Theorem 3.1.1808ED9D7CF1CF

Proof. If $d=1$ and $B=(a,b)$, then $H_ Bf$ is the affine function with the values $f(a)$ and $f(b)$ at the endpoints, as noted above. It is harmonic on $B$ and extends continuously to $\overline B$ with boundary values $f$, and the kernel is positive with $P_ B(x,a)+P_ B(x,b)=1$. The remaining assertions are proved below in all dimensions.

Now let $d\geq 2$; after a translation we may assume that $c=0$. Fix $\zeta \in \partial B$ and put $A(x)=R^2-|x|^2$ and $t(x)=|x-\zeta |$ for $x\neq \zeta $. We have $\nabla A=-2x$, $\Delta A=-2d$ and $\nabla t^{-d}=-dt^{-d-2}(x-\zeta )$, and (3.1.15), applied to $F(t)=t^{-d}$ after translating $\zeta $ to the origin, gives $\Delta t^{-d}=\bigl (d(d+1)-d(d-1)\bigr )t^{-d-2}=2dt^{-d-2}$. By the product rule $\Delta (At^{-d})=(\Delta A)t^{-d}+2\nabla A\cdot \nabla t^{-d}+A\Delta t^{-d}$, and since $t^2=|x|^2-2x\cdot \zeta +|\zeta |^2$,

Thus $P_ B(\cdot ,\zeta )$ is harmonic on $B$ for every $\zeta \in \partial B$. Every partial derivative $\partial _ x^\alpha P_ B(x,\zeta )$ is continuous on $B\times \partial B$, hence bounded on $K\times \partial B$ for each compact set $K\subset B$, where $|x-\zeta |\geq \operatorname {dist}(K,\partial B)>0$. By the mean value theorem and dominated convergence on the finite measure space $(\partial B,S)$, we may therefore differentiate under the integral sign any number of times, so $H_ Bf$ is smooth on $B$ and $\Delta H_ Bf(x)=\int _{\partial B}\Delta _ xP_ B(x,\zeta )f(\zeta )\, \mathrm dS(\zeta )=0$.

Next we show that the kernel has integral one. Taking $f=1$, we see that $I(x)=\int _{\partial B}P_ B(x,\zeta )\, \mathrm dS(\zeta )$ is harmonic on $B$. It is also radial. Indeed, $P_ B(Qx,Q\zeta )=P_ B(x,\zeta )$ for every orthogonal transformation $Q$, and surface measure on $\partial B$ is invariant under $Q$, so $I(Qx)=I(x)$. Moreover, if $|x|=|x'|$ and $x\neq x'$, the orthogonal reflection $y\mapsto y-2(y\cdot e)e$ with $e=(x-x')/|x-x'|$ maps $x$ to $x'$. Hence $I(x)=F(|x|)$, where $F(r)=I(re_1)$ is smooth on $(-R,R)$ and $e_1$ is the first coordinate vector. By (3.1.15), $(r^{d-1}F'(r))'=0$ on $(0,R)$, so $r^{d-1}F'(r)$ is constant there. Since $F'$ is bounded near $0$ and $d\geq 2$, this constant is $\lim _{r\to 0}r^{d-1}F'(r)=0$. Hence $F$ is constant on $(0,R)$, and since $I$ is continuous at the origin, $I$ is constant on $B$, with value $I(0)=R^2\cdot s_{d-1}R^{d-1}/(s_{d-1}R\cdot R^ d)=1$.

For the boundary behavior, fix $\zeta _0\in \partial B$ and $\varepsilon >0$, and choose $\delta >0$ such that $|f(\zeta )-f(\zeta _0)|<\varepsilon $ for $\zeta \in \partial B$ with $|\zeta -\zeta _0|<\delta $. Since the kernel is positive with integral one,

If $|x-\zeta _0|<\delta /2$, then $|x-\zeta |\geq \delta /2$ in the last integral, so $P_ B(x,\zeta )\leq (R^2-|x-c|^2)/(s_{d-1}R(\delta /2)^ d)$ there, and the integral is at most $R^{d-2}(2/\delta )^ d(R^2-|x-c|^2)$. This tends to zero as $x\to \zeta _0$, because $|\zeta _0-c|=R$. Hence $\varlimsup _{x\to \zeta _0}|H_ Bf(x)-f(\zeta _0)|\leq \varepsilon $ for every $\varepsilon >0$. Since $f$ is continuous on $\partial B$, it follows that the function equal to $H_ Bf$ on $B$ and to $f$ on $\partial B$ is continuous on $\overline B$.

Uniqueness follows from the last assertion of proposition 3.1.10. The operator $H_ B$ is linear by definition, and since the kernel is positive with integral one, $H_ B$ preserves order and $|H_ Bf(x)|\leq \int _{\partial B}P_ B(x,\zeta )|f(\zeta )|\, \mathrm dS(\zeta )\leq \max _{\partial B}|f|$. Finally, if $h$ is harmonic on a neighborhood of $\overline B$, then $h$ and $H_ B(h|_{\partial B})$ are harmonic on $B$, continuous on $\overline B$ and equal on $\partial B$, so they coincide. $\square$