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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F(z)$ be an inner function in the upper half plane, (i.e. $F$ is bounded analytic function such that $\lim_{y\to 0^{+}}|F(x+iy)|=1$ for almost all $x\in \mathbb R$ with respect to the Lebesgue measure). I need to prove that:

If $F$ admits an analytic extension across the real axis (hence is meromorphic in the whole complex plane) then there is a well-defined branch of the argument of $F$ on the line, i.e., an increasing differentiable function $\psi$ such that $F(x)=\exp(i \psi(x)), x\in \mathbb R$.

share|cite|improve this question
@all: I mistakenly identified Analytic Functions in the Upper Half Plane (which is closely related) as a duplicate thread. Please ignore my vote to close. – t.b. Jan 3 '12 at 19:47

Your Answer


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

Browse other questions tagged or ask your own question.