Proposition 4.1.164A7451727417

Proof. (1) Since $K$ is a compact subset of $B$, there is $c>0$ such that $q(x)=c(|x|^2-R^2)\leq -1$ on $K$. The function $q$ is smooth and subharmonic, since $\Delta q=2dc>0$, and it is nonpositive on $B$; hence it is admissible, and $q\leq f_ K\leq u_ K\leq 0$. In particular $f_ K$ and $u_ K$ tend to zero at $\partial B$.

We show that $f_ K$ is the supremum of the continuous admissible functions. Let $v$ be admissible. We first extend $\max \{ v,q\} $ to a subharmonic function on $\mathbb R^ d$: put $w=\max \{ v,q\} $ on $B$ and $w=q$ on $\mathbb R^ d\setminus B$. At $\zeta \in \partial B$ we have $w(\zeta )=0$ and $\varlimsup _{x\to \zeta }w(x)\leq 0$, because $v\leq 0$ on $B$ and $q$ is continuous with $q(\zeta )=0$; so $w$ is upper semicontinuous. We verify the comparison principle of theorem 3.4.5 for $w$ on an arbitrary ball $D$. Let $h$ be harmonic on $D$ and continuous on $\overline D$, with $\varlimsup _{D\ni x\to \zeta }w(x)\leq h(\zeta )$ for $\zeta \in \partial D$. Since $w\geq q$ and $q$ is continuous, $q\leq h$ on $\partial D$, hence on $D$ by the comparison principle for $q$; in particular $w=q\leq h$ on $D\setminus B$. On the bounded open set $D\cap B$, the function $v-h$ has nonpositive boundary upper limits: at points of $\partial D\cap B$ because $v\leq w$, and at points of $\partial B$ because $\varlimsup v\leq 0=q\leq h$ there. By the comparison principle on bounded open sets (the remark after theorem 3.4.5), $v\leq h$ on $D\cap B$. Thus $w\leq h$ on $D$, and the converse part of theorem 3.4.5 shows that $w$ is subharmonic on $\mathbb R^ d$.

Since $w\geq q$, it is locally integrable. Its mollifications $w*\rho _\varepsilon $ are smooth and satisfy $\Delta (w*\rho _\varepsilon )=(\Delta w)*\rho _\varepsilon \geq 0$, so they are subharmonic, and they decrease to $w$ as $\varepsilon \downarrow 0$ by (3.4.13). Fix $\delta >0$. On $B$ we have $w*\rho _\varepsilon \leq \sup _{B(0,R+\varepsilon )}w\leq c((R+\varepsilon )^2-R^2)$, because $w\leq 0$ on $B$ and $w=q$ outside $B$; hence $w*\rho _\varepsilon -\delta \leq 0$ on $B$ for small $\varepsilon $. On $K$ we have $w\leq -1$. Since the continuous functions $w*\rho _\varepsilon $ decrease to $w$, the open sets $\{ w*\rho _\varepsilon <-1+\delta \} $ increase and cover the compact set $K$, so $w*\rho _\varepsilon <-1+\delta $ on $K$ for small $\varepsilon $. For such $\varepsilon $ the function $w*\rho _\varepsilon -\delta $ is continuous and admissible, and it is at least $w-\delta \geq v-\delta $. Since $v$ and $\delta $ are arbitrary, $f_ K$ is the supremum of the continuous admissible functions; in particular $f_ K$ is lower semicontinuous. Every admissible function is at most $-1$ on $K$, so $f_ K=-1$ on $K$.

Let $D$ be an open ball with $\overline D\subset B\setminus K$; we show that $f_ K=u_ K$ on $D$ and that $u_ K$ is harmonic there. For a continuous admissible function $v$, let $v^ D$ be its harmonic lifting on $D$, which equals the Poisson integral of $v|_{\partial D}$ on $D$ and $v$ elsewhere. As noted in subsection 3.1.23, the continuous subsolutions defined there are exactly the real-valued continuous subharmonic functions. Hence lemma 3.1.24 shows that $v^ D$ is continuous and subharmonic and that $v^ D\geq v$. It is admissible: $v^ D\leq 0$ because the Poisson integral preserves order, and $v^ D=v$ on $K$.

Apply Choquet's lemma (lemma 3.4.22) to the family of continuous admissible functions, whose supremum is $f_ K$: there are continuous admissible functions $v_1,v_2,\ldots $ with $(\sup _ jv_ j)^*=u_ K$. The functions $w_ j=\max \{ -1,v_1,\ldots ,v_ j\} $ are continuous, admissible and increasing, and so are the functions $w_ j^ D$. On $D$, the harmonic functions $w_ j^ D$ take values in $[-1,0]$ and increase, so by corollary 3.1.21 they converge locally uniformly to a harmonic function $H$. On the one hand, each $w_ j^ D$ is admissible, so $H\leq f_ K\leq u_ K$ on $D$. On the other hand, $H\geq \sup _ jw_ j$. The upper semicontinuous regularization of $\sup _ jw_ j=\max \{ -1,\sup _ jv_ j\} $ is $\max \{ -1,u_ K\} =u_ K$, so $\sup _ jw_ j=u_ K$ almost everywhere by theorem 3.4.23. Hence $H=u_ K$ almost everywhere on $D$; since both functions are subharmonic and locally integrable, $H=u_ K$ on $D$ by theorem 3.4.12. Therefore $f_ K=u_ K=H$ on $D$. As $D$ is arbitrary, $f_ K=u_ K$ on $B\setminus K$, and $u_ K$ is harmonic there.

(2) Fix $0<r<R$ with $\operatorname{Supp}\lambda \subset B(0,r)$. By the definition of $G_ B$ in theorem 3.1.34 and the symmetry in lemma 4.1.14,

By Fubini's theorem the last integral is the Poisson integral of the continuous function $\zeta \mapsto E_ d*\lambda (\zeta )$ on $\partial B$, so it is harmonic on $B$. Hence $u$ is subharmonic with Riesz measure $\lambda $, by proposition 3.4.11. By lemma 4.1.14, for $y\in \operatorname{Supp}\lambda $ the formula given there for $G_ B(x,y)$ defines a harmonic function of $x$ on the annulus $r<|x|<R^2/r$, which vanishes for $|x|=R$. Since this function is jointly smooth in $(x,y)$, its derivatives in $x$ are bounded uniformly in $y\in \operatorname{Supp}\lambda $ on compact subsets of the annulus. Differentiating under the integral sign, we see that $-\int G_ B(x,y)\, \mathrm d\lambda (y)$ defines a harmonic function on this annulus that extends $u$ across $\partial B$ and vanishes on $\partial B$; we denote the extension again by $u$.

Choose $r<r'<R$ and $\chi \in C_ c^\infty (B)$ with $\chi =1$ on a neighborhood of $\overline{B(0,r')}$. Then $\lambda (B)=\int \chi \, \mathrm d\lambda =\int _ Bu\Delta \chi \, \mathrm d x$. The integrand vanishes outside the annulus $r'<|x|<R$, near whose closure $u$ is smooth and harmonic. The divergence identity (3.1.4) on this annulus, applied to the field $u\nabla \chi -\chi \nabla u$, whose divergence there is $u\Delta \chi $, gives

because $\chi $ and its derivatives vanish near $|x|=R$, while $\chi =1$ near $|x|=r'$; here $\partial _ r$ is the radial derivative. Applying the divergence identity on the same annulus to $\nabla u$, whose divergence vanishes there, we see that this integral equals $\int _{\partial B}\partial _ nu\, \mathrm dS$. This proves the flux identity.

For the comparison of masses, let $u'=-G_ B\lambda '$ be a second function as above with $u\leq u'$ on $B$. Then $u-u'\leq 0$ on $B$ and $u-u'=0$ on $\partial B$, and since $u-u'$ is smooth near $\partial B$, its derivative at a point of $\partial B$ along the outward normal is the limit of difference quotients $((u-u')(\zeta -sn)-0)/(-s)\geq 0$ as $s\downarrow 0$. Hence $\partial _ nu\geq \partial _ nu'$ on $\partial B$, and integration gives $\lambda (B)\geq \lambda '(B)$.

For the converse, let $v$ and $\lambda $ be as stated, extend $\lambda $ by zero to $\mathbb R^ d$, and put $p=E_ d*\lambda $. By proposition 3.4.11, $\Delta (v-p)=0$ on $B$ in the sense of distributions, so by theorem 3.1.11 the locally integrable function $v-p$ agrees almost everywhere on $B$ with a harmonic function $h$. Then $p+h$ is a locally integrable subharmonic function on $B$ that agrees with $v$ almost everywhere, hence everywhere by theorem 3.4.12. The function $p$ is finite and continuous on the neighborhood $\mathbb R^ d\setminus \operatorname{Supp}\lambda $ of $\partial B$: if $x\notin \operatorname{Supp}\lambda $, then $E_ d(w-y)\to E_ d(x-y)$ uniformly in $y\in \operatorname{Supp}\lambda $ as $w\to x$, because $E_ d$ is uniformly continuous on compact subsets of $\mathbb R^ d\setminus \{ 0\} $. As $v$ tends to zero at $\partial B$, the harmonic function $h=v-p$ extends continuously to $\overline B$ with boundary values $-p|_{\partial B}$, and the uniqueness in theorem 3.1.18 gives $h=-H_ B(p|_{\partial B})$. By the formula for $-G_ B\lambda $ proved at the beginning of (2), $v=p-H_ B(p|_{\partial B})=-G_ B\lambda $.

(3) By (1), $u_ K$ is harmonic on $B\setminus K$, so its Riesz measure $\lambda _ K$, a positive Radon measure on $B$ by theorem 3.4.12, is supported on the compact set $K$; in particular it is finite. Moreover $-1\leq u_ K\leq 0$, so $u_ K$ is locally integrable, and $u_ K$ tends to zero at $\partial B$ by (1). The converse in (2) therefore gives $u_ K=-G_ B\lambda _ K$. The flux formula for $C_ B(K)$ is now (2) with $\lambda =\lambda _ K$. $\square$