3.2 Kelvin transforms and half-space potentials9728BD930155

Proof. Translate $a$ to $0$, and set $r=|x|$, $e=x/r$, $Q=I-2ee^{\mathsf T}$, $y=T(x)=\rho ^2x/r^2$ and $w(x)=\rho ^{d-2}r^{2-d}$. The matrix $Q$ is symmetric and orthogonal, and $Qx=-x$. Differentiating $y_ j=\rho ^2x_ jr^{-2}$ gives

For completeness, $\Delta r^\alpha =\alpha (\alpha +d-2)r^{\alpha -2}$ follows by differentiating $\partial _ jr^\alpha =\alpha x_ jr^{\alpha -2}$ and summing. Applying the product rule to $x_ jr^{-2}$ proves the second identity above. The same radial formula yields

The chain rule and the orthogonality of $Q$ now give

In $\Delta (w(u\circ T))$, the terms containing $x\cdot \nabla u(y)$ therefore cancel, leaving $\rho ^{d+2}r^{-d-2}(\Delta u)(y)$. This proves (3.2.3), including $d=2$.

Since $|T(x)|=\rho ^2/|x|$, substitution gives $T(T(x))=x$ and $w(x)w(T(x))=1$. In particular $T$ is a smooth diffeomorphism of $\mathbb R^ d\setminus \{ 0\} $ onto itself, so $T(U)$ is open and $K^2u=u$. The positive weight $w$ preserves signs. The Laplacian identity and involutivity prove the assertion about harmonicity. $\square$

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. $\square$

Proof. We first verify the normalization

Translation and the substitution $z=x'+tw$ reduce the integral to $\frac{2}{s_{d-1}}\int _{\mathbb R^{d-1}}(1+|w|^2)^{-d/2}\, dw$. The map

parametrizes the upper hemisphere. With $m=d-1$, its metric matrix is $DF^{\mathsf T}DF=(1+|w|^2)^{-1}I-(1+|w|^2)^{-2}ww^{\mathsf T}$. Its eigenvalues are $(1+|w|^2)^{-1}$ in the $m-1$ perpendicular directions and $(1+|w|^2)^{-2}$ in the direction of $w$. The square root of its determinant is $(1+|w|^2)^{-d/2}$. At $w=0$ this formula holds by continuity. The surface area formula and reflection in the equatorial plane show that the last integral is $s_{d-1}/2$, proving (3.2.13). This also treats $d=2$, when $m=1$ and there are no perpendicular directions.

The weighted measure $|f(z)|\, dz$ satisfies (3.2.8), by the same integral at $(0,1)$. Applying the differentiation estimates in the proof of Theorem 3.2.6 to the positive and negative parts of $f$ proves harmonicity. The positivity and normalization of $P_ H$ give the stated bound.

Fix $\varepsilon >0$. At a continuity point $z_0$, choose $\delta >0$ so that $|f(z)-f(z_0)|<\varepsilon $ for $|z-z_0|<\delta $. The contribution of this set to $|\int P_ H(x,z)(f(z)-f(z_0))\, dz|$ is at most $\varepsilon $. If $|x'-z_0|<\delta /2$ and $|z-z_0|\geq \delta $, then $|z-x'|\geq \delta /2$. Writing $s_{d-2}$ for the area of the unit sphere in $\mathbb R^{d-1}$ (in particular $s_0=2$), polar integration gives

Thus

near $(z_0,0)$. Letting $(x',t)\to (z_0,0)$ and then $\varepsilon \downarrow 0$ proves the boundary limit. Finally $|u(0,t)/t-c|\leq \| f\| _\infty /t\to 0$. $\square$