Use a version of the Schwarz lemma to prove Liouville’s theorem.
Solution
Suppose \(f\) is entire and bounded, we’ll show \(f\) is constant. If \(f\) is bounded by \(M\), then \(f(\CC) \subseteq \DD_M(0)\). Without loss of generality, replace \(f\) with \(g(z) \da f(z) - f(0)\), so \(g(0) = 0\) and is still bounded by \(M' \da M + \abs{f(0)}\) by the triangle inequality. This is still finite since \(0\) is not a singularity since \(f\) is entire.
By the radius \(R\) variant of the Schwarz lemma, for every \(\DD_R(0)\), \[ \abs{g(z)} \leq {M\over R}\abs{z} \qquad \text{for } z\in \DD_R(0) .\] Using that \(g(0) = 0\), \[ \abs{g(z) - g(0) \over z} \leq {M\over R}\convergesto{R\to\infty}0 ,\] where dividing by \(z\) is not an issue since \(z=0\) is a zero of \(g\) of at least order one. This forces \(g(z) = g(0) = 0\) for all \(z\), so \(f(z) = f(z_0)\) is constant.