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I'm stuck on the following problem:

Let $\Omega$ be a bounded domain, and $f: \Omega \rightarrow \Omega$ analytic such that $f(z_0) = z_0$. Show that $|f'(z_0)| \leq 1$.

What I did so far was suppose that $|f'(z_0)| > 1$, and define $g_n := f \circ \cdots \circ f$ ($n$ times). Then $g: \Omega \rightarrow \Omega$ is analytic with $g_n(z_0) = z_0$ and $g_n'(z_0) = f'(z_0)^n$. It follows that $|g_n'(z_0)|$ is large for $n$ large.

So now we are reduced to the case where $g: \Omega \rightarrow \Omega$ is analytic, fixes a point $z_0$, and $g'(z_0)$ is very large. Somehow I want to show that as a result of $g'(z_0)$ being so large, it must map some point in $\Omega$ outside of $\Omega$, contradiction. Can anyone give a hint?

D_S
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1 Answers1

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Let $h(z)=\frac{g(z)-z_0}{z-z_0}$; since $g(z)$ is analytic in $\Omega$, $h(z)$ has a removable singularity at $z_0$ and is analytic in $\Omega$ as well. Thus $$ g'(z_0)=\lim_{z\to z_0}\frac{g(z)-g(z_0)}{z-z_0}=\lim_{z\to z_0}\frac{g(z)-z_0}{z-z_0}=h(z_0) $$ Since $\Omega$ is open, we can choose an open disc $U$ centered at $z_0$ of radius $r>0$ contained in $\Omega$. If $M>0$ is a bound for $\Omega$, then on $U\cap\{z\in\mathbb{C}:|z-z_0|\geq\frac{r}{2}\}$, $$ |h(z)|=\left|\frac{g(z)-z_0}{z-z_0}\right|\leq\frac{|g(z)|+|z_0|}{\left|z-z_0\right|}<\frac{M+|z_0|}{\frac{r}{2}}<\frac{4M}{r} $$ Consider the closed disc $V=\{z\in\mathbb{C}:|z-z_0|\leq\frac{r}{2}\}$. By the above, we know that $|h(z)|<\frac{4M}{r}$ on $\partial V$ and so the inequality holds on $\text{int}(V)\ni z_0$ by the maximum modulus principle. You can then use your idea of making $n$ large and setting $g=g_n$ to reach a contradiction.

Michael M
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