Suppose \(f\) is entire and there exist \(A, R >0\) and natural number \(N\) such that \[|f(z)| \geq A |z|^N\ \text{for}\ |z| \geq R.\] Show that (i) \(f\) is a polynomial and (ii) the degree of \(f\) is at least \(N\).
Solution
Goal: Suppose \(f\) is entire and \(\abs{f(z)} \ge A\abs{z}^N\) for \(\abs{z} \ge R\). Show (i) \(f\) is a polynomial and (ii) \(\deg f \ge N\).
1. All zeros of \(f\) lie in \(\abs{z} \le R\).
For \(\abs{z} \ge R\), \(\abs{f(z)} \ge A\abs{z}^N > 0\), so no zeros outside the disk; finitely many inside by isolation of zeros on a compact set.
2. \(f\) is a polynomial.
Let \(p\) be the polynomial with exactly the zeros of \(f\) (with multiplicity); \(g = f/p\) is entire and zero-free.
Since \(\abs{f(z)} \ge A\abs{z}^N\) and \(\abs{p(z)} \le C\abs{z}^{\deg p}\) for large \(\abs{z}\), we get \(\abs{1/g(z)} = \abs{p(z)/f(z)} \le \frac{C}{A}\abs{z}^{\deg p - N} \le C'\abs{z}^{\deg p}\) for large \(\abs{z}\) (also bounded on the compact disk \(\abs{z} \le R\)). So \(1/g\) is an entire function of polynomial growth, hence a polynomial (extended Liouville / Cauchy estimates). As \(1/g\) has no zeros, it is constant, so \(g\) is constant and \(f\) is a polynomial.
3. \(\deg f \ge N\).
If \(\deg f = d < N\), then \(\abs{f(z)}/\abs{z}^N \to 0\) as \(\abs{z} \to \infty\), contradicting \(\abs{f(z)} \ge A\abs{z}^N\), i.e. \(\abs{f(z)}/\abs{z}^N \ge A > 0\).
4. Q.E.D.
2 and 3 prove (i) and (ii).