7.12 Mixed products for a common measure07DF7DC201F0

Proof. Let $n\geq 2$, and put $\lambda =(\prod _ if_ i^{1/n})\mu $. We first prove the theorem when $\Omega $ is a ball $B=B(a_0,R)$ and all $u_ i$ belong to $\mathcal E_0(B)$ (Steps 1 and 2), and then reduce the general case to this one by modifying the potentials near the boundary of a ball (Step 3). For a ball $B=B(a_0,R)$ we write $\rho (z)=|z-a_0|^2-R^2$.

We use the generalized Hölder inequality

for a positive measure $\nu $, a measurable set $A$ and nonnegative measurable functions $g_ i$. If one of the integrals $c_ i=\int _ Ag_ i\, \mathrm d\nu $ vanishes, then $g_ i=0$ $\nu $-almost everywhere on $A$ and both sides vanish, with the convention $0\cdot \infty =0$; if none vanishes and one is infinite, the right side is infinite. Otherwise, integrating the arithmetic–geometric mean inequality $\prod _ i(g_ i/c_ i)^{1/n}\leq \frac1n\sum _ ig_ i/c_ i$ over $A$ gives $\int _ A\prod _ i(g_ i/c_ i)^{1/n}\, \mathrm d\nu \leq 1$, which is (7.12.3). For every compact $K\subset \Omega $ we have $\int _ Kf_ i\, \mathrm d\mu \leq (\mathrm{dd}^{\mathrm{c}}u_ i)^ n(K)<\infty $ by hypothesis, so (7.12.3) shows that $\lambda $ is a positive Radon measure.

Step 1. Exhaustion by compactly supported measures. Choose closed balls $K_ l\subset B$ concentric with $B$, increasing with $l$ and exhausting $B$, and set

These measures are finite since $u_ i\in \mathcal E_0(B)$, they have compact support in $B$, and they are dominated by the Monge–Ampère measure of the bounded plurisubharmonic function $u_ i$. By theorem 7.11.30, the equation $(\mathrm{dd}^{\mathrm{c}}A)^ n=\alpha _{i,l}$ has a bounded solution $A_{i,l}\in \mathcal E_0(B)$ with zero boundary values, and $A_{i,l}$ is the largest bounded plurisubharmonic function $v$ on $B$ with $(\mathrm{dd}^{\mathrm{c}}v)^ n\geq \alpha _{i,l}$ and upper boundary limits at most zero. The function $u_ i$ has these properties for every $l$, and so has $A_{i,l+1}$, because $\alpha _{i,l+1}\geq \alpha _{i,l}$. Therefore

We claim that $A_{i,l}\downarrow u_ i$ as $l\to \infty $. The decreasing limit $A_{i,\infty }$ satisfies $u_ i\leq A_{i,\infty }\leq 0$, so it is a bounded plurisubharmonic function (theorem 5.10.1) with zero boundary values. By theorem 7.3.7, $(\mathrm{dd}^{\mathrm{c}}A_{i,\infty })^ n$ is the weak limit of $\alpha _{i,l}$, that is, $(\mathrm{dd}^{\mathrm{c}}A_{i,\infty })^ n=(\mathrm{dd}^{\mathrm{c}}u_ i)^ n$, and this measure has finite total mass. Let $\varepsilon >0$ and $S=\{ u_ i<A_{i,\infty }+\varepsilon \rho \} $. Since $u_ i-(A_{i,\infty }+\varepsilon \rho )$ tends to zero at $\partial B$, the comparison theorem (theorem 7.3.11) applies. Expanding $(\mathrm{dd}^{\mathrm{c}}(A_{i,\infty }+\varepsilon \rho ))^ n$ and dropping the mixed terms, which are positive, we obtain

The integrals of $(\mathrm{dd}^{\mathrm{c}}u_ i)^ n$ are finite, so $S$ is Lebesgue null. Hence $u_ i\geq A_{i,\infty }+\varepsilon \rho $ almost everywhere, and therefore everywhere on $B$, by the uniqueness assertion of theorem 5.9.1 applied to $\max \{ u_ i,A_{i,\infty }+\varepsilon \rho \} $ and $u_ i$. Letting $\varepsilon \downarrow 0$ gives $A_{i,\infty }=u_ i$, as claimed.

Step 2. A common grid. Fix $l$. Choose a sequence of partitions of $\mathbb C^ n=\mathbb R^{2n}$ into half-open cubes whose side lengths decrease to zero, the same for all $i$. Let $\alpha _{i,l,j}$ be the corresponding cube averages of $\alpha _{i,l}$, as in proposition 7.11.5, and let $A_{i,l,j}\in \mathcal E_0(B)$ be the corresponding canonical approximants of $\alpha _{i,l}$, that is, the zero-boundary solutions of $(\mathrm{dd}^{\mathrm{c}}A_{i,l,j})^ n=\alpha _{i,l,j}$; they are defined for all large $j$. Discarding finitely many indices $j$, we may assume that they are defined for all $j$ and all $i$, and that every cube meeting $K_ l$ has its closure in $B$. By theorem 7.11.30, $A_{i,l,j}\to A_{i,l}$ locally in capacity as $j\to \infty $. By proposition 7.11.5, applied with the bounded subsolution $u_ i$, the energies $e_1(A_{i,l,j})$ are bounded uniformly in $j$; since there are only finitely many indices $i$, they have a common bound. (If $\alpha _{i,l}=0$, all these solutions vanish.) Therefore corollary 7.11.38 gives

weakly on $B$.

Let $\lambda _ l=\mathbf1_{K_ l}\lambda $, a finite measure, and let $\lambda _{l,j}$ be its cube averages. For every cube $Q$, the hypothesis gives $\alpha _{i,l}(Q)\geq \int _{Q\cap K_ l}f_ i\, \mathrm d\mu $, and (7.12.3) then gives

Let $Q$ be a cube with $\lambda _ l(Q)>0$; it meets $K_ l$, so its interior lies in $B$. There the measures $(\mathrm{dd}^{\mathrm{c}}A_{i,l,j})^ n=\alpha _{i,l,j}$ have the constant densities $\alpha _{i,l}(Q)/|Q|$ with respect to Lebesgue measure $\mathrm dV$. Recall that $\eta ^ n=c_ n\, \mathrm dV$ with $\eta =\mathrm{dd}^{\mathrm{c}}|z|^2$ and $c_ n=n!\pi ^{-n}$, by (7.2.4). Applying corollary 7.8.8 on the open set $\operatorname{Int}Q$ to the constant densities $c_ n^{-1}\alpha _{i,l}(Q)/|Q|$ with respect to $\eta ^ n$, and using (7.12.5), we obtain on $\operatorname{Int}Q$

The factor $c_ n$ cancels because the $n$ exponents $1/n$ sum to one. The measure $\lambda _{l,j}$ vanishes on the cubes with $\lambda _ l(Q)=0$ and does not charge the cube boundaries, which are Lebesgue null. Since the mixed product is positive, we conclude that $\bigwedge _ i\mathrm{dd}^{\mathrm{c}}A_{i,l,j}\geq \lambda _{l,j}$ on $B$.

As $j\to \infty $, $\lambda _{l,j}\to \lambda _ l$ weakly. Indeed, for $\chi \in C_ c(B)$, replacing the values of $\chi $ by its cube averages changes $\int \chi \, \mathrm d\lambda _ l$ by at most $\lambda _ l(B)$ times the modulus of continuity of $\chi $ at the cube diameter. Letting $j\to \infty $ and using (7.12.4), we obtain $\bigwedge _ i\mathrm{dd}^{\mathrm{c}}A_{i,l}\geq \lambda _ l$. Finally let $l\to \infty $. By Step 1, $A_{i,l}\downarrow u_ i$ with $u_ i\leq A_{i,l}\leq 0$, so theorem 7.3.7 gives $\bigwedge _ i\mathrm{dd}^{\mathrm{c}}A_{i,l}\to \bigwedge _ i\mathrm{dd}^{\mathrm{c}}u_ i$ weakly, while $\lambda _ l$ increases to $\lambda $. Testing with nonnegative functions in $C_ c(B)$ and using monotone convergence on the right, we obtain (7.12.2) for potentials in $\mathcal E_0(B)$.

Step 3. Localization. Now let $u_1,\ldots ,u_ n$ be locally bounded on $\Omega $. Choose a ball $B=B(a_0,R)$ with $\overline B\subset \Omega $ and a compact set $K\subset B$. Since adding constants does not change the products, we may assume that $u_ i\leq -1$ on $B$. With constants $C_ i>0$ so large that $C_ i\rho <\inf _ Bu_ i$ on a neighborhood of $K$, put

Then $\widetilde u_ i=u_ i$ on a neighborhood of $K$, and $\widetilde u_ i=C_ i\rho $ near $\partial B$, because $C_ i\rho \to 0$ at $\partial B$ while $u_ i\leq -1$. Thus $\widetilde u_ i\in \mathcal E_0(B)$: its Monge–Ampère measure is smooth near $\partial B$ and locally finite on the rest of $B$. By locality of the products on the open neighborhood of $K$ where $\widetilde u_ i=u_ i$, we have $(\mathrm{dd}^{\mathrm{c}}\widetilde u_ i)^ n\geq f_ i\mathbf1_ K\mu $ on $B$. The case already proved, applied to the measure $\mathbf1_ K\mu $, gives $\bigwedge _ i\mathrm{dd}^{\mathrm{c}}\widetilde u_ i\geq \mathbf1_ K\lambda $, and on a neighborhood of $K$ the left side equals $\bigwedge _ i\mathrm{dd}^{\mathrm{c}}u_ i$ by locality. Thus $P=\bigwedge _ i\mathrm{dd}^{\mathrm{c}}u_ i$ satisfies $P(E)\geq \lambda (E)$ for every Borel set $E\subset K$. The interiors of such compact sets $K$ cover $\Omega $, and since $\Omega $ has a countable basis, countably many of them, say $\operatorname{Int}L_1,\operatorname{Int}L_2,\ldots $, already cover $\Omega $. A Borel set $E\subset \Omega $ is the disjoint union of the Borel sets $E_ m=(E\cap L_ m)\setminus \bigcup _{m'<m}L_{m'}$, and therefore $P(E)=\sum _ mP(E_ m)\geq \sum _ m\lambda (E_ m)=\lambda (E)$. $\square$

Proof. Apply theorem 7.12.1 to $k$ copies of $u$ with density $f$ and $n-k$ copies of $v$ with density $g$. $\square$

Proof. By additivity and symmetry of the products, $(\mathrm{dd}^{\mathrm{c}}(u+v))^ n=\sum _{k=0}^ n\binom nk(\mathrm{dd}^{\mathrm{c}}u)^ k\wedge (\mathrm{dd}^{\mathrm{c}}v)^{n-k}$. Bound the terms with $1\leq k\leq n-1$ by corollary 7.12.6 and the terms with $k=0,n$ by the hypotheses; the binomial formula then gives the assertion. The density on the right is locally $\mu $-integrable, since $f^{k/n}g^{(n-k)/n}\leq f+g$ by the weighted arithmetic–geometric mean inequality, and $f$ and $g$ are locally $\mu $-integrable because $f\mu \leq (\mathrm{dd}^{\mathrm{c}}u)^ n$ and $g\mu \leq (\mathrm{dd}^{\mathrm{c}}v)^ n$. $\square$