Actually, I'm solving the following problem. there are some steps I can't understand. Can you guys help me to understand?

The problem is: Find all entire functions that map the unit circle to itself. (problem from Rudin's real & complex analysis chapter 12 ex.4)

excluding the constant function, I first showed that $f$ should have zero inside unit disk and $f$ should map open unit disk into itself by using maximum modulus theorem.

Lots of proof I found then say that its zero should locate at only origin. That's the first part that I cannot understand.

and then they consider $ g(z) = [\bar f(\frac{1}{\bar z})]^{-1} $ and showed that $g(z) = f(z)$ on unit circle. which has a limit point in $C - \{0\}$. so by identity theorem, (that's the second part; I'm not sure $g(z)$ is even analytic except some singular point, since it involves conjugation.) $f(z) = g(z)$ on $C-\{0\}$. Then by considering the order of pole at $0$, we can conclude that.

To study further, I tried to find lots of materials and above discussion may due to identity theorem for meromorphic functions, analytic continuation, etc. But we never learned this.

I think there may be an easier way by just using elementary properties of analytic function or schwarz lemma. Can you help me please?


Hint: $f(z) \overline{f(1/\overline{z})}$ is analytic on ${\mathbb C} \backslash \{0\}$. What is it on the unit circle?

  • $\begingroup$ I think it is 1. But we must priori prove that 0 is the only zero for f inside unit disk?? $\endgroup$ – Detectives Nov 8 '12 at 6:25
  • $\begingroup$ If it is $1$ on the unit circle, it is $1$ everywhere in ${\mathbb C}\backslash \{0\}$. But how could it be $1$ at a point where $f(z) = 0$? $\endgroup$ – Robert Israel Nov 8 '12 at 7:00
  • $\begingroup$ I'm not sure about the analyticity of given function on C/{0} since it involves conjugation, which is not analytic in general.. I got the point about using identity theorem to conclude that f==1 on C/{0} so the only zero of f is 0 $\endgroup$ – Detectives Nov 8 '12 at 7:38
  • $\begingroup$ It's a simple exercise to prove that if $f$ is differentiable at $z=p$, then $z \to \overline{f(\overline{z})}$ is differentiable at $z = \overline{p}$. $\endgroup$ – Robert Israel Nov 8 '12 at 22:20
  • $\begingroup$ Or if you prefer you can use power series: if $f(z) = \sum_n a_n (z - p)^n$ near $z=p$, then $\overline{f(\overline{z})} = \sum_n \overline{a_n} (z - \overline{p})^n$ near $z=\overline{p}$. $\endgroup$ – Robert Israel Nov 8 '12 at 22:24

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