Let \(f\) be entire and suppose that \(\lim_{z \rightarrow \infty} f(z) = \infty\). Show that \(f\) is a polynomial.
Solution
Note that \(f\) has finitely many zeros: since \(f\) is unbounded, there is some \(R\) such that \(f(\DD_R^c) \subseteq \DD^c\), so in particular \(f\) is nonvanishing on \(\DD_R^c\). So \(Z_f\) is a closed subset of a compact set, so is either finite or has an accumulation point. In the latter case, \(f\equiv 0\) by the identity principle, so suppose not.
Write \(Z_f = \ts{z_k}_{k\leq n}\) for the \(n\) many zeros of \(f\), included with multiplicity, and set \[ \Phi(z) \da \prod_{k\leq n} (z-z_k), \qquad F(z) \da {\Phi(z) \over f(z) } .\] Now \(F\) is a nonvanishing entire function.
\(F\) is bounded on \(\CC\).
(of claim)
Choose \(R\gg 1\) so that all of \(z_k\) are in \(\DD_R\), so \(\abs{\xi - z_k} < R\) for all \(\xi \in \DD_R\) and all \(k\). By Cauchy’s integral formula, \[ \abs{F(z)} &\leq {1\over 2\pi} \oint_{\abs{\xi} = R} \abs{F(\xi) \over \xi} \dxi \\ &={1\over 2\pi} \oint_{\abs{\xi} = R} \abs{\Psi(\xi) \over f(\xi) \cdot \xi} \dxi \\ &\leq {1\over 2\pi} \oint_{\abs{\xi} = R} {R^m \over \abs{ f(\xi) } R } \dxi \\ &\leq {1\over 2\pi} \oint_{\abs{\xi} = R} R^{m-1} \dxi \\ &= R^m ,\] where \(R\) is increased if necessary to ensure \(\abs{1\over f(z)} < 1\), which can be done since \(\abs{f(z)}\to \infty\) as \(R\to \infty\). So \(\abs{F(z)} \leq C\abs{z}^m\) in \(\bar{\DD_R}^c\) for \(R\) large enough, making \(F\) a polynomial of degree at most \(m\). Since \(F\) is continuous in \(\bar{\DD_R}\) which is compact, \(F\) is bounded in here as well, making \(F\) bounded on all of \(\CC\).
Given this, \(F\) is entire and bounded and thus constant by Liouville. So \(F(z) = c\), and as a result \(f(z) = c\Phi(z)\) which is a polynomial of degree \(n\).