Analysis Problem of the Day 80

Today’s problem appeared as Problem 6 on the UCLA Fall 2009 Analysis Qual:

Problem 6. Recall that for a continuous function $h$ on the closed unit disc $|z| \leq 1$ that is harmonic on the open unit disc, for the Poisson kernel

$P_\rho(\theta) := \text{Re}\left(\frac{1+\rho e^{i\theta}}{1-\rho e^{-i\theta}}\right),$

one has the representation formula

$h(re^{i\eta}) = \frac{1}{2\pi}\int_0^1 P_{\frac{r}{R}}(\eta-\theta) h(Re^{i\theta}) d\theta$

for $r<1, \eta \in [0,2\pi).$ Show that if $f$ is harmonic positive function on the open unit disc, there exists a positive Borel measure on $[0,2\pi]$ such that

$f(re^{i\eta}) = \frac{1}{2\pi}\int_0^1 P_{r}(\eta-\theta) d\mu(\theta).$

Solution: Note that a harmonic function is defined by the mean value property, i.e. $\frac{1}{2\pi}\int_0^{2\pi} f(r\theta)d\theta = f(0)>0.$ Without loss of generality, suppose $f(0)=1,$ so that $\frac{1}{2\pi} f(r\theta) d\theta:= d\mu_r$ for a probability Borel measure $\mu_r$ on $[0,2\pi].$ Now, by Prokhorov’s theorem, $\{\mu_r\}$ is a collection of probability measures on a compact space, and is therefore weak $-^*$ sequentially compact, i.e. there exists a sequence $\mu_{r_n} \to \mu$ for some Borel measure $\mu$ in the weak topology. We now show $\mu$ satisfies the required representation formula. Indeed, by the definition of the topology on the space of measure,