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

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.

• Did you mean $f^n(z)-(f(z))^n$? or $f^n(z)=(f^{n-1}\circ f)(z)$, $f^1=f$, In the former, isn't it trivial? $|f^n(z)|=|f(z)|^n\xrightarrow{n\rightarrow\infty}0$ since $|f(z)|<1$. – Oliver Diaz May 28 at 3:29
• oh no this is composition. – T C May 28 at 3:30
• That's what I thought. I edited your question to add composition symbols to avoid confusion. – Oliver Diaz May 28 at 3:39
• This seems to be related to what is known as normal families. See pages 281-282 of Rudin's real and complex analysis. In particular Theorem14.6 – Oliver Diaz May 28 at 3:51
• Do you mean that $f$ is a holomorphic function on the unit disc in $\mathbb{C}$? – WimC Jun 2 at 11:00

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 $$m \to \infty$$.)
• @MichaelBurr Yes, the Taylor coefficients at $z=0$. – WimC Jun 2 at 10:56
• The family described in the problem is a normal family since $f^{n}=f\circ\ldots\circ f$ ($n$ times) is in fact uniformly bound. One can either used a theorem that states that families of holomorphic functions in a region that are uniformly bounded in compact subsets are normal (meaning every subsequence has a uniform convergent subsequence etc) or use Cauchy estimates to prove Lipschitz condition and some versions of the Ascolli-Arzela theorem. – Oliver Diaz Jun 2 at 14:46