Balayage

Remarks

Write $U^\mu(y) = \int \abs{y - x}^{2-n}\, d\mu(x)$ for the Newtonian potential. Then $\mu'(\partial D) = \mu(D)$ and $U^{\mu'} = U^\mu$ on $\R^n \setminus D$. Indeed, for $y \notin \overline{D}$ the function $x \mapsto \abs{y - x}^{2-n}$ is harmonic on $D$ and continuous on $\overline{D}$, so it is the integral of its boundary values against $\omega_x$. Inside $D$ one has $U^{\mu'} \le U^\mu$. Example: let $D$ be the ball of radius $R$ about $0$ and $\mu = \delta_0$. Then $\omega_0$ is the normalised surface measure on the sphere, so $\mu'$ is that measure. Its potential is $\min(R^{2-n}, \abs{y}^{2-n})$, which equals $\abs{y}^{2-n} = U^{\delta_0}(y)$ for $\abs{y} \ge R$. Non-example: $\mu$ itself is not a balayage of $\mu$ onto $\partial D$, since it is not carried by $\partial D$. More generally, for any set $E \subseteq \R^n$ the balayage $\mu^E$ is the measure whose potential is the regularised reduced function $\widehat{R}^E_{U^\mu}$.