4.2.9 Green energy and compact reduction27012C9B879C

Proof. We first prove the gradient identity for the partial sums of a decomposition of $\mu $ and $\nu $ into pieces with bounded Green potentials, to which proposition 4.1.29 applies, and then pass to the limit. Translating, we may assume that $B=B(0,R)$ for some $R>0$, and we choose $r<R$ such that the supports of $\mu $ and $\nu $ lie in $B(0,r)$. By lemma 4.1.14, the function $\kappa (x,y)=G_ B(x,y)-\Gamma _ d(x-y)$ extends to a continuous function on the compact set $\overline B\times \overline{B(0,r)}$; let $C$ be a bound for $|\kappa |$ there. Since $\Gamma _ d=-E_ d$, every finite positive measure $\lambda $ supported in $B(0,r)$ satisfies

and the last integral is a continuous function of $x\in \overline B$ bounded by $C\lambda (B)$.

Step 1: decomposition. Since $\int G_ B\mu \, \mathrm d\mu =I_ B(\mu )<\infty $, the Green potential $G_ B\mu $ is finite $\mu $-almost everywhere, and by the display so is $E_ d*\mu $. By part (2) of theorem 4.1.10, $\mu $ is therefore the sum of its restrictions to countably many disjoint compact sets, each of which has a finite and continuous fundamental potential. Let $\mu _ n$ be the sum of the first $n$ of these restrictions. Since $E_ d*\mu _ n$ is finite and continuous, the display shows that $G_ B\mu _ n$ is finite, continuous and bounded on $B$; this proves the last assertion of the proposition. By part (2) of proposition 4.1.16, $-G_ B\mu _ n$ is a subharmonic function on $B$ with Riesz measure $\mu _ n$, which extends harmonically across $\partial B$ and vanishes on $\partial B$; in particular, it tends to zero at $\partial B$. Thus proposition 4.1.29 applies to the functions $-G_ B\mu _ n$, $n\geq 1$. Hence $\nabla G_ B\mu _ n\in L^2(B)$ and

Step 2: the gradient of $G_ B\mu $. For $m>n$, expanding the square with the last display gives

where the inequality holds because $0\leq \mu _ m-\mu _ n\leq \mu -\mu _ n$ and $G_ B\geq 0$. The measure $\mu -\mu _ n$ is the restriction of $\mu $ to a decreasing sequence of sets whose intersection is $\mu $-null. Since $G_ B$ is nonnegative and integrable for $\mu \otimes \mu $, dominated convergence gives $I_ B(\mu -\mu _ n)\to 0$. Thus $(\nabla G_ B\mu _ n)$ is a Cauchy sequence in $L^2(B)$; let $g$ be its limit.

On the other hand, $G_ B\mu _ n\uparrow G_ B\mu $ pointwise by monotone convergence, and $G_ B\mu \in L^1(B)$. Indeed, for $y\in \operatorname{Supp}\mu $, the bound $|\kappa |\leq C$ and the local integrability of $\Gamma _ d$ (proposition 3.1.13) give

and integrating against $\mu $, by Tonelli's theorem, shows that $\int _ BG_ B\mu \, \mathrm d x$ is at most $\mu (B)$ times the right side. Hence $G_ B\mu _ n\to G_ B\mu $ in $L^1(B)$ by monotone convergence. For a smooth compactly supported vector field $\varphi $ on $B$ we obtain

so $g=\nabla G_ B\mu $. Since $\| \nabla G_ B\mu _ n\| _2^2=I_ B(\mu _ n)\uparrow I_ B(\mu )$ by monotone convergence, we get $\| \nabla G_ B\mu \| _2^2=I_ B(\mu )$.

Step 3: the mutual energy. Construct the partial sums $\nu _ n$ of $\nu $ in the same way. By proposition 4.1.29 and the Cauchy–Schwarz inequality in $L^2(B)$,

As $n\to \infty $, the left side increases to $I_ B(\mu ,\nu )$ by monotone convergence, so the mutual energy is finite, while the integral converges to $\int _ B\nabla G_ B\mu \mathbin {\cdot }\nabla G_ B\nu \, \mathrm d x$ by convergence in $L^2(B)$. This proves the identity for positive measures. For signed measures, we expand both sides bilinearly in the positive and negative parts.

Step 4: positive definiteness. Let $\sigma $ be a signed measure as in the statement, and put $G_ B\sigma =G_ B\sigma ^+-G_ B\sigma ^-$, a function in $L^1(B)$ by Step 2. Then $I_ B(\sigma )=\| \nabla G_ B\sigma \| _2^2\geq 0$. Suppose that $I_ B(\sigma )=0$, so that $\nabla G_ B\sigma =0$ as a distribution on $B$. Since $B$ is connected, corollary 2.5.5 shows that the distribution defined by $G_ B\sigma $ is a constant. On the other hand, $-\Delta G_ B\sigma =\sigma $ in the sense of distributions, by part (2) of proposition 4.1.16. Hence $\sigma ^+$ and $\sigma ^-$, which are Radon measures by proposition 2.2.2, define the same distribution, so $\sigma ^+=\sigma ^-$ by part (4) of proposition 2.5.3, that is, $\sigma =0$. Finally, the Cauchy–Schwarz inequality for $I_ B$ follows from the gradient identity and the Cauchy–Schwarz inequality in $L^2(B)$. $\square$

Proof. By part (3) of proposition 4.1.16, the function $w_ K=-u_ K$ is the Green potential $G_ B\lambda _ K$, and $0\leq w_ K\leq 1$ because $-1\leq u_ K\leq 0$. Since $u_ K$ is the upper semicontinuous regularization of the envelope $f_ K$, which equals $-1$ on $K$, part (2) of theorem 4.1.26 shows that $w_ K=1$ quasi-everywhere on $K$. In particular $I_ B(\lambda _ K)=\int w_ K\, \mathrm d\lambda _ K\leq \lambda _ K(K)$, so $\lambda _ K$ has finite Green energy. As measures of finite Green energy do not charge polar sets, we obtain, with $m=C_ B(K)$,

for every probability measure $\tau $ on $K$ with finite Green energy; in the second identity we used the symmetry of $G_ B$.

Suppose that $m>0$, and let $\tau $ be a probability measure on $K$. If $I_ B(\tau )<\infty $, the Cauchy–Schwarz inequality of proposition 4.2.11 gives $1=I_ B(\lambda _ K,\tau )^2\leq mI_ B(\tau )$; thus $I_ B(\tau )\geq 1/m$ in all cases. Equality holds for $\tau =\lambda _ K/m$, whose energy is $m/m^2=1/m$. Hence the infimum equals $1/m$, which proves the formula. If $\tau _1\neq \tau _2$ were two minimizers, their midpoint would again be a probability measure on $K$, and by positive definiteness

a contradiction. Suppose next that $m=0$. Then $K$ is polar by theorem 4.1.17, so it carries no probability measure of finite energy, and the infimum is $+\infty $ (if $K=\varnothing $, the family is empty).

For the last assertion, recall that balls are regular domains (theorem 3.1.29). The monotonicity of Green functions in the domain (theorem 3.1.34) gives $G_ D\leq G_ B$ on $D\times D$. Hence $I_ B(\tau )\geq I_ D(\tau )$ for every measure $\tau $ supported on a compact set $K\subset D$, and the variational formula, applied in $B$ and in $D$, gives $C_ B(K)\leq C_ D(K)$. Taking suprema over compact subsets, we obtain $C_ B(O)\leq C_ D(O)$ for every open $O\Subset D$. Every such $O$ containing $A$ is also allowed in the infimum defining $C_ B^*(A)$, so $C_ B^*(A)\leq C_ D^*(A)$. $\square$

Proof. Enlarging $A$ or $s$ shrinks the class of admissible functions, which proves the two monotonicity assertions. For $c>0$, multiplication by $c$ is a bijection between the admissible functions for $s$ and those for $cs$; for $c=0$, both sides vanish (with $0\cdot \infty =0$), since the zero function is admissible. If $h_1$ and $h_2$ are admissible for $s$ and $t$ respectively, then $h_1+h_2$ is admissible for $s+t$, so $R_{s+t}^ A\leq R_ s^ A+R_ t^ A$. Monotonicity and homogeneity are preserved by lower semicontinuous regularization. For subadditivity, recall that each reduction agrees with its regularization outside a polar set, hence almost everywhere. Therefore $\widehat R_{s+t}^ A\leq \widehat R_ s^ A+\widehat R_ t^ A$ almost everywhere. Both sides are superharmonic, so the inequality holds everywhere by the ball-average formula above.

Now let $U$ be bounded and $A$ polar. Then $A$ has compact closure, so proposition 4.1.11 provides a finite positive measure $\nu $ of compact support with $E_ d*\nu =-\infty $ on $A$. The kernel $E_ d$ is bounded above on the bounded set of differences $x-y$ with $x\in U$ and $y\in \operatorname{Supp}\nu $, so adding a constant to $-E_ d*\nu $ gives a nonnegative superharmonic function $p$ on $U$ with $p=+\infty $ on $A$; by proposition 3.4.11, $p$ is locally integrable. For every $\varepsilon >0$, the function $\varepsilon p$ is admissible, even at points of $A$ where $s=+\infty $. Hence $R_ s^ A\leq \varepsilon p$ for every $\varepsilon >0$, and so $R_ s^ A=0$ on the set $\{ p<\infty \} $. This set has full Lebesgue measure, because $p$ is locally integrable, and is therefore dense in $U$. Consequently $\widehat R_ s^ A=0$. $\square$

Proof. We show that $r$ is harmonic off $A$ by replacing admissible functions with their Poisson integrals on balls in $B\setminus A$ (Steps 1–3), and then identify $r$ with the Green potential of its Riesz measure (Step 4).

Step 1: capped admissible functions. Choose a closed ball $L\subset B$ whose interior contains $A$, and a constant $M\geq 0$ with $s\leq M$ on $A$. The relative extremal function $u_ L$ takes values in $[-1,0]$ and tends to zero at $\partial B$ by part (1) of proposition 4.1.16; it equals $-1$ on $\operatorname{Int}L$, because it is the upper semicontinuous regularization of the envelope $f_ L$, which equals $-1$ on $L$. Hence $h_0=-Mu_ L$ is a nonnegative superharmonic function on $B$, bounded by $M$, which majorizes $s$ on $A$ and tends to zero at $\partial B$. If $h$ is admissible, so is $\min \{ h,h_0\} $. Therefore $R_ s^ A$ is also the infimum of the admissible functions $h$ with $h\leq h_0$; we call these capped. In particular $0\leq r\leq R_ s^ A\leq h_0\leq M$, so $r$ is bounded and tends to zero at $\partial B$.

Step 2: Poisson modification. Let $D$ be an open ball with $\overline D\subset B\setminus A$, and let $h$ be a capped admissible function. Its restriction to $\partial D$ is bounded and lower semicontinuous, so by lemma 3.4.4, applied to $-h$, it is the increasing limit of continuous functions $g_ k$ on $\partial D$. By monotone convergence, the Poisson integrals $H_ Dg_ k$ of theorem 3.1.18 increase to

Since $0\leq h\leq M$ and the Poisson kernel has integral one, $0\leq H\leq M$, and $H$ is harmonic by part (2) of corollary 3.1.21. For $\zeta \in \partial D$, lower semicontinuity gives $\varliminf _{y\to \zeta }h(y)\geq h(\zeta )\geq g_ k(\zeta )$, so the comparison principle for the superharmonic function $h$ (theorem 3.4.5) gives $H_ Dg_ k\leq h$ on $D$; hence $H\leq h$ on $D$. On the other hand, $H\geq H_ Dg_ k$, and $H_ Dg_ k$ extends continuously to $\overline D$ with boundary values $g_ k$; letting $k\to \infty $, we get

Define $h^ D=H$ on $D$ and $h^ D=h$ on $B\setminus D$. The last inequality and the lower semicontinuity of $h$ show that $h^ D$ is lower semicontinuous, and $0\leq h^ D\leq h$.

We check that $h^ D$ is superharmonic, using the converse part of theorem 3.4.5. Let $E$ be a ball with $\overline E\subset B$, and let $q$ be harmonic on $E$ and continuous on $\overline E$ with $\varliminf _{E\ni y\to \zeta }h^ D(y)\geq q(\zeta )$ for $\zeta \in \partial E$. Since $h\geq h^ D$, the function $h$ satisfies the same boundary inequality, so $h\geq q$ on $E$. In particular $h^ D=h\geq q$ on $E\setminus D$. On the open set $E\cap D$ consider the harmonic function $H-q$. At boundary points of $E\cap D$ lying on $\partial E$, its lower limits are nonnegative by the hypothesis on $q$. At boundary points $\zeta \in E\cap \partial D$, they are at least $h(\zeta )-q(\zeta )\geq 0$ by the inequality displayed above. The minimum principle (proposition 3.1.10) gives $H\geq q$ on $E\cap D$. Hence $h^ D\geq q$ on $E$, and $h^ D$ is superharmonic. Since $D\cap A=\varnothing $, the function $h^ D$ is again a capped admissible function. Finally, $h^ D$ depends monotonically on $h$, by the integral formula for $H$.

Step 3: harmonicity off $A$. By Choquet's lemma (lemma 3.4.22), applied to the negatives of the capped admissible functions, there is a sequence of capped admissible functions whose infimum has the same lower semicontinuous regularization $r$ as $R_ s^ A$. By theorem 3.4.23, applied to the negatives of this sequence, the infimum equals $r$ almost everywhere. Replacing the functions of the sequence by their successive minima, we obtain a decreasing sequence $(h_ j)$ of capped admissible functions whose limit equals $r$ almost everywhere.

Now fix a ball $D$ as in Step 2. On $D$, the modified functions $h_ j^ D$ are harmonic, decrease, and lie between $0$ and $M$; by part (3) of corollary 3.1.21, they converge on $D$ to a harmonic function $\Phi $. Each $h_ j^ D$ is admissible, so $h_ j^ D\geq R_ s^ A\geq r$; moreover $h_ j^ D\leq h_ j$. Thus $r\leq \Phi \leq \lim _ jh_ j=r$ almost everywhere on $D$. The bounded superharmonic functions $r$ and $\Phi $ therefore agree almost everywhere, hence everywhere on $D$ by theorem 3.4.12. Since such balls cover $B\setminus A$, the function $r$ is harmonic on $B\setminus A$.

Step 4: Green representation. The function $-r$ is bounded and subharmonic on $B$, so $\nu =-\Delta r$ is a positive measure by part (1) of theorem 3.4.12. As $r$ is harmonic on $B\setminus A$, the measure $\nu $ is supported on the compact set $A$ and is therefore finite. The function $-r$ is bounded, hence locally integrable, its Riesz measure $\nu $ has compact support in $B$, and it tends to zero at $\partial B$; so the converse in part (2) of proposition 4.1.16 gives $-r=-G_ B\nu $, that is, $r=G_ B\nu $ on $B$. The reduction $R_ s^ A$ equals $s$ on $A$, and $r=R_ s^ A$ outside a polar set, so $r=s$ quasi-everywhere on $A$. Finally $I_ B(\nu )=\int r\, \mathrm d\nu \leq M\nu (A)<\infty $. $\square$