Let \(f(z)\) be entire and assume that \(\abs{f(z)} \leq M |z|^2\) outside of some disk for some constant \(M\). Show that \(f(z)\) is a polynomial in \(z\) of degree \(\leq 2\).
Solution
Prove a more general statement: if \(\abs{f(z)} \leq M\abs{z}^n\), then \(f\) is a polynomial of degree at most \(n\).
Since \(f\) is entire, it is analytic everywhere, so \(f(z) = \sum_{k\geq 0}c_k z^k\) where \(c_k = f^{(k)}(0)/n!\) is given by the coefficient of its Taylor expansion about \(z=0\).
Applying Cauchy’s estimate, on a circle of radius \(R\), \[ \abs{f^{(k)}(0)} \leq { \sup_{\gamma}\abs{f(z)} n! \over R^k} \leq {M\abs{z}^n n! \over R^k} = {M R^n n! \over R^k} .\]
So for \(k \geq n+1\), this goes to zero as \(R\to \infty\), so \(\abs{f^{k}(0)} = 0\) for all such \(k\).
But then \(f\) is a power series annihilated by taking \(n+1\) derivatives, so it is a polynomial of degree at most \(n\).