I have been reading Harmonic Measure by John Garnett and Donald Marshall. Their formulation of Beurling theorem (page 105) is:
If $E\subset\overline{\mathbb{D}}\setminus\{0\}$, and $E^*=\{|z|: z\in E\}$ is the circular projection of $E$, then $$\omega(z,E,\mathbb{D}\setminus E)\geqslant\omega(-|z|,E^*,\mathbb{D}\setminus E^*),$$
where $\omega$ is harmonic measure.
First, they claim that by Green's theorem $$\omega(z)=\omega(z,E,\mathbb{D}\setminus E) = \frac{1}{2\pi}\int_{\partial E} g(z,\zeta)\frac{\partial \omega(\zeta)}{\partial n_\zeta}ds,$$
where $g$ is Green's function. I think that this statement is true when Green's function is defined on region $\mathbb{D}\setminus E$.
1.question : Is it true that $g$ can be defined on $\mathbb{D}$ and previous equatation is valid for that $g$? (I think it is not, but I'm not sure, because I've seen some authors to assume that.)
Then, they claim that $$g(-|z|,|\zeta|)\leqslant g(z,\zeta)\leqslant g(|z|,|\zeta|),$$ which is trivial when region is $\mathbb{D}$.
2.question: I don't see why is this inequality true when $g$ is defined on $\mathbb{D}\setminus E$. If it is valid, please can you help me to understand why?
If I suppose that something of my previous questions is true, then the rest of the proof is clear to me.
If these hypothesis are not true, can you please refere me to another proof of this theorem, or try to explain me this proof.
Sorry for my bad english.
Here, $g(z,\zeta)$ isn't the Green function for $\mathbb{D}\setminus{E}$ or some other complicated domain. Rather, $g(z,\zeta)=-\log{|z-\zeta|}$.
This is well- defined on $\mathbb{D}\setminus{\{z\}}$, as a function of $\zeta$ for fixed $z$, and the inequality
$$g(-|z|,|\zeta|)\leq{g(z,\zeta)}\leq{g(|z|,|\zeta|)}$$
does indeed hold trivially.
Edit: sorry, what I meant to say is that $g(z,\zeta)=-\log{\big|\frac{z-\zeta}{1-\overline{\zeta}z}\big|}$ which is the Green function for the disk from the perspective of $\zeta$. In this case, again, the desired inequality holds trivially. As for the formula you asked about, this follows by applying Green's formula. Namely, since $g(z,\zeta)$ and $\omega(z)$ are both harmonic on $\mathbb{D}\setminus{\{\zeta\}}$, for any sufficiently small $\varepsilon>0$ we have:
$$ \int_{\mathbb{D}\setminus{(E\cup{B(\zeta,\varepsilon)})}}g(z,\zeta)\Delta{\omega{(z)}}-\omega(z)\Delta{g(z,\zeta)}dA(z)=0=\int_{\partial{\mathbb{D}}}g(z,\zeta)\partial_{n}\omega{(z)}-\omega{(z)}\partial_{n}g(z,\zeta)d\sigma(z)+\int_{\partial{E}}g(z,\zeta)\partial_{n}\omega{(z)}-\omega{(z)}\partial_{n}g(z,\zeta)d\sigma(z)+\int_{\partial{B(\zeta,\varepsilon)}}g(z,\zeta)\partial_{n}\omega{(z)}-\omega{(z)}\partial_{n}g(z,\zeta)d\sigma(z)$$
where $dA(z)$ is the usual 2- dimensional area measure and $d\sigma{(z)}$ refers to integration with respect to arclength. Since $g(z,\zeta), \omega{(z)}=0$ for $z\in{\partial{\mathbb{D}}}$, the integral over $\partial{\mathbb{D}}$ is identically $0$. Also, we note that $\omega{(z)}=1$ for $z\in{\partial{E}}$ which is relevant for computing the integral over $\partial{E}$. In summa, we see that:
$$ \int_{\partial{E}}g(z,\zeta)\partial_{n}\omega{(z)}-\partial_{n}g(z,\zeta)d\sigma(z)= -\int_{\partial{B(\zeta,\varepsilon)}}g(z,\zeta)\partial_{n}\omega{(z)}-\omega{(z)}\partial_{n}g(z,\zeta)d\sigma(z)$$
Next, observe that:
$$ \int_{\partial{E}}\partial_{n}g(z,\zeta)d\sigma(z)= \int_{\partial{E}}\nabla_{z}g(z,\zeta)\cdot{n_{z}}d\sigma{(z)}=0$$
by the divergence theorem. This leaves us with:
$$ \int_{\partial{E}}g(z,\zeta)\partial_{n}\omega{(z)}d\sigma(z)= -\int_{\partial{B(\zeta,\varepsilon)}}g(z,\zeta)\partial_{n}\omega{(z)}-\omega{(z)}\partial_{n}g(z,\zeta)d\sigma(z)$$
Finally, using our explicit formula for $g(z,\zeta)$ along with the fact that $\omega(z)$ is smooth in a neighbourhood of $\zeta$ we get that:
$$ \int_{\partial{B(\zeta,\varepsilon)}}g(z,\zeta)\partial_{n}\omega{(z)}-\omega{(z)}\partial_{n}g(z,\zeta)d\sigma(z)=2\pi\omega{(\zeta)}+O(\varepsilon\log\big(\frac{1}{\varepsilon}\big))$$
Letting $\varepsilon$ tend to $0$ we get that:
$$ \omega(\zeta)=\frac{-1}{2\pi}\int_{\partial{E}}g(z,\zeta)\partial_{n}\omega{(z)}d\sigma(z)$$
Which is exactly what we want.