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Let $f$ be a holomorphic function on an open, connected set $\Omega\subset \mathbb{C}$ with $z_0\in \Omega$ a fixed point, and $\{f^n\}_{n\in \mathbb{N}}$ the sequence of iterates.

I want to prove that if $|f'(z_0)|<1$ then there is a neighborhood of $z_0$ such that $\{f^n\}$ is normal in it.

I don't really know what to do, because I don't know how to handle the iterates. I think the fact that $|f(z)| < |z-z_0| + |z_0| + \epsilon |z-z_0|$ might be useful to apply Montel's theorem, but I didn't get very far.

This is not homework, it's a problem I'm trying to solve as preparation for a complex analysis exam.

Also: I'd be grateful to have a geometrical interpretation of $|f'(z_0)|<1$ or of $|f'|$ in general. I don't really understand what does it mean for $|f'(z_0)|<1$, for example in Schwarz's lemma.

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1 Answer 1

up vote 2 down vote accepted

I think I got it. Correct me if I'm wrong.

There exists a neighborhood $U$ of $z_0$ such that $|f(z)-z_0|\leq \rho |z-z_0|$, where $|f'(z_0)|<\rho<1$.

Then for all $z\in U$, $|f^n(z_0)-z_0| = |f(f^{n-1}(z))-z_0| \leq \rho |f^{n-1}(z)-z_0| \leq \dots \leq \rho^n |z-z_0|$.

This means the sequence $\{f^n\}$ converges uniformly to $z_0$ in $U$, because $\rho^n\to 0$ as $n\to \infty$. In particular, it converges uniformly in compacts of $U$, whence $\{f^n\}$ is normal in $U$.

There is still the last part of the question to be answered, though.

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The fact that the absolute value is less than one, implies that a small disk containing z0, will be mapped to something disk-like, but with less area, since the mapping shrinks with the same factor as the derivative. –  Per Alexandersson May 30 '11 at 19:47

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