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Let $\Delta(a, r)$ be the open disk of radius $r$ centered at the point $a$ in the complex plane, and $\operatorname{Aut}(\Delta(0, 1))$ be the set of linear fractional transformations that preserves the open unit disk, i.e. transformations of the form $z\mapsto e^{i\theta}(z-a)/(1-\bar az)$, where $a\in\Delta(0, 1)$ and $\theta\in\mathbb R$.

  1. I want to show $\operatorname{Aut}(\Delta(0, 1))$ is equicontinuous on every compact subset of $\Delta(0, 1)$. Does it suffice to show it is equicontinuous on $\overline{\Delta(0,r)}$ for any $r<1$? I think it surely suffices to show that it is continuous on every closed disk contained in the unit disk, but I'm unsure about the former.

  2. I want to show $\operatorname{Aut}(\Delta(0, 1))$ is equicontinuous on $\overline{\Delta(b,r)}$ which is contained in the unit disk. To do this, I evaluated $|f(z)-f(w)|$ for an arbitrary $f$ in $\operatorname{Aut}(\Delta(0, 1))$ and got $|f(z)-f(w)| \le |z-w|/(|1-\bar az||1-\bar aw|)$. But I can't go further. How do you get an upper bound for this?

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up vote 2 down vote accepted
  1. A compact subset $K$ of $\Delta(0,1)$ is contained on a set of the form $\overline{\Delta(0,r)}$ for some $r\in (0,1)$. Indeed, for each $x\in K$ we can find $r_x$ such that $\Delta(x,2r_x)\subset \Delta(0,1)$, so $K\subset \bigcup_{j=1}^N\Delta(x_j,r_{x_j})$ for some $N$ and $x_1,\ldots,x_N\in K$. Then put $r:=\max_{1\leq j\leq N}(1+r_{x_j})|x_j|$.

So it's enough to show equi-continuity on the sets of the form $\overline{\Delta(0,r)}$. 2. We have for $z_1,z_2\in\overline{\Delta(0,r)}$ that $$|f(z_1)-f(z_2)|\leq\frac{|z_1-z_2|}{|1-\bar az_1|\cdot |1-\bar az_2|}$$ and using triangular inequality and the fact that $|a|<1$ $$|1-\bar az_1|\cdot |1-\bar az_2|\geq (1-|z_1|)(1-|z_2|)\geq (1-r)^2$$ so $$|f(z_1)-f(z_2)|\leq \frac{|z_1-z_2|}{(1-r)^2},$$ which proves equi-continuity.

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For 1., why is $K$ contained for some $\overline{\Delta(0, r)}$? –  Pteromys Mar 17 '12 at 11:55
    
I will edit the answer in order to add it. –  Davide Giraudo Mar 17 '12 at 12:11
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