3.5.10 Continuity of potentials and bounded densities7AECB61AA15F

Proof. Near $K$ the Green potential differs from its fundamental potential by a harmonic function. Both potentials are locally integrable and have the same distributional Laplacian, so Weyl's lemma gives a harmonic difference. Adding that harmonic function preserves superharmonicity; canonical representatives then identify the equality everywhere, including infinite values. No joint continuity of a cancelled diagonal kernel is assumed.

It suffices to prove the first assertion for $U\mu (x)=\int \Phi (x-y)d\mu (y)$. If $U\mu (a)=\infty $, lsc suffices. Otherwise choose a small ball $B$ about $a$ and split $\mu =\mu '+\mu ''$, with $\mu '=\mu |_ B$, so that the local singular integral $U\mu '(a)<\eta $. In the planar case take all relevant distances less than one so this local kernel is nonnegative. The remainder $U\mu ''$ is harmonic near $a$. If $z_ x\in K$ is a nearest point to $x$, then $z_ x\to a$ as $x\to a$, and $|z_ x-y|\le 2|x-y|$ for $y\in K$. For $d\ge 3$ this gives $U\mu '(x)\le 2^{d-2}U\mu '(z_ x)$. For $d=2$ it gives $U\mu '(x)\le U\mu '(z_ x)+(\log 2)\mu '(B)$. The restriction of $U\mu '$ to $K$ is continuous at $a$, being the difference of the assumed continuous restriction of $U\mu $ and the harmonic remainder. Thus the limsup of the local contribution is at most $2^{d-2}\eta $, or $\eta +(\log 2)\mu (B)$ in dimension two. In that case $\mu (\{ a\} )=0$ by finiteness of the potential, so $\mu (B)\to 0$ as the radius shrinks. The harmonic remainder is continuous; letting $\eta \to 0$ proves continuity.

For the second assertion, Lusin's theorem gives compact $C\subset K$ with lost mass less than $\varepsilon $ and $G\mu |_ C$ finite continuous. On $C$, $G(\mu |_ C)=G\mu -G(\mu |_{K\setminus C})$ is upper semicontinuous, since the second term is lsc; it is also lsc. It is therefore continuous on $C$, and the first assertion gives continuity at every point of $C$. Off $C$ it is harmonic. This proves global finite continuity. $\square$

Proof. Separate the positive and negative parts of $f$. They are finite measures, so have superharmonic potentials. On an interior ball, the contribution from its complement is harmonic; the local contribution differs from a fundamental potential by a harmonic function. Its singular integral over a ball of radius $r$ is uniformly at most $\| f\| _\infty s_{d-1}r^2/2$ for $d\ge 3$, and $2\pi \| f\| _\infty \int _0^ r s\log (1/s)ds$ in the plane, after adding a harmless local constant. These bounds tend to zero. To justify that a ball centered elsewhere has no larger integral of the radial decreasing positive kernel, use its layer-cake representation: intersection with every centered superlevel ball has volume at most the smaller of the two ball volumes. The nonsingular part is continuous by dominated convergence, proving the first assertion and finiteness.

For the boundary assertions, Harnack chains in the pole variable make $G(x,\cdot )\to 0$ uniformly on any fixed compact subset when it tends to zero at one pole. The explicit ball kernel supplies this condition at every ball boundary point, so compactly supported potentials tend to zero. Exhaust $\Omega $ by compact $K_ j$. In dimension at least three, $G(x,y)\le |x-y|^{2-d}$, and the layer-cake estimate just proved yields

For a bounded planar domain choose $D>\operatorname {diam}\Omega $; $G(x,y)\le \log (D/|x-y|)$ by minimality, and the same estimate is the integral of this radial kernel over a disk of the tail's area, tending to zero. Thus the compact-support integrals approximate $Gf$ uniformly in $x$, and their boundary limits are zero. $\square$