# Entire function that is a bijection on the unit disk is a rotation

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.

• $f(z)= 2z$ is entire but is a bijection of $\Delta(0,1)$ onto its image without being a rotation. Is there any further asumption? – DIdier_ May 27 '20 at 18:17
• 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 – Conrad May 27 '20 at 18:17
• @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 – Conrad May 27 '20 at 18:21

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 $$z= 1/\overline a$$).
• @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. – Martin R May 27 '20 at 19:39