1
$\begingroup$

Let $U$ be an open unit disk and $f: U \rightarrow U$ is holomorphic mapping, such that $f(1/2)=1/3$. How to find the maximum value of $|f'(1/2)|$?

I tried to use Schwarz lemma but I didn’t succeed.

$\endgroup$
1
$\begingroup$

$$|f'(a)|\leq \frac{1-|f(a)|^2}{1-|a|^2},\forall a\in U. $$

To prove this, Fix $z\in U$. Apply Schwarz's lemma to $\phi_{f(z)}\circ f\circ \phi_{z}^{-1}:U\to U$ which sends $0$ to $0$.

Here for $w\in U$ we have bijective holomorphic map $\phi_w:z\mapsto \frac{z-w}{1-\overline wz},\forall z\in U.$

$\endgroup$
3
  • $\begingroup$ Why is the maximum reached? $\endgroup$ – Shorty12319 Nov 6 '19 at 14:54
  • $\begingroup$ Consider suitable $\beta\cdot \phi_w$ for some $\beta,w$ with $|\beta|,|w|\leq1$. $\endgroup$ – Sumanta Nov 6 '19 at 15:01
  • $\begingroup$ Sorry, I thought for a long time, but I can’t understand how this will help us. Why will equality be achieved? $\endgroup$ – Shorty12319 Nov 6 '19 at 16:21
1
$\begingroup$

Hint: Consider the function $g(z)=\frac{f-1/3}{1-f/3}\left(\frac{1/2-z}{1-z/2}\right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.