2
$\begingroup$

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 $F(x)=\exp(i \psi(x)), x\in \mathbb R$.

  • 0
    @all: I mistakenly identified [Analytic Functions in the Upper Half Plane](http://math.stackexchange.com/questions/93614/analytic-functions-in-the-upper-half-plane) (which is closely related) as a duplicate thread. Please ignore my vote to close.2012-01-03

1 Answers 1

0

If you can extend the inner function across the real line then the singular measure associated to the inner function can have support only at the point at infinity. The inner function is hence F(z) = B(z)*exp(iaz) with B a Blaschke product and a >= 0.

Now every zero of the Blaschke product contributes an increasing function of finite variation to the phase of F on the line, and the exponent also contributes a linear increasing function. The phase of F is hence an increasing function.