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Let $f: \Omega \rightarrow \Omega$ be holomorphic where $\Omega \subset \mathbb{C}$ is a bounded region containing 0. If $f(0)=0$ and $f'(0)=1$, does $f(z)=z$?

This is true if $\Omega$ is a disk centered at 0 but does it hold if $D$ is only bounded?

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up vote 5 down vote accepted

There is a substantial generalization of Schwarz' Lemma valid for arbitrary maps $f:\ \Omega\to\Omega$ of a Riemann surface into itself. It says that any such map which is not a conformal automorphism of $\Omega$ actually decreases the hyperbolic distance between points. This implies that if $|f'(z)|=1$ at some fixed point $z$ then $f$ is a conformal automorphism. This can be proven by lifting the map $f$ to a map $\tilde f:\ \tilde\Omega\to\tilde\Omega$ of the universal cover of $\Omega$. The latter is ("in most cases") conformally equivalent to the unit disk, where Pick's invariant version of Schwarz' Lemma can be applied. See here, p. 27:

Huber, Heinz: Analytische Abbildungen Riemannscher Flächen in sich. Comm. Math. Helv. 27 (1953), 1–73.

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Find a conformal map $h$ that maps $\Omega$ to the unit disc $D$, fixing $0$. Then $g:= h\circ f\circ h^{-1}:D\to D$ is holomorphic, with $g(0)=0$ and $g'(0)=1$.

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but $\Omega$ is just a bounded region containing zero. It might not be simply connected. If $\Omega$ was simply connected and open then I agree that such a conformal map exists by Riemann mapping theorem. I am not sure such a map exists. – Shankara Pailoor Sep 26 '12 at 14:29
you are right, I missed this condition.. – Berci Sep 27 '12 at 9:44
@Berci Well with this we can conclude that $ f(h(z))=h(az) $ for some $a \in \mathbb{C} $ with $|a|=1$. How we can conclude that $a=1$ ? – Daniel Oct 8 '12 at 6:29

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