Let \(f\) be entire and define a function \[ m: [0, \infty) &\to \mathbb{R} \\ r &\mapsto \min_{\abs{z} = r} \abs{f(z)} .\] Suppose \(\tilde m := \lim_{r\to \infty}m(r)\) exists and is a finite positive real number, and show that \(f\) is constant.
Hint: consider \(g(z) := f(1/z)\).
Solution
Let \[ m(r)=\min_{|z|=r}|f(z)|, \qquad m(r)\longrightarrow \widetilde m\in(0,\infty). \]
1. Since \(m(r)\to\widetilde m>0\), there is \(R>0\) such that\[ |f(z)|\ge \widetilde m/2 \qquad(|z|\ge R). \]Thus\[ g(w):=f(1/w) \]is holomorphic on \(0<|w|<1/R\) and bounded away from \(0\) there.
2. The singularity of \(g\) at \(0\) is not essential. Indeed, Casorati–Weierstrass would force values of \(g\) arbitrarily close to \(0\), contradicting the lower bound.
3. The singularity is not a pole either. If \(g\) had a pole of order \(k\ge1\), then\[ w^k g(w)\longrightarrow a\ne0 \qquad(w\to0). \]Hence uniformly on \(|w|=1/r\),\[ |g(w)|\sim |a|r^k, \]so\[ m(r)=\min_{|w|=1/r}|g(w)|\longrightarrow\infty, \]contrary to the assumed finite limit.
4. Therefore \(0\) is removable for \(g\). After extending \(g\) holomorphically across \(0\), it is bounded on a small closed disk about \(0\). Translating back, \(f\) is bounded for all sufficiently large \(|z|\). It is also bounded on the remaining compact disk, so \(f\) is bounded on \(\mathbb C\).
By Liouville’s theorem, \(f\) is constant.