- Let U be a bounded domain in C, and let \(f : U \to U\) such that f is holomorphic. Let \(P \in U\) and suppose that \(f ( P ) = P\) . Prove that \(| f ^ { \prime } ( P ) | \leq 1\)
Hint
Consider the iterates \(f_n=f\circ\cdots\circ f\) with \(n\) factors.
Solution
1. The derivatives of all iterates at \(P\) have a common upper bound.
Since \(U\) is open and contains \(P\), choose \(r>0\) with \(\overline{D(P,r)}\subset U\). Boundedness of \(U\) gives a constant \(M>0\) such that \(|z|\leq M\) for every \(z\in U\). For \(n\geq1\), the iterate \(f_n=f^{\circ n}\) is holomorphic on \(U\), maps \(U\) to itself and fixes \(P\), by induction using \(f(U)\subset U\) and \(f(P)=P\). Thus \(|f_n|\leq M\) on the same circle \(|z-P|=r\) for every \(n\). Cauchy’s derivative formula gives \[ f_n'(P)=\frac{1}{2\pi i}\int_{|z-P|=r} \frac{f_n(z)}{(z-P)^2}\,dz, \qquad |f_n'(P)|\leq\frac Mr \] [@SS03]. Neither the radius nor this bound depends on \(n\).
2. The chain rule forces \(|f'(P)|\leq1\).
Let \(a=f'(P)\). Because every iterate fixes \(P\), the chain rule gives \(f_{n+1}'(P)=f'(P)f_n'(P)\). Starting with \(f_1'(P)=a\), induction yields \(f_n'(P)=a^n\). Step 1 therefore implies \[ |a|^n\leq M/r,\qquad |a|\leq(M/r)^{1/n} \quad(n\geq1). \] Since \(M/r\) is a fixed positive number, its \(n\)th root tends to one. Passing to the limit proves \(|f'(P)|\leq1\).