# Property of a composed holomorphic function

I have a question about the following problem.

Let $D$ be the unit disc in $\mathbb{C}$. Suppose $f: D \to D$ is a holomorphic function with $f(0) = 0$ and $f'(0) = 1$. By applying Cauchy's estimates to $$f_k = f \circ f \circ \cdots \circ f\;\; \text{k times}$$ for large $k$, show that $f$ is a linear function.

Someone suggested using Schwarz lemma, but I haven't learned it in class yet. I'm having trouble because I don't have an idea what result I want when I want to show a function is a linear function. Thank you!

• @user1952009 If $f=z^2/2$ then $f'(0)\ne 1$. The problem is correct as stated. – David C. Ullrich Apr 20 '16 at 13:20
• The Schwarz Lemma says precisely that $f$ is linear. But whether you've covered that in class or not you can't use that - you're supposed to give a solution by a specified method, not by any method at all. – David C. Ullrich Apr 20 '16 at 13:21
• @DavidC.Ullrich How can I apply Cauchy's estimate when we don't know $f_k$ is bounded by some $M$? – user12994 Apr 20 '16 at 13:40
• @user12994, $|f_k| \le 1$ because $|f| \le 1$. – lhf Apr 20 '16 at 13:41
• @DavidC.Ullrich Oh yes, I forgot about that condition. So if we let $z_0 \in D$ we have $|f_k(z_0)| \leq 1$ for $|z-z_0| = r$ where $0 < r < 1$. Then we can use Cauchy's estimate $\forall \; n \geq 0$, $|f_k^{(k)}(z_0) | \leq \frac{k!}{r^k}$. We need $k$th derevative to be a constant for $f$ to be a linear function right? – user12994 Apr 20 '16 at 13:48

The Cauchy estimates $|a_n|\le r^{-n}\sup_{|z|=r}|f(z)|$ show that any function that takes values in $D$ has Taylor coefficients (about $z=0$) that are $\le 1$ in absolute value (take $r\to 1$).
The $k$-fold composition $f_k$ is such a function. Now suppose that $f(z)=z+az^n+\ldots$, with $a\not= 0$. Then $f_k(z)=z+kaz^n+\ldots$, by an easy induction. For large $k$, this contradicts what we just observed.