Theorem 3.2.6 (Half-space representation)6BC86EEF03A9

Proof. We first derive the ball representation needed here from the ball Poisson formula and compactness of bounded positive measures on a compact set. Let $v\geq 0$ be harmonic on $B=B(b,R)$ and let $0<r<R$. On $\partial B$ define the measure

The mean-value property gives $\mu _ r(\partial B)=s_{d-1}R^{d-1}v(b)$. For a sequence $r_ j\uparrow R$, compactness therefore gives a subsequence converging weakly to a finite positive measure $\mu $ on $\partial B$. To apply the Euclidean measure-selection theorem, extend $\mu _{r_ j}$ by zero to $\mathbb R^ d$; their common compact support gives uniform tightness, and their weak limit is supported on that same closed sphere. For $|x-b|<r$, the ball Poisson formula, followed by the dilation from $\partial B$ to $\partial B(b,r)$, reads

For fixed $x\in B$ the integrands converge uniformly on $\partial B$ as $r\uparrow R$. The masses are uniformly bounded, so weak convergence gives

This argument also covers $v(b)=0$, in which case all measures and $v$ vanish.

Take $a=-e_ d$, $b=-e_ d/2$, $R=1/2$, and $T(x)=a+(x-a)/|x-a|^2$. Write $q=T(x)$ and $L=|x-a|$. Direct expansion gives

For the last identity, square both sides and expand the inner products; the numerator on the left becomes $|x-a|^2+|y-a|^2-2(x-a)\cdot (y-a)=|x-y|^2$ after multiplying by $|x-a|^2|y-a|^2$. The middle identity and $T^2=1$ show that $T$ maps $H$ onto $B(b,1/2)$ and maps $\partial H$ onto $\partial B\setminus \{ a\} $.

By Theorem 3.2.2, $v(q)=|q-a|^{2-d}h(T(q))$ is nonnegative and harmonic on $B$. Apply (3.2.9) and split $\mu =m\delta _ a+\mu _0$, where $\mu _0(\{ a\} )=0$. Since $h(x)=L^{2-d}v(T(x))$, the atom contributes $2mt/s_{d-1}$. Let $\sigma =T_*\mu _0$ on $\partial H$, and put

The measure $\sigma $ is finite and positive, and its continuous positive weight is bounded on compact sets. Hence $\nu $ is a positive Radon measure. The distance identity in (3.2.10) transforms the remaining integral into

Also $|(z,0)-a|^ d=(1+|z|^2)^{d/2}$, so the integral in (3.2.8) equals $\sigma (\mathbb R^{d-1})<\infty $. This proves the required representation.

For the converse, each kernel is smooth and harmonic as a function of $x\in H$. Indeed, with $r=|x-(z,0)|$, the radial differentiation formula used above gives $\Delta r^{-d}=2d r^{-d-2}$ and $\partial _ t r^{-d}=-dt r^{-d-2}$. Consequently $\Delta (t r^{-d})=2dt r^{-d-2}-2dt r^{-d-2}=0$. On a compact subset $K\subset H$, $t$ is bounded above and away from $0$, and $|x'|$ is bounded. Writing $v=x-(z,0)$, differentiation gives

For $|z|$ large, $r\geq |z|/2$ and $|v_ i|\leq r$, so these formulas bound the kernel and all its $x$ derivatives of orders at most two by $C_ K(1+|z|^2)^{-d/2}$. For bounded $z$, the lower bound on $t$ supplies the same estimate after increasing $C_ K$. These estimates also hold on a compact neighborhood of $K$ inside $H$, so dominated convergence justifies two differentiations under the integral. Condition (3.2.8) gives finiteness and harmonicity; positivity is immediate. $\square$