I'm reading a proof about when two annuli cannot be conformally equivalent and I got stuck at some point.

The proof goes like this.

Let $f:D(0,1,R_1) \to D(0,1,R_2)$ be a conformal mapping and let $R_2 \lt R_1$

We constructa sequence $f_n$ of $1-1$ functions like this:

$f_1=f$ and $f_{n+1}=f \circ f_n$

If $Ω=D(0,R_2,R_1)$ we can show by induction that $f_n(Ω) \cap f_m(Ω)= \emptyset$

Since $f_n$ are uniformly bounded, using Montel we can find a subsequence $f_{n_k} \to h$ uniformly on every compact subspace of $D(0,1,R_1)$

So far, so good.

Here's the part I don't get:

We want to show that $h$ is constant. By assuming the opposite we take some $z_0 \in Ω$ so that $f(z_0)=w_0$. By the Uniqueness theorem there exists an $r \gt0$ so that $0 \notin h(D(z_0,r)- \ {z_0})$. Then by Hurwitz there exists some $k_0 \in \Bbb N$ so that for every $k\ge k_0$ there is a $z_k \in D(z_0,r)$ so that $f_{n_k}(z_k)=w_0$. By an earlier statement this is a contradiction.

I can understand why this is a contradiction, what I don't seem to get is what does that got to do with $h$ being non-constant? Why is $w_0$ in the image of $f_{n_k}$ for large $k$?

I know that $h$ must be $1-1$ if it's non-constant, also from Hurwitz, but I can't make anything of it.

Please help!


1 Answer 1


Hurwitz's theorem says that if $g_k\to g$, $g(z_0)=0$, and $g$ is not constant in a neighborhood of $z_0$, then for any sufficiently small ball $B$ around $z_0$ and all $k$ sufficiently large (depending on $B$), there exists a $z_k\in B$ such that $g_k(z_k)=0$. (In fact, it says more than that, but that's all we need here.) Apply this with $g_k(z)=f_{n_k}(z)-w_0$ and $g(z)=h(z)-w_0$ and you get that $w_0$ is in the image of $f_{n_k}$ for all sufficiently large $k$.

  • $\begingroup$ I'm sorry but it's still unclear to me why $w_0$ is in the image of $f_{n_k}$ for large $k$. Could you please give me more details? $\endgroup$ Jun 9, 2016 at 3:51
  • 1
    $\begingroup$ Oh, I didn't realize that was what you were asking about. $\endgroup$ Jun 9, 2016 at 3:55
  • $\begingroup$ Yes, sorry if it wasn't clear. I edit the question so it would be more obvious. $\endgroup$ Jun 9, 2016 at 5:08

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