4
$\begingroup$

$f$ is analytic mapping of the unit disk into itself such that $f(a) = 0$. Show $|f(z)| \leq |\frac{z-a}{1-\bar{a}z}|$

I considered $F(z) =f(\phi_{-a}(z))$ where $\phi_a(z) = \frac{z-a}{1-\bar{a}z}$ maps unit disk to unit disk and $\phi_{-a}$ is the inverse of $\phi_a$

I get F is analytic mapping unit disk to unit disk, and $F(0) = 0$, so by Schwarz lemma I get $|F(z)|\leq |z|$ and $|F'(0)|\leq1$. And then I get stuck. I am not sure what I can do to get the final inequality. Or maybe I am thinking about the problem in a wrong way.

$\endgroup$

1 Answer 1

2
$\begingroup$

You're almost there. Now, let $\xi=\phi_{-a}(z)$. We get $F(z)=f(\phi_{-a}(z))=f(\xi)$, for So the inequality would be, in this terms,

$$|f(\xi)|<|\phi_{a}(\xi)|=\left|\frac{\xi-a}{1-\bar a\xi}\right|$$

$\endgroup$
2
  • $\begingroup$ hehe, that was simple! Thanks a lot! $\endgroup$
    – aregak
    Commented Mar 5, 2017 at 19:36
  • $\begingroup$ You can obtain another condition if you write $F'(0)$ using the chain rule, I think. $\endgroup$ Commented Mar 5, 2017 at 19:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .