Theorem 3.1.34476CFD04AAC6

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\} $. $\square$