# Thread: Schwartz's lemma related problem

1. ## Schwartz's lemma related problem

Let $f$ be analytic and copies the open unit disc $D$ onto itself. assume $f$ is 1-1 and $f(0)=0$. Show that $f=e^{it}z, t\in R$ for $|z|<1$.

OK, so according to Schwartz's lemma all I need to show is that there exists $0\neq z\in D$ such that $|f(z)|=|z|$ or that $f'(0)=1$, how do I use the fact that $f$ is 1-1 to show this? I was trying to define a new function such as $\frac{f(z)}{a}$ where $a$ is a point in the unit disc, but didnt get anywhere.

2. Since $f$ is one-to-one, it is invertible. So consider the function $g(z)=f^{-1}(z)$, which maps the disk onto the disk and fixes the origin. By the Schwarz lemma, we have $|f'(0)|\leq 1$ and $|g'(0)|\leq 1$. We also have $(f\circ g)(z)=z$ and $(f\circ g)'(z)=1$. Using the chain rule gives $1=(f\circ g)'(0)=f'(0)g'(0)$. This forces $|f'(0)|=|g'(0)|=1$. This allows us to apply the Schwarz lemma to conclude that $f(z)=e^{it}z$ for some $t\in \mathbb{R}$.

3. I have one question, to use the schwartz lemma on g we need to know g is analytic in the open unit disc. But for this we need to know that f does not zero in the open unit disc? how can we be sure of that

4. Originally Posted by skyking
I have one question, to use the schwartz lemma on g we need to know g is analytic in the open unit disc. But for this we need to know that f does not zero in the open unit disc? how can we be sure of that

f is 1-1 so only zero is mapped into zero and Schwartz lemma is appliable.

Tonio