$f:D\to D$ is analytic then $f^{n_i}(z)$ converges pointwise for all $z$

Question

This is a problem from my past Qual.

"Let $D$ denote the unit disk and $f:D\to D$ be analytic. Show that there exists a sequence $n_i$ s.t. $f^{n_i}(z)$ converges pointwise for all $z\in D$. Here $f^n=f\circ f\circ\ldots\circ f$ ($n$ times)."

I have no idea how to start. I have an analytic function, so I have its Taylor series in a small neighborhood, I know the Cauchy-Riemann equations. That's it. I mean usually when I deal with $f^n$, I study $f$. In this case it seems I don't have a lot of information to study $f^n$. So I'm stuck here.

Answer 1

Note that all coefficients of all functions $f^n$ have modulus at most $1$, by their Cauchy integral formula. Since the closed unit disc is compact, it follows that there is a subsequence $f^{n_k}$ such that their coefficients (viewed as functions $\mathbb{N}\to \overline{\mathbb{D}}$) converge pointwise to a sequence $a_0, a_1, \ldots$ in the closed unit disc. Set $g(z)=\sum_ka_kz^k$. This is a holomorphic function on the unit disc. Then show that the sequence of functions $f^{n_k}$ converges pointwise to $g$ on the unit disc. (For example, consider only the first $m$ terms in their series and derive a pointwise bound depending om $m$, then take $m \to \infty$.)