Let \(G \subset \mathbb{C}\) be a bounded, simply connected region and let \(f\) be an analytic self-map of \(G\) (i.e. \(f(G) \subset G\)). Assume that \(f\) has two fixed points. Show that \(f(z) = z\).
Solution
By the Riemann mapping theorem choose a biholomorphism \(\phi:G\to\mathbb D\). Then \(F=\phi\circ f\circ\phi^{-1}\) is a holomorphic self-map of \(\mathbb D\) with two distinct fixed points \(\alpha,\beta\in\mathbb D\).
Let \(T\) be a disk automorphism with \(T(\alpha)=0\). Then \(H=T\circ F\circ T^{-1}\) fixes \(0\) and the nonzero point \(T(\beta)\). Schwarz’s lemma gives \(|H(z)|\le |z|\). Since equality holds at the nonzero fixed point, the equality case gives \(H(z)=e^{i\theta}z\). That same fixed point forces \(e^{i\theta}=1\). Hence \(H\), then \(F\), then \(f\), is the identity.