Let \(f\) be an entire function. Assume that for some \(k \in \mathbb{N}\), and sufficiently large \(|z|\), we have that \(|f(z)| \leq A+B|z|^{k}\). Prove that \(f\) is a polynomal of degree at most \(k\).
Solution
Induct on \(k\ge0\). For \(k=0\), \(\abs f\le A+B\) outside a disk and \(f\) is bounded on the disk by continuity, so \(f\) is constant by Liouville.
For \(k\geq1\), assume the statement for \(k-1\) and let \(\abs{f(z)}\le A+B\abs z^k\) for \(\abs z\ge R\). Consider \[ g(z) \da \begin{cases} {f(z) - f(0) \over z} & z\neq 0, \\ f'(0) & z=0, \end{cases} \] which is entire. For \(\abs z\geq\max(R,1)\), \[ \abs{g(z)} \leq {A+\abs{f(0)}+B\abs{z}^k \over \abs{z}} \leq \qty{A+\abs{f(0)}}+B\abs{z}^{k-1} .\] By the induction hypothesis \(g\) is a polynomial of degree at most \(k-1\), so \(f(z)=f(0)+zg(z)\) is a polynomial of degree at most \(k\).