Let $\{f_n\}_{n=1}^{\infty}$ be such that $f_1(z)=\sin(z)$, and $f_{n+1}(z)=\sin(f_n(z))$. Show $\{f_n\}$ is not locally bounded near the origin.

Question

Let $\{f_n\}_{n=1}^{\infty}$ be the sequence of iterates of the sine function: namely, $f_1(z)=\sin(z)$, and $f_{n+1}(z)=\sin(f_n(z))$ when $n\geq 1$. Show that this sequence $\{f_n\}$ is not locally bounded in any neighborhood of the origin.

I'm having trouble with this. This is my attempt:

Suppose to the contrary that $\{f_n\}$ is locally bounded on some $D=\{z\in\mathbb{C}\,:\,|z|<\varepsilon\}$. Then Montel's theorem shows that there exists a subsequence $\{f_{n_k}\}$ converging to some holomorphic $f$ uniformly on compact subsets of $D$. Since $f(0)=0$ we can choose (through continuity) $\varepsilon$ small enough that $f(D)\subseteq\mathbb{D}$. Now (I think) f'(0)=1$... something something something sine behaves weirdly causing a contradiction.

Is Schwarz' lemma applicable here? I feel like I'm close but I have a few technical details I need to work out. Any help is greatly appreciated. Thank you

Answer 1

Here are a couple different possible approaches to the problem.

1. If $f_{n_k}$ converges to $f$ uniformly, then the derivatives $f^{(m)}_{n_k}$ must converge uniformly to $f^{(m)}$ as well. Try computing $f_n^{(m)}(0)$ for different small values of $m$ until you can find a contradiction.

2. If $x$ is small and real, how does $\sin(x)$ compare to $x$? What does this tell you about the behavior of $f_n(x)$ as $n\to\infty$? If $f_{n_k}$ converges uniformly to $f$, what does this tell you about $f$? Now compute $f'(0)$ as suggested above to get a contradiction.