4

Studying for a preliminary exam, I came across the following question:

Let $D = \{z \in \Bbb C : \text{Re}( z )> 0\}$ and $f : D \to D $ be a holomorphic function. Prove that $$|f'(z)|\leq\frac{\text{Re}(f(z))}{\text{Re}(z)}\quad \text{for all }z\in D.$$

I thought the best way might be to use the fact that harmonic functions satisfy the maximum modulus principle, too, along with the limit definition of the derivative, but I wasn't able to get it worked out. How should I approach this problem?

Clayton
  • 24,751
  • 3
    This is a part of Schwarz-Pick theorem (the half-plane case); see e.g. Wikipedia (there you'll find a proof and some context) – user8268 Aug 26 '13 at 19:46
  • The outline of a possible proof can be found here: http://anhngq.wordpress.com/2009/12/17/schwarzs-lemma-schwarz-pick-theorem-and-some-applications-involving-inequalities/ ,too. – Jack D'Aurizio Aug 26 '13 at 22:44

1 Answers1

1

Let's put this into wider context.

Fact 1. If $D$ is a simply connected domain and $f:D\to D$ is a holomorphic map with fixed point $a$, then $|f'(a)|\le 1$.

Proof: apply the Schwarz lemma to $g^{-1}\circ f\circ g$ where $g$ maps the unit disk onto $D$ so that $g(0)=a$. $\quad \Box$

Remark: one can drop "simply connected" in Fact 1, but then the proof requires more tools.

To apply Fact 1 here, fix $a\in D$ and consider $g=\phi\circ f$ where $\phi$ is an automorphism $\phi$ of $D$ that sends $f(a)$ back to $a$. Since $|g'(a)|\le 1$, it follows that $|f'(a)|\le |\phi'(f(a))|^{-1}$.

Notice that $D$ can still be any simply connected domain above. Its shape comes into play only when we construct $\phi$. Which for the right half-plane turns out to be
$$\phi(z)=\frac{\operatorname{Re}a}{\operatorname{Re}f(a)}\, (z-\operatorname{Im}f(a)) +i \operatorname{Im}a $$

user98130
  • 2,707