Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
I think it is 1. But we must priori prove that 0 is the only zero for f inside unit disk?? –  Detectives Nov 8 '12 at 6:25
    
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 $f(z) = 0$? –  Robert Israel Nov 8 '12 at 7:00
    
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 –  Detectives Nov 8 '12 at 7:38
    
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 $z = \overline{p}$. –  Robert Israel Nov 8 '12 at 22:20
    
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 $z=\overline{p}$. –  Robert Israel Nov 8 '12 at 22:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.