3.1.16 Poisson kernels and Harnack inequalities9C3CB4CF29A2

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. $\square$

Proof. (1) Let $r<R'<R$. By theorem 3.1.18 on $B(0,R')$, both $h(x)$ and $h(0)$ are integrals of the nonnegative boundary values of $h$ against the Poisson kernel. For $|\zeta |=R'$ we have $R'-r\leq |x-\zeta |\leq R'+r$, so the ratio

lies between the two coefficients in (3.1.20) with $R'$ in place of $R$. Integrating and letting $R'\uparrow R$ gives (3.1.20).

(2) If a nonnegative harmonic function on a domain vanishes at a point, it attains its minimum there and is identically zero by proposition 3.1.10.

(3) We first apply (1), after a translation, to a ball $B(a,2t)\subset \Omega $: for $y\in B(a,t)$ we have $r=|y-a|<t$, and the right inequality in (3.1.20) on $B(a,2t)$ gives

Now let $y\in \Omega $. Since $\Omega $ is connected, $y$ can be joined to $x_0$ by a polygonal path in $\Omega $: the set of points that can be so joined is open, and so is its complement in $\Omega $, because balls are convex. The path is compact, so there is $t>0$ with $B(z,2t)\subset \Omega $ for every point $z$ of the path. Choose points $x_0=p_0,p_1,\ldots ,p_ N=y$ along the path with $|p_{i+1}-p_ i|<t$. Applying the ball estimate on $B(p_ i,2t)$ for $i=0,\ldots ,N-1$ gives $h(y)\leq C_0^ Nh(x_0)$, and once more on $B(y,2t)$ gives $h\leq C_0^{N+1}h(x_0)$ on $B(y,t)$. Here $t$ and $N$ depend on $y$, $x_0$ and $\Omega $, but not on $h$. Finitely many of the balls $B(y,t)$, $y\in K$, cover $K$, and the largest of the corresponding constants $C_0^{N+1}$ is the required $C$. $\square$

Proof. Let $h_ j\geq 0$ be harmonic on $D$ with $h_ j(x_0)\leq M$ for some $x_0\in D$. Let $K\subset D$ be compact and choose $\delta >0$ such that $K_\delta =\{ y:\operatorname {dist}(y,K)\leq \delta \} $ is contained in $D$. By proposition 3.1.19(3), $h_ j\leq C(K_\delta ,x_0,D)M$ on $K_\delta $ for all $j$, and (3.1.8) with radius $\delta $ then bounds each derivative $\partial ^\alpha h_ j$ on $K$ uniformly in $j$. Taking for $K$ a closed ball in $D$ and applying this bound to the first derivatives of $\partial ^\alpha h_ j$, we see that the functions $\partial ^\alpha h_ j$ are uniformly Lipschitz on every closed ball in $D$, and hence equicontinuous on compact subsets of $D$.

Let $K_ m$ be the sets of the compact exhaustion of $D$ constructed in section 2.2; every compact subset of $D$ lies in some $K_ m$. Applying proposition 2.7.4 on each $K_ m$ and making a diagonal extraction over all $m$ and all multi-indices $\alpha $, we obtain a subsequence along which every $\partial ^\alpha h_ j$ converges uniformly on compact subsets of $D$, say to $g_\alpha $. To identify the limits, let $e_ k$ be the $k$th standard basis vector. Passing to the uniform limit in

which holds whenever the segment from $y$ to $y+te_ k$ lies in $D$, we find that $\partial _ kg_\alpha =g_{\alpha +e_ k}$. By induction, $g=g_0$ is smooth with $\partial ^\alpha g=g_\alpha $, so the subsequence converges to $g$ in $C^\infty $ on compact subsets, and $\Delta g=\lim _ j\Delta h_ j=0$. This proves the first assertion.

We also need a lower Harnack bound: for compact $K\subset D$ and $x_0\in D$ there is $c>0$ such that $h\geq c\, h(x_0)$ on $K$ for every nonnegative harmonic $h$ on $D$. Indeed, if $B(a,2t)\subset D$ and $y\in B(a,t)$, then $r=|y-a|<t$, and the left inequality in (3.1.20) on $B(a,2t)$ gives $h(y)\geq (2t)^{d-2}(2t-r)(2t+r)^{1-d}h(a)\geq c_0h(a)$ with $c_0=2^{d-2}3^{1-d}$. Cover $K$ by finitely many balls $B(a_ k,t_ k)$ with $B(a_ k,2t_ k)\subset D$. Proposition 3.1.19(3), applied to the compact set $\{ x_0\} $ and the base point $a_ k$, gives $h(x_0)\leq C_ kh(a_ k)$. Hence $h\geq (c_0/\max _ kC_ k)h(x_0)$ on $K$.

Now let $(h_ j)$ be increasing and fix $x_0\in D$. The functions $h_ j-h_1$ are nonnegative and harmonic. If $h_ j(x_0)$ is bounded, the first assertion applies to every subsequence of $(h_ j-h_1)$. Since the sequence increases and, by proposition 3.1.19(3), is bounded on compact sets, it has a finite pointwise limit, and every subsequence has a further subsequence converging in $C^\infty $ on compact subsets to this limit. Hence the whole sequence converges in $C^\infty $ on compact subsets, and the limit is a finite harmonic function. If $h_ j(x_0)\to +\infty $, the lower Harnack bound gives $h_ j-h_1\geq c(h_ j(x_0)-h_1(x_0))$ on each compact $K$, and since $h_1$ is bounded on $K$, $h_ j\to +\infty $ uniformly on $K$. For a decreasing sequence, apply this to $(-h_ j)$. $\square$

Proof. After changing the sign and adding a constant, we may assume that the function $h$ is nonnegative. Fix $x\in \mathbb R^ d$ and put $r=|x|$. For $R>r$, (3.1.20) on $B(0,R)$ gives $c_ Rh(0)\leq h(x)\leq C_ Rh(0)$, where the coefficients $c_ R$ and $C_ R$ of (3.1.20) tend to one as $R\to \infty $. Letting $R\to \infty $ gives $h(x)=h(0)$. Hence $h$ is constant. $\square$