Iteration of analytic function Suppose $f$ is analytic on the unit disc $D$ with $f(0)=0$ and $f(D)\subset D$. Define $f_n=f\circ f\circ\dots\circ f$.
If $f$ is not a rotation, can we say $f_n\to 0$ uniformly on compact subsets of $D$?
I know the Schwarz Lemma shows the sequence is a normal family, but I don't know how to continue, any suggestions? Thanks!
 A: Consider $g(x)=\frac{f(x)}{x}$, that is a holomorphic function on the unit disk.
By the maximum modulus principle:
$$ |g(x)|\leq \sup_{x\in\partial D}\left|\frac{f(x)}{x}\right|=\sup_{x\in D}|f(x)|\leq c <1 \tag{1}$$
where the last inequality follows from $f(D)\subset D$. However, $(1)$ gives:
$$ |f(x)| \leq c\,|x|\tag{2} $$
hence:
$$ \sup_{x\in D} |f_n(x)| \leq c^n \to 0.\tag{3}$$
A: Not as slick as Jack's answer but another way...
Since $f_n$ is a uniformly bounded family of holomorphic functions it is a normal family. Thus on compact subsets we have a subsequence converging uniformly. 
Since any compact subset is contained in a closed disc of radius $r<1$ we can just consider closed discs. The maximum modulus occurs on the boundary, so let $z_0\in \partial D_r(0)$. If we can show that $|f_n(z_0)|$ converges to 0, we are done. 
By assumption, $$|f_{n+1}(z_0)|<|f_n(z_0)|.$$ So suppose for the sake of contradiction that $|f_n(z_o)|> c>0$ for all $n$. Since these norms form a decreasing sequence bounded below, they converge to the inf, say $c$.
By the uniform convergence above, there is a subsequence $\{f_{n_k}\}$ converging to $g$ on $D_r(0)$. Thus $|g(z_0)|= c$. And by Schwarz lemma, $|f(g(z_0))|< c$. By continuity of $f$ there exists a neighborhood $N$ about $g(z_0)$ such that $|f(y)|<c$ for all $y\in N$. But now by the uniform convergence we can choose $K>0$ such that $f_{n_k}(z_0)\in N$ for all $k>K$. Thus, $|f_{n_k+1}(z_0)|< c$ which is contradiction.
