3.1 Harmonic functions and the Dirichlet problem952EDF2CE146

Proof. Fix $0<r<R$. For $\omega \in S^{d-1}$ and small $h\neq 0$, the fundamental theorem of calculus gives

Since $\nabla u$ is uniformly continuous on the compact set $\overline{B(x,R)}$, these difference quotients converge to $\nabla u(x+r\omega )\cdot \omega $ uniformly in $\omega $ as $h\to 0$. Integrating over the finite measure space $(S^{d-1},S)$ and using the definition of surface measure on $\partial B(x,r)$, where $\nu (x+r\omega )=\omega $, we obtain

Applying (3.1.3) on $B(x,r)$ to $F=\nabla u$, which is $C^1$ on $\Omega $ with divergence $\Delta u$, gives (3.1.6).

Now let $u$ be harmonic. Then $M_ u(x,\cdot )$ has zero derivative, and hence is constant, on $(0,R)$. Since $|M_ u(x,r)-u(x)|\leq \max _{|y-x|=r}|u(y)-u(x)|$, which tends to zero as $r\downarrow 0$ because $u$ is continuous at $x$, the constant is $u(x)$. If $\overline{B(x,r)}\subset \Omega $, then the compact set $\overline{B(x,r)}$ has positive distance from the closed set $\mathbb R^ d\setminus \Omega $, so there is $R>r$ with $\overline{B(x,R)}\subset \Omega $, and therefore $u(x)=M_ u(x,r)$; this is the sphere mean-value property. Since $M_ u(x,t)=u(x)$ for $0<t\leq r$, (3.1.2) gives $\int _{B(x,r)}u\, \mathrm dy=s_{d-1}r^ du(x)/d=|B(x,r)|\, u(x)$, which is the ball mean-value property. $\square$

Proof. As in section 2.5, we write $u_\varepsilon =u*\rho _\varepsilon $. Let $u$ be continuous with the sphere mean-value property, and let $\overline{B(x,\varepsilon )}\subset \Omega $. The compact set $\overline{B(x,\varepsilon )}$ has positive distance from $\mathbb R^ d\setminus \Omega $, so $x\in \Omega _\varepsilon $, and $M_ u(x,r)=u(x)$ for $0<r\leq \varepsilon $. Since $\rho $ is radial, we may write $\rho _\varepsilon (y)=\phi _\varepsilon (|y|)$; as $\rho _\varepsilon $ is even and vanishes outside $B(0,\varepsilon )$, (3.1.2) gives

Every point of $\Omega _\varepsilon $ is the center of such a closed ball, so $u=u_\varepsilon $ on the open set $\Omega _\varepsilon $, and $u_\varepsilon $ is smooth there (section 2.5). As $\Omega $ is the union of the sets $\Omega _\varepsilon $, the function $u$ is smooth.

Next we show that $\Delta u=0$. Fix $x\in \Omega $. Taylor's formula gives

uniformly in $\omega \in S^{d-1}$. We average over the sphere. The reflections $\omega \mapsto -\omega $ and $\omega _ j\mapsto -\omega _ j$ preserve $S$, so $\int \omega _ j\, \mathrm dS=0$ and $\int \omega _ j\omega _ k\, \mathrm dS=0$ for $j\neq k$. Permutations of the coordinates also preserve $S$, so the integrals $\int \omega _ j^2\, \mathrm dS$ are all equal, and their sum is $s_{d-1}$ because $\sum _ j\omega _ j^2=1$. Hence

Since $M_ u(x,r)=u(x)$ for all small $r$, it follows that $\Delta u(x)=0$. Thus $u$ is harmonic. Conversely, harmonic functions have the sphere mean-value property by proposition 3.1.5; in particular they are smooth, and they satisfy $h=h_\varepsilon =h*\rho _\varepsilon $ on $\Omega _\varepsilon $ by the first part of the proof.

Now let $h$ be harmonic and $\overline{B(x,r)}\subset \Omega $. Then $x$ lies in the open set $\Omega _ r$, and $h=h_ r$ on $\Omega _ r$ by the first part of the proof. By the differentiation formula $\partial ^\alpha h_ r=h*\partial ^\alpha \rho _ r$ of section 2.5, and since $\partial ^\alpha \rho _ r$ vanishes outside $B(0,r)$, we get

Since $\partial ^\alpha \rho _ r(z)=r^{-d-|\alpha |}(\partial ^\alpha \rho )(z/r)$, this proves (3.1.8) with $C_{d,\alpha }=\sup |\partial ^\alpha \rho |$, where $\rho $ is the mollifier on $\mathbb R^ d$ fixed in section 2.5. $\square$

Proof. For the first assertion, suppose that $h$ attains its maximum $m$ at a point of $D$, and let $E=\{ x\in D:h(x)=m\} $. This set is nonempty and closed in $D$. If $x\in E$ and $\overline{B(x,r)}\subset D$, the ball mean-value property gives $\int _{B(x,r)}(m-h)\, \mathrm dy=0$. The integrand is continuous and nonnegative, so it vanishes on $B(x,r)$. Thus $E$ is open, and $E=D$ by connectedness. For the minimum, apply this to $-h$.

For the second assertion, let $\varepsilon >0$, and suppose that $h>\varepsilon $ somewhere in $D$; we derive a contradiction. Each $\zeta \in \partial D$ has a ball $B(\zeta ,r_\zeta )$ such that $h<\varepsilon $ on $D\cap B(\zeta ,r_\zeta )$. Since $D$ is bounded, $\partial D$ is compact, and finitely many of these balls cover it; let $W$ be their union. Then $K=\overline D\setminus W=D\setminus W$ is compact. Since $h<\varepsilon $ on $D\setminus K=D\cap W$, the set $K$ is nonempty. The maximum of $h$ over $K$ is attained at some point $x_0$, and $h(x_0)>\varepsilon $; hence $h(x_0)$ is also the maximum of $h$ over $D$. Let $D_0$ be the component of $D$ containing $x_0$; it is open, because every point of $D$ lies in a ball contained in $D$, and balls are connected. By the first assertion, $h=h(x_0)$ on $D_0$.

Since $\mathbb R^ d$ is connected and $D_0$ is a nonempty, bounded, open subset, $D_0$ is not closed, so it has a boundary point $\zeta $. This point does not lie in $D$: otherwise the component of $D$ containing $\zeta $ would be an open neighborhood of $\zeta $, so it would meet $D_0$ and hence equal $D_0$, which does not contain its boundary point $\zeta $. Thus $\zeta \in \partial D$ and $\varlimsup _{D\ni x\to \zeta }h(x)\geq h(x_0)>0$, contradicting the hypothesis. Hence $h\leq \varepsilon $ on $D$, and letting $\varepsilon \downarrow 0$ gives $h\leq 0$.

Finally, if two harmonic functions on a bounded open set are continuous on its closure and agree on its boundary, apply the second assertion to the real and imaginary parts of their difference and of its negative. $\square$

Proof. Recall from section 2.5 that $(T*\phi )(y)=\langle T,\phi (y-\cdot )\rangle $ for a test function $\phi $ on $\mathbb R^ d$ supported in $\overline{B(0,s)}$ and $y\in \Omega _ s$, and that $T_\varepsilon =T*\rho _\varepsilon $. By part (1) of proposition 2.5.3, $T*\phi $ is smooth on $\Omega _ s$ and $\partial ^\alpha (T*\phi )=T*\partial ^\alpha \phi $; moreover $\Delta T_\varepsilon =(\Delta T)_\varepsilon =0$, so $T_\varepsilon $ is a smooth harmonic function on $\Omega _\varepsilon $. We show that $T_\varepsilon $ converges to the required $h$, locally uniformly with all derivatives, by writing $T_\varepsilon =T*(\rho _\varepsilon *\rho _ r)$ near each point for a fixed $r>0$.

Fix $x\in \Omega $ and $r>0$ with $\overline{B(x,3r)}\subset \Omega $, and let $0<\varepsilon <r$. Since $\operatorname {dist}(x,\mathbb R^ d\setminus \Omega )>3r$, every point of $B(x,2r)$ has distance greater than $r$ from $\mathbb R^ d\setminus \Omega $, and every point of $B(x,r)$ has distance greater than $2r$; thus $B(x,2r)\subset \Omega _\varepsilon $ and $B(x,r)\subset \Omega _{2r}$. Let $y\in B(x,r)$. The closed ball $\overline{B(y,r)}$ lies in $B(x,2r)$, where $T_\varepsilon $ is harmonic, so the last assertion of theorem 3.1.7, applied to $T_\varepsilon $ on $B(x,2r)$, gives $T_\varepsilon (y)=\int T_\varepsilon (z)\rho _ r(y-z)\, \mathrm dz$. The test function $\chi =\rho _ r(y-\cdot )$ is supported in $\overline{B(y,r)}\subset \Omega _\varepsilon $, so (2.5.4) gives $T_\varepsilon (y)=\langle T,\chi *\rho _\varepsilon \rangle $. Substituting $z=w+v$ and using that $\rho _\varepsilon $ is even, we find

The function $\rho _\varepsilon *\rho _ r$ is a test function supported in $\overline{B(0,\varepsilon +r)}\subset \overline{B(0,2r)}$, so we have shown that

Since $\rho _\varepsilon *\rho _ r=\rho _ r*\rho _\varepsilon $, part (3) of lemma 2.5.1, applied on $\mathbb R^ d$ to $\rho _ r$, shows that $\partial ^\alpha (\rho _\varepsilon *\rho _ r)\to \partial ^\alpha \rho _ r$ uniformly for every multi-index $\alpha $ as $\varepsilon \to 0$. Let $C$ and $m$ be the constants of definition 2.4.1 for $T$ and the compact set $\overline{B(x,3r)}$, which contains the supports of $(\rho _\varepsilon *\rho _ r)(y-\cdot )$ and $\rho _ r(y-\cdot )$ for $y\in B(x,r)$. For $y\in B(x,r)$ and a multi-index $\alpha $, the difference $\partial ^\alpha T_\varepsilon (y)-\partial ^\alpha (T*\rho _ r)(y)$ is the pairing of $T$ with $w\mapsto \psi (y-w)$, where $\psi =\partial ^\alpha (\rho _\varepsilon *\rho _ r-\rho _ r)$, and the derivatives of order $\beta $ of this test function are $\pm (\partial ^\beta \psi )(y-w)$. Hence

which tends to zero. Thus $T_\varepsilon \to T*\rho _ r$ on $B(x,r)$, uniformly with all derivatives.

Every point of $\Omega $ is the center of such a ball. Therefore $h(y)=\lim _{\varepsilon \to 0}T_\varepsilon (y)$ exists for every $y\in \Omega $, and on each ball $B(x,r)$ as above it equals $T*\rho _ r$ and is the limit of $T_\varepsilon $ uniformly with all derivatives. Thus $h$ is smooth and $\Delta h=\lim _{\varepsilon \to 0}\Delta T_\varepsilon =0$. Every compact subset of $\Omega $ is covered by finitely many of these balls, so $T_\varepsilon \to h$ uniformly on compact sets, and part (2) of proposition 2.5.3 gives $\int h\chi \, \mathrm dx=\lim _{\varepsilon \to 0}\int T_\varepsilon \chi \, \mathrm dx=\langle T,\chi \rangle $ for every test function $\chi $. If two continuous functions define the same distribution, they agree almost everywhere by part (4) of proposition 2.5.3, and hence everywhere, since nonempty open sets have positive measure. This proves uniqueness. Finally, if $T$ is given by a locally integrable function $u$, then $u$ and $h$ define the same distribution, so $u=h$ almost everywhere by the same proposition. $\square$

Proof. We begin with (3.1.15). Let $F\in C^2(a,b)$. The function $x\mapsto |x|$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, so $F(|x|)$ is of class $C^2$ on the annulus. For $a<|x|<b$ the chain rule gives $\partial _ jF(|x|)=F'(|x|)x_ j/|x|$ and

summing over $j$ gives the first expression in (3.1.15), and the product rule gives the second.

Write $E_ d(x)=F_ d(|x|)$ for $x\neq 0$. Differentiating the three cases of (3.1.14) and using $s_0=2$ and $s_1=2\pi $, we find

in every dimension $d\geq 1$. Hence $t^{d-1}F_ d'(t)$ is constant, and $E_ d$ is harmonic on $\mathbb R^ d\setminus \{ 0\} $ by (3.1.15). The functions $E_ d$ are continuous off the origin, and they are locally integrable. For $d\geq 2$ this follows from (3.1.2): $\int _{B(0,1)}|E_ d|\, \mathrm dx=s_{d-1}\int _0^1r^{d-1}|F_ d(r)|\, \mathrm dr$, which is a constant multiple of $\int _0^1r|\log r|\, \mathrm dr$ when $d=2$ and of $\int _0^1r\, \mathrm dr$ when $d\geq 3$, hence finite, while $E_ d$ is bounded on every compact set disjoint from $B(0,1)$. In dimension one, $E_1$ is continuous, and for every test function $\chi $ on $\mathbb R$ an integration by parts on each half-line gives

Thus $\langle E_1'',\chi \rangle =\int E_1\chi ''\, \mathrm dx=\chi (0)$, that is, $E_1''=\delta _0$.

Let $d\geq 2$, let $\chi $ be a test function, and choose $R$ with $\operatorname{Supp}\chi \subset B(0,R)$. The vector field $E_ d\nabla \chi -\chi \nabla E_ d$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, which is an open neighborhood of every closed annulus $\{ \varepsilon \leq |x|\leq R\} $, and its divergence is

because $\Delta E_ d=0$ off the origin. The vector field vanishes near the sphere $|x|=R$, since $\operatorname{Supp}\chi $ is a compact subset of $B(0,R)$, and $\Delta \chi =0$ for $|x|\geq R$. Hence (3.1.4), with $c=0$ and $\nu =x/|x|$, gives

where $\partial _ r$ denotes the derivative in the direction of $x/|x|$. The first boundary term tends to zero as $\varepsilon \to 0$: its absolute value is at most $\sup |\nabla \chi |\, |F_ d(\varepsilon )|\, s_{d-1}\varepsilon ^{d-1}$, which is $\sup |\nabla \chi |$ times $\varepsilon |\log \varepsilon |$ if $d=2$, and times $\varepsilon /(d-2)$ if $d\geq 3$. On the sphere $|x|=\varepsilon $ we have $\partial _ rE_ d=F_ d'(\varepsilon )=(s_{d-1}\varepsilon ^{d-1})^{-1}$, so the second term equals $s_{d-1}^{-1}\int _{S^{d-1}}\chi (\varepsilon \omega )\, \mathrm dS(\omega )=M_\chi (0,\varepsilon )$, which tends to $\chi (0)$ because $\chi $ is continuous at $0$. On the left, $E_ d\Delta \chi $ is integrable, so dominated convergence gives the limit $\int E_ d\Delta \chi \, \mathrm dx$. Hence $\langle \Delta E_ d,\chi \rangle =\langle E_ d,\Delta \chi \rangle =\chi (0)$, that is, $\Delta E_ d=\delta _0$.

Finally, let $f\in C_ c^\infty (\mathbb R^ d)$ and write $(E_ d*f)(x)=\int E_ d(y)f(x-y)\, \mathrm dy$. Fix a compact set $K\subset \mathbb R^ d$, and let $L$ be the compact set of points $x'-z$ with $\operatorname {dist}(x',K)\leq 1$ and $z\in \operatorname{Supp}f$. If $g\in C_ c(\mathbb R^ d)$ is supported in $\operatorname{Supp}f$, then for $x\in K$ the function $y\mapsto E_ d(y)g(x-y)$ vanishes outside $L$ and is bounded by $\sup |g|\, |E_ d|\mathbf1_ L$, which is integrable; by dominated convergence, $E_ d*g$ is continuous on $K$. Moreover, for $x\in K$, a coordinate vector $e_ k$ and $0<|t|\leq 1$, the mean value theorem gives

so dominated convergence shows that $\partial _ k(E_ d*f)=E_ d*\partial _ kf$ on $K$. Since $K$ is arbitrary and $\partial _ kf$ is again a test function supported in $\operatorname{Supp}f$, induction on $|\alpha |$ shows that $E_ d*f$ is smooth and $\partial ^\alpha (E_ d*f)=E_ d*\partial ^\alpha f$. In particular, since $\Delta _ y[f(x-y)]=(\Delta f)(x-y)$,

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. $\square$

Proof. For the first assertion we may assume that $D\neq \varnothing $. Let $\varepsilon >0$, and put $w(y)=v(y)+\varepsilon |y|^2$ for $y\in D$. If $\overline{B(c,r)}\subset D$, then $v(c)\leq M_ v(c,r)$, and the identity $M_{|\cdot |^2}(c,r)=|c|^2+r^2$ gives $w(c)\leq M_ w(c,r)-\varepsilon r^2$. Consequently $w$ does not attain its supremum over $D$ at any point $c\in D$, for we would then have $M_ w(c,r)\leq w(c)$ for small $r$. Choose $x_ k\in D$ with $w(x_ k)\to \sup _ Dw$. Since $D$ is bounded, we may assume, after passing to a subsequence, that $x_ k$ converges to a point $p\in \overline D$. If $p$ were in $D$, the continuity of $w$ would give $w(p)=\sup _ Dw$, which we have excluded. Hence $p\in \partial D$. Since $\sup _ Dw=\lim _ k\bigl (v(x_ k)+\varepsilon |x_ k|^2\bigr )$, the hypothesis gives $\sup _ Dw\leq \varlimsup _{D\ni x\to p}v(x)+\varepsilon |p|^2\leq \varepsilon \max _{\overline D}|y|^2$. Thus $v\leq w\leq \varepsilon \max _{\overline D}|y|^2$ on $D$, and letting $\varepsilon \downarrow 0$ gives $v\leq 0$.

For the second assertion, $v^ B$ is harmonic on $B$ and equals $v$ on $D\setminus B$; it is continuous at the points of $\partial B$ because $H_ B(v|_{\partial B})$ extends continuously to $v$ on $\partial B$ (theorem 3.1.18). Moreover $v^ B\geq v$, because $v\leq H_ B(v|_{\partial B})$ on $B$. To verify the subsolution inequality, let $A$ be a ball with $\overline A\subset D$, and let $H=H_ A(v^ B|_{\partial A})$, regarded as a continuous function on $\overline A$. Since $v\leq v^ B$ and $H_ A$ preserves order, $v\leq H_ A(v|_{\partial A})\leq H$ on $A$; the same inequality holds on $\partial A$, where $H=v^ B\geq v$. On the open set $A\cap B$, the function $v^ B-H$ is harmonic, and it is continuous on the closure of $A\cap B$. A boundary point $\zeta $ of $A\cap B$ lies either on $\partial A$, where $v^ B(\zeta )=H(\zeta )$, or in $A\cap \partial B$, where $v^ B(\zeta )=v(\zeta )\leq H(\zeta )$. By part (2) of proposition 3.1.10, applied to $v^ B-H$ on the bounded open set $A\cap B$, we get $v^ B\leq H$ on $A\cap B$. On $A\setminus B$ we have $v^ B=v\leq H$. Hence $v^ B\leq H_ A(v^ B|_{\partial A})$ on $A$, which is the subsolution inequality for $v^ B$ on $A$. $\square$

Proof. The bounds for $U_ f$ were proved above. To prove harmonicity, fix a ball $B$ with $\overline B\subset \Omega $ and a point $x_0\in B$; we show that $U_ f$ agrees on $B$ with a harmonic function. If $v\in \mathcal S_ f$, its harmonic lifting $v^ B$ on $B$ is a continuous subsolution by lemma 3.1.24, and it coincides with $v$ near $\partial \Omega $ because $\overline B$ is a compact subset of $\Omega $; hence $v^ B\in \mathcal S_ f$.

Choose $v_ j\in \mathcal S_ f$ with $v_ j(x_0)\to U_ f(x_0)$, and put $w_ j=\max \{ v_1,\ldots ,v_ j,\min _{\partial \Omega }f\} $. Then $(w_ j)$ is an increasing sequence in $\mathcal S_ f$ with values in $[\min _{\partial \Omega }f,\max _{\partial \Omega }f]$, and $w_ j(x_0)\to U_ f(x_0)$ because $v_ j\leq w_ j\leq U_ f$. The liftings $w_ j^ B$ belong to $\mathcal S_ f$ and satisfy $w_ j\leq w_ j^ B\leq U_ f$. On $B$ they are harmonic, increase with $j$ because $H_ B$ preserves order, and are bounded by $\max _{\partial \Omega }f$. By part (2) of corollary 3.1.21, applied on the domain $B$ to this sequence, which is bounded above, they converge on $B$ to a harmonic function $h$, and $h\leq U_ f$ on $B$, while $h(x_0)=U_ f(x_0)$.

Now let $v\in \mathcal S_ f$ be arbitrary. The same argument applied to the increasing sequence $\max \{ w_ j,v\} \in \mathcal S_ f$ shows that the liftings $(\max \{ w_ j,v\} )^ B$ increase on $B$ to a harmonic function $k$. Since these liftings dominate $w_ j^ B$ and $v$ and lie in $\mathcal S_ f$, we have $h\leq k\leq U_ f$ and $k\geq v$ on $B$. In particular $k(x_0)\leq U_ f(x_0)=h(x_0)$, so the nonnegative harmonic function $k-h$ on $B$ vanishes at $x_0$ and therefore vanishes identically, by proposition 3.1.19(2). Hence $v\leq h$ on $B$. Taking the supremum over $v\in \mathcal S_ f$ gives $U_ f\leq h$, so $U_ f=h$ on $B$. Since $B$ was arbitrary, $U_ f$ is harmonic on $\Omega $.

Finally, let $H$ be harmonic on $\Omega $ and continuous on $\overline\Omega $ with $H=f$ on $\partial \Omega $. Then $H\in \mathcal S_ f$, so $H\leq U_ f$. Conversely, for $v\in \mathcal S_ f$ the function $v-H$ is a continuous subsolution with $\varlimsup _{x\to \zeta }(v-H)(x)\leq 0$ at every $\zeta \in \partial \Omega $, so $v\leq H$ by lemma 3.1.24. Hence $U_ f=H$. $\square$

Proof. We may assume that $K\neq \varnothing $. Let $\delta >0$. By compactness there are points $\zeta _1,\ldots ,\zeta _ N\in K$ with $K\subset \bigcup _ iB(\zeta _ i,\delta )$. Put $\chi _ i(x)=\rho ((x-\zeta _ i)/(2\delta ))$, where $\rho $ is the mollifier of section 2.5; this function is smooth, positive on $B(\zeta _ i,2\delta )$ and zero outside it. The function $\sigma =\sum _ i\chi _ i$ is positive on the open set $W=\bigcup _ iB(\zeta _ i,2\delta )\supset K$. Choose a smooth function $\eta $ with compact support in $W$ which equals one near $K$ (lemma 2.5.1(4)), and put

extended by zero outside $W$. This is a smooth function with compact support on $\mathbb R^ d$, because it is smooth on $W$, where $\sigma >0$, and vanishes on the open set $\mathbb R^ d\setminus \operatorname{Supp}\eta $; it is real-valued if $f$ is. For $\zeta \in K$ we have $\eta (\zeta )=1$, so $F(\zeta )$ is a convex combination of values $f(\zeta _ i)$ with $|\zeta _ i-\zeta |<2\delta $, and therefore $|F(\zeta )-f(\zeta )|\leq \sup \{ |f(y)-f(y')|:y,y'\in K,\ |y-y'|<2\delta \} $. By the uniform continuity of $f$ on the compact set $K$, this bound tends to zero as $\delta \to 0$. $\square$

Proof. Write $\| f\| =\max _{\partial \Omega }|f|$, and let $\mathcal R$ be the set of $f\in C(\partial \Omega )$ with $U_ f=V_ f$. We show that $\mathcal R$ is a closed linear subspace of $C(\partial \Omega )$ containing a dense subset.

Step 1: $\mathcal R$ is a closed linear subspace, and $f\mapsto U_ f$ is linear on it. If $v\in \mathcal S_ f$ and $w\in \mathcal S_ g$, then $v+w\in \mathcal S_{f+g}$; taking suprema over $v$ and $w$ at a fixed point gives $U_ f+U_ g\leq U_{f+g}$. Applying this to $-f$ and $-g$ and changing signs gives $V_{f+g}\leq V_ f+V_ g$. Together with the inequality $U_{f+g}\leq V_{f+g}$ proved above, we obtain

Moreover $\mathcal S_{tf}=t\, \mathcal S_ f$ for $t>0$, so $U_{tf}=tU_ f$ and $V_{tf}=tV_ f$, while $U_{-f}=-V_ f$ and $V_{-f}=-U_ f$ by definition. Also $U_0=0$, because $0\in \mathcal S_0$ and every member of $\mathcal S_0$ is nonpositive by lemma 3.1.24; hence $V_0=-U_0=0$ and $0\in \mathcal R$. Consequently, if $f,g\in \mathcal R$, the display forces $U_{f+g}=V_{f+g}=U_ f+U_ g$, and $tf\in \mathcal R$ for every real $t$. Thus $\mathcal R$ is a linear subspace, and $f\mapsto U_ f$ is linear on it.

To see that $\mathcal R$ is closed, we first show that $U_ f$ and $V_ f$ depend $1$-Lipschitz continuously on $f$ for the supremum norm. If $f\leq g$, then $\mathcal S_ f\subset \mathcal S_ g$ and hence $U_ f\leq U_ g$; and $\mathcal S_{f+t}=\mathcal S_ f+t$ for constants $t$, so $U_{f+t}=U_ f+t$. Since $f\leq g+\| f-g\| $, we get $U_ f\leq U_ g+\| f-g\| $, and by symmetry $|U_ f-U_ g|\leq \| f-g\| $; as $V_ f=-U_{-f}$ and $V_ g=-U_{-g}$, also $|V_ f-V_ g|\leq \| f-g\| $. If $f_ n\in \mathcal R$ converge uniformly to $f$, then $0\leq V_ f-U_ f\leq |V_ f-V_{f_ n}|+|U_{f_ n}-U_ f|\leq 2\| f-f_ n\| \to 0$, so $f\in \mathcal R$. Hence $\mathcal R$ is closed.

Step 2: $\mathcal R$ contains the restriction to $\partial \Omega $ of every real smooth function on a neighborhood of $\overline\Omega $. Suppose first that $F$ is a real smooth function on a neighborhood of $\overline\Omega $ with $\Delta F\geq 0$ there. Then $F$ is a continuous subsolution on $\Omega $, and $F(x)\to F(\zeta )$ as $\Omega \ni x\to \zeta \in \partial \Omega $, so $F\in \mathcal S_{F|_{\partial \Omega }}$ and $U_{F|_{\partial \Omega }}\geq F$. Consequently the harmonic function $U_{F|_{\partial \Omega }}$ satisfies $\varliminf _{x\to \zeta }U_{F|_{\partial \Omega }}(x)\geq F(\zeta )$ at every $\zeta \in \partial \Omega $, so it belongs to $-\mathcal S_{-F|_{\partial \Omega }}$ and $V_{F|_{\partial \Omega }}\leq U_{F|_{\partial \Omega }}$. Thus $F|_{\partial \Omega }\in \mathcal R$.

Now let $F$ be an arbitrary real smooth function on an open neighborhood $W$ of $\overline\Omega $, and choose $\delta >0$ such that the compact set $L=\{ x:\operatorname {dist}(x,\overline\Omega )\leq \delta \} $ lies in $W$. Since $\Delta F$ is continuous, hence bounded below, on $L$, there is a constant $A\geq 0$ with $\Delta F+2dA\geq 0$ on $L$. As $\Delta (A|x|^2)=2dA$, both $F+A|x|^2$ and $A|x|^2$ are smooth with nonnegative Laplacian on the open neighborhood $\{ x:\operatorname {dist}(x,\overline\Omega )<\delta \} $ of $\overline\Omega $. By the first case their restrictions to $\partial \Omega $ lie in $\mathcal R$, and by Step 1 so does the restriction of their difference $F$.

Step 3: these restrictions are dense in $C(\partial \Omega )$. This is lemma 3.1.26, applied to real-valued functions on the compact set $\partial \Omega $.

By Steps 1–3, $\mathcal R=C(\partial \Omega )$, and $f\mapsto U_ f$ is linear. By Step 1 it preserves order and satisfies $|U_ f-U_ g|\leq \| f-g\| $. Since $U_0=0$, it follows that $U_ f\geq 0$ whenever $f\geq 0$, and that $|U_ f|\leq \| f\| $. $\square$

Proof. Suppose that $b$ is a barrier at $\xi $, and let $f\in C(\partial \Omega )$ and $\varepsilon >0$. We show that $|U_ f-f(\xi )|\leq \varepsilon +Cb$ on $\Omega $ for a suitable constant $C$. Choose $\delta >0$ such that $|f(\zeta )-f(\xi )|<\varepsilon $ for all $\zeta \in \partial \Omega $ with $|\zeta -\xi |<2\delta $, and put $c=\inf _{\Omega \setminus B(\xi ,\delta )}b>0$; if $\Omega \subset B(\xi ,\delta )$, then $|\zeta -\xi |\leq \delta $ for every $\zeta \in \partial \Omega $, and we put $c=1$. If $\zeta \in \partial \Omega $ and $|\zeta -\xi |\geq 2\delta $, then $\Omega \cap B(\zeta ,\delta )$ lies outside $B(\xi ,\delta )$, so $\varliminf _{x\to \zeta }b(x)\geq c$. Let $M=\max _{\partial \Omega }|f|$ and fix $C\geq 2M/c$, which depends on $f$, $\varepsilon $ and $b$. We claim that, for every $\zeta \in \partial \Omega $,

If $|\zeta -\xi |<2\delta $, this holds because $b>0$ and $f(\xi )-\varepsilon <f(\zeta )<f(\xi )+\varepsilon $. If $|\zeta -\xi |\geq 2\delta $, it holds because $f(\xi )-Cc\leq M-2M\leq f(\zeta )$ and $f(\xi )+Cc\geq -M+2M\geq f(\zeta )$. Hence the continuous subsolution $f(\xi )-\varepsilon -Cb$ belongs to $\mathcal S_ f$, and the continuous supersolution $f(\xi )+\varepsilon +Cb$ belongs to $-\mathcal S_{-f}$, so that it dominates every member of $\mathcal S_ f$, as shown after theorem 3.1.25. Therefore $|U_ f-f(\xi )|\leq \varepsilon +Cb$ on $\Omega $. Letting $x\to \xi $ and then $\varepsilon \downarrow 0$ shows that $U_ f(x)\to f(\xi )$. Thus $\xi $ is regular.

Conversely, let $\xi $ be regular, and take the boundary function $f(\zeta )=|\zeta -\xi |$. The smooth function $v(x)=|x-\xi |^2/\operatorname {diam}\Omega $ satisfies $\Delta v\geq 0$, and $v\leq f$ on $\partial \Omega $ because $|\zeta -\xi |\leq \operatorname {diam}\overline\Omega =\operatorname {diam}\Omega $ for $\zeta \in \partial \Omega $; hence $v\in \mathcal S_ f$. Thus $b=U_ f$ is harmonic by theorem 3.1.25, and hence a continuous supersolution, with $b\geq |x-\xi |^2/\operatorname {diam}\Omega $. In particular $b$ is positive on $\Omega $, and if $N$ is a neighborhood of $\xi $ containing a ball $B(\xi ,t)$, then $\inf _{\Omega \setminus N}b\geq t^2/\operatorname {diam}\Omega >0$. By regularity, $b(x)\to f(\xi )=0$ as $x\to \xi $. Hence $b$ is a barrier.

Now suppose that every boundary point is regular, and let $f\in C(\partial \Omega )$. Then $U_ f$ is harmonic on $\Omega $ by theorem 3.1.25. By regularity and the continuity of $f$ on $\partial \Omega $, the function equal to $U_ f$ on $\Omega $ and to $f$ on $\partial \Omega $ is continuous on $\overline\Omega $. Uniqueness follows from proposition 3.1.10.

Finally, suppose that $\overline{B(a,r)}\cap \overline\Omega =\{ \xi \} $. Then $|\xi -a|=r$, since otherwise the open ball $B(a,r)$ would be a neighborhood of $\xi $ and would therefore contain points of $\Omega $, which differ from $\xi \notin \Omega $. In particular $a\notin \overline\Omega $, and every point of $\overline\Omega \setminus \{ \xi \} $ satisfies $|x-a|>r$. Put

For $d\geq 2$ this is harmonic on $\mathbb R^ d\setminus \{ a\} $ by proposition 3.1.13, being a positive multiple of $E_ d(x-a)$ plus a constant. In dimension one, each component of $\Omega $ is an open interval disjoint from $[a-r,a+r]$, so it lies on one side of this interval; hence $|x-a|$ is affine on each component, and $b$ is harmonic on $\Omega $ as well. In all dimensions $b$ is an increasing function of $|x-a|$ which vanishes when $|x-a|=r$, so it is continuous on the compact set $\overline\Omega $, positive on $\overline\Omega \setminus \{ \xi \} $ and zero at $\xi $. Hence $b(x)\to 0$ as $x\to \xi $, and if $N$ is a neighborhood of $\xi $ with $\Omega \setminus N\neq \varnothing $, the infimum of $b$ over the nonempty compact set $\overline\Omega \setminus \operatorname{Int}N$ is attained and positive. Thus $b$ is a barrier at $\xi $. $\square$

Proof. Choose $r>0$ with $\overline{B(\xi ,2r)}\subset W$ and $0<t<r$. The set $L=\overline\Omega \cap \{ r-t\leq |x-\xi |\leq r+t\} $ is compact, and $b$ is positive and continuous on it, so there is $c>0$ with $b\geq 2c$ on $L$ (if $L$ is empty, any $c>0$ will do). Define $\tilde b=\min \{ b,c\} $ on $\Omega \cap B(\xi ,r)$ and $\tilde b=c$ on $\Omega \setminus B(\xi ,r)$. Since $b\geq 2c$ on $L$, we have $\tilde b=\min \{ b,c\} $ on $\Omega \cap B(\xi ,r+t)$ and $\tilde b=c$ on $\Omega \setminus \overline{B(\xi ,r)}$. These two open sets cover $\Omega $, and on each of them $\tilde b$ is a continuous supersolution, being a minimum of two continuous supersolutions or a constant. By the locality of the subsolution property, noted after the definition of continuous subsolutions, $\tilde b$ is a continuous supersolution on $\Omega $.

The function $\tilde b$ is positive, and it tends to zero at $\xi $, because $0<\tilde b\leq b$ on $\Omega \cap B(\xi ,r)$ and the extension of $b$ vanishes at $\xi $. If $N$ is a neighborhood of $\xi $, then $\tilde b\geq c$ on $\Omega \setminus B(\xi ,r)$, while on $(\Omega \cap B(\xi ,r))\setminus N$ it is at least the minimum of $c$ and of $b$ over the compact set $(\overline\Omega \cap \overline{B(\xi ,r)})\setminus \operatorname{Int}N$, on which $b$ is positive. Hence $\tilde b$ is a barrier at $\xi $, and theorem 3.1.29 shows that $\xi $ is regular. $\square$

Proof. The map $g\mapsto H_\Omega (g|_{\partial \Omega })(x)$ is a positive linear functional on $C_ c(\mathbb R^ d,\mathbb R)$, so by theorem 2.3.3 it is integration against a Radon measure $\mu $ on $\mathbb R^ d$. If $g\in C_ c(\mathbb R^ d)$ is supported in the open set $\mathbb R^ d\setminus \partial \Omega $, then $g|_{\partial \Omega }=0$ and $\int g\, \mathrm d\mu =0$; by (2.3.4), $\mu (\mathbb R^ d\setminus \partial \Omega )=0$. If $\eta \in C_ c(\mathbb R^ d)$ equals one on the compact set $\partial \Omega $, then $\mu (\mathbb R^ d)=\mu (\partial \Omega )=\int \eta \, \mathrm d\mu =H_\Omega 1(x)=1$. Let $\omega _ x$ be the restriction of $\mu $ to the Borel subsets of $\partial \Omega $; it is a probability measure.

Every $f\in C(\partial \Omega )$ is a uniform limit of restrictions of functions in $C_ c(\mathbb R^ d)$, by lemma 3.1.26. Both sides of the asserted identity are continuous for uniform convergence on $\partial \Omega $, the right side because $\omega _ x$ is finite and the left side because $|H_\Omega f|\leq \max _{\partial \Omega }|f|$. Hence the identity holds for real $f$, and by linearity for complex $f$. If $\omega '$ is another Borel probability measure on $\partial \Omega $ with the same property, then $\omega _ x$ and $\omega '$, extended by zero to $\mathbb R^ d$, are finite Borel measures, hence Radon measures by proposition 2.2.2, with the same integrals against every $g\in C_ c(\mathbb R^ d)$. The uniqueness assertion of theorem 2.3.3 gives $\omega '=\omega _ x$. The function $x\mapsto \int f\, \mathrm d\omega _ x=H_\Omega f(x)$ is harmonic by the definition of $H_\Omega $. If $\Omega $ is a ball, $P_\Omega (x,\cdot )\, \mathrm dS$ is a Borel probability measure on $\partial \Omega $ with the required property, by theorem 3.1.18, so it equals $\omega _ x$. $\square$

Proof. (1) Fix $x\in \Omega $ and let $\gamma =\Gamma _ d(\cdot -x)|_{\partial \Omega }$, which is continuous because $\Gamma _ d(\cdot -x)$ is continuous on $\mathbb R^ d\setminus \{ x\} $. The function $H_\Omega \gamma $ is harmonic on $\Omega $, and $\Gamma _ d(\cdot -x)$ is harmonic on $\mathbb R^ d\setminus \{ x\} $ by proposition 3.1.13, so $G_\Omega (x,\cdot )$ is harmonic on $\Omega \setminus \{ x\} $. By the same proposition, $\Gamma _ d(\cdot -x)$ is locally integrable, and for every test function $\chi $ on $\Omega $, which we extend by zero to $\mathbb R^ d$,

Since $H_\Omega \gamma $ is smooth on $\Omega $ with $\Delta H_\Omega \gamma =0$, integration by parts gives $\int H_\Omega \gamma \, \Delta \chi \, \mathrm dy=0$. Hence $G_\Omega (x,\cdot )$ is locally integrable on $\Omega $ and $\langle -\Delta _ yG_\Omega (x,\cdot ),\chi \rangle =\chi (x)$, that is, $-\Delta _ yG_\Omega (x,\cdot )=\delta _ x$. As $y\to \zeta \in \partial \Omega $, both $\Gamma _ d(y-x)$ and $H_\Omega \gamma (y)$ tend to $\gamma (\zeta )$, so $G_\Omega (x,y)\to 0$.

We prove positivity when $d\geq 2$; for $d=1$ it follows from (2) below. First we show that $G_\Omega (x,\cdot )\geq 0$. Let $v\in \mathcal S_\gamma $; we claim that $v\leq \Gamma _ d(\cdot -x)$ on $\Omega \setminus \{ x\} $. Since $v\leq \max _{\partial \Omega }\gamma $ and $\Gamma _ d(y-x)\to +\infty $ as $y\to x$, there is $\varepsilon _0>0$ such that $\overline{B(x,\varepsilon _0)}\subset \Omega $ and $v\leq \Gamma _ d(\cdot -x)$ on $\partial B(x,\varepsilon )$ for $0<\varepsilon <\varepsilon _0$. On the bounded open set $\Omega \setminus \overline{B(x,\varepsilon )}$, whose boundary lies in $\partial \Omega \cup \partial B(x,\varepsilon )$, the function $v-\Gamma _ d(\cdot -x)$ is a continuous subsolution. At a point $\zeta \in \partial \Omega $ its upper limit is at most $\gamma (\zeta )-\gamma (\zeta )=0$, and on $\partial B(x,\varepsilon )$ it is continuous and nonpositive. By lemma 3.1.24, $v\leq \Gamma _ d(\cdot -x)$ on $\Omega \setminus \overline{B(x,\varepsilon )}$, and letting $\varepsilon \downarrow 0$ gives the claim. Taking the supremum over $v$ and using $H_\Omega \gamma =U_\gamma $ (theorem 3.1.29), we obtain $H_\Omega \gamma \leq \Gamma _ d(\cdot -x)$ on $\Omega \setminus \{ x\} $, that is, $G_\Omega (x,\cdot )\geq 0$.

Strict positivity follows from the minimum principle, because $\Omega \setminus \{ x\} $ is connected when $d\geq 2$. To see this, let $y_0,y_1\in \Omega \setminus \{ x\} $. They can be joined by a polygonal path $p\colon [0,1]\to \Omega $, because the points of the domain $\Omega $ that can be joined to $y_0$ by a polygonal path in $\Omega $ form an open set whose complement in $\Omega $ is also open, balls being convex. Suppose that $p$ passes through $x$, and choose $r>0$ with $\overline{B(x,r)}\subset \Omega $ and $r<\min \{ |y_0-x|,|y_1-x|\} $. By the intermediate value theorem, the closed set of times $t$ with $|p(t)-x|=r$ is nonempty; let $t_1$ and $t_2$ be its smallest and largest elements. Then $|p(t)-x|>r$ for $t<t_1$ and for $t>t_2$, and replacing $p$ on $[t_1,t_2]$ by a path from $p(t_1)$ to $p(t_2)$ in the sphere $\partial B(x,r)$, which is path-connected because $d\geq 2$ (example 3.1.31), we obtain a path from $y_0$ to $y_1$ in $\Omega \setminus \{ x\} $. Now if $G_\Omega (x,\cdot )$ vanished at some point of the domain $\Omega \setminus \{ x\} $, it would attain its minimum there, and part (1) of proposition 3.1.10 would make it identically zero on $\Omega \setminus \{ x\} $, contradicting $G_\Omega (x,y)\to +\infty $ as $y\to x$. Hence $G_\Omega (x,\cdot )>0$.

For uniqueness, let $G'$ be another function with the stated properties. The locally integrable function $G_\Omega (x,\cdot )-G'$ has vanishing distributional Laplacian on $\Omega $, so by theorem 3.1.11 it agrees almost everywhere with a harmonic function $h$ on $\Omega $. Away from $x$, the functions $G_\Omega (x,\cdot )$ and $G'$ are harmonic, so their difference is continuous there and therefore equals $h$ everywhere on $\Omega \setminus \{ x\} $. Then $h$ is real and has zero boundary limits, because $G_\Omega (x,\cdot )$ and $G'$ do, and $h=0$ by proposition 3.1.10, applied to $h$ and to $-h$. This argument does not use positivity, and it applies in every dimension.

(2) Let $d=1$, $\Omega =(a,b)$ and $x\in (a,b)$, and let $g(y)$ denote the right side of the formula in (2). The function $g$ is continuous, positive on $(a,b)$, and tends to zero at $a$ and $b$. It is affine on each side of $x$, with slope $g'_-=(b-x)/(b-a)$ on $(a,x)$ and $g'_+=-(x-a)/(b-a)$ on $(x,b)$, so $g'_+-g'_-=-1$. For a test function $\chi $ on $(a,b)$, an integration by parts on each of the intervals $(a,x)$ and $(x,b)$, on which $g'$ is constant, gives

so $-g''=\delta _ x$ in the sense of distributions. Moreover $g$ is harmonic on $(a,b)\setminus \{ x\} $. By the uniqueness argument just given, $g$ coincides with $G_{(a,b)}(x,\cdot )$ on $(a,b)\setminus \{ x\} $, and also at $x$, since both functions are continuous there. This proves (2), and with it the positivity in (1) when $d=1$.

(3) Let $D'\subset D''$ be bounded regular domains and $x\in D'$. On $D'\setminus \{ x\} $ the fundamental solutions cancel, and

which is harmonic on all of $D'$. At every $\zeta \in \partial D'$ its lower limit is nonnegative, because $G_{D'}(x,y)\to 0$ as $y\to \zeta $ and $G_{D''}(x,y)\geq 0$. By proposition 3.1.10, applied to its negative, this harmonic function is nonnegative on $D'$, so $G_{D'}(x,y)\leq G_{D''}(x,y)$ for $y\in D'\setminus \{ x\} $. For $y=x$, both sides are $+\infty $ if $d\geq 2$; if $d=1$, the display also holds at $y=x$, so the inequality holds there as well.

(4) Let $D_ j$ and $D$ be as in the statement, fix $x\in D$, and consider only $j$ large enough that $x\in D_ j$. Put $h_ j=H_{D_ j}(\Gamma _ d(\cdot -x)|_{\partial D_ j})$ and $h=H_ D(\Gamma _ d(\cdot -x)|_{\partial D})$; we show that $h_ j$ decreases to $h$ in $C^\infty $ on compact subsets of $D$. By (3), $G_{D_ j}(x,\cdot )\leq G_{D_{j+1}}(x,\cdot )\leq G_ D(x,\cdot )$ on $D_ j\setminus \{ x\} $, that is, $h_ j\geq h_{j+1}\geq h$ there, and also at $x$, since these functions are continuous at $x$. Let $B$ be a ball with $\overline B\subset D$. The compact set $\overline B$ is covered by the increasing open sets $D_ j$, so $\overline B\subset D_ j$ for all large $j$. On $B$ the functions $h_ j$ then form a decreasing sequence of harmonic functions bounded below by $h$, and part (3) of corollary 3.1.21 shows that they converge in $C^\infty $ on compact subsets of $B$ to a finite harmonic function. These local limits are pointwise limits, so they define a harmonic function $h_\infty $ on $D$. Every compact subset of $D$ is covered by finitely many balls $B(a,s)$ with $\overline{B(a,2s)}\subset D$, and $\overline{B(a,s)}$ is a compact subset of $B(a,2s)$; hence $h_ j\to h_\infty $ in $C^\infty $ on compact subsets of $D$.

It remains to show that $h_\infty =h$. The function $\Gamma _ d(\cdot -x)-h_\infty $ is the pointwise limit of $G_{D_ j}(x,\cdot )$, so it lies between $0$ and $G_ D(x,\cdot )$. Therefore the harmonic function $h_\infty -h=G_ D(x,\cdot )-(\Gamma _ d(\cdot -x)-h_\infty )$ satisfies $0\leq h_\infty -h\leq G_ D(x,\cdot )$ on $D\setminus \{ x\} $, so it has zero boundary limits and vanishes by proposition 3.1.10. Thus $h_ j\to h$, which means that $G_{D_ j}(x,y)$ increases to $G_ D(x,y)$ for every $y\neq x$. Moreover $h_ j-h\to 0$ in $C^\infty $ on compact subsets of $D$, and $G_{D_ j}(x,\cdot )-G_ D(x,\cdot )=h-h_ j$ on $D_ j\setminus \{ x\} $; since every compact subset of $D\setminus \{ x\} $ lies in $D_ j$ for large $j$, $G_{D_ j}(x,\cdot )\to G_ D(x,\cdot )$ in $C^\infty $ on compact subsets of $D\setminus \{ x\} $. $\square$