Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Related: math.stackexchange.com/q/39000 –  Jonas Meyer Dec 24 '12 at 2:56

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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. –  Deepak Dec 24 '12 at 4:33
    
@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. –  Jonas Meyer Dec 24 '12 at 4:40
    
I see your point. Thanks @Jonas Meyer. Appreciated. Happy holidays. –  Deepak Dec 24 '12 at 4:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.