Let \(f:D\to D\) be holomorphic on the unit disk, with \(f(0)=0\), and suppose \(f\) is not a rotation \(z\mapsto e^{i\theta}z\). For \(w\in D\), set \(w_0=w\) and \(w_{n+1}=f(w_n)\). Prove that \(w_n\to0\).
Solution
1. The quotient \(f(z)/z\) is strictly bounded by one on each compact subdisk.
Because \(f(0)=0\), the quotient extends holomorphically to \(h\) on \(D\), with \(h(0)=f'(0)\). Schwarz’s lemma and continuity give \(|h|\leq1\) on all of \(D\) [@SS03]. If equality held at any point, the maximum modulus principle would make \(h\) a constant of modulus one, so \(f\) would be a rotation. Therefore \(|h(z)|<1\) at every point. On \(|z|\leq r<1\), continuity and compactness give a maximum \(M_r<1\).
2. The iterates decay geometrically to zero.
If \(w=0\), all iterates are zero. Otherwise let \(r=|w|<1\) and choose \(q=(1+M_r)/2\), so \(0<q<1\). Schwarz’s lemma gives \(|w_{n+1}|\leq|w_n|\), so the whole orbit stays in \(|z|\leq r\). On this disk step 1 implies \[ |w_{n+1}|=|w_n|\,|h(w_n)|\leq q|w_n|. \] Induction yields \(|w_n|\leq q^n|w|\) for all \(n\geq0\). Since \(q^n\to0\), the required limit follows.