# Maximum value of derivative of a holomorphic function

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.

## 2 Answers

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

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 $$\phi_w:z\mapsto \frac{z-w}{1-\overline wz},\forall z\in U.$$

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

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