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.

this is an exercise I came across in Rudin's "Real and complex analysis" Chapter 16.

Suppose $\Omega$ is the complement set of $E$ in $\mathbb{C}$, where $E$ is a compact set with positive Lebesgue measure in the real line.

Does there exist a non-constant bounded holomorphic function on $\Omega$?

Especially, do this for $\Omega=[-1,1]$.


Some observations:

Suppose there exists such function $f$, then WLOG, we may assume $f$ has no zeros points in $\Omega$ by adding a large enough positive constant, then, $\Omega$ is simply-connected implies $\int_{\gamma}fdz=0$, for any closed curve $\gamma\subset \Omega$, how to deduce any contradiction?

share|improve this question
    
Why do you say that $\Omega$ is simply connected? // You are going in the wrong direction: such a function exists on any domain with complement of positive measure on the line. Think of the Cauchy integral formula. // In the case of $E=[-1,1]$ (you have a typo there in the question) it's possible to find it explicitly using conformal maps. –  user53153 Jan 3 '13 at 0:16
    
@Pavel M, the function constructed using Cauchy integral formula is not bounded on $\Omega$, this is the first several questions before this one on the book, you can check it. That is why Rudin post this question on the book. –  ougao Jan 3 '13 at 0:23
    
Maybe it depends on how you use the Cauchy integral. Anyway, here's an example for $E=[-1,1]$: the conformal map $g(z)=\frac12(z+z^{-1})$ sends the unit disk onto the complement of $E$. Therefore, its inverse $f=g^{-1}$ is bounded by $1$ on the complement of $E$. // At least this shows that your attempt to find a contradiction could not work. –  user53153 Jan 3 '13 at 0:37
    
You are correct, I double checked the problem, and found I have made a silly typo, in fact, I want to know whether there exists a non-constant bounded holomorphic(not just analytic) function, sorry for the confusion. –  ougao Jan 3 '13 at 1:06
    
Does not change anything in my replies. I assumed you wanted a holomorphic function from the beginning. –  user53153 Jan 3 '13 at 1:34

1 Answer 1

up vote 6 down vote accepted

Reading Exercise 8 of Chapter 16, I imagine Rudin interrogating the reader.

Let $E\subset\mathbb R$ be a compact set of positive measure, let $\Omega=\mathbb C\setminus E$, and define $f(z)=\int_E \frac{dt}{t-z}$. Now answer me!
a) Is $f$ constant?
b) Can $f$ be extended to an entire function?
c) Does $zf(z)$ have a limit at $\infty$, and if so, what is it?
d) Is $\sqrt{f}$ holomorphic in $\Omega$?
e) Is $\operatorname{Re}f$ bounded in $\Omega$? (If yes, give a bound)
f) Is $\operatorname{Im}f$ bounded in $\Omega$? (If yes, give a bound)
g) What is $\int_\gamma f(z)\,dz$ if $\gamma$ is a positively oriented loop around $E$?
h) Does there exist a nonconstant bounded holomorphic function on $\Omega$?

Part h) appears to come out of the blue, especially since $f$ is not bounded: we found that in part (e). But it is part (f) that's relevant here: $\operatorname{Im}f$ is indeed bounded in $\Omega$ (Hint: write it as a real integral, notice that the integrand has constant sign, extend the region of integration to $\mathbb R$, and evaluate directly). Therefore, $f$ maps $\Omega$ to a horizontal strip. It's a standard exercise to map this strip onto a disk by some conformal map $g$, thus obtaining a bounded function $g\circ f$.

share|improve this answer
    
thanks, I should be more careful to deal with part (f). –  ougao Jan 3 '13 at 14:20

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.