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Thread: Schwartz's lemma related problem

  1. #1
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    Schwartz's lemma related problem

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

    OK, so according to Schwartz's lemma all I need to show is that there exists $\displaystyle 0\neq z\in D$ such that $\displaystyle |f(z)|=|z|$ or that $\displaystyle f'(0)=1$, how do I use the fact that $\displaystyle f$ is 1-1 to show this? I was trying to define a new function such as $\displaystyle \frac{f(z)}{a}$ where $\displaystyle a$ is a point in the unit disc, but didnt get anywhere.
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  2. #2
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    Since $\displaystyle f$ is one-to-one, it is invertible. So consider the function $\displaystyle g(z)=f^{-1}(z)$, which maps the disk onto the disk and fixes the origin. By the Schwarz lemma, we have $\displaystyle |f'(0)|\leq 1$ and $\displaystyle |g'(0)|\leq 1$. We also have $\displaystyle (f\circ g)(z)=z$ and $\displaystyle (f\circ g)'(z)=1$. Using the chain rule gives $\displaystyle 1=(f\circ g)'(0)=f'(0)g'(0)$. This forces $\displaystyle |f'(0)|=|g'(0)|=1$. This allows us to apply the Schwarz lemma to conclude that $\displaystyle f(z)=e^{it}z$ for some $\displaystyle t\in \mathbb{R}$.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by skyking View Post
    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
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