Let \(\Omega\) be a simply connected open set and let \(\gamma\) be a simple closed contour in \(\Omega\) and enclosing a bounded region \(U\) anticlockwise. Let \(f: \ \Omega \to {\mathbb C}\) be a holomorphic function and \(|f(z)|\leq M\) for all \(z\in \gamma\). Prove that \(|f(z)|\leq M\) for all \(z\in U\).
Solution
Theorem.
If \(f:\Omega\to\mathbb C\) is
holomorphic, \(\gamma\) is a simple
closed contour in \(\Omega\), and \(|f|\le M\) on \(\gamma\), then \(|f(z)|\le M\) for every \(z\in U\).
Proof.
- Let \(\overline U\) be the closure of \(U\) together with \(\gamma\). Since \(\gamma\subset\Omega\) is a simple closed contour, \(\overline U\subset\Omega\) and \(f\) is holomorphic on an open neighborhood of \(\overline U\).
- Assume for contradiction that there exists \(z_0\in U\) with \(|f(z_0)|>M\).
- Define \(g(z)=\dfrac{f(z)}{f(z_0)}\). Then \(g\) is holomorphic on a neighborhood of \(\overline U\), and \(|g(z)|\le 1\) for all \(z\in\gamma\).
- By the maximum modulus principle, \(|g(z)|\le 1\) for all \(z\in U\).
- We have \(|g(z_0)|=1\), so \(|g|\) attains its maximum at an interior point. By the strong form of the maximum modulus principle, \(g\) is constant on \(U\), hence on \(\overline U\).
- Then \(f\equiv f(z_0)\) on \(U\), so \(|f|\equiv |f(z_0)|>M\) on \(\gamma\), contradicting the boundary bound.
Therefore no such \(z_0\) exists and \(|f(z)|\le M\) for all \(z\in U\).