1
$\begingroup$

A singular inner function $M$ (an analytic function on the open unit disk without zeros which takes on unimodular boundary values almost everywhere) can be written as $$M(z)=c \exp\left(\int_0^{2\pi} \frac{z+e^{it}}{z-e^{it}} d\mu(t)\right)$$ for $|z|<1$ where $|c|=1$ and $\mu$ is a positive Borel measure that is singular w.r.t. Lebesgue (cf. Rudin, Real and Complex Analysis, ch. 17).

Is $\mu$ uniquely determined? How do I prove it?

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes, $\mu$ is unique. The function $-\log M$ is holomorphic with positive real part in the disk. The measure $\mu$ gives the Herglotz representation of $-\log M$. Why is such $\mu$ unique? Because subtracting two such measures $\mu_1,\mu_2$ we would have a nonzero measure $\nu=\mu_1-\mu_2$ on the boundary whose Poisson integral is identically zero. But the Poisson integral is $\sum r^{|n|}c_n e^{in\theta}$ where $c_n$ are the Fourier coefficients of $\nu$. So, all the coefficients must be zero. Therefore, all trigonometric polynomials have zero integral with respect to $\nu$. By their density, all continuous functions on the circle have zero integral with respect to $\nu$. Thus $\nu$ is the zero measure.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .