# Immediate consequence of standard Schwarz-Pick lemma! How to interpret?

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|

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!

• Just typo! Sorry. – Riemann Apr 12 at 6:18
• $D(0,1)=\{\|z|<1\}$. – Riemann Apr 12 at 6:25

Let $$|z|. Define $$H(z)=\frac {z-a} {1-az}$$. Verify that $$G=H^{-1}$$. We have to show that $$g(z)=G(\zeta)$$ with $$|\zeta| . Define $$\zeta$$ as $$H(g(z))$$. Then $$g(z)=G(\zeta)$$. To show that $$|\zeta| apply Schwarz Pick Lemma with $$f$$ changed to $$g$$ and $$\alpha =0$$.
• But $g(z)=G(\zeta)=G(H(z))=z$!! – Riemann Apr 12 at 6:31