Lemma 3.5.167C49056207BE

Proof. Shrink $R$ while keeping $r<R$, and subtract a lower bound so $u\ge 0$ on a neighborhood of the new closed ball. Reduce $u$ onto $K=\overline{B(a,r)}$ relative to $B(a,R)$. The regularized reduction $p$ equals $u$ inside $K$, is a potential with measure supported in $K$, and is harmonic outside $K$. Its measure is finite by compact support, so $p\to 0$ on the outer sphere. Harmonic reflection extends it to a slightly larger annulus; denote the extension again by $p$. Let $F(x)=\alpha (\Phi (x-a)-\Phi (Re_1))$. It is radial harmonic outside $a$ and zero on the outer sphere. The normal derivative of $\Phi $ is strictly negative there. Since $p$ is smooth on a compact collar of that sphere and zero on it, a sufficiently large $\alpha >0$ makes $F\ge p$ on the inner half of a thinner collar and $F\le p$ on its outer half: divide the radial differences by $|x-a|-R$ and use uniform continuity of the radial derivatives. Define $u^*=p$ inside the sphere and $u^*=F$ outside. In the collar it is the minimum of two harmonic functions, hence superharmonic, so locality proves it is entire superharmonic. It is finite outside $K$. Finally restore the subtracted constant. $\square$