One of the past qualifying question is as follows,

Let $D$ be a bounded domain, and f analytic from $D$ on $D$.

Let $z_{0}$ be a fixed point of $f$ and $|f^`(z_{0})|< 1$.

Now I am trying to show that nth iterate of $f$ converges uniformly on compact subsets of D to $z_{0.}$

I have a hunch that it is an application of Mittag- Leffler theorem. But this is getting nowhere with that thinking. I would love to see rigorous proof of this problem.


Let $\phi:D\to D$ be defined by $\phi(z)=\frac{z_0-z}{1-\overline{z_0}z}$, so that $\phi$ is a bijective analytic map such that $\phi(z_0)=0$ and $\phi(\phi(z))=z$ for all $z\in D$. Let $g=\phi\circ f\circ \phi$. Note that $g^n=\phi\circ f^n\circ\phi$ and $f^n=\phi\circ g^n\circ \phi$ (where the exponents denote iteration).

Since $g:D\to D$ is analytic, $g(0)=0$, and $|g'(0)|<1$, Schwarz's Lemma implies that $|g(z)|<|z|$ for all $z\in D\setminus\{0\}$. It follows that if $0<r<1$, then there exists $c$ with $0<c<1$ such that $|g(z)|\leq c|z|$ for all $z$ with $|z|\leq r$ (using compactness of the closed disk of radius $r$). Hence, $|g^n(z)|\leq c^n|z|$ for all $z$ with $|z|\leq r$. This implies that $g^n\to 0$ uniformly on compact sets, which in turn implies that $f^n=\phi\circ g^n\circ\phi\to z_0$ uniformly on compact sets.

  • $\begingroup$ Could you please justify this one, "using compactness of the closed disk of radius r". How did you get that? Rest of the proof I can follow. Thanks a lot. $\endgroup$ – Deepak Dec 24 '12 at 4:33
  • $\begingroup$ @Deepak: One way to get that is to use the fact that $\left|\frac{g(z)}{z}\right|$, taking out the removable singularity at $0$, is continuous, hence takes on a maximum on each compact subset of the disk. $\endgroup$ – Jonas Meyer Dec 24 '12 at 4:40
  • $\begingroup$ I see your point. Thanks @Jonas Meyer. Appreciated. Happy holidays. $\endgroup$ – Deepak Dec 24 '12 at 4:43

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