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.

Let $G$ be a bounded domain, $f$ and $g$ holomorphic and continuous functions on whole $\overline{G}$ and let $|f(z)|=|g(z)|$ for all $z\in \partial G$. What can we say about $f$ and $g$?

Define: $h(z) = \frac{f(z)}{g(z)}$, if g has zero-points, then we cant say much because of the singularities. If g has no zero points, then with the maximum principle for bounded domains it follows that h takes its maximum on the edge. Because |f(z)|=|g(z)| for all $z\in \partial G$ it follows that the modulus is 1. If now f has zero points, then we cant say anything more about it. If f has no zero points, then we can use the minimum principle for bounded domains. With that it follows that also the modulus of the minimum must be 1.

From that we can conclude that : $|h(z)|=\frac{|f(z)|}{|g(z)|}=1$ for all $z\in G$. So if f and g have no zero points, $f(z)=\lambda g(z)$

Is this correct?

share|improve this question
    
What do you expect? With $f(z)=z$ and $g(z)=z^2$, $G=\{z,|z|<1\}$, we can see we can't have equality in the whole $G$ in general. –  Davide Giraudo Sep 10 '12 at 20:13

4 Answers 4

If $f$ and $g$ doesn't have any $0$ in $G$ we can conclude, by maximum modulus principle, that $|f(z)|=|g(z)|$ for all $z\in G$.

We can't get that when there is a zero. For example, $G$ is the open unit disk and $f(z)=z$, $g(z)=z^2$. Then $|g(z)|=|f(z)|(=1)$ on $\partial G$, but of course not on $g$.

share|improve this answer

In general, we cannot say much about $f$ and $g$. For example, suppose that $G$ is a Jordan domain. Then $G$ is simply connected, so there is a conformal map $f$ of $G$ onto the unit disk. Moreover, it is a well known theorem that in this case, $f$ extends to a homeomorphism of $\overline{G}$ onto the closed unit disk, mapping the boundary of $G$ onto the unit circle. Hence $|f(z)|=|g(z)|$ on $\partial G$, where $g \equiv 1$...

share|improve this answer

If $G$ is a Jordan domain (that is, $\partial G$ is a simple closed curve), we can reduce the problem to the case when $G$ is the unit disk $D=\{z:|z|<1\}$ by a conformal change of variable. Then the Schwarz reflection principle applies, allowing us to extend $f/g$ (originally a meromorphic function on $D$) to a meromorphic function on $\overline{\mathbb C}$. It follows that $f/g$ is a rational function, more precisely the ratio of two finite Blaschke products.

The argument does not seem to work in multiply connected domains. I don't know how bad $f/g$ can be in general.

share|improve this answer

Define: $h(z) = \frac{f(z)}{g(z)}$, if g has zero-points, then we cant say much because of the singularities. If g has no zero points, then with the maximum principle for bounded domains it follows that h takes its maximum on the edge. Because |f(z)|=|g(z)| for all $z\in \partial G$ it follows that the modulus is 1. If now f has zero points, then we cant say anything more about it. If f has no zero points, then we can use the minimum principle for bounded domains. With that it follows that also the modulus of the minimum must be 1.

From that we can conclude that : $|h(z)|=\frac{|f(z)|}{|g(z)|}=1$ for all $z\in G$. So if f and g have no zero points, $f(z)=\lambda g(z)$

Is this correct?

share|improve this answer

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.