Let $g:D(0,1)\to D(0,1)$ be analytic function and $g(0)=a\in[0,1)$;
suppose $$G(z)=\frac{z+a}{1+az},\text{for}\ z\in D(0,1).$$
we want to prove that:$$g(D(0,r))\subset G(D(0,r))\ \forall \ r\in(0,1).$$
In the origin paper, it says that this is an immediate consequence of the standard Schwarz-Pick lemma. My question is why this is right and how to interpret.
Here
$$D(0,1)=\{z \in \mathbb{C}||z|<1\}$$ $$D(0,r)=\{z \in \mathbb{C}||z|<r\}.$$
Standard Schwarz-Pick lemma: if $f:D(0,1)\to D(0,1)$ is analytic and
$\alpha \in D(0,1)$, then
$$\left|\frac{f(z)-f(\alpha)}{1-\overline{f(\alpha)}f(z)}\right|
\leq \left|\frac{z-\alpha}{1-\overline{\alpha}z}\right|, \forall \ z\in D(0,1).$$
Any hints and help will welcome!
Best Answer
Let $|z|<r$. Define $H(z)=\frac {z-a} {1-az}$. Verify that $G=H^{-1}$. We have to show that $g(z)=G(\zeta)$ with $|\zeta| <r$. Define $\zeta$ as $H(g(z))$. Then $g(z)=G(\zeta)$. To show that $|\zeta| <r$ apply Schwarz Pick Lemma with $f$ changed to $g$ and $\alpha =0$.