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!

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

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 $\alpha =0$.

  • $\begingroup$ But $g(z)=G(\zeta)=G(H(z))=z$!! $\endgroup$ – Riemann Apr 12 at 6:31
  • $\begingroup$ Sorry, there was a typo. I have corrected it. $\endgroup$ – Kavi Rama Murthy Apr 12 at 6:34
  • $\begingroup$ Can you explain to me in some detail. Thank you a lot! @Kavi Rama Murthy $\endgroup$ – Riemann Apr 12 at 6:59
  • $\begingroup$ Thank you very,I see now! $\endgroup$ – Riemann Apr 12 at 7:05

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