Compact convergence of a series of functions

Question

Let $D$ be the unit disk on the complex plane. Suppose $f:D\to D$ is holomorphic and $f(0)=0$.

Let $f_1=f$ and $f_{n+1}=f\circ f_n$ for $n=1,\:2,...$ If $|f'(0)|<1$, show that $\sum_{n=1}^{\infty}f_n(z)$ converges uniformly on compact sets of $U$.

My attempt: By Schwarz lemma, I got $|f(z)|\le |z|$, and so, by induction, $|f_n(z)|\le|z|$ for all $n$.

However, I cannot apply Weierstrass M-test directly here to prove uniform convergence on a compact set. Can I refine the bound further using the condition $|f'(0)|<1$? Or should I take different approach?

Any hints or advices will help a lot! Thanks!

Answer 1

Hint: Consider $D(0, r)$ where $r<1$. By Schwarz's lemma, we see that \begin{align} \sup_{z\in \overline{D(0, r)}} \frac{|f(z)|}{|z|}=\alpha<1. \end{align} Next, observe \begin{align} \frac{|f_n(z)|}{|z|} = \frac{|f_n(z)|}{|f_{n-1}(z)|}\cdot \frac{|f_{n-1}(z)|}{|f_{n-2}(z)|}\cdots \frac{|f(z)|}{|z|} \leq \alpha^n \ \ \implies \ \ |f_n(z)| \leq \alpha^n r. \end{align}

Answer 2

Let $0 < r < 1$ and let $\alpha_r = \max_{|z| \le r} \frac{|f(z)|}{|z|}$. By Schwarz lemma, using $|f'(0)| < 1$ and $\{z : |z| \le r\}$ is compact we have $\alpha_r < 1$. We can now prove by induction that $|f_n(z)| \le (\alpha_r)^n |z|$. this proves the normal convergence and the uniform convergence.