Show that if \(f\) is entire and \(\abs{f(z)} \in \bigo(\abs{z}^p)\) for \(\abs{z}\) sufficiently large, then \(f\) is a polynomial of degree at most \(\floor{p}\).
Solution
Write \(f(z)=\sum_{k\geq0}c_kz^k\), and choose \(C\) and \(R_0\) with \(\abs{f(\xi)}\le C\abs\xi^p\) for \(\abs\xi\ge R_0\). For \(R\ge R_0\), Cauchy’s integral formula for \(c_k=f^{(k)}(0)/k!\) gives \[ \abs{c_k} &\leq {1\over 2\pi}\int_{\abs{\xi} = R} {\abs{f(\xi)} \over \abs{\xi}^{k+1}}\abs{\dxi}\\ &\leq {1 \over 2\pi}\int_{\abs \xi = R}{ C\abs{\xi}^{p} \over \abs{\xi}^{k+1} }\abs{\dxi} \\ &= {C \over 2\pi} {1\over R^{k+1-p}} \cdot 2\pi R \\ &= C R^{p-k} ,\] which converges to \(0\) as \(R\to \infty\) provided that \(k>p\). So every coefficient \(c_k\) with \(k> p\), that is, \(k\geq \floor{p}+1\), vanishes, and \(f\) is a polynomial of degree at most \(\floor p\).