Let \(f : D \to D\) be a holomorphic function where \(D = \{z \in \mathbb{C} : |z| < 1\}\) is the unit disk. Prove that if \(f\) has at least 2 fixed points then \(f\) is the identity map. (Note: A point \(a\) is said to be a fixed point of \(f\) if \(f(a) = a\).)
Solution
Choose distinct fixed points \(a,b\in D\).
1. Conjugation moves the two fixed points to zero and a nonzero point.
Set \[ \phi_a(z)=\frac{z-a}{1-\overline a z}, \qquad \phi_a^{-1}(w)=\frac{w+a}{1+\overline a w}. \] The denominators do not vanish on \(D\). Direct substitution verifies that these maps are inverse, and \[ 1-|\phi_a(z)|^2 =\frac{(1-|a|^2)(1-|z|^2)}{|1-\overline a z|^2}>0 \] for \(z\in D\). Applying the same formula to \(-a\) shows that the inverse also maps \(D\) into itself. Thus \(\phi_a\) is an automorphism of \(D\) with \(\phi_a(a)=0\).
The function \(F=\phi_a\circ f\circ\phi_a^{-1}\) maps \(D\) holomorphically into \(D\). It fixes zero and also \(\beta=\phi_a(b)\), which is nonzero because \(b\ne a\).
2. The second fixed point forces \(F\), and hence \(f\), to be the identity.
Schwarz’s lemma gives \(|F(z)|\leq|z|\) [@SS03]. The quotient \(H(z)=F(z)/z\) extends holomorphically to zero with value \(F'(0)\), by the Taylor expansion of \(F\). It satisfies \(|H|\leq1\) on \(D\), including zero by continuity. At the nonzero fixed point \(\beta\), \(H(\beta)=F(\beta)/\beta=1\). The maximum modulus principle makes \(H\) constant, and its value there makes that constant one [@SS03]. Thus \(F(z)=z\) throughout \(D\). Undoing the conjugation gives \(f=\phi_a^{-1}\circ\operatorname{id}_D\circ\phi_a =\operatorname{id}_D\).