Suppose that \(f: \mathbb{D} \rightarrow \mathbb{D}\) is holomorphic and \(f(0)=0\). Let \(n \geq 1\), and define the function \(f_{n}(z)\) to be the \(n\)-th composition of \(f\) with itself; more precisely, let
\[ f_{1}(z):=f(z), f_{2}(z):=f(f(z)), \text { in general } f_{n}(z):=f\left(f_{n-1}(z)\right) . \]
Suppose that for each \(z \in \mathbb{D}, \lim _{n \rightarrow \infty} f_{n}(z)\) exists and equals to \(g(z)\). Prove that either \(g(z) \equiv 0\) or \(g(z)=z\) for all \(z \in D\).
Solution
Note that there is a unique fixed point. We have \(f(0) = 0\), so there is at least one, so suppose \(a\) is another fixed point with \(f(a) = a\). By Schwarz, \(\abs{f(z)}\leq \abs{z}\) with equality at any nonzero point implying \(f\) is a rotation, and \(f(a) = a\implies \abs{f(a)} = \abs{a}\), so write \(f(z) = e^{i\theta}z\). Now \(f(a) = a = e^{i\theta }a\) forces \(\theta = 0\), so \(f(z) = z\) is the identity.
Since \(f(0) = 0\), the Schwarz lemma applies and either
\(f(z) = e^{i\theta} z\) is a rotation, or
\(\abs{f'(0)} < 1\) and \(\abs{f(z)} < z\) for all \(z\in \DD\).
Supposing the latter, \(f\) is a contraction, and \(\abs{f_{n+1}(z)} < \abs{f_{n}(z)}\) for all \(n\) and all \(z\), so \(\abs{f_n(z)} \convergesto{n\to\infty} 0\) for all \(z\). Since \(f_n\to g\) pointwise, this means \(g(z) = 0\) for all \(z\), making \(g\equiv 0\).
Otherwise, suppose \(f\) is a rotation. Then if \(f(z) = e^{i\theta}z\), \(f_n(z) = e^{in\theta}z\). The pointwise limit \(\lim_{n\to\infty}e^{in\theta}z\) can only exist if \(\theta = 0\), otherwise this is periodic when \(\theta\) is rational or the points \(e^{i\theta}z, e^{2i\theta }z,\cdots\) form form a countably infinite set of distinct points. So \(f(z) = z\), making \(\lim_{n\to \infty}f_n(z) = z\) as well.