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I am seeking help on a complex analysis qualifying exam problem.

Let $D$ be a bounded open connected subset of $\mathbf{C}$ containing $0$ and let $f \colon D \to D$ be an analytic function satisfying $f(0) = 0$ and $\left| f^\prime \right|(0) < 1$. Define $f_n = f \circ f \circ \dots \circ f$ ($n$ times). Prove that $f_n \longrightarrow 0$ uniformly on compact sets.

The hint is to start locally around zero. I was able to prove that there exists a neighborhood $U$ contained within the radius of convergence of $f$ about $0$ such that $f_n \longrightarrow 0$ uniformly on compact sets contained in $U$. I note that the proof did not use the boundedness of $D$.

I am having trouble extending the result to the entirety of $D$. Perhaps I am supposed to exploit the connectedness of $D$, considering something like $E = \left\{ z \in D \colon \text{ the result is true locally around } z \right\}$ and showing this set is open and closed. If this is true, then for an arbitrary compact subset of $D$ we can take a finite cover applying the local result and be done with it. It is obvious that $E$ is open, but I am having trouble showing it is closed.

I don't have an idea where the boundedness of $D$ comes into play...

Many thanks in advance for your help.

Doug
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  • I’m an outsider to complex stuff, but I don’t think that boundedness of $D$ is needed. After all, you take a compact subset of $D$, and prove something about that, right? And that’s the whole game, isn’t it? – Lubin Jan 17 '16 at 05:31
  • The only reason I could see possibly using that $D$ is bounded is because it gives trivially that $f_n(Z)$ is uniformally bounded, hence the Arzela theorem can be applied to say that a subsequence $f_{n_k}$ converges to some analytic function $f$ uniformally on compact sets. In fact, now that I think about it, we know that $f$ equals 0 since this is true locally around 0. So we have actually proved that a subsequence of $f_n$ converges uniformally on compact sets to 0, and the whole sequence converges uniformally on compact sets to 0 on an open subset of $D$. – Doug Jan 17 '16 at 06:05
  • So...let $K$ be a compact set. Then there exists $N$ such that $|f(K)| \leq \epsilon$. Take $\epsilon$ small enough such that this image $K'$, which is compact, lies in $U$. Choose $N$ such that $|f_n(K') \leq \epsilon$ for all $n \geq N$. Then for all $n \geq N + M$, $|f_n(K)| \leq \epsilon$. I think that is a solution. – Doug Jan 17 '16 at 06:46

1 Answers1

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Assume we have proved there exists $U \ni 0$ open on which $f_n \longrightarrow 0$ uniformly on compact sets. (This is not too hard. Start with $U$ small enough so that the power series for $f$ at $0$ converges on $U$. Then by assumption, for $z \in U$ we have $$ \left| f(z) \right| = \left|z\right| \left| f^\prime(0) + c_2 z + \dots \right|. $$ Since $f^\prime(0) < 1$, upon shrinking $U$ further, by continuity there exists $\alpha < 1$ such that $$ \left| f(z) \right| < \alpha |z|. $$ If $K \subset U$ is compact, then $$ \left| f_n(z) \right| \leq \alpha^n \sup_{w \in K} \left| w \right| \longrightarrow 0.)$$

Now, since $D$ is bounded and $f$ maps into $D$, $f_n(z)$ is uniformly bounded, hence by the Arzelà-Ascoli theorem for analytic functions (i.e. Montel's theorem) there is a subsequence $f_{n_k}$ that converges uniformly on compact sets to some analytic function $f$. In particular, since $f_n \longrightarrow 0$ on $U$, $f(U) = 0$, hence $f = 0$.

Fix $\epsilon > 0$ and let $K$ be a compact subset of $D$. By what we just showed, there exists $N$ such that $\left| f_N(K) \right| \leq \epsilon$. Take $\epsilon$ sufficiently small so that $K^\prime \subset U$, where $K^\prime = f_N(K)$. Since $f_n \longrightarrow 0$ uniformly on $K^\prime$, take $M$ so that for all $n \geq M$, $\left| f_M(K^\prime) \right| \leq \epsilon$. But $f_M \circ f_N = f_{N + M}$, hence for all $n \geq N + M$ we conclude $$ \left| f_n(K) \right| \leq \epsilon. $$

Doug
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    Arzela-Ascoli for analytic functions? Do you mean Montel's Theorem? – zhw. Jan 17 '16 at 18:54
  • I just checked, and that is correct. In class we just called it Arzelà-Ascoli for analytic functions since it was a corollary of the more general Arzelà-Ascoli theorem. – Doug Jan 17 '16 at 19:11