7.11 Local energy moments867F8B92120A

Proof. We prove the tail bound by applying Błocki's estimate (proposition 7.4.5) to truncations of $\varphi $, and then integrate it. For $s>0$ let

By the remark following definition 7.11.1, there is a compact set $K_ s\subset B$ with $\varphi >-s$ on $B\setminus K_ s$, and since $K_ s$ is compact there is a ball $B_ s\Subset B$, concentric with $B$, which contains $K_ s$. Thus $h_ s=\varphi $ on $B_ s\setminus K_ s$, which is a collar of $\partial B_ s$, and $d_ s$ vanishes outside $K_ s$. The functions $\varphi $, $h_ s$ and $w$ are bounded and plurisubharmonic near $\overline{B_ s}$, $\varphi \le h_ s$, and the oscillation of $w$ on $B_ s$ is at most $M$. Błocki's estimate on $B_ s$, together with $\mu \leq (\mathrm{dd}^{\mathrm{c}}w)^ n$, therefore gives

Now take $s=t/2$. On $\{ \varphi <-t\} $ we have $d_ s>t/2$. On the other hand, $0\le d_ s\le -\varphi $, and $d_ s=0$ where $-\varphi \le t/2$; since $p\geq 1$, we have $-\varphi \le (t/2)^{1-p}(-\varphi )^ p$ where $-\varphi >t/2$, and hence

Combining the two estimates and dividing by $(t/2)^{n+1}$, we obtain

Since also $\mu (\{ \varphi <-t\} )\le m$, this proves (7.11.3).

If $m=0$, both sides of (7.11.4) vanish. If $\kappa _{n,p}e_ p(\varphi )=0$, then (7.11.3) shows that $\mu (\{ \varphi <-t\} )=0$ for every $t>0$, so the left side of (7.11.4) vanishes. Otherwise put

The first equality below is the layer-cake formula, which follows from Tonelli's theorem applied to the nonnegative function $(z,t)\mapsto pt^{p-1}\mathbf1_{\{ \varphi (z)<-t\} }$ on $B\times (0,\infty )$. Using (7.11.3) with the bound $m$ for $t<T$ and the second bound for $t\ge T$, we obtain

By the choice of $T$,

so the right-hand side equals $(1+p/n)mT^ p=A_ pe_ p(\varphi )^{p/(n+p)}$. This proves (7.11.4). $\square$

Proof. If $m=0$, then $\mu _ j=0$ and $U_ j=0$ for all $j$, and all assertions hold. Assume that $m>0$, put $d=\operatorname {dist}(K,\mathbb C^ n\setminus B)>0$, and choose $j_0$ such that $h_ j<d/(6\sqrt{2n})$ for $j\geq j_0$; in the rest of the proof we consider only $j\geq j_0$. Since $\mu $ is a finite Borel measure, it is a Radon measure (proposition 2.2.2), so $\mu $ vanishes outside its support $K$. Hence a cube $Q$ with $\mu (Q)>0$ meets $K$, and as its diameter is $\sqrt{2n}\, h_ j<d/6$, it lies in the $(d/6)$-neighborhood of $K$, which is contained in $B$. These cubes are disjoint, each has volume $h_ j^{2n}$, and they lie in a bounded set; so there are only finitely many of them. Therefore $\mu _ j$ is a finite sum of measures with constant densities on cubes contained in $B$, so it has a bounded density and compact support in $B$, and $\mu _ j(B)=\sum _ Q\mu (Q)=m$. This proves (1).

For the weak convergence, let $g$ be a continuous function on $\mathbb C^ n$. Then

and on every cube with $\mu (Q)>0$ the integrand is bounded by the modulus of continuity of $g$ at $\sqrt{2n}\, h_ j$ on the closed $(d/6)$-neighborhood of $K$. Hence the difference tends to zero.

To prove (2), fix a cube $Q\in \mathcal Q_ j$ which meets $K$ and a point $z\in Q$, and put $\varrho =4\sqrt2\, h_ j$. Every point of $Q$ differs from $z$ by less than $\sqrt2\, h_ j$ in each complex coordinate, so $Q\subset P(z,\varrho /2)$. Moreover $P(z,\varrho )$ lies in the Euclidean ball of radius $4\sqrt{2n}\, h_ j$ about $z$, and $\operatorname {dist}(z,K)\le \sqrt{2n}\, h_ j$, so this polydisc lies in the $(5d/6)$-neighborhood of $K$ and hence in $B$. Let $v\le 0$ be a plurisubharmonic function on $P(z,\varrho )$; if $v(z)=-\infty $, there is nothing to prove, so we assume that $v(z)>-\infty $. The function $\zeta \mapsto v(z+\varrho \zeta )$ is plurisubharmonic and nonpositive on the unit polydisc (proposition 5.9.3), with the finite value $v(z)$ at the origin. The first estimate of proposition 5.10.8, with its constant $C_{n,p}$, and the change of variables $\zeta \mapsto z+\varrho \zeta $, whose Jacobian is $\varrho ^{2n}$, give

Since $|Q|=h_ j^{2n}$ and $(\varrho /h_ j)^{2n}=32^ n$, we obtain (7.11.6) with $D_{n,p}=32^ nC_{n,p}^ p$.

For (3), let $\varphi \in \mathcal E_0(B)$. Since $\varphi \le 0$ is plurisubharmonic on $B$, (7.11.6) applies to $v=\varphi $ at every point $z$ of every cube with $\mu (Q)>0$. Integrating over $z\in Q$ with respect to $\mu $ and summing over the cubes, we obtain

and (7.11.7) follows from (7.11.4).

For (4), note that $\mu _ j$ equals $f_ j\eta ^ n$ for a bounded density $f_ j\ge 0$, so the solution $U_ j$ exists by theorem 7.10.2, applied with zero boundary values. Since $U_ j$ is subharmonic (theorem 5.9.1) with boundary limits $0$, the maximum principle (proposition 3.4.7) gives $U_ j\le 0$, and its Monge–Ampère measure has finite mass $\mu _ j(B)=m$; hence $U_ j\in \mathcal E_0(B)$. Taking $\varphi =U_ j$ in (7.11.7), we get

If $e_ p(U_ j)>0$, dividing by $e_ p(U_ j)^{p/(n+p)}$ gives the asserted bound; if $e_ p(U_ j)=0$, there is nothing to prove.

Finally write $B=B(a,R)$ and $\rho (z)=|z-a|^2-R^2$. The function $\rho $ is bounded and plurisubharmonic on $B$, and $(\mathrm{dd}^{\mathrm{c}}\rho )^ n=\eta ^ n=n!\pi ^{-n}\, \mathrm dV$. Applying theorem 7.11.2 to the finite measure $(\mathrm{dd}^{\mathrm{c}}\rho )^ n$ on $B$, with $p=2$, and using the uniform bound for $e_2(U_ j)$, we obtain $S=\sup _ j\int _ BU_ j^2\, \mathrm dV<\infty $. Since $U_ j\le 0$, theorem 5.10.4 shows that every subsequence has a further subsequence converging in $L^1_{\mathrm{loc}}(B)$ to a plurisubharmonic function $u$; convergence to $-\infty $ is excluded by the $L^2$ bound. Exhausting $B$ by compact sets and passing to a diagonal subsequence, we may assume that $U_ j\to u$ almost everywhere on $B$, so Fatou's lemma gives $\int _ Bu^2\, \mathrm dV\le S$. For a compact set $L\subset B$, the Cauchy–Schwarz inequality gives

which is small when $|B\setminus L|$ is small. Since $U_ j\to u$ in $L^1(L)$, it follows that $U_ j\to u$ in $L^1(B)$ along this subsequence. In particular, if a subsequence of $(U_ j)$ converges in $L^1(B)$, its limit is represented by such a function $u$. Finally $u\le 0$ almost everywhere, hence everywhere, by the uniqueness assertion of theorem 5.9.1 applied to $\max \{ u,0\} $ and $0$. $\square$

Proof. We show that $\varlimsup _ jv_ j=v$ $\nu $-almost everywhere (Step 1) and that $\int v_ j\, \mathrm d\nu \to \int v\, \mathrm d\nu $ (Step 2), and combine the two facts (Step 3). First, the bounds $-L\le v\le L$ hold almost everywhere: for every compact set $K\subset D$ we have $(|v|-L)_+\le |v-v_ j|$, so $\int _ K(|v|-L)_+\, \mathrm dV\le \| v_ j-v\| _{L^1(K)}\to 0$. They hold everywhere by the uniqueness assertion of theorem 5.9.1, applied on each connected component of $D$ to the plurisubharmonic functions $\max \{ v,-L\} $ and $v$, and to $\max \{ v,L\} $ and the constant $L$.

Step 1. The upper limit of $v_ j$. We show that

For $\ell \ge 1$ let $b_\ell =(\sup _{j\geq \ell }v_ j)^*$ be the upper semicontinuous regularization of the supremum. By theorem 5.10.2, these are plurisubharmonic functions with values in $[-L,L]$, and they decrease as $\ell $ increases. We claim that $b_\ell \downarrow v$ everywhere. Let $z\in D$ and $\varepsilon >0$. By upper semicontinuity, $v<v(z)+\varepsilon $ on a neighborhood of a closed ball $\overline{B'}\subset D$ centered at $z$, and Hartogs' lemma (theorem 5.10.4), applied on the connected component of $D$ containing $z$ with the constant majorant $v(z)+\varepsilon $, gives $v_ j<v(z)+\varepsilon $ on $\overline{B'}$ for all large $j$. Hence $\sup _{j\ge \ell }v_ j\le v(z)+\varepsilon $ on $B'$ for all large $\ell $, so $b_\ell (z)\le v(z)+\varepsilon $, and the decreasing limit of $b_\ell $ is at most $v$.

Conversely, $b_\ell \ge v_ j$ for all $j\ge \ell $, so $(v-b_\ell )_+\le |v-v_ j|$ for $j\ge \ell $; integrating over a compact set $K\subset D$ and letting $j\to \infty $, we find that $b_\ell \ge v$ almost everywhere on $K$, hence almost everywhere on $D$. The decreasing limit of the $b_\ell $ is bounded below by $-L$, so it is plurisubharmonic by theorem 5.10.1; it is at most $v$ and equals $v$ almost everywhere, hence everywhere by the uniqueness of plurisubharmonic representatives, applied on each connected component of $D$.

The sets $N_\ell =\{ \sup _{j\ge \ell }v_ j<b_\ell \} $ are Borel, and by corollary 7.6.2 they satisfy $\operatorname {Cap}_ D^*(N_\ell \cap K)=0$ for every compact $K\subset D$. By countable subadditivity of outer capacity (proposition 7.4.3), the union $N=\bigcup _\ell N_\ell $ has the same property, so $\nu (N)=0$ by hypothesis. For $z\notin N$ we have $\sup _{j\ge \ell }v_ j(z)=b_\ell (z)$ for every $\ell $, and therefore $\varlimsup _ jv_ j(z)=\lim _\ell b_\ell (z)=v(z)$. This proves (7.11.10). The argument applies to every sequence of plurisubharmonic functions with values in $[-L,L]$ that converges to $v$ in $L^1_{\mathrm{loc}}(D)$.

Step 2. Convergence of $\int v_ j\, \mathrm d\nu $. We first construct convex combinations of the $v_ j$ whose integrals converge to $\int v\, \mathrm d\nu $. Let $\mathcal C_\ell $ be the set of finite convex combinations of $v_\ell ,v_{\ell +1},\ldots $; its elements are plurisubharmonic functions with values in $[-L,L]$ (proposition 5.9.5). Let

Since $\mathcal C_{\ell +1}\subset \mathcal C_\ell $, the numbers $d_\ell $ increase, and they are bounded by $L\, \nu (D)^{1/2}$; let $d$ be their limit. Choose $r_\ell \in \mathcal C_\ell $ with

For $k\geq \ell $ the midpoint $(r_ k+r_\ell )/2$ belongs to the convex set $\mathcal C_\ell $, so its norm is at least $d_\ell $, and the parallelogram identity gives

The right-hand side tends to zero as $\ell \to \infty $, so we can choose $\ell _1<\ell _2<\cdots $ with $\| r_{\ell _{i+1}}-r_{\ell _ i}\| _{L^2(\nu )}\le 2^{-i}$. By the triangle inequality in $L^2(\nu )$ and the monotone convergence theorem, the function $g=\sum _ i|r_{\ell _{i+1}}-r_{\ell _ i}|$ satisfies $\| g\| _{L^2(\nu )}\le 1$. Hence $g<\infty $ $\nu $-almost everywhere, and at every point where $g$ is finite the sequence $(r_{\ell _ i})$ converges.

On the other hand, the $r_{\ell _ i}$ are plurisubharmonic with values in $[-L,L]$, and $r_\ell \to v$ in $L^1_{\mathrm{loc}}(D)$, since $\| r_\ell -v\| _{L^1(K)}\le \sup _{j\ge \ell }\| v_ j-v\| _{L^1(K)}$ for every compact $K\subset D$. By Step 1, applied to the sequence $(r_{\ell _ i})$, we have $\varlimsup _ ir_{\ell _ i}=v$ $\nu $-almost everywhere. Therefore $r_{\ell _ i}\to v$ $\nu $-almost everywhere, and dominated convergence, with the constant bound $L$ and the finite measure $\nu $, gives $\int r_{\ell _ i}\, \mathrm d\nu \to \int v\, \mathrm d\nu $.

We now show that $\int v_ j\, \mathrm d\nu \to \int v\, \mathrm d\nu $. Otherwise there are $c>0$ and a subsequence $(v_{j_ k})$ such that either $\int v_{j_ k}\, \mathrm d\nu \ge \int v\, \mathrm d\nu +c$ for all $k$, or $\int v_{j_ k}\, \mathrm d\nu \le \int v\, \mathrm d\nu -c$ for all $k$. The subsequence satisfies the hypotheses of the lemma, so the construction above, applied to it, provides convex combinations of its terms whose integrals converge to $\int v\, \mathrm d\nu $. This is impossible, since every convex combination of the $v_{j_ k}$ satisfies the same inequality as the $v_{j_ k}$ themselves.

Step 3. Conclusion. By Step 1, $\varlimsup _ jv_ j\le v$ everywhere, so $(v_ j-v)_+\to 0$ pointwise. These functions are bounded by $2L$, so $\int (v_ j-v)_+\, \mathrm d\nu \to 0$ by dominated convergence. Since

Step 2 proves the assertion.

Finally, suppose that $\nu \le (\mathrm{dd}^{\mathrm{c}}w)^ n$ with $w$ bounded and plurisubharmonic on $D$, and put $M=\sup _ Dw-\inf _ Dw$. If $M=0$, then $w$ is constant and $\nu =0$. Otherwise $(w-\inf _ Dw)/M$ is a competitor for $\operatorname {Cap}_ D$ (definition 7.4.2), whose Monge–Ampère measure is $M^{-n}(\mathrm{dd}^{\mathrm{c}}w)^ n$. Hence $\nu (A)\leq M^ n\operatorname {Cap}_ D(A)$ for every Borel set $A\subset D$, and $\operatorname {Cap}_ D(A)\le \operatorname {Cap}_ D^*(A)$ because $\operatorname {Cap}_ D$ is increasing. If $E$ is as in the hypothesis, then $\nu (E\cap K)=0$ for every compact $K\subset D$, and exhausting $D$ by countably many compact sets gives $\nu (E)=0$. $\square$

Proof. If $m=0$, then $\mu =\mu _ j=0$ and there is nothing to prove. We assume that $m>0$, and use the cubes $Q\in \mathcal Q_ j$, the polydiscs $P(z,r)$ and the estimate (7.11.6) of proposition 7.11.5; recall that $K$ denotes the support of $\mu $.

Step 1. Convergence against the fixed measure $\mu $. By theorem 7.11.2 with $p=2$ and the uniform bound for $e_2(U_ j)$ in proposition 7.11.5, we have $S=\sup _ j\int _ BU_ j^2\, \mathrm d\mu <\infty $. Fix $T>0$. The truncations $\max \{ U_ j,-T\} $ are plurisubharmonic with values in $[-T,0]$, and they converge to $\max \{ u,-T\} $ in $L^1_{\mathrm{loc}}(B)$, since $s\mapsto \max \{ s,-T\} $ is $1$-Lipschitz. The measure $\mu $ satisfies the hypothesis of lemma 7.11.9, because $\mu \le (\mathrm{dd}^{\mathrm{c}}w)^ n$, so the truncations converge in $L^1(\mu )$. Since $|a^2-b^2|\le 2T|a-b|$ for $a,b\in [-T,0]$, it follows that $\int \max \{ u,-T\} ^2\, \mathrm d\mu \le S$. Letting $T\to \infty $, monotone convergence gives $\int u^2\, \mathrm d\mu \le S$; in particular $u\in L^2(\mu )$, and $u$ is finite $\mu $-almost everywhere. For every measurable $v\le 0$ we have $\int _ B|v-\max \{ v,-T\} |\, \mathrm d\mu \le T^{-1}\int _ Bv^2\, \mathrm d\mu $, because $0\le \max \{ v,-T\} -v\le v^2/T$. Applying this to $v=U_ j$ and $v=u$, and using the convergence of the truncations in $L^1(\mu )$, we get $\varlimsup _ j\int _ B|U_ j-u|\, \mathrm d\mu \le 2S/T$, and letting $T\to \infty $ proves the first convergence in (7.11.12).

Step 2. Cube averages. For $z\in K$ let $Q_ j(z)$ be the cube of $\mathcal Q_ j$ containing $z$, and write

for the average over this cube of a function $v$ integrable on it. If $v$ is $\mu _ j$-integrable, then $v$ is integrable on every cube $Q$ with $\mu (Q)>0$; on a cube with $\mu (Q)=0$ on which $v$ is not integrable we set $\mathcal A_ jv=0$, which does not affect integrals against $\mu $. Since $\mu $ is carried by $K$, the definition of $\mu _ j$ gives

For a plurisubharmonic function $v\le 0$ on $B$ and $z\in K$, the Cauchy–Schwarz inequality for the probability measure $\mathrm dV/|Q_ j(z)|$ on $Q_ j(z)$ and (7.11.6) with $p=2$ give

if $v(z)=-\infty $, the right side is infinite. This applies in particular to the possibly unbounded function $u$.

Fix $z\in K$ with $u(z)>-\infty $ and $\varepsilon >0$, and put $c=u(z)+2\varepsilon $. By upper semicontinuity of $u$ and Hartogs' lemma (theorem 5.10.4), there is a neighborhood of $z$ on which $u\le c$ and $U_ j\le c$ for all large $j$. Since the polydiscs $P(z,4\sqrt2\, h_ j)$ shrink to $z$, for $v=u$ or $v=U_ j$ and all large $j$ the function $v-c$ is plurisubharmonic and nonpositive on $P(z,4\sqrt2\, h_ j)$. Then (7.11.6) with $p=1$, applied to $v-c$, gives $\mathcal A_ j(c-v)(z)\le D_{n,1}(c-v(z))$, while $\mathcal A_ jv(z)\le c$ because $v\le c$ on $Q_ j(z)$. Thus, for all large $j$,

Consequently $\mathcal A_ ju(z)\to u(z)$, and $\mathcal A_ jU_ j(z)\to u(z)$ whenever $U_ j(z)\to u(z)$. Indeed, in both cases the lower bound tends to $u(z)+2\varepsilon (1-D_{n,1})$ as $j\to \infty $, the upper bound is $u(z)+2\varepsilon $, and $\varepsilon >0$ is arbitrary.

We use the following elementary fact: if Borel functions $F_ j$ on $K$ converge to zero $\mu $-almost everywhere and are bounded in $L^2(\mu )$, then $\int _ K|F_ j|\, \mathrm d\mu \to 0$. Indeed, for $\varepsilon >0$ the Cauchy–Schwarz inequality gives

and $\mu (\{ |F_ j|>\varepsilon \} )=\int _ K\mathbf1_{\{ |F_ j|>\varepsilon \} }\, \mathrm d\mu \to 0$ by dominated convergence; we let first $j\to \infty $ and then $\varepsilon \downarrow 0$.

By (7.11.14) and Step 1, the sequences $(\mathcal A_ jU_ j-u)$ and $(\mathcal A_ ju-u)$ are bounded in $L^2(\mu )$. Since $u$ is finite $\mu $-almost everywhere, we have shown that $\mathcal A_ ju\to u$ $\mu $-almost everywhere, so $\mathcal A_ ju\to u$ in $L^1(\mu )$ by the elementary fact. By the first convergence in (7.11.12), every subsequence of $(U_ j)$ has a further subsequence $(U_{j_ k})$ with $\sum _ k\int _ B|U_{j_ k}-u|\, \mathrm d\mu <\infty $. By the monotone convergence theorem, $\sum _ k|U_{j_ k}-u|$ is $\mu $-integrable, hence finite $\mu $-almost everywhere, so $U_{j_ k}\to u$ $\mu $-almost everywhere. By the pointwise convergence proved above, applied along this subsequence, $\mathcal A_{j_ k}U_{j_ k}\to u$ $\mu $-almost everywhere, and hence in $L^1(\mu )$. Consequently the whole sequence $\mathcal A_ jU_ j$ converges to $u$ in $L^1(\mu )$. The function $U_ j-u$ is $\mu _ j$-integrable, since $u$ is locally integrable and $\mu _ j$ has a bounded density with compact support. Therefore (7.11.13) gives

Step 3. The absolute difference. Put $g_ j=\mathcal A_ j((U_ j-u)_+)$ on $K$. Let $z\in K$ with $u(z)>-\infty $, and let $c$ be as in Step 2. For all large $j$ we have $U_ j\le c$ and $u\le c$ on $Q_ j(z)$, so $(U_ j-u)_+\le c-u$ there, and hence $0\le g_ j(z)\le c-\mathcal A_ ju(z)$. Since $\mathcal A_ ju(z)\to u(z)$, letting $j\to \infty $ and then $\varepsilon \downarrow 0$ shows that $g_ j\to 0$ $\mu $-almost everywhere. The Cauchy–Schwarz inequality and (7.11.14) give

so the $g_ j$ are bounded in $L^2(\mu )$ by Step 1. The elementary fact of Step 2 gives $\int _ Kg_ j\, \mathrm d\mu \to 0$, and by (7.11.13) this integral equals $\int _ B(U_ j-u)_+\, \mathrm d\mu _ j$. Combining this with (7.11.15) and the identity $|U_ j-u|=2(U_ j-u)_+-(U_ j-u)$, we obtain the second convergence in (7.11.12).

Finally, since $(\mathrm{dd}^{\mathrm{c}}U_ j)^ n=\mu _ j$, we have $\int _ BU_ j(\mathrm{dd}^{\mathrm{c}}U_ j)^ n=\int _ B(U_ j-u)\, \mathrm d\mu _ j+\int _ Bu\, \mathrm d\mu _ j$. The first term tends to zero by the second convergence in (7.11.12), and by (7.11.13) and Step 2 the second term equals $\int _ K\mathcal A_ ju\, \mathrm d\mu $, which tends to $\int _ Bu\, \mathrm d\mu $. This proves the last assertion. $\square$

Proof. The two measures agree on the open set $\Omega \setminus K$, because the products are local (proposition 7.2.2). For $0\leq k\leq n-1$ put $S_ k=(\mathrm{dd}^{\mathrm{c}}\varphi )^ k\wedge (\mathrm{dd}^{\mathrm{c}}\psi )^{n-1-k}$. By the symmetry of the products (theorem 7.3.7) and definition 7.2.1,

so the sum over $k$ telescopes and

The coefficient measures of the current $R$ of order zero vanish on $\Omega \setminus K$, so $\operatorname{Supp}R\subset K$. By lemma 2.5.1 there is $\chi \in C^\infty _ c(\Omega )$ with $0\leq \chi \leq 1$ which equals one on a neighborhood of $K$. Then $\mathrm{dd}^{\mathrm{c}}\chi $ vanishes on a neighborhood of $\operatorname{Supp}R$, and therefore

where both integrals are finite. Since $\chi =1$ on $K$ and the two measures agree on $\Omega \setminus K$, it follows that $(\mathrm{dd}^{\mathrm{c}}\varphi )^ n(K)=(\mathrm{dd}^{\mathrm{c}}\psi )^ n(K)<\infty $, and adding the equal masses of $\Omega \setminus K$ gives the last assertion. $\square$

Proof. Since $\varphi $ tends to zero uniformly at $\partial B$, there is an open neighborhood $V$ of $\partial B$ such that $\varphi >-T$ on $V\cap B$; there $\varphi _ T=\varphi $. Thus $\varphi _ T$ is bounded, nonpositive and plurisubharmonic, it has zero boundary values, and it coincides with $\varphi $ outside the compact set $\overline B\setminus V\subset B$. By lemma 7.11.18, the measures $(\mathrm{dd}^{\mathrm{c}}\varphi _ T)^ n$ and $(\mathrm{dd}^{\mathrm{c}}\varphi )^ n$ have the same total mass, which is finite; in particular $\varphi _ T\in \mathcal E_0(B)$.

Put $F_ T=\{ \varphi \leq -T\} $. By maximum locality (proposition 7.3.10), the two measures agree on the complement $\{ \varphi >-T\} $. Since their total masses are equal and finite, they also assign the same mass $m_ T$ to $F_ T$, and Chebyshev's inequality gives

As $\varphi _ T=-T$ on $F_ T$, we get

The two measures differ only on $F_ T$, where each has mass $m_ T$; this proves (7.11.20). Finally, $(\varphi _ T+T)/T$ is plurisubharmonic with values in $[0,1]$, so it is a competitor for $\operatorname {Cap}_ B$ (definition 7.4.2), and therefore $\int _ E(\mathrm{dd}^{\mathrm{c}}\varphi _ T)^ n\leq T^ n\operatorname {Cap}_ B(E)$. Since the two measures agree on $E\setminus F_ T$, we conclude

Proof. (1) Let $h$ be a competitor for $\operatorname {Cap}_ B$, that is, $h\in \mathrm{PSH}(B)$ and $0\leq h\leq 1$. The set $\{ \varphi <-T\} $ is open, because $\varphi $ is upper semicontinuous, and it is relatively compact in $B$, because $\varphi $ has zero boundary values. Hence $\nu =\mathbf1_{\{ \varphi <-T\} }(\mathrm{dd}^{\mathrm{c}}h)^ n$ is a finite measure, and it is dominated by $(\mathrm{dd}^{\mathrm{c}}h)^ n$. The tail estimate (7.11.3) of theorem 7.11.2, applied to $\nu $ and $w=h$, involves the constant $\kappa _{n,p}=2^{n+p}(n+1)!M^ n$, where $M\leq 1$ is the oscillation of $h$ on $B$, but not the mass of $\nu $; it gives

Taking the supremum over $h$ proves (1).

(2) Choose an open set $U$ and compact sets $L,L'\subset B$ with $K\subset U\Subset \operatorname{Int}L$ and $L\subset \operatorname{Int}L'$. By the Chern–Levine–Nirenberg inequality (theorem 7.2.5, with the base current $1$ and the pair $L\subset \operatorname{Int}L'$), there is a constant $C_ L$ with $\int _ L(\mathrm{dd}^{\mathrm{c}}h)^ n\leq C_ L$ for every competitor $h$, since $\| h\| _{L^\infty (L')}\leq 1$. The finite measure $\mathbf1_ L(\mathrm{dd}^{\mathrm{c}}h)^ n$ is dominated by $(\mathrm{dd}^{\mathrm{c}}h)^ n$ and has mass at most $C_ L$, so the constant $A_ p$ of theorem 7.11.2 for this measure and $w=h$ is at most

Hence the moment estimate (7.11.4), applied to $\varphi _ j$, gives

for all $j$ and all competitors $h$.

We pass this bound to the limit $u$. Since the $\varphi _ j$ are nonpositive, $u\leq 0$ almost everywhere, and hence everywhere by the submean inequality. Fix $S>0$. The truncations $\max \{ \varphi _ j,-S\} $ take values in $[-S,0]$ and converge to the plurisubharmonic function $\max \{ u,-S\} $ in $L^1_{\mathrm{loc}}(B)$, since $t\mapsto \max \{ t,-S\} $ is $1$-Lipschitz. Since $\mathbf1_ L(\mathrm{dd}^{\mathrm{c}}h)^ n\leq (\mathrm{dd}^{\mathrm{c}}h)^ n$, lemma 7.11.9 shows that they also converge in $L^1(\mathbf1_ L(\mathrm{dd}^{\mathrm{c}}h)^ n)$. As $t\mapsto t^ p$ is Lipschitz on $[0,S]$ and $(-\max \{ \varphi _ j,-S\} )^ p\leq (-\varphi _ j)^ p$, it follows that

Letting $S\to \infty $, monotone convergence gives the same bound for $\int _ L(-u)^ p(\mathrm{dd}^{\mathrm{c}}h)^ n$. The set $U\cap \{ u<-T\} $ is open, because $u$ is upper semicontinuous, and it contains $K\cap \{ u<-T\} $. By Chebyshev's inequality its $(\mathrm{dd}^{\mathrm{c}}h)^ n$-measure is at most $C_{K,p}E_ p^{p/(n+p)}T^{-p}$. Taking the supremum over $h$ bounds $\operatorname {Cap}_ B(U\cap \{ u<-T\} )$, and hence the outer capacity in (7.11.23). Finally, $K\cap \{ u=-\infty \} $ is contained in $K\cap \{ u<-T\} $ for every $T$, so letting $T\to \infty $ shows that it has zero outer capacity. $\square$

Proof. We estimate the capacities of the sets where the truncation $\max \{ U_ j,-T\} $ lies far below or far above $u_ T$ (Steps 1 and 2) and deduce convergence in capacity (Step 3). The estimate for $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ then follows from the convergence of the truncated measures (Step 4). Fix $T>0$ and write $a_ j=\max \{ U_ j,-T\} $ and $b=u_ T$. These plurisubharmonic functions take values in $[-T,0]$, and $a_ j\to b$ in $L^1_{\mathrm{loc}}(B)$ because truncation is $1$-Lipschitz. By theorem 7.11.11, $\varepsilon _ j=\int _ B|U_ j-u|\, \mathrm d\mu _ j\to 0$.

Step 1. The sets where $a_ j<b-\delta $. Let $\delta >0$. The set $\{ a_ j<b-\delta /2\} $ is relatively compact in $B$, because $b\leq 0$ while $a_ j$ tends to zero at $\partial B$. Applying lemma 7.10.1 with $s=t=\delta /2$ and then Chebyshev's inequality, we obtain

The weight $(b-a_ j)_+$ takes values in $[0,T]$, and it is at most $|U_ j-u|$, since truncation does not increase distances. By (7.11.20) with $p=2$, $\| (\mathrm{dd}^{\mathrm{c}}a_ j)^ n-\mu _ j\| _{\mathrm{TV}}\leq 2E_2T^{-2}$, where $\mu _ j=(\mathrm{dd}^{\mathrm{c}}U_ j)^ n$. Hence

and therefore $\operatorname {Cap}_ B(\{ a_ j<b-\delta \} )\leq 2^{n+1}\delta ^{-(n+1)}(\varepsilon _ j+2E_2/T)$.

Step 2. The sets where $a_ j>b+\delta $. Let $L\subset B$ be compact. By theorem 7.5.2, for every $\sigma >0$ there is an open set $G\subset B$ with $\operatorname {Cap}_ B(G)<\sigma $ such that $b$ is continuous on $B\setminus G$. We claim that $L\cap \{ a_ j>b+\delta \} \subset G$ for all sufficiently large $j$. Let $z\in L\setminus G$. Since $b$ is upper semicontinuous, $b<b(z)+\delta /3$ on a closed ball $\overline{B(z,r)}\subset B$. Hartogs' lemma (theorem 5.10.4), applied to the sequence $a_ j$ with the constant majorant $b(z)+\delta /3$, gives $a_ j<b(z)+\delta /3$ on $\overline{B(z,r)}$ for all large $j$. Since $b$ is continuous on $B\setminus G$, after shrinking $r$ we also have $b>b(z)-\delta /3$ on $(L\setminus G)\cap B(z,r)$. Hence $a_ j-b<2\delta /3$ on $(L\setminus G)\cap B(z,r)$ for all large $j$. As $L\setminus G$ is compact, finitely many such balls cover it, which proves the claim. Consequently $\varlimsup _ j\operatorname {Cap}_ B(L\cap \{ a_ j>b+\delta \} )\leq \sigma $ for every $\sigma >0$, that is, $\operatorname {Cap}_ B(L\cap \{ a_ j>b+\delta \} )\to 0$.

Step 3. Convergence in capacity. Let $K\subset B$ be compact, and apply Step 2 to a compact set $L\subset B$ with $K\subset \operatorname{Int}L$. The bounded functions $a_ j-b$ are quasi-continuous by theorem 7.5.2 and the remark following definition 7.5.1. Therefore lemma 7.5.4, together with Steps 1 and 2 applied with $\delta /2$ in place of $\delta $, gives

It remains to remove the truncation. Outside $\{ U_ j<-T\} \cup \{ u<-T\} $ we have $a_ j=U_ j$ and $b=u$. Hence both sets $K\cap \{ U_ j>u+\delta \} $ and $K\cap \{ U_ j<u-\delta \} $ in definition 7.5.5 are contained in

By proposition 7.11.22 with $p=2$, applied in part (1) to $U_ j$, whose energy is at most $E_2$, and in part (2) to the sequence $(U_ j)$,

and the capacity of the open set $\{ U_ j<-T\} $ equals its outer capacity. By subadditivity of the outer capacity, the upper limits as $j\to \infty $ of the outer capacities of the two sets in definition 7.5.5 are bounded by the sum of three terms, each of which tends to zero as $T\to \infty $. Since these upper limits do not depend on $T$, they vanish. Thus $U_ j\to u$ locally in capacity.

Step 4. The truncated limit measure. Fix $T>0$ again. By Step 3 and (7.11.17), the uniformly bounded functions $a_ j$ converge to $b$ locally in capacity, so $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n\to (\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ weakly on $B$ by theorem 7.5.8. By lemma 7.11.19, each $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n$ has total mass $\mu _ j(B)=m$. Testing with functions $\chi \in C_ c(B)$, $0\leq \chi \leq 1$, and letting $\chi $ increase to one, we see that $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ has total mass at most $m$. Moreover, (7.11.20) gives

for every real $\chi \in C_ c(B)$ with $|\chi |\leq 1$. Since $\mu _ j\to \mu $ weakly by proposition 7.11.5, letting $j\to \infty $ yields the same bound with $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ and $\mu $ in place of $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n$ and $\mu _ j$. Both limit measures are finite, so by the remark following the definition of the total variation distance, the supremum over such $\chi $ equals $\| (\mathrm{dd}^{\mathrm{c}}u_ T)^ n-\mu \| _{\mathrm{TV}}$. This proves (7.11.25). $\square$

Proof. Let $C$ be a common bound for $|f_ j|$ and $|f|$. Since $\omega $ is nonnegative and nondecreasing with $\omega (t)\to 0$ as $t\downarrow 0$, we have $\omega (0)=0$ and $\omega (t_ j)\to 0$ whenever $t_ j\geq 0$ and $t_ j\to 0$. Choose $\chi \in C_ c(B)$ with $0\leq \chi \leq 1$ and $\chi =1$ on $K_0$. Since the measures are carried by $K_0$, we have $\nu _ j(B)=\int \chi \, \mathrm d\nu _ j\to \int \chi \, \mathrm d\nu =\nu (B)$, so the masses of the $\nu _ j$ and of $\nu $ have a common bound $M$. The capacity bound passes to $\nu $ on open sets: if $G\subset B$ is open, then $\operatorname {Cap}^*_ B(G)=\operatorname {Cap}_ B(G)$, and proposition 2.6.2 gives $\nu (G)\leq \varliminf _ j\nu _ j(G)\leq \omega (\operatorname {Cap}_ B(G))$.

We first consider the fixed weight $f$. Let $L\subset B$ be a compact neighborhood of $K_0$ on which $f$ is quasi-continuous, and let $\kappa >0$. Choose an open set $G\subset B$ with $\operatorname {Cap}_ B(G)<\kappa $ such that $f$ is continuous on the compact set $L\setminus G$. By lemma 7.5.3, this restriction extends to a continuous function on $\mathbb C^ n$ bounded by $C$. Multiplying the extension by a cutoff in $C_ c(\operatorname{Int}L)$ with values in $[0,1]$ and equal to one on $K_0$, we obtain $g\in C_ c(B)$ with $|g|\leq C$ and $g=f$ on $K_0\setminus G$. Since $\nu _ j$ and $\nu $ are carried by $K_0$,

and similarly for $\nu $, since $\nu (G)\leq \omega (\operatorname {Cap}_ B(G))\leq \omega (\kappa )$. Since $\int g\, \mathrm d\nu _ j\to \int g\, \mathrm d\nu $, we get $\varlimsup _ j|\int f\, \mathrm d\nu _ j-\int f\, \mathrm d\nu |\leq 4C\omega (\kappa )$. Letting $\kappa \downarrow 0$ proves the assertion for the fixed weight.

For the varying weights, let $\delta >0$. Since $|f_ j-f|\leq 2C$ and $\nu _ j$ is carried by $K_0$,

As $j\to \infty $, the capacity on the right tends to zero, and hence so does the second term. Letting $\delta \downarrow 0$ gives $\int |f_ j-f|\, \mathrm d\nu _ j\to 0$, which together with the fixed-weight case proves the first assertion.

For the level sets, fix $s>0$ and let $\sigma ^-_ s,\sigma ^+_ s\colon \mathbb R\to [0,1]$ be the continuous piecewise linear functions such that $\sigma ^-_ s$ equals one on $(-\infty ,c-s]$, zero on $[c,\infty )$ and is linear in between, and $\sigma ^+_ s$ equals one on $(-\infty ,c]$, zero on $[c+s,\infty )$ and is linear in between. Then

Since $\sigma ^\pm _ s$ are Lipschitz, $\sigma ^\pm _ s\circ f_ j$ converge to $\sigma ^\pm _ s\circ f$ in the same sense as $f_ j\to f$, and $\sigma ^\pm _ s\circ f$ is quasi-continuous on a neighborhood of $K_0$ (remark following definition 7.5.1). The first assertion therefore gives

As $s\downarrow 0$, the two outer terms tend to $\nu (\{ f<c\} )$ and $\nu (\{ f\leq c\} )$, which are equal since $\nu (\{ f=c\} )=0$. This proves (7.11.28). $\square$

Proof. By the moment estimate (7.11.4) with $p=2$, the integrals $\int _ B\varphi _ j^2\, \mathrm d\nu $ are bounded uniformly in $j$. Since the $\varphi _ j$ are nonpositive, so is $u$, by the submean inequality. Fix $T>0$. By lemma 7.11.9, the truncations $\max \{ \varphi _ j,-T\} $ converge to $\max \{ u,-T\} $ in $L^1(\nu )$, and so do their squares, since squaring is Lipschitz on $[-T,0]$. Hence $\int _ B\max \{ u,-T\} ^2\, \mathrm d\nu \leq \sup _ j\int _ B\varphi _ j^2\, \mathrm d\nu $, and letting $T\to \infty $ shows that $u\in L^2(\nu )$ with the same bound. For a function $v\leq 0$ we have $0\leq \max \{ v,-T\} -v\leq v^2/T$, so the $L^1(\nu )$ distances between $\varphi _ j,u$ and their truncations are at most $T^{-1}$ times a bound independent of $j$. Letting first $j\to \infty $ and then $T\to \infty $ proves that $\varphi _ j\to u$ in $L^1(\nu )$. $\square$

Proof. If $\mu =0$, then all $\mu _ j$ and all $U_ j$ vanish, and $U=0$ is a solution. By the maximum principle (proposition 3.4.7), every plurisubharmonic function on $B$ whose upper boundary limits are at most zero is nonpositive, so $U=0$ is the stated supremum, and the last assertion of theorem 7.3.11, applied in both directions, shows that any two bounded solutions coincide. We may therefore assume that $m=\mu (B)>0$. We construct a barrier $W$ (Step 1) and, along a convergent subsequence of canonical approximants, prove that the masses of certain sublevel sets converge (Step 2). The comparison theorem then shows that the limit lies above $W$ and is a solution (Step 3), and uniqueness gives the convergence of the whole sequence (Step 4).

Step 1. A barrier with zero boundary values. Write $B=B(a,R)$ and $\rho (z)=|z-a|^2-R^2$, let $M=\sup _ Bw-\inf _ Bw$, and put $w_0=w-\sup _ Bw-1$, so that $-M-1\leq w_0\leq -1$. Since $\operatorname{Supp}\mu $ is a compact subset of $B$, the function $\rho $ is bounded away from zero on an open neighborhood $N\Subset B$ of $\operatorname{Supp}\mu $, and we can choose $A>0$ with $A\rho <-M-2$ on $N$. Put

a plurisubharmonic function on $B$ (proposition 5.9.5). Then $W=w_0$ on $N$, while $W=A\rho $ near $\partial B$, because $A\rho \to 0$ at $\partial B$ and $w_0\leq -1$. Thus $W$ is bounded and negative and has zero boundary values. By locality on the open set $N$ (proposition 7.2.2), $(\mathrm{dd}^{\mathrm{c}}W)^ n=(\mathrm{dd}^{\mathrm{c}}w)^ n\geq \mu $ there; as $\mu $ is carried by $N$, we get $(\mathrm{dd}^{\mathrm{c}}W)^ n\geq \mu $ on $B$.

For $\varepsilon >0$ put $W_\varepsilon =W+\varepsilon \rho $. It equals $(A+\varepsilon )\rho $ near $\partial B$, so its Monge–Ampère measure is a smooth measure there and is finite on the remaining compact part of $B$; hence it has finite total mass. Expanding $(\mathrm{dd}^{\mathrm{c}}W+\varepsilon \mathrm{dd}^{\mathrm{c}}\rho )^ n$ and dropping the mixed terms, which are positive, we obtain

Step 2. Convergence of the masses of sublevel sets. By proposition 7.11.5, a subsequence of the canonical approximants, still denoted $U_ j$ (together with the corresponding measures $\mu _ j$), converges in $L^1(B)$; let $u$ be the nonpositive plurisubharmonic representative of the limit. By theorem 7.11.24, $U_ j\to u$ locally in capacity, and $E_2=\sup _ je_2(U_ j)<\infty $. By corollary 7.11.29 with $\nu =\nu _\varepsilon $, we have $U_ j\to u$ in $L^1(\nu _\varepsilon )$ for every $\varepsilon >0$.

We check that lemma 7.11.27 applies to the measures $\mu _ j$ and $\mu $. Let $K$ be the support of $\mu $ and $d=\operatorname {dist}(K,\mathbb C^ n\setminus B)>0$, and let $h_ j$ be the side lengths of the cubes defining $\mu _ j$. A cube $Q$ with $\mu _ j(Q)>0$ satisfies $\mu (Q)>0$, so it meets $K$. Since $h_ j\to 0$, after discarding finitely many indices every such cube has diameter $\sqrt{2n}\, h_ j<d/2$, and then $\mu _ j$ is supported in the compact set $K_0=\{ z:\operatorname {dist}(z,K)\leq d/2\} \subset B$, which also contains the support of $\mu $. By proposition 7.11.5, $\mu _ j\to \mu $ weakly. By (7.11.21) with $p=2$, and since $\operatorname {Cap}_ B(E)\leq \operatorname {Cap}_ B^*(E)$, we have $\mu _ j(E)\leq T^ n\operatorname {Cap}_ B^*(E)+E_2T^{-2}$ for every Borel set $E\subset B$ and every $T>0$. Let $\omega (t)$ be the supremum of $\mu _ j(E)$ over all $j$ and all Borel sets $E\subset B$ with $\operatorname {Cap}_ B^*(E)\leq t$. Then $\omega $ is nondecreasing with values in $[0,m]$, and $\omega (t)\leq T^ nt+E_2T^{-2}$ for every $T>0$. Choosing first $T$ large and then $t$ small shows that $\omega (t)\to 0$ as $t\downarrow 0$.

Fix $\varepsilon >0$. The level sets $\{ u-W_\varepsilon =-\delta \} $, $0<\delta <1$, are pairwise disjoint Borel sets, and $\mu +\nu _\varepsilon $ is a finite measure, so by lemma 2.2.5 all but countably many of these sets are null for $\mu +\nu _\varepsilon $, hence for both measures. Choose $\delta \in (0,1)$ such that $\{ u-W_\varepsilon =-\delta \} $ is null for both $\mu $ and $\nu _\varepsilon $, and fix $T>\| W_\varepsilon \| _\infty +2$. Since $W_\varepsilon -\delta >-T$, the sets $\{ U_ j<W_\varepsilon -\delta \} $ and $\{ u<W_\varepsilon -\delta \} $ do not change when $U_ j$ and $u$ are replaced by $a_ j=\max \{ U_ j,-T\} $ and $b=\max \{ u,-T\} $.

We apply lemma 7.11.27 to the functions $f_ j=a_ j-W_\varepsilon $ and $f=b-W_\varepsilon $, which are bounded by $T+\| W_\varepsilon \| _\infty $ and quasi-continuous by theorem 7.5.2 and the remark following definition 7.5.1. Since $U_ j\to u$ locally in capacity, (7.11.17) shows that $a_ j\to b$ locally in capacity, and hence $f_ j\to f$, because $W_\varepsilon $ is finite. Moreover, $\mu (\{ f=-\delta \} )=0$ by the choice of $\delta $, since $\{ f=-\delta \} =\{ u-W_\varepsilon =-\delta \} $: indeed, $b=u$ wherever $u\geq -T$, while $u-W_\varepsilon <-2$ and $b-W_\varepsilon \leq -T+\| W_\varepsilon \| _\infty <-2$ wherever $u<-T$. Thus (7.11.28), with $c=-\delta $, gives

The analogous convergence holds for the fixed measure $\nu _\varepsilon $. Indeed, for $s>0$, outside $\{ |U_ j-u|>s\} $, the indicators of $\{ U_ j<W_\varepsilon -\delta \} $ and $\{ u<W_\varepsilon -\delta \} $ can differ only on $\{ |u-W_\varepsilon +\delta |\leq s\} $. The $\nu _\varepsilon $-mass of the first set tends to zero as $j\to \infty $, by the convergence in $L^1(\nu _\varepsilon )$, and that of the second set tends to $\nu _\varepsilon (\{ u-W_\varepsilon =-\delta \} )=0$ as $s\downarrow 0$. Hence

Step 3. The limit is a bounded solution. Since $U_ j$ and $W_\varepsilon $ have zero boundary values, the difference $U_ j-(W_\varepsilon -\delta )$ has boundary limits $\delta >0$. The comparison theorem (theorem 7.3.11) therefore gives

Let $j\to \infty $ and put $S=\{ u<W_\varepsilon -\delta \} $. By (7.11.31),

Since $\mu (S)$ is finite, $S$ has Lebesgue measure zero. Thus $u\geq W_\varepsilon -\delta $ almost everywhere, and hence everywhere, by the uniqueness assertion of theorem 5.9.1 applied to $\max \{ u,W_\varepsilon -\delta \} $ and $u$. Letting $\delta \downarrow 0$ through admissible values and then $\varepsilon \downarrow 0$, we obtain $W\leq u\leq 0$.

As $W$ has zero boundary values, $u$ is bounded and has zero boundary values. For $T>\| u\| _\infty $ the truncation $\max \{ u,-T\} $ is $u$ itself, so (7.11.25) gives $\| (\mathrm{dd}^{\mathrm{c}}u)^ n-\mu \| _{\mathrm{TV}}\leq 2E_2T^{-2}$ for all large $T$. Hence $(\mathrm{dd}^{\mathrm{c}}u)^ n=\mu $. In particular $(\mathrm{dd}^{\mathrm{c}}u)^ n$ has finite mass $m$, and $u\in \mathcal E_0(B)$.

Step 4. Uniqueness and convergence of the whole sequence. Let $v$ be a bounded plurisubharmonic function on $B$ with $(\mathrm{dd}^{\mathrm{c}}v)^ n\geq \mu $ and upper boundary limits at most zero, and put $v_\varepsilon =v+\varepsilon \rho $ for $\varepsilon >0$. As in Step 1, $(\mathrm{dd}^{\mathrm{c}}v_\varepsilon )^ n\geq \mu +\varepsilon ^ n(\mathrm{dd}^{\mathrm{c}}\rho )^ n$. Since $u-v_\varepsilon $ has nonnegative lower boundary limits, the comparison theorem, applied to $u$ and $v_\varepsilon $, gives

The two $\mu $-masses are finite, so the set $\{ u<v_\varepsilon \} $ is Lebesgue null. Therefore $u\geq v_\varepsilon $ everywhere, by the uniqueness assertion of theorem 5.9.1 applied to $\max \{ u,v_\varepsilon \} $ and $u$. Letting $\varepsilon \downarrow 0$ gives $u\geq v$. Since $u$ is itself such a function, it is the pointwise supremum in the statement. Any bounded solution with zero boundary values is such a function, hence is at most $u$; exchanging the roles of the two solutions proves uniqueness. We denote the solution by $U$.

Finally, by proposition 7.11.5, every subsequence of the canonical approximants has a further subsequence converging in $L^1(B)$. By Step 3 and uniqueness, its limit is $U$, and theorem 7.11.11, theorem 7.11.24 give the asserted convergence of the integrals and convergence in capacity along it. If one of the asserted convergences failed for the whole sequence, some subsequence would stay away from the stated limit, and no further subsequence could converge as just described. This contradiction proves all the assertions. $\square$

Proof. Write $B=B(a,R)$ and $\rho (z)=|z-a|^2-R^2$. We say that $v\in \mathcal E_0(B)$ has a quadratic collar if $v=c\rho $ on a neighborhood of $\partial B$ in $B$ for some constant $c>0$. We first prove the inequality with constant one for potentials with quadratic collars (Steps 1 and 2), and then approximate arbitrary elements of $\mathcal E_0(B)$ by such potentials while keeping their energies under control (Steps 3 and 4). For functions $b_0,\ldots ,b_ n$ which either all have quadratic collars, or are smooth real functions on a neighborhood of $\overline B$ with $b_2,\ldots ,b_ n$ plurisubharmonic, put

The integral is finite: in the first case every mixed product of the $b_ l$ is smooth near $\partial B$ and locally finite on the rest of $B$, and in the second the integrand is a smooth form on a neighborhood of $\overline B$.

Step 1. Smooth functions. Let $b_0,\ldots ,b_ n$ be smooth real functions on a neighborhood of $\overline B$, with $b_2,\ldots ,b_ n$ plurisubharmonic and $b_0=0$ on $\partial B$, and put $S=\mathrm{dd}^{\mathrm{c}}b_2\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}b_ n$, with $S=1$ if $n=1$. With $\mathrm d^{\mathrm c}$ as in (5.2.10), we claim that

Fix a smooth function $\theta \colon \mathbb R\to [0,1]$ with $\theta =1$ on $(-\infty ,-2]$ and $\theta =0$ on $[-1,\infty )$, and put $\chi _ k=\theta (k\rho )$ for $k\geq 1$; then $\chi _ k\in C^\infty _ c(B)$. The $(2n-1)$-form $\chi _ kb_0\, \mathrm d^{\mathrm c}b_1\wedge S$ is smooth with compact support in $B$, and $S$ is closed, so proposition 5.2.6 and the Leibniz rule give

As $k\to \infty $, $\chi _ k\to 1$ pointwise on $B$, and the forms $\mathrm db_0\wedge \mathrm d^{\mathrm c}b_1\wedge S$ and $b_0\, \mathrm{dd}^{\mathrm{c}}b_1\wedge S$ have bounded coefficients on $B$; by dominated convergence, the first and the last integral tend to $\int _ B\mathrm db_0\wedge \mathrm d^{\mathrm c}b_1\wedge S$ and $-I(b_0,\ldots ,b_ n)$. In the middle integral, $\mathrm d\chi _ k=k\theta '(k\rho )\, \mathrm d\rho $ vanishes outside the set where $-2\leq k\rho \leq -1$. Let $\Lambda $ be the maximum of $|\nabla b_0|$ on $\overline B$. If $z\in B$ and $\zeta \in \partial B$ is a nearest boundary point, then $b_0(\zeta )=0$ and $|z-\zeta |=R-|z-a|$, so the mean value theorem on the segment from $z$ to $\zeta $ gives

Hence $|b_0\, k\theta '(k\rho )|\leq (\Lambda /R)|k\rho |\, |\theta '(k\rho )|\leq 2\Lambda \sup |\theta '|/R$ on $B$, uniformly in $k$, and $b_0\, k\theta '(k\rho )\to 0$ pointwise on $B$. Since $\mathrm d\rho \wedge \mathrm d^{\mathrm c}b_1\wedge S$ has bounded coefficients on $B$, dominated convergence shows that the middle integral tends to zero. This proves (7.11.35).

As observed at the beginning of section 7.3, $\mathrm db\wedge \mathrm d^{\mathrm c}c\wedge S=\mathrm dc\wedge \mathrm d^{\mathrm c}b\wedge S$ for smooth real $b,c$. Moreover, $\mathrm db\wedge \mathrm d^{\mathrm c}b=(\mathrm i/2\pi )\partial b\wedge \bar\partial b$ is a positive $(1,1)$-form, and so are the forms $\mathrm{dd}^{\mathrm{c}}b_ l$, $l\geq 2$, since the complex Hessian of a smooth plurisubharmonic function is positive semidefinite (proposition 5.9.2). Positive $(1,1)$-forms are strongly positive (section 5.3), so by proposition 5.3.2 the $(n,n)$-form $\mathrm db\wedge \mathrm d^{\mathrm c}b\wedge S$ is a nonnegative multiple of the volume form. Consequently the real bilinear form

defined for smooth real functions $b,c$ on a neighborhood of $\overline B$ vanishing on $\partial B$, is symmetric and positive semidefinite, and by (7.11.35) we have $Q(b,c)=I(b,c,b_2,\ldots ,b_ n)$. The Cauchy–Schwarz inequality for $Q$, which follows from the nonnegativity of the quadratic polynomial $t\mapsto Q(b+tc,b+tc)$, therefore gives

The symmetry of $Q$ allows us to interchange the first two arguments of $I$, and the symmetry of the products allows us to permute the last $n$ arguments. Hence $I$ is symmetric in all $n+1$ arguments when they are smooth plurisubharmonic functions on a neighborhood of $\overline B$ vanishing on $\partial B$.

Step 2. Quadratic collars. Let $u_0,\ldots ,u_ n$ have quadratic collars, say $u_ l=c_ l\rho $ near $\partial B$. We first show that the assertions of Step 1 persist for them. Extend each $u_ l$ by $c_ l\rho $ to a plurisubharmonic function on a neighborhood of $\overline B$. Let $\vartheta _\varepsilon $ be the radial mollifiers of section 2.5, denoted $\rho _\varepsilon $ there; we change the letter because $\rho $ is now the quadratic function above. Because $\rho $ is a quadratic polynomial and $\vartheta _\varepsilon $ is radial, $\rho *\vartheta _\varepsilon =\rho +\gamma \varepsilon ^2$ with the constant $\gamma =\int |y|^2\vartheta _1(y)\, \mathrm dV(y)$. Hence

equals $c_ l\rho $ on a fixed neighborhood of $\partial B$ for all small $\varepsilon $. By proposition 5.9.2, the convolutions $u_ l*\vartheta _\varepsilon $ are smooth and plurisubharmonic on a neighborhood of $\overline B$ for small $\varepsilon $, and they decrease to $u_ l$ as $\varepsilon \downarrow 0$. Hence each $u_{l,\varepsilon }$, which differs from $u_ l*\vartheta _\varepsilon $ by a constant, is smooth and plurisubharmonic on a neighborhood of $\overline B$ and vanishes on $\partial B$, although the $u_{l,\varepsilon }$ need not decrease. By theorem 7.3.7,

weakly, and the constant $c_0\gamma \varepsilon ^2$ changes the integral against a fixed test function only by an amount tending to zero, by the local mass bounds of theorem 7.2.5. Near $\partial B$, all the $u_{l,\varepsilon }$ and $u_ l$ coincide with $c_ l\rho $. Splitting the integrals over $B$ by a cutoff in $C_ c(B)$ which equals one outside this neighborhood of $\partial B$, we conclude that $I(u_{0,\varepsilon },\ldots ,u_{n,\varepsilon })\to I(u_0,\ldots ,u_ n)$, and likewise for every rearrangement of the potentials. Passing to the limit in the smooth case proves that $I$ is symmetric and satisfies (7.11.36) for potentials with quadratic collars.

We now prove the inequality with constant one for these potentials. Since $u_ l=c_ l\rho $ near $\partial B$ with $c_ l>0$, the energy $e_1(u_ l)$ is positive: already its integral over a neighborhood of $\partial B$ is positive. Since $I$ is positively homogeneous of degree one in each argument and $e_1(tv)=t^{n+1}e_1(v)$ for $t>0$, both sides of the inequality are multiplied by the same factor when each $u_ l$ is replaced by $e_1(u_ l)^{-1/(n+1)}u_ l$. We may therefore assume that $e_1(u_ l)=1$ for every $l$. Consider the finitely many ordered $(n+1)$-tuples whose entries are taken from the functions $u_0,\ldots ,u_ n$, and let $F$ be the largest value of $I$ on these tuples. Since $I(u_ l,\ldots ,u_ l)=e_1(u_ l)=1$, we have $F\geq 1$. Among the tuples on which $I$ equals $F$, choose one for which the sum of the squares of the multiplicities of its distinct entries is largest.

Suppose that this tuple contains two distinct functions $b$ and $c$, with multiplicities $r$ and $s$. By symmetry we may place them in the first two slots. In (7.11.36), the left side is $F^2$, while the two factors on the right are nonnegative values of $I$ on tuples of our family, hence at most $F$. Therefore both factors equal $F$. Replacing one entry $c$ by $b$ changes the sum of the squares of the multiplicities by $(r+1)^2+(s-1)^2-r^2-s^2=2(r-s)+2$, and replacing one entry $b$ by $c$ changes it by $2(s-r)+2$. One of these numbers is positive, which contradicts the choice of the tuple. Hence the chosen tuple consists of a single function repeated $n+1$ times, and $F=1$. In particular $I(u_0,\ldots ,u_ n)\leq 1$, which is the inequality with constant one.

Step 3. Approximation with controlled energy. Let $u\in \mathcal E_0(B)$, put $m_\rho =\int _ B(\mathrm{dd}^{\mathrm{c}}\rho )^ n$, and for $j\geq 1$ let

Since $u-\delta _ j\leq -\delta _ j$ while $j\rho >-\delta _ j$ near $\partial B$, the function $v_ j$ equals $j\rho $ on a neighborhood of $\partial B$ in $B$. Since $v_ j$ and $j\rho $ coincide outside a compact subset of $B$, lemma 7.11.18 gives $\int _ B(\mathrm{dd}^{\mathrm{c}}v_ j)^ n=j^ nm_\rho $, and hence $v_ j\in \mathcal E_0(B)$. Moreover, $v_ j\to u$ uniformly on $B$. Indeed, let $s>0$, and choose a neighborhood $V$ of $\partial B$ with $u>-s$ on $V\cap B$. Since $u-\delta _ j\leq v_ j\leq 0$, we have $|v_ j-u|\leq s+\delta _ j$ on $V\cap B$. On the set $B\setminus V$, whose closure is a compact subset of $B$, the function $\rho $ is bounded away from zero and $u$ is bounded, so $j\rho <u-\delta _ j$, and hence $v_ j=u-\delta _ j$, there for all large $j$.

Next we estimate $e_1(v_ j)$. Let $t>4\delta _ j$. Since $u\leq v_ j+\delta _ j$ and $v_ j\leq 0$,

for the first inclusion note that on $\{ v_ j<-t\} $ we have $v_ j/2<-t/2<-t/4-\delta _ j$, hence $u\leq v_ j+\delta _ j<v_ j/2-t/4$. As $u$ tends to $0$ and $v_ j/2-t/4$ tends to $-t/4$ at $\partial B$, the comparison theorem (theorem 7.3.11) applies to $u$ and $v_ j/2-t/4$ and gives

For $0<t\leq 4\delta _ j$ we use the total mass $j^ nm_\rho $ instead. Integrating over $t$ by the layer-cake formula $e_1(v)=\int _0^\infty (\mathrm{dd}^{\mathrm{c}}v)^ n(\{ v<-t\} )\, \mathrm dt$, we obtain

Step 4. The general case. Apply the construction of Step 3 to each $u_ l$, obtaining functions $v_{l,j}\in \mathcal E_0(B)$ which equal $j\rho $ near $\partial B$. They are uniformly bounded, since $u_ l-1\leq v_{l,j}\leq 0$, and they converge uniformly, hence locally in capacity, to $u_ l$. Let $\chi \in C_ c(B)$ with $0\leq \chi \leq 1$. The weight $-u_0$ is bounded and quasi-continuous (theorem 7.5.2), so the weighted form of theorem 7.5.8 and Step 2 give

By (7.11.37), the right side is at most

Letting $\chi $ increase to one proves (7.11.34). Vanishing energies need no separate treatment, since the energies of the $v_{l,j}$ are positive. $\square$

Proof. We compare the products with those of the truncations at a level $-T$: the latter converge by theorem 7.5.8, and by theorem 7.11.33 the two differ by a term of order $1/T$. For $T>0$ put $a_{l,j}=\max \{ u_{l,j},-T\} $, and write

By lemma 7.11.19, $a_{l,j}\in \mathcal E_0(B)$ and $e_1(a_{l,j})\leq E$. For each $l$, theorem 7.11.33 with $u_0=u_{l,j}$, respectively $u_0=a_{l,j}$, gives

Let $D_{j,T}=\bigcup _{l=1}^ n\{ u_{l,j}\leq -T\} $. On $\{ u_{l,j}\leq -T\} $ both $-u_{l,j}$ and $-a_{l,j}$ are at least $T$, so Chebyshev's inequality gives

On the complement of $D_{j,T}$, every $u_{l,j}$ is greater than $-T$. For $0\leq k\leq n$ put

so that $P_0=R_ j$ and $P_ n=R_{j,T}$. By the symmetry of the products, $P_ k=\mathrm{dd}^{\mathrm{c}}a_{k,j}\wedge \Theta _ k$ and $P_{k-1}=\mathrm{dd}^{\mathrm{c}}u_{k,j}\wedge \Theta _ k$ with the same closed positive current $\Theta _ k$ formed by the other $n-1$ factors, and maximum locality (proposition 7.3.10 with $p=1$, $u=u_{k,j}$, $v=-T$ and base current $\Theta _ k$) gives $P_ k=P_{k-1}$ on $\{ u_{k,j}>-T\} $. Hence $R_ j=R_{j,T}$ on the complement of $D_{j,T}$, and for every $\chi \in C_ c(B)$,

Now fix $\chi \in C_ c(B)$, choose an open set $B'\Subset B$ containing $\operatorname{Supp}\chi $, and let $T>\max _ l\sup _{B'}|u_ l|$. By (7.11.17), the uniformly bounded functions $a_{l,j}$ converge to the locally bounded plurisubharmonic function $\max \{ u_ l,-T\} $ locally in capacity, and theorem 7.5.8 shows that $R_{j,T}\to \bigwedge _ l\mathrm{dd}^{\mathrm{c}}\max \{ u_ l,-T\} $ weakly on $B$. On $B'$ we have $\max \{ u_ l,-T\} =u_ l$, so by locality (proposition 7.2.2) this limit equals $\mathrm{dd}^{\mathrm{c}}u_1\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}u_ n$ there. Consequently

for all large $T$, and letting $T\to \infty $ completes the proof. $\square$