For a map $F: \mathbb{H} \rightarrow \mathbb{D}$, prove that $|F(z)| \leq |z-i|/|z+i|$ for all $z \in \mathbb{H}$.

922 Views Asked by At

For notation let $\mathbb{H}$ denote the upper-half plane and $\mathbb{D}$ the open unit disk. There was exercise on which I was stuck on:

Let $F: \mathbb{H} \rightarrow \mathbb{C}$ be a holomorphic function that satisfies $$|F(z)| \leq 1$$ and $$F(i)=0.$$ Prove that $|F(z)| \leq |z-i|/|z+i|$ for all $z \in \mathbb{H}$.

So far:

The condition $|F(z)| \leq 1$ is basically saying that we are mapping $\mathbb{H}$ to the disk $\mathbb{D}$. We also have that the map $\varphi(z) = (z-i)/(z+i)$ is a map from $\mathbb{H}$ to $\mathbb{D}$ that sends $i \mapsto 0$. Suppose we consider the map $F(\varphi^{-1}(z))$; then this is self-analytic map of the disk that fixes the origin. We can therefore invoke the Schwarz-lemma that tells us that $|F(\varphi^{-1}(z))| \leq |z|$.

I'm having trouble seeing how to finish the problem from here, I was wondering if I could get recommendations on either a different approach or on how to finish.

1

There are 1 best solutions below

1
On BEST ANSWER

Thus $$|F(z)| =|F(\varphi^{-1} (\varphi (z)) )|\leq |\varphi (z) |$$