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Let $\Omega$ be a bounded connected open subset of $\mathbb{C}$ containing $0$. Let $f: \Omega \rightarrow \Omega$ be holomorphic and $f(0) = 0$, $f'(0) = 1$. The problem I am working on is to show that $f(z) = z$.

If $\Omega = \mathbb{D}$, then this follows from the Schwarz Lemma. I also know of a solution (posted here) which involves looking at the power series coefficients of $f^n := f\circ f \circ f \circ \cdots \circ f$ ($n$ times) and using the Cauchy estimates, but is there a different way of doing this problem that doesn't involve taking the $k$th derivative of $f^n$?

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A solution which I find more natural than looking at iterations of $f$ involves a holomorphic covering map $\phi:\mathbb D\to\Omega$. Such a map exists for every domain $\Omega$ provided that $\mathbb C\setminus \Omega$ has at least two points. This is Theorem 16.5.1 in Conway's book Functions of one complex variable, vol. II.

Normalize the covering map so that $\phi(0)=0$. Then $\phi^{-1}\circ f\circ \phi$ is a holomorphic map of disk to itself which fixes the origin and has derivative $1$ there. (Note: although $\phi$ is not globally invertible, it is locally invertible; therefore $\phi^{-1}\circ f\circ \phi$ is an analytic function, and since $\mathbb D$ is simply-connected, it has a single-valued holomorphic branch in $\mathbb D$.)

By the Schwarz lemma the map $\phi^{-1}\circ f\circ \phi$ is the identity, and therefore $f$ is the identity.

Key words for additional reading: uniformization theorem, hyperbolic metric.

  • I did not get to reformulate the proof because I have a problem with the word 'locally invertible': and I do not see exactly the application of Schwartz theorem: in other words for me " $\phi^{-1} $ is a collection of maps "; here's the problem: Let $z\in \Bbb D$ , $z'=\phi(z)\in \Omega$, and there is and "elementary" neighborhood of $z'$ such that, $\phi^{-1}(U)=\cup_i V_i$ (disjoint union ) with $\phi_{/V_i}:V_i\to U$ is biholomorphic. Now $V_i\to \phi(V_i)=U\to f(U) \to \phi_{/V_i}^{-1}(f(U))$? – Hamou Aug 23 '14 at 20:40
  • @Hamou Think of $\phi^{-1}$ in terms of analytic continuation: it can be analytically continued along every path in $\Omega$ (starting at $0$, to be concrete). Therefore, $\phi^{-1}\circ f\circ \phi$ can be continued along every path in $\mathbb D$. But since $\mathbb D$ is simply connected, the result of continuation does not depend on the path, only on the endpoint. Thus, it defines a holomorphic function. –  Aug 23 '14 at 20:45