Suppose that \(U\) is a bounded, open and simply connected domain in \(\mathbb{C}\) and that \(f(z)\) is a complex-valued non-constant continuous function on \(\bar{U}\) whose restriction to \(U\) is holomorphic.
- Prove the maximum modulus principle by showing that if \(z_{0} \in U\), then
\[ \left|f\left(z_{0}\right)\right|<\sup \{|f(z)|: z \in \partial U\} . \]
- Show furthermore that if \(|f(z)|\) is constant on \(\partial U\), then \(f(z)\) has a zero in \(U\) (i.e., there exists \(z_{0} \in U\) for which \(f\left(z_{0}\right)=0\) ).
Solution
Let \(M\da \sup_{z\in \bd U}\abs{f(z)}\). If \(M=0\), then \(f\) must be the constant zero function, so assume \(M>0\).
Suppose toward a contradiction that there exists a \(z_0 \in U\) with \(\abs{f(z_0)} = M\). Note that the map \(z\mapsto \abs{z}\) is an open in discs that don’t intersect \(z=0\). Since \(f\) is holomorphic, by the open mapping theorem \(f\) is an open map, so consider \(D_\eps(z_0)\) a small disk not containing \(0\). Then \(f(D_\eps(z_0))\) is open, and the composition \(z\mapsto f(z) \mapsto \abs{f(z)}\) is an open map \(D_\eps(z_0)\to \RR\). Now if \(f\) is nonconstant, \(\abs{f(D_\eps(z_0))} \contains (M-\eps, M+\eps)\) contains some open interval about \(M\), which contradicts maximality of \(f\) at \(z_0\).
See notes for a proof using the mean value theorem.
Suppose toward a contradiction that \(f\) has no zeros in \(U\). Then \(g(z) \da 1/f(z)\) is holomorphic in \(U\). Now if \(\abs{f(z)} = C\) on \(\bd U\), we have \(\abs{g(z)} = \abs{1\over f(z)} = {1\over C}\) on \(\bd U\), so \(\max_{z\in U} \abs{f(z)} = C\) and \(\min_{\abs{f(z)}} = {1\over C}\). Since \(\abs{f(z)}\) is constant on the boundary, we must have \(\max \abs{f(z)} = \min \abs{f(z)} = C\), so \(f\) is constant on \(\bd U\). By the identity principle, \(f\) is constant on \(U\), a contradiction.