- A complex number \(w \in \mathbb { D }\) is a fixed point for the map \(f : \mathbb { D } \to \mathbb { D } { \mathrm { i f ~ } } f ( w ) = w\)
Prove that if \(f : \mathbb { D } \to \mathbb { D }\) is analytic and has two distinct fixed points, then f is the identity, that is, \(f ( z ) = z { \mathrm { ~ f o r } }\) all \(z \in \mathbb { D }\)
Must every holomorphic function \(f : \mathbb { D } \to \mathbb { D }\) have a fixed point? [Hint: Consider the upper half-plane.]
Solution
For part (a), let \(a,b\in\mathbb D\) be distinct fixed points of \(f\). Choose a disc automorphism \(\phi\) with \(\phi(a)=0\) and define \[ g=\phi\circ f\circ\phi^{-1}. \] Then \(g:\mathbb D\to\mathbb D\) is holomorphic, \(g(0)=0\), and \(c=\phi(b)\ne0\) is another fixed point. Schwarz’s lemma gives \(|g(z)|\le|z|\). Since \(|g(c)|=|c|\) for \(c\ne0\), the equality case of Schwarz’s lemma implies \[ g(z)=e^{i\theta}z \] for some real \(\theta\). But \(g(c)=c\) and \(c\ne0\), so \(e^{i\theta}=1\). Hence \(g\) is the identity, and therefore so is \(f\).
For part (b), no. Let \[ \phi(z)=\frac{z-i}{z+i} \] map the upper half-plane biholomorphically onto \(\mathbb D\), and let \(T(z)=z+1\), an automorphism of the upper half-plane with no fixed point there. Then \[ F=\phi\circ T\circ\phi^{-1}:\mathbb D\to\mathbb D \] is a holomorphic self-map. If \(F(w)=w\), then applying \(\phi^{-1}\) would give a fixed point of \(T\), impossible. Thus a holomorphic self-map of the disc need not have a fixed point.