Theorem 7.11.306C02DFB33B7C

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. $\square$