PomeTheorem 3.5.12.9CF388990FE4 If $f\in L^1(\Omega )\cap L^\infty (\Omega )$, then $G_\Omega f$ is finite continuous on $\Omega $. For a ball and compactly supported measure its potential tends to zero at every boundary point. More generally suppose $\Omega $ is connected, has finite volume for $d\ge 3$, and is bounded for $d=2$. If $G_\Omega (x,y_0)\to 0$ as $x\to a\in \partial \Omega $ for one $y_0\in \Omega $, then $G_\Omega f(x)\to 0$ for every bounded measurable $f$.
Theorem 3.5.129CF388990FE4
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
\[ \sup _ x\int _{\Omega \setminus K_ j}G(x,y)|f(y)|dy \le \tfrac 12s_{d-1}\| f\| _\infty (|\Omega \setminus K_ j|/v_ d)^{2/d}\longrightarrow 0. \]
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. $\square$
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