Iterated Complex Mappings

Make sure the Symbol font is enabled on your browser...I'll use ¶ for the complex differentiation operator.

Let G be a bounded connected (open) region in the complex plane, and let f:G®G be an analytic mapping. Suppose there exists z₀ÎG with f(z₀)=z₀, and that |¶f(z₀)| is less than 1. Define f₁(z)=f(z), and inductively, f_n+1(z)=f(f_n(z)). Note that this makes sense, since the range of f_n is contained in the domain of f, and that all the f_n are analytic, and, by induction, f_n(z₀)=z₀ for all n. The goal is to show that f_n converges to the constant function z₀ uniformly on compact subsets of G.

Differentiate:

¶f_n+1(z)=¶[f(f_n(z))]=¶f(f_n(z))¶f_n(z)

By induction, we have

¶f_n(z)=Õ¶f(f_k(z)), k=0..n-1

Now, since G is bounded, and f_n maps G into G, the functions G are all uniformly bounded by some constant. Therefore, since the f_n are all analytic, the family {f_n} is normal: applying the Arzela-Ascoli Theorem, we learn that {f_n} is equicontinuous on each compact subset of G. (Incidentally, I've heard the term 'compacta' used for 'compact subsets', and I like it.) Now, the fun begins. Since |¶f(z₀)| is less than 1, and |¶f| is continuous, $e,h, both positive, with h less than 1, with |¶f(z)| less than h for all z with |z-z₀| less than e. Since {f_n} is equicontinuous in a neighbourhood of z₀, $d positive with |f_n(z)-f_n(z₀)|=|f_n(z)-z₀| less than e for all z with |z-z₀| less than d. Therefore, in this d-disk about z₀, |¶f(f_n(z)| is less than h, for all n. Therefore |¶f_n(z)| is less than hⁿ, which tends to 0 as n approaches infinity. Therefore ¶f_n tends to 0 uniformly on this disk.

Must be off now, but I'll finish later.

UPDATE: Okay, I solved it in the shower just now. Since {f_n} is a normal family, it contains a subsequence {f_nk} converging uniformly on compacta to some function g. By Weierstrass' Theorem, g is analytic, and ¶f_nk converges uniformly to ¶g on compacta. But these derivatives are converging to 0 on an open neighbourhood of z₀, so, since nonzero analytic functions have only isolated zeroes, these derivatives are converging to 0, and hence ¶g=0: since G is connected, and g(z₀)=z₀, g is the constant function z₀. Now, fix a compact subset K of G. {f_n} is equicontinuous on K, so "e $d so that if |z-z₀| is less than d, then for all n, |f_n(z)-f_n(z₀)|=|f_n(z)-z₀| is less than e. Furthermore, there exists n_K such that |f_nK(z)-z₀| is less than d for all z in K. Thus, for all z in K, for all m, |f_m(f_nK(z))-z₀|=|f_nK+m(z)-z₀| is less than e. Therefore, for all large n, |f_n(z)-z₀| is less than e for all z in K: f_n converges uniformly to z₀ on all compact subsets of G. QED

Posted by aloysius at July 23, 2003 12:31 PM | TrackBack |