Let \(f\) be a holomorphic self-map of the unit disk which is not the identity. Show that \(f\) has at most one fixed point.
Solution
Suppose, for contradiction, that \(f\) has two distinct fixed points \[ a,b\in\mathbb D. \] Let \[ \phi_a(z)=\frac{z-a}{1-\overline a z}, \] the disk automorphism sending \(a\) to \(0\). Define \[ g=\phi_a\circ f\circ\phi_a^{-1}. \] Then \(g:\mathbb D\to\mathbb D\) is holomorphic and \[ g(0)=0. \]
1. The second fixed point gives a nonzero fixed point of \(g\).
Set \[ c=\phi_a(b). \] Since \(a\ne b\), we have \(c\ne0\). Also \[ \begin{aligned} g(c) &=\phi_a\bigl(f(\phi_a^{-1}(c))\bigr)\\ &=\phi_a(f(b))\\ &=\phi_a(b)\\ &=c. \end{aligned} \] Thus \(g\) fixes both \(0\) and the nonzero point \(c\).
2. Schwarz’s lemma forces \(g\) to be the identity.
Since \(g\) is a holomorphic self-map of the disk with \(g(0)=0\), Schwarz’s lemma gives \[ |g(z)|\le|z| \] for every \(z\in\mathbb D\). At the nonzero point \(c\) from step 1, \[ |g(c)|=|c|. \] The equality case of Schwarz’s lemma therefore implies \[ g(z)=e^{i\theta}z \] for some real \(\theta\). Since \(g(c)=c\) and \(c\ne0\), \[ e^{i\theta}=1. \] Hence \[ g(z)=z \] throughout \(\mathbb D\).
3. Conclude the contradiction.
From \[ g=\phi_a\circ f\circ\phi_a^{-1}=\operatorname{id}_{\mathbb D}, \] we obtain \[ f=\phi_a^{-1}\circ g\circ\phi_a =\operatorname{id}_{\mathbb D}, \] contrary to the hypothesis. Therefore a nonidentity holomorphic self-map of the unit disk has at most one fixed point.