Proposition 3.1.33C5D6FBB32967

Proof. The map $g\mapsto H_\Omega (g|_{\partial \Omega })(x)$ is a positive linear functional on $C_ c(\mathbb R^ d,\mathbb R)$, so by theorem 2.3.3 it is integration against a Radon measure $\mu $ on $\mathbb R^ d$. If $g\in C_ c(\mathbb R^ d)$ is supported in the open set $\mathbb R^ d\setminus \partial \Omega $, then $g|_{\partial \Omega }=0$ and $\int g\, \mathrm d\mu =0$; by (2.3.4), $\mu (\mathbb R^ d\setminus \partial \Omega )=0$. If $\eta \in C_ c(\mathbb R^ d)$ equals one on the compact set $\partial \Omega $, then $\mu (\mathbb R^ d)=\mu (\partial \Omega )=\int \eta \, \mathrm d\mu =H_\Omega 1(x)=1$. Let $\omega _ x$ be the restriction of $\mu $ to the Borel subsets of $\partial \Omega $; it is a probability measure.

Every $f\in C(\partial \Omega )$ is a uniform limit of restrictions of functions in $C_ c(\mathbb R^ d)$, by lemma 3.1.26. Both sides of the asserted identity are continuous for uniform convergence on $\partial \Omega $, the right side because $\omega _ x$ is finite and the left side because $|H_\Omega f|\leq \max _{\partial \Omega }|f|$. Hence the identity holds for real $f$, and by linearity for complex $f$. If $\omega '$ is another Borel probability measure on $\partial \Omega $ with the same property, then $\omega _ x$ and $\omega '$, extended by zero to $\mathbb R^ d$, are finite Borel measures, hence Radon measures by proposition 2.2.2, with the same integrals against every $g\in C_ c(\mathbb R^ d)$. The uniqueness assertion of theorem 2.3.3 gives $\omega '=\omega _ x$. The function $x\mapsto \int f\, \mathrm d\omega _ x=H_\Omega f(x)$ is harmonic by the definition of $H_\Omega $. If $\Omega $ is a ball, $P_\Omega (x,\cdot )\, \mathrm dS$ is a Borel probability measure on $\partial \Omega $ with the required property, by theorem 3.1.18, so it equals $\omega _ x$. $\square$