Suppose \(f\) is entire and there exist \(A, R >0\) and natural number \(N\) such that \(|f(z)| \leq A |z|^N\) for \(|z| \geq R\). Show that (i) \(f\) is a polynomial and (ii) the degree of \(f\) is at most \(N\).
Solution
Write the Taylor series of \(f\) at the origin as \[ f(z)=\sum_{k=0}^\infty a_kz^k. \] For any \(\rho\ge R\), Cauchy’s estimate gives \[ |a_k|=\frac{|f^{(k)}(0)|}{k!} \le \frac{M(\rho)}{\rho^k}, \qquad M(\rho):=\max_{|z|=\rho}|f(z)|. \] The hypothesis gives \(M(\rho)\le A\rho^N\), hence \[ |a_k|\le A\rho^{N-k}. \] If \(k>N\), letting \(\rho\to\infty\) yields \(a_k=0\). Thus all Taylor coefficients above degree \(N\) vanish, so \[ f(z)=a_0+a_1z+\cdots+a_Nz^N. \] Therefore \(f\) is a polynomial and \[ \boxed{\deg f\le N} \] unless \(f\equiv0\), in which case the conclusion is immediate.