial of degree at most $n$.
 A: Hints:


*

*We have by Cauchy's integral formula that 
$$|f^{(d)}(0)|=\frac{d!}{2\pi R}\left|\int_{C(0,R)}\frac{f(z)}{z^{d+1}}dz\right|.$$

*What about $f^{(d)}(0)$ if $d\geq n+1$?

*Use the fact that $f$ is analytic at $0$ to get that $f(z)=\sum_{j=0}^n\frac{f^{(j)}(0)}{j!}z^j$ in a neighborhood of $0$. 

*Show that the last formula is in fact true for all $z\in\Bbb C$. 

A: Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $\infty$, which must match its Taylor series there. 
$$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$$
Since $|f(z)|\leq k|z|^m$, Cauchy's estimate gives
$$|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}$$
for all $|z|=R$. For $n>m$, letting $R\rightarrow\infty$, we see that $|f^{(n)}|=0$. It follows that $f$ is a polynomial of degree $\leq m$.
A: Look at closed discs centered at the origin, use maximum modulus principle to show that the function obtains its maximum value on the boundary, show that if you take a larger disc, you obtain a higher value, and thus use Liouville's Theorem to get that $\lim_{|z| \to \infty} |f(z)| = \infty$.  Then show that such a function is a polynomial.
A: A really dirty way to do this:  


*

*Theorem 1: Jensen's Formula Corollary
Suppose $f$ has order of growth $\rho$.
Then there is a $C$ that for large enough $R$, $n(R) \le C R^{\rho} $ where $n(R)$ is the number of zeros whose magnitude is less than $R$. 

*Theorem 2: Hadamard Factorization Theorem
Suppose $f$ has order of growth $k \le \rho \lt k+1$ where $k$ is an integer. Then $f(z)$ can be written $z^m e^{g(z)} \prod_n E_k(z/a_n)$ where $E_k$ is the kth canonical Weirerstrass factor and $a_n$ is the nth zero of $f$ and $g(z)$ is a polynomial of degree $k$.  
Then note that by assumption $f$ has zero order of growth. Theorem 1 it follows that $f$ has finitely many zeros. From Theorem 2 it follows that $f$ is a polynomial. Then we need to put in a tiny bit of work to show that the degree of this polynomial is the one we need. (Just argue about $|f(z)/z^n|$ as $z$ grows large)
A: Here's an attempt by contradiction. Suppose $f(z)$ is a polynomial of degree $n+1$ and satisfies $|f(z)| \leq k|z|^n$ for some constant $k > 0$.
$$ f(z) = a_0 + a_1 z + \dots + a_{n+1} z^{n+1}$$
By the Fundamental Theorem of Algebra, $f(z)$ has $n+1$ roots and can be written as
$$ f(z) = (z-z_1)(z-z_2)\dots(z-z_{n+1}) g(z)$$
where $z_i, i = 1, 2, \dots, n+1$, are roots of $f(z)$ and $g(z) \neq 0$ for all $z$.
Consider $w \in \mathbb{C}$ which is not a root of $f(z)$.
\begin{align}f(w) &= (w-z_1)\dots(w-z_{n+1})g(w) \\
&= (w^{n+1} + c_1 w^{n} + \dots + c_n \prod_{i=1}^{n+1} z_i)g(w)\\
\end{align}
Then
\begin{align}
|f(w)| &= |w^{n+1} + \dots c_n \prod_{i=1}^{n+1}z_i||g(w)|\\
&= |w^{n+1} g(w) + \dots + c_n \prod_{i=1}^{n+1}z_i g(w)|\\
&\leq k|w|^n
\end{align}
This implies $k \geq |w g(w) + \dots + c_n \prod_{i=1}^{n+1}z_i|$. In other words, $k$ is dependent on $w$ and not a constant which contradicts the assumption, so $f(z)$ is a polynomial of at most degree $n$.
