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I'm working on this problem

"Let $f$ be an entire function. Suppose $f$ restricted to the unit disk is a bijection. Prove that $f$ is a rotation."

My attempt: It is tempting to use Schwarz lemma. Let $T$ be the linear transformation that maps the unit disk to itself and $T(f(0))=0$ (If I remember it right, $T=\frac{z-f(0)}{1-\overline{f(0)}z}$). Then $|Tf(z)|\leq 1$ in the unit disk and $T(f(0))=0$.

So, if I can show that $|Tf(z)|=|z|$ for some nonzero $z$, we are done by the lemma. I want to say $Tf$ achieves maximum on the unit circle. So if the maximum is 1, $|T(f(z))|=1=|z|$ for some $z$ there. However I am stuck here.

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  • $\begingroup$ $f(z)= 2z$ is entire but is a bijection of $\Delta(0,1)$ onto its image without being a rotation. Is there any further asumption? $\endgroup$ – Didier May 27 '20 at 18:17
  • $\begingroup$ an idea is to use $T \circ T^{-1}=id$ to get that $|T'(0)(T^{-1})'(0)|=1$ and conclude from there and Schwarz $\endgroup$ – Conrad May 27 '20 at 18:17
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    $\begingroup$ @Dl - bijection of the unit disc onto itself if I understand it right - otherwise one needs to classify the univalent functions on the disc that extend to entire functions and those are many $\endgroup$ – Conrad May 27 '20 at 18:21
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All holomorphic bijections of the unit disk onto itself are Möbius transformations of the form $$ T(z) = e^{i\lambda} \frac{z-a}{1-\overline a z} $$ for some $a \in \Bbb D$ and $\lambda \in \Bbb R$. (See for example Can we characterize the Möbius transformations that maps the unit disk into itself?.)

If $T$ is the restriction of an entire function $f$ then necessarily $a=0$ (otherwise $f$ would have a pole at $z= 1/\overline a$).

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  • $\begingroup$ Thank you. I will look into that site $\endgroup$ – T C May 27 '20 at 18:57
  • $\begingroup$ it's amazing how you don't use any theorem. Just computations $\endgroup$ – T C May 27 '20 at 18:58
  • $\begingroup$ in the site you cited, they proe that Mobius trnasformation which is a bijection on unit disk must have your form. However, in my problem this function is not a Mobius transformation, right? $\endgroup$ – T C May 27 '20 at 19:31
  • $\begingroup$ @TC: All holomorphic bijections of the unit disk are Möbius transformations. I'll see if I can find a reference. $\endgroup$ – Martin R May 27 '20 at 19:32
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    $\begingroup$ @TC: See for example Theorem 2.5 here wwwf.imperial.ac.uk/~dcheragh/Teaching/2016-F-GCA/…. The idea is that you apply the Schwarz Lemma to both $f$ and its inverse, and conclude that equality holds. $\endgroup$ – Martin R May 27 '20 at 19:39

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