Example 4.2.20372946152BF9

Proof. We first check that $\partial W$ is the disjoint union of $\partial A$ and $\partial B$. Since $W=B\cap (\mathbb R^ d\setminus A)$ is the intersection of two open sets, $\partial W\subset \partial B\cup \partial A$, and $\partial A\subset A\subset B$ does not meet $\partial B$. Conversely, no point of $\partial B\cup \partial A$ lies in $W$. A point of $\partial B$ is a limit of points of $B\setminus A=W$, because the compact set $A\subset B$ has positive distance from $\partial B$. Every small ball around a point of $\partial A$ lies in $B$ and contains points outside $A$, that is, points of $W$. Consequently $\partial A$ and $\partial B$ are disjoint compact sets whose union is $\partial W$, and $f_ x$ is continuous on $\partial W$, because $G_ B(x,\cdot )$ is continuous on $B\setminus \{ x\} \supset A$.

Let $H=H_ Wf_ x$ be the solution of the Dirichlet problem on $W$ with boundary values $f_ x$; it is continuous on $\overline W$ since $W$ is regular, and $H\geq 0$ because $f_ x\geq 0$ and $H_ W$ preserves order. The function $G_ B(x,\cdot )=G_ B\delta _ x$ is superharmonic on $B$ by part (2) of proposition 4.1.16, it is continuous on $B\setminus \{ x\} $, and it tends to zero at $\partial B$ (theorem 3.1.34); hence its boundary limits on $\partial W$ are the values of $f_ x$. Therefore the maximum principle (proposition 3.4.7), applied to the subharmonic function $H-G_ B(x,\cdot )$ on the bounded open set $W$, gives $H\leq G_ B(x,\cdot )$ on $W$. Define $r=H$ on $W$ and $r=G_ B(x,\cdot )$ on $A$. Then $r$ is continuous on $B$, because $H$ and $G_ B(x,\cdot )$ have the same values on $\partial A$, and $0\leq r\leq G_ B(x,\cdot )$.

To see that $r$ is superharmonic, we argue as in Step 2 of the proof of theorem 4.2.14. Let $E$ be a ball with $\overline E\subset B$, and let $q$ be harmonic on $E$ and continuous on $\overline E$ with $q\leq r$ on $\partial E$. Then $q\leq G_ B(x,\cdot )$ on $\partial E$, hence on $E$ by superharmonicity of $G_ B(x,\cdot )$; in particular $r\geq q$ on $E\cap A$. On $E\cap W$ the harmonic function $H-q$ has nonnegative boundary limits: at points of $\partial E$ because $H$ extends continuously to $r\geq q$ there, and at points of $E\cap \partial A$ because $H$ tends to $G_ B(x,\cdot )\geq q$ there. The minimum principle gives $H\geq q$ on $E\cap W$. Hence $r\geq q$ on $E$, and $r$ is superharmonic by theorem 3.4.5.

The function $r$ is admissible for $R_{G_ B(x,\cdot )}^ A$: it is nonnegative and superharmonic on $B$ and equals $G_ B(x,\cdot )$ on $A$. Let $h$ be any admissible function. Then $h\geq r$ on $A$. On $W$, the superharmonic function $h$ satisfies $\varliminf _{y\to \zeta }h(y)\geq h(\zeta )\geq f_ x(\zeta )$ at points $\zeta \in \partial A$, by lower semicontinuity, and $\varliminf _{y\to \zeta }h(y)\geq 0=f_ x(\zeta )$ at points $\zeta \in \partial B$, since $h\geq 0$. Applying proposition 3.4.7 to the subharmonic function $H-h$ on $W$, we get $h\geq H$ there. Hence $h\geq r$ on $B$, and $R_{G_ B(x,\cdot )}^ A=r$ pointwise. Since $r$ is continuous, $\widehat R_{G_ B(x,\cdot )}^ A=r$, that is, $r_ x=r$.

On $W$, both $G_ B(x,\cdot )$ and $G_ W(x,\cdot )$ are locally integrable and satisfy $-\Delta =\delta _ x$ there (theorem 3.1.34). Their difference is therefore distributionally harmonic on $W$; by Weyl's lemma (theorem 3.1.11) it has a harmonic representative, which coincides with the difference off $x$. This harmonic function tends to $f_ x$ at $\partial W$, because $G_ W(x,\cdot )$ tends to zero there. By uniqueness of the solution of the Dirichlet problem (proposition 3.1.10), it equals $H$. The harmonic-measure formula is proposition 3.1.33 applied to $H=H_ Wf_ x$.

Finally, the measure $\delta _ x^ A=-\Delta r_ x$ vanishes on $W$, where $r_ x=H$ is harmonic, and on $\operatorname{Int}A$, where $r_ x=G_ B(x,\cdot )$ is harmonic because $x\notin A$. It is therefore supported on $\partial A$, and its mass is at most $\delta _ x(B)=1$ by theorem 4.2.16.

For the last assertion, write $B=B(c,R)$ and let $A=\overline{B(c,\rho )}$ with $0<\rho <R$, so that $W=\{ y:\rho <|y-c|<R\} $ is a bounded open annulus. It is connected: every point of $W$ is joined by a radial segment in $W$ to the sphere $\partial B(c,(\rho +R)/2)$, which is path-connected because $d\geq 2$. By theorem 3.1.29, it suffices to find, for every $\zeta \in \partial W$, a closed ball that meets $\overline W$ only at $\zeta $. If $|\zeta -c|=R$, the closed ball $\overline{B(2\zeta -c,R)}$ meets $\overline{B(c,R)}\supset \overline W$ only at $\zeta $, as noted at the beginning of subsection 3.1.32. If $|\zeta -c|=\rho $, let $m=(c+\zeta )/2$. Every $y\in \overline{B(m,\rho /2)}$ satisfies $|y-c|\leq |y-m|+|m-c|\leq \rho $, with equality only if $y-m=(\zeta -c)/2$, that is, $y=\zeta $. Since $|y-c|\geq \rho $ on $\overline W$, the closed ball $\overline{B(m,\rho /2)}$ meets $\overline W$ only at $\zeta $. Hence every boundary point of $W$ is regular, and $W$ is a regular domain. $\square$