3.5.13 The generalized Laplacian589DA36A9C17

Proof. Near a fixed $x$, the contribution of the measure outside a small closed ball is harmonic, and the local Green potential differs from its fundamental potential by a harmonic function. Harmonic functions contribute zero to the generalized Laplacian. The sphere average of $\Phi (\cdot -y)$ at center $x$ is $\Phi (r)$ when $|x-y|\le r$ and $\Phi (x-y)$ otherwise. For the second case this is the harmonic mean property. For the first, rotation invariance gives a radial harmonic function of $y$ inside the sphere, smooth there by differentiation, hence constant, with its value at the center equal to $\Phi (r)$. Here is uniform control at the sphere: its surface measure in any ball of radius $s$ centered on or near it is at most $Cs^{d-1}$, by finitely many bounded graph charts. Layer-cake integration therefore bounds its fundamental singular integral on such a cap by $Cs$ for $d\ge 3$, and by $Cs(1+|\log s|)$ for $d=2$. These bounds tend to zero uniformly in the pole. The remaining kernel is uniformly continuous, proving continuity across the sphere. Consequently

The bracket is nonpositive. Its absolute integral is $\pi r^2/2$ for $d=2$, since $2\pi \int _0^ r s\log (r/s)ds=\pi r^2/2$; for $d\ge 3$ it is $s_{d-1}r^2(1/2-1/d)$. Thus the expression for constant $f(x)$ is exactly $-c_ df(x)$, while the error is at most $c_ d\sup _{|y-x|<r}|f(y)-f(x)|$, tending to zero.

For the converse fix a ball $B\Subset \Omega $, let $H$ be the continuous Poisson solution with boundary values $u$, and let $w=G_ B1$, continuous, zero at the boundary, with $\widetilde\Delta w=-c_ d$ by the first part. For $t>0$, both $u-H+tw$ and $H-u+tw$ have strictly negative generalized Laplacian. Hence at every point they satisfy the strict local supermean inequality for all sufficiently small radii. The elementary minimum principle for continuous locally supermean functions, proved by the open-and-closed minimum set argument, implies they are nonnegative on $B$ because their boundary values are zero. Let $t\downarrow 0$; then $u=H$. As $B$ was arbitrary, $u$ is harmonic. $\square$