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Math Help - Schwartz's lemma related problem

  1. #1
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    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.
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  2. #2
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    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}.
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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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