Equicontinuity of automorphism on compact subsets of the unit disk

Question

Is the family of the functions $Aut(∆(1))$ equicontinuous on compact subsets of $ ∆(1)$? Here $∆(1)$ is the disk of radius $1$ centered at the origin on the complex plane.

I am lost where to start. How to go about proving/disproving this? Thank you.

Answer 1

There might be an easier proof but here is what I came up with:

Take any $f\in Aut(\Delta(1))$, so $f(z)=e^{i\theta}\frac{z-a}{1-\overline{a}z}$ with $a\in\Delta(1)$ and $\theta\in\mathbb{R}$. Now take a compact subset $K\subset\Delta(1)$ and $z,w\in K$. Then

$|f(z)-f(w)|=|e^{i\theta}\frac{z-a}{1-\overline{a}z}-e^{i\theta}\frac{w-a}{1-\overline{a}w}|=|e^{i\theta}||\frac{(z-a)(1-\overline{a}w)-(w-a)(1-\overline{a}z)}{(1-\overline{a}z)(1-\overline{a}w)}|$ $\\$ $=\frac{|z-\overline{a}wz-a+|a|^2w-w+\overline{a}wz+a-|a|^2z|}{|1-\overline{a}z||1-\overline{a}w|}$ $=\frac{|1-|a|^2||z-w|}{|1-\overline{a}z||1-\overline{a}w|}$.

Also observe that since $|a|<1$ we have that $|1-|a|^2|\leq1$ and $|\overline{a}z|=|az|=|a||z|\leq|z|$. Thus $0<1-|z|\leq1-|\overline{a}z|=|1-|\overline{a}z||\leq|1-\overline{a}z|$. Hence

$\frac{1}{|1-\overline{a}z|}\leq\frac{1}{1-|z|}=\frac{1}{|1-|z||}$ and $\frac{1}{|1-\overline{a}w|}\leq\frac{1}{|1-|w||}$. $\\$

Since $K$ is compact and $0<|1-|z||$ for any $z\in K$, there is $M>0$ such that $M\leq|1-|z||$. Hence $\frac{1}{|1-|z||}\leq\frac{1}{M}$ for any $z\in K$. It follows that

$|f(z)-f(w)|=\frac{|1-|a|^2||z-w|}{|1-\overline{a}z||1-\overline{a}w|} \leq\frac{|z-w|}{|1-\overline{a}z||1-\overline{a}w|}\leq\frac{|z-w|}{|1-|z|||1-|w||}\leq\frac{|z-w|}{M^2}$.

Now for $\epsilon>0$ take $\delta=\epsilon M^2>0$, then we have $|f(z)-f(w)|\leq\frac{|z-w|}{M^2}<\epsilon$ for any $z,w\in K$ with $|z-w|<\delta$ and for any $f\in Aut(\Delta(1))$. Therefore, the family of functions $Aut(\Delta(1))$ is equicontinuous on compact subsets of $\Delta(1)$.