Let $U\subset\mathbb C$ be a bounded region (i.e., open and connected) and $f$ be a holomorphic function on $U$, with $f(U)\subset U$. Denote by $f_{n} = f \circ f \circ · · · \circ f $.
Suppose $z_{0} \in U$, be such that $f (z_{0} ) = z_{0}$ and $\lvert\, f' (z_{0} )\rvert \lt 1$.
Show that the sequence $\{f_{n}\}_{n\in\mathbb N}$ converges uniformly on compact subsets of $U$ to the constant function $h(z)=z_{0}$.
It this an application of Schwartz lemma or Hurwitz theorem?
Any hint is welcome.
Without loss of generality, and for simplicity, set $z_0=0$.
If we set $$ g(z)=\frac{f(z)}{z}, \quad z\in U\smallsetminus\{0\}, $$ then $g$ extends as a holomorphic function in $U$, and $g(0)=f'(0)$, and as $\lvert\, f'(0)\rvert<1$, then for $$ a=\frac{\lvert\, f'(0)\rvert+1}{2}<1, $$ there is an $r_0>0$, such that $\lvert g(z)\rvert\le a$, and hence $$ \lvert\,f(z)\rvert\le a\lvert z\rvert, $$ for all $|z|\le r_0$, which in turn implies that $$ \lvert\,f_{n}(z)\rvert\le a^n \lvert z\rvert\le a^n r_0\longrightarrow 0 \tag{1} $$ uniformly in the closed disc $\overline D_{r_0}=\{z: |z|\le r_0\}$.
In order to show that $f_{n}\to 0$, locally uniformly in $U$, it suffices to show that every subsequence of $\{f_n\}$ contains a sub-subsequence which converges to $0$ locally uniformly in $U$.
Let $\{h_n\}$ a subsequence of $\{f_n\}$ and $K\subset U$ compact.
Then, as $U$ is bounded we have that $U\subset D(0,R)$, for some $R>0$, and hence $\lvert \,f(z)\rvert,\,\lvert\,f_n(z)\rvert\le R$, and let $d=\mathrm{dist}(K,\partial U)>0$. Then for every $z\in K$ and $d_1\in(0,d)$ we have $$ h_n'(z)=\frac{1}{2\pi i}\int_{|z-\zeta|=d_1}\frac{h_n(\zeta)\,d\zeta}{(\zeta-z)^2}, $$ and thus $$ \lvert h_n'(z)\rvert \le\frac{1}{2\pi}\cdot 2\pi d_1 \cdot \frac{M}{d_1^2}=\frac{M}{d_1}, $$ and as this holds for every $d_1<d$ and $n\in\mathbb N$, then $$ \sup_{z\in K}\lvert h_n'(z)\rvert\le \frac{M}{d}. $$ This implies that the sequence $\{h_n\}_{n\in\mathbb N}$, when restricted to $K$, is equicontinuous, and due to Arzelà–Ascoli theorem it is precompact, and thus $\{h_n\}$ possesses a convergent subsequence. Repeating this procedure for $$ K_j=\big\{z\in D : \mathrm{dist}(z,\partial D)\le 2^{-j}\big\}, $$ and using a suitable diagonal argument we can pick a subsequence, call it also $\{h_n\}$ which converges locally uniformly in $U$ to an analytic function $h$. But $h_n=f_{k_n}$, for some $k_n\ge n$, and due to $(1)$, $h_n\to 0$ uniformly in $\overline D_{r_0}$. Thus $h\equiv 0$ in $\overline D_{r_0}$, and consequently $h\equiv 0$ in $U$.