Uniform convergence of n-fold composition using Schwarz lemma

Question

Let $f$ be an analytic function mapping the unit disk $\mathbb D$ to itself with $f(0) = 0$ and $|f'(0)| < 1$. Let $f^{n} = f \circ f \circ \dots \circ f$ be the function obtained by composing $f$ with itself $n$-times. Prove that $f^n \rightarrow 0$ uniformly on compact subsets of $\mathbb D$.

My attempt:

Every compact subset will be contained in a closed ball $\overline B(0, R)$ with $R < 1$, so it suffices to show uniform convergence on $\overline B(0,R)$. By the maximum modulus principle and Schwarz lemma,

$|f(z)| \leq R$

for every $z \in \overline B(0, R)$. Moreover, by Schwarz lemma, the sequence $f^n(z)$ is decreasing for every $z \in \overline B(0,R)$. The issue is that the bound of the sequence $R$ does not decrease to 0 as $n \rightarrow \infty$ . I'm guessing I need to use the condition $|f'(0)| < 1$ to get a tighter bound, but I don't know where to go from here.

Answer 1

Hint: In fact, the Schwarz Lemma implies that $|f(z)| < |z|$ for $z\ne 0,$ unless $f$ is a rotation.

Answer 2

As you already observed yourself, $|f^n(z)|$ decreases in $n$ for each fixed $z\in D$. Moreover, $f^n(z)$ will approach a limit along a subsequence, and this limit can only be zero, by the same argument (and thus passing to a subsequence was unnecessary).

So $f^n\to 0$ pointwise, but these functions also form a normal family, so in fact $f^n\to 0$ locally uniformly.