6
$\begingroup$

$f: D \rightarrow D$ is an analytic function on a bounded domain $D$ with $f(0)=0$ and $|f'(0)| < 1$. If $F_n(z) := f \circ \dots \circ f(z)$, show that $F_n(z) \rightarrow 0$ as $n \rightarrow \infty$ uniformly on compact subsets of $D$.

My first instinct was to use Schwarz somehow, but turns out that $D$ is not necessarily simply connected, so I cannot use the Riemann mapping theorem to map $D$ to the unit disc. So, I was thinking about using Arzela-Ascoli somehow, but not sure how...

There also is a suggestion to consider the behavior of $F_n$ on a small neighborhood of 0. In a neighborhood of 0, I assume that something similar to contraction mapping happens. Help!

$\endgroup$

2 Answers 2

3
$\begingroup$

Since $D$ is bounded, all iterations of $f$ are uniformly bounded; thus they form a normal family. Suppose $F_n$ fail to converge to $0$ on compact subsets. Then there is a compact set $K$ and a subsequence $F_{n_j}$ such that $\sup_{K}|F_{n_j}|$ is bounded below by a positive constant independent of $j$. Using the normality, pick a convergent sequence $F_{n_j}\to g$; since the convergence is uniform on $K$, the function $g$ is not identically zero.

On the other hand, pick a number $\gamma$ strictly between $|f'(0)| $ and $1$ and use the definition of $f'(0)$ to find a circular neighborhood of $0$, i.e., $U=\{z:|z|<r\}$, in which $|f(z)|\le \gamma|z|$. Clearly, $F_n\to 0$ on $U$.

Putting the conclusions of the two paragraphs together, we see that $g$ is a nonconstant holomorphic function which vanishes on $U$. This is a contradiction.

$\endgroup$
2
  • $\begingroup$ $U$ and $K$ are not necessarily related, right? So, the said subsequence in the first part might not converge to anything outside of $K$? I may be missing something obvious here though... $\endgroup$ Jan 31, 2014 at 6:06
  • $\begingroup$ Never mind. So you take the convergent subsequence and then you can use Arzela-Ascoli again on that sequence in $U$. So, $g$ is $0$ in $U$. Woo! $\endgroup$ Jan 31, 2014 at 6:20
2
$\begingroup$

If we set $g(z)=f(z)/z$, then $g$ is analytic in $D$, and $g(0)=f'(0)$, and as $|f'(0)|<1$, then for $$a=\frac{|f'(0)|+1}{2}<1,$$ there is an $r_0\in(0,1]$, such that $|g(z)|\le a$, and hence $$ |f(z)|\le a|z|, $$ for all $|z|\le r_0$, which in turn implies that $$ |f^{n\circ}(z)|\le a^n |z|\le a^n r_0\to 0 \tag{1} $$ uniformly in $\overline D_{r_0}=\{z: |z|\le r_0\}$, where $f^{n\circ}=\underbrace{f\circ f\circ \cdots \circ f}_{n\,\,\,\text{times}}$.

In order to show that $f^{n\circ}\to 0$, locally uniformly in $D$, it suffices to show every subsequence of $\{f^{n\circ}\}$ contains a sub-subsequence which converges to $0$ locally uniformly in $D$.

Let $\{h_n\}$ a subsequence of $\{f^{n\circ}\}$ and $K\subset D$ compact. Then $K\subset \bar D_r$, for some $r<1$. Let $r_1\in(r,1)$. Then $|h_n(z)|\le 1$, for $|z|=r_1$, and hence since $$ h_n'(z)=\frac{1}{2\pi i}\int_{|z|=r_1}\frac{h_n(\zeta)\,d\zeta}{(\zeta-z)^2}, $$ for every $z\in K$, then $$ |h_n'(z)|\le\frac{1}{2\pi}\cdot 2\pi r_1 \cdot 1 \cdot\frac{1}{(r_1-r)^2}=\frac{r}{(r_1-r)^2}, $$ for every $z\in K$. This implies that $\{h_n\}$, when restricted to $K$, is equicontinuous, and due to Arzelà–Ascoli theorem it is precompact, and thus $\{h_n\}$ possesses a convergent subsequence. Repeating this procedure for $K_j=\overline{D}_{1-2^{-j}}$, and using a suitable diagonal argument we can pick a subsequence, call it also $\{h_n\}$ which converges locally uniformly in $D$ to an analytic function $h$. But $h_n=f^{k_n\circ}$, for some $k_n\ge n$, and due to $(1)$, $h_n\to 0$ uniformly in $\overline D_{r_0}$. Thus $h\equiv 0$ in $\overline D_{r_0}$, and consequently $h\equiv 0$ in $D$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .