Suppose \(f:\DD\to \DD\) with \(f(a) = a\) a fixed point where \(\abs{f'(a)} < 1\). Show that for any initial point \(z_0\), the sequence \(z_k \da f(z_{k-1})\) converges to \(a\).
Solution
First suppose \(a=0\) – then Schwarz applies, and since \(\abs{f'(a)} < 1\) is strict, \(f\) is not a rotation.
For any choice of \(z_0\in \DD\), there is an \(r\) with \(0< \abs{z_0} < r < 1\) and a constant \(C<1\) such that \(\abs{f(z)} \leq C\abs{z}\) for \(\abs{z} < r\).
With such an \(r\) and \(C<1\) in hand,
\[ \abs{z_k} = \abs{f(z_{k-1})} \leq C\abs{z_{k-1}} = C\abs{f(z_{k-2})} \leq C^2 \abs{z_{k-2}} \cdots \implies \abs{z_k} \leq C^k\abs{z_0} \convergesto{k\to\infty}0 ,\] which proves the \(a=0\) case.
(That $f$ is a contraction)
The claim is that for any given \(r\), the constant \(C\da M/r\) works, where \(M\da \max_{\abs{z} = r} \abs{f(z)}>0\). The scaled Schwarz lemma gives \(\abs{f(z)}\leq {M\over r}\abs{z} = C\abs{z}\), and \(\abs{C} \leq 1\) since \(\abs{M} \leq r\), which follows because \(\abs{f(z)}\leq \abs{z}\) on \(\DD\) itself.
For \(a\neq 0\), take a Blaschke factor \(\psi_a(z)\) and consider \(F \da \inverseof{\psi_a} \circ f\circ \psi_a\). The claim is that this reduces to the case \(a=0\).
Note \(F(0) = 0\), so \(0\) is a fixed point of \(F\). Moreover, a clever calculation shows \[ F'(0) &= (\inverseof{\psi_a})'(f(\psi_a(0))) \cdot f'(\psi_a(0)) \cdot \psi_a'(0) \\ \\ &= (\inverseof{\psi_a})'(f( a )) \cdot f'(a ) \cdot \psi_a'(0) \\ \\ &= (\inverseof{\psi_a})'(a) \cdot f'(a ) \cdot \psi_a'(0) \qquad \text{since } f(a) = a \\ \\ &= (\psi_a)'(a) \cdot \psi_a'(0) \cdot f'(a) \qquad \text{since } \inverseof{\psi_a} = \psi_a \\ \\ &= (\psi_a)'( \psi_a(0) ) \cdot \psi_a'(0) \cdot f'(a) \\ \\ &= (\psi_a \circ \psi_a)'(0) \cdot f'(a) \\ &= 1 \cdot f'(a) ,\] so \(\abs{F'(0)} = \abs{f'(a)} < 1\). Now setting \(w_k \da \psi_a(z_n)\) and writing \(f = \psi_a \circ F \circ \inverseof{\psi_a}\), by continuity we have \[ f(z_k) = \psi_a(F(w_k)) \convergesto{k\to\infty} \psi_a(0) = a .\]