17.5 Variational Monge–Ampère equations16614E7E6F9C

Proof. (1) This is lemma 16.12.1(2) with $W=X$; the second assertion follows by applying the first in both directions.

(2) Since $X$ is compact, weak convergence of measures on $X$ is convergence of the integrals of all continuous functions (see the discussion following definition 5.3.5). Hence $\int _ Xg\, \mathrm d\lambda =\lim _ j\int _ Xg\, \mathrm d\lambda _ j\geq \int _ Xg\, \mathrm d\sigma $ for every continuous function $g\geq 0$ on $X$, and (1) gives $\lambda \geq \sigma $. $\square$

Proof. Let $B=\sup _ X|f|$. The potential $u-|t|B$ is a competitor for the obstacle $u+tf$, and every competitor lies below $u+|t|B$, so $u-|t|B\leq u_ t\leq u+|t|B$. Hence $u_ t\sim u$, and $E(u_ t)\geq E(u)-|t|Bm>-\infty $, so $u_ t\in \mathcal E^1(X,\theta ,\phi )$ by theorem 17.3.14. The obstacle $u+tf$ is upper semicontinuous, so $u_ t\leq u+tf$ everywhere, by proposition 16.11.4(1),(2). In particular $u_0\leq u$; since $u$ is itself a competitor for the obstacle $u$, also $u\leq u_0$, so $u_0=u$.

Next we show that $\mu _{u_{t+s}}\to \mu _{u_ t}$ weakly as $s\to 0$. The obstacles $u+tf$ and $u+(t+s)f$ differ by at most $|s|B$, so proposition 16.11.4(4) gives $u_ t-|s|B\leq u_{t+s}\leq u_ t+|s|B$. In particular all $u_ t$ have the pole set of $u$, and $|u_{t+s}-u_ t|\leq |s|B$ outside it. Since $|\arctan a-\arctan b|\leq |a-b|$ for real $a,b$, and adding the same smooth local potential of $\theta $ to $u_{t+s}$ and $u_ t$ does not change their difference, the distance $d_*$ of definition 13.7.2 between the local potentials of $u_{t+s}$ and $u_ t$ is at most $|s|B$ everywhere. Hence, for every $\delta >0$, the sets where it exceeds $\delta $ are empty once $|s|B<\delta $, and $u_{t+s}\to u_ t$ in extended capacity as $s\to 0$. These potentials have full mass relative to $\phi $, so corollary 13.9.4, applied along every sequence $s_ k\to 0$, gives the asserted weak convergence; in particular $\int _ Xf\, \mu _{u_{t+s}}\to \int _ Xf\, \mu _{u_ t}$.

The obstacle $u+tf$ is bounded above, and it is quasi-continuous, because $u$ is (section 16.11) and $f$ is continuous. Hence theorem 16.13.4 shows that $\mu _{u_ t}$ is carried by the contact set $\{ u_ t=u+tf\} $. On the contact set of $u_{t+s}$ we have $u_{t+s}=u+(t+s)f$ and $u_ t\leq u+tf$, so $u_{t+s}-u_ t\geq sf$ there. On the contact set of $u_ t$ we have $u_ t=u+tf$ and $u_{t+s}\leq u+(t+s)f$, so $u_{t+s}-u_ t\leq sf$ there. The differences are bounded and the pole sets carry no mass, so the tangent bounds of lemma 17.4.10 give, for every real $s$,

Dividing by $s>0$, respectively $s<0$, and letting $s\to 0$, we obtain the right and left derivatives from the weak convergence just proved. Both equal $\int _ Xf\, \mu _{u_ t}$, which depends continuously on $t$ by the same weak convergence. $\square$

Proof. Let $v\in \mathcal D_\nu $. Then $v\leq \phi +C$ for a constant $C$, and $\phi \leq 0$. Thus the negative part of $\phi $ is at most that of $v$ plus $C$, and the positive part of $\phi $ vanishes; hence $\phi \in L^1(\nu )$. If $\mu _ u=\nu $, the tangent bound of lemma 17.4.10 gives

that is, $F_\nu (v)\leq F_\nu (u)$.

Conversely, let $u$ be a maximizer and $f\in C(X,\mathbb R)$. By proposition 17.5.4, the potentials $u_ t=P_\theta (u+tf)$ have bounded difference from $u$ and finite energy, so they belong to $\mathcal D_\nu $; moreover $u_0=u$ and $u_ t\leq u+tf$ everywhere, hence $\nu $-almost everywhere. Therefore

The left side vanishes at $t=0$ and is differentiable there by proposition 17.5.4; since it attains its maximum at $t=0$, its derivative vanishes there. Thus $\int _ Xf\, \mu _ u=\int _ Xf\, \mathrm d\nu $ for every $f\in C(X,\mathbb R)$, and lemma 17.5.3(1) gives $\mu _ u=\nu $. Since $\mu _ u(X)=m$, a maximizer can exist only if $\nu (X)=m$. $\square$

Proof. Put $\nu =g\, \mathrm dV$, a non-pluripolar measure of mass $m$. Every $\theta $-psh potential is integrable for $\mathrm dV$, hence for $\nu $, so $\mathcal D_\nu =\mathcal E^1(X,\theta ,\phi )$; we write $F=F_\nu $. By the shift formula for the energy and $\nu (X)=m$, the functional $F$ is invariant under adding constants, so we may restrict it to potentials $v$ with $\sup _ Xv=0$. For such $v$, lemma 16.22.2 gives $v\leq \phi $, hence $E(v)\leq 0$. Since $\mathrm dV\leq c_ V\, \omega ^ n$ for a constant $c_ V$, proposition 10.3.1 shows that $\int _ X|v|\, \mathrm dV$ is bounded independently of $v$, and $\phi \in L^1(\mathrm dV)$, so

with $C$ independent of $v$. Consequently $\sup F\in [0,C]$, since $F(\phi )=0$ and $F(v)\leq 0+C$.

Let $(v_ j)$ be a normalized maximizing sequence with $F(v_ j)\geq -1$. Then $E(v_ j)=F(v_ j)+\int _ X(v_ j-\phi )g\, \mathrm dV\geq -C-1$. By corollary 17.4.5 a subsequence converges in $L^1(X)$ to some $u\in \mathcal E^1(X,\theta ,\phi )$, and $\sup _ Xu=0$, because the supremum is continuous for $L^1$ convergence of potentials (proposition 10.3.1). The linear term $v\mapsto \int _ X(v-\phi )g\, \mathrm dV$ is continuous for $L^1$ convergence because $g$ is bounded, and the energy is upper semicontinuous by proposition 17.4.4. Hence $F(u)\geq \varlimsup _ jF(v_ j)=\sup F$, so $u$ is a maximizer, and theorem 17.5.5 gives $\mu _ u=\nu $. $\square$

Proof. Let $A=\{ u<v\} $. The comparison principle of proposition 16.25.4 gives

Both potentials are finite $\nu $-almost everywhere, and on $A$ we have $\mathrm e^{\lambda v}>\mathrm e^{\lambda u}$. The difference $\mathrm e^{\lambda v}-\mathrm e^{\lambda u}$ is therefore positive on $A$ and has nonpositive integral over $A$, so $\nu (A)=0$. Hence $\mu _ u(A)=0$, and the domination principle (theorem 16.22.10) gives $u\geq v$ everywhere. If $u$ and $v$ are two solutions, applying this in both directions gives $u=v$. $\square$

Proof. Let $f\in C(X,\mathbb R)$ and $u_ t=P_\theta (u+tf)$ for $t\in \mathbb R$. By proposition 17.5.4, $u_ t\in \mathcal E^1(X,\theta ,\phi )$, $u_0=u$ and $u_ t\leq u+tf$ everywhere. Hence $\mathrm e^{\lambda u_ t}\leq \mathrm e^{\lambda (u+tf)}$, and $F_\lambda (u_ t)\leq F_\lambda (u)$ gives

The left side vanishes at $t=0$. Its energy term is differentiable, with derivative $\int _ Xf\, \mu _{u_ t}$, by proposition 17.5.4. For $|t|\leq 1$ the $t$-derivative $\lambda f\mathrm e^{\lambda (u+tf)}$ of the integrand is bounded by $\lambda \sup _ X|f|\, \mathrm e^{\lambda \sup _ Xu+\lambda \sup _ X|f|}$, which is integrable for the finite measure $\nu $; by the mean value theorem and dominated convergence, the integral is differentiable as well, with derivative $\lambda \int _ Xf\mathrm e^{\lambda (u+tf)}\, \mathrm d\nu $. Since the left side is maximal at $t=0$, its derivative vanishes there: $\int _ Xf\, \mu _ u=\int _ Xf\mathrm e^{\lambda u}\, \mathrm d\nu $. Since $f$ is arbitrary and $\mathrm e^{\lambda u}\nu $ is a finite measure, lemma 17.5.3(1) gives $\mu _ u=\mathrm e^{\lambda u}\nu $. $\square$

Proof. By lemma 17.4.3, the tail envelopes $a_ j=(\sup _{k\geq j}q_ k)^*$ are $\theta $-psh and decrease pointwise to $q$. For each $j$, corollary 5.47.11, applied after adding smooth local potentials of $\theta $, gives $a_ j=\sup _{k\geq j}q_ k$ outside a pluripolar set. The countable union of these sets is pluripolar (theorem 5.47.8), and outside it $q=\lim _ ja_ j=\varlimsup _ jq_ j$. $\square$

Proof. By lemma 17.5.10, $\varlimsup _ jq_ j=q$ outside a pluripolar set, hence $\nu $-almost everywhere. Thus $(q_ j-q)_+\to 0$ $\nu $-almost everywhere, and $\int _ X(q_ j-q)_+\, \mathrm d\nu \to 0$ by bounded convergence, since $0\leq (q_ j-q)_+\leq 2B$ and $\nu $ is finite. Since $|q_ j-q|=2(q_ j-q)_+-(q_ j-q)$, it remains to show that $\int _ Xq_ j\, \mathrm d\nu \to \int _ Xq\, \mathrm d\nu $. We have just seen that $\varlimsup _ j\int _ X(q_ j-q)\, \mathrm d\nu \leq 0$, so it suffices to prove that $L=\varliminf _ j\int _ Xq_ j\, \mathrm d\nu $ is at least $\int _ Xq\, \mathrm d\nu $.

Choose a subsequence $(q_{j_ k})$ with $\int _ Xq_{j_ k}\, \mathrm d\nu \to L$. For bounded Borel functions $y$ on $X$ put $\| y\| =(\int _ Xy^2\, \mathrm d\nu )^{1/2}$. For $l\geq 1$ let $C_ l$ be the set of finite convex combinations of the functions $q_{j_ k}$ with $k\geq l$, and let $d_ l=\inf \{ \| y\| :y\in C_ l\} $. Since $C_{l+1}\subset C_ l$ and $\| y\| \leq B\nu (X)^{1/2}$ for $y\in C_ l$, the numbers $d_ l$ increase to a finite limit. Choose $y_ l\in C_ l$ with $\| y_ l\| ^2\leq d_ l^2+1/l$. For $l\leq l'$ the function $\frac12(y_ l+y_{l'})$ belongs to the convex set $C_ l$, so its norm is at least $d_ l$. The parallelogram identity, obtained by integrating the pointwise identity $(a-b)^2+(a+b)^2=2a^2+2b^2$, therefore gives

and the right side tends to zero as $l\to \infty $, uniformly in $l'\geq l$.

By the Cauchy–Schwarz inequality, $\int _ X|y_ l-y_{l'}|\, \mathrm d\nu \leq \nu (X)^{1/2}\| y_ l-y_{l'}\| $, so we can choose $l_1<l_2<\cdots $ with $\int _ X|y_{l_{i+1}}-y_{l_ i}|\, \mathrm d\nu \leq 2^{-i}$. By monotone convergence the series $\sum _ i|y_{l_{i+1}}-y_{l_ i}|$ has finite integral, hence converges $\nu $-almost everywhere, and therefore $y_{l_ i}$ converges $\nu $-almost everywhere; let $h=\varlimsup _ iy_{l_ i}$, a Borel function with $|h|\leq B$. By bounded convergence, $\int _ Xy_{l_ i}\, \mathrm d\nu \to \int _ Xh\, \mathrm d\nu $. On the other hand, $\int _ Xy_ l\, \mathrm d\nu $ is a convex combination of the numbers $\int _ Xq_{j_ k}\, \mathrm d\nu $ with $k\geq l$, which tend to $L$, so $\int _ Xy_ l\, \mathrm d\nu \to L$. Hence $L=\int _ Xh\, \mathrm d\nu $.

It remains to show that $h=q$ $\nu $-almost everywhere. Each $y_{l_ i}$ is a convex combination of $\theta $-psh functions, hence $\theta $-psh, and $\| y_{l_ i}-q\| _{L^1(X)}\leq \sup _{k\geq l_ i}\| q_{j_ k}-q\| _{L^1(X)}$, which tends to zero. By lemma 17.5.10, applied to the sequence $(y_{l_ i})_ i$, the function $h=\varlimsup _ iy_{l_ i}$ equals $q$ outside a pluripolar set, hence $\nu $-almost everywhere. Thus $L=\int _ Xq\, \mathrm d\nu $, as required. $\square$

Proof. By proposition 10.3.1 there is a constant $C$ with $q_ j,q\leq C$. We show that the functions $\mathrm e^{\lambda q_ j}$ and $\mathrm e^{\lambda q}$, with value zero on the pole sets, are $C_1\omega $-psh for a constant $C_1$ independent of $j$; the preceding lemma then applies to them.

Choose $C_0\geq 0$ with $\theta \leq C_0\omega $, and a finite cover of $X$ by coordinate balls $W'\Subset W$ such that the smaller balls $W'$ still cover $X$, together with smooth functions $h_ W$ near $\overline W$ with $\mathrm{dd}^{\mathrm{c}}h_ W=\theta $. The function $q_ j+h_ W$ is plurisubharmonic on $W$, so for every $\delta >0$ smaller than the distance from $W'$ to the boundary of $W$, its regularization $g_\delta =(q_ j+h_ W)*\rho _\delta $ by the radial mollifier is a smooth plurisubharmonic function on the ball $W'$, and $g_\delta $ decreases to $q_ j+h_ W$ as $\delta \downarrow 0$, by proposition 5.9.2(2). Put $q_{j,\delta }=g_\delta -h_ W$. Then $\mathrm{dd}^{\mathrm{c}}q_{j,\delta }\geq -\theta \geq -C_0\omega $. Moreover $g_\delta (x)$ is an average of values of $q_ j+h_ W\leq C+h_ W$ at points within distance $\delta $ of $x$, so by uniform continuity of $h_ W$ on a compact neighborhood of $\overline{W'}$ in $W$ we have $q_{j,\delta }\leq C+1$ on $W'$ for all small $\delta $, independently of $j$.

For a smooth real function $s$, the chain rule $\partial \bar\partial \, \mathrm e^{\lambda s}=\lambda \mathrm e^{\lambda s}\partial \bar\partial s+\lambda ^2\mathrm e^{\lambda s}\partial s\wedge \bar\partial s$ and the identity $\frac{\mathrm i}{2\pi }\partial s\wedge \bar\partial s=\mathrm ds\wedge \mathrm d^{\mathrm c}s$, which follows from (5.2.10), give

because $\mathrm dq_{j,\delta }\wedge \mathrm d^{\mathrm c}q_{j,\delta }$ is a semipositive $(1,1)$-form. Thus $\mathrm e^{\lambda q_{j,\delta }}$ is $C_1\omega $-psh on $W'$ with $C_1=\lambda C_0\mathrm e^{\lambda (C+1)}$, and so is its decreasing limit $\mathrm e^{\lambda q_ j}$ as $\delta \downarrow 0$, by theorem 5.10.1 applied after adding a smooth local potential of $C_1\omega $; the limit is not identically $-\infty $, since it is nonnegative. Since the balls $W'$ cover $X$, the function $\mathrm e^{\lambda q_ j}$ belongs to $\mathrm{PSH}(X,C_1\omega )$, and the same argument applies to $\mathrm e^{\lambda q}$.

The exponential is Lipschitz with constant $\lambda \mathrm e^{\lambda C}$ on $(-\infty ,C]$, so $\mathrm e^{\lambda q_ j}\to \mathrm e^{\lambda q}$ in $L^1(X)$. These functions take values in $[0,\mathrm e^{\lambda C}]$, and lemma 17.5.11, with the background $C_1\omega $, gives the assertion. $\square$

Proof. By corollary 17.4.5, the set $K_\phi $ of potentials $w\in \mathrm{PSH}(X,\theta )$ with $w\leq \phi $ and $\sup _ Xw=0$ is compact in $L^1(X)$; it contains $\phi $. By corollary 17.5.12, the function $w\mapsto \int _ X\mathrm e^{\lambda w}\, \mathrm d\nu $ is continuous on $K_\phi $. Its values are positive, since the pole set of $w$ is $\nu $-null and $\nu (X)>0$. Hence it has a positive minimum $a$ on $K_\phi $.

Any $v\in \mathcal E^1(X,\theta ,\phi )$ can be written $v=w+c$ with $c=\sup _ Xv$ and $w\in K_\phi $, by lemma 16.22.2. Since $E(w)\leq 0$, the shift formula for the energy gives

The right side tends to $-\infty $ as $c\to \pm \infty $. Since $F_\lambda (\phi )$ is finite, a maximizing sequence $v_ j=w_ j+c_ j$ with $F_\lambda (v_ j)\geq F_\lambda (\phi )-1$ therefore has bounded constants $c_ j$. Moreover $E(v_ j)\geq F_\lambda (v_ j)$, because the exponential term is nonnegative, so $E(w_ j)=E(v_ j)-mc_ j$ is bounded below. By corollary 17.4.5, after passing to a subsequence, $c_ j$ converges and $w_ j$ converges in $L^1(X)$; hence $v_ j\to u$ in $L^1(X)$ for some $u\in \mathcal E^1(X,\theta ,\phi )$. The energy is upper semicontinuous (proposition 17.4.4), and $\int _ X\mathrm e^{\lambda v_ j}\, \mathrm d\nu \to \int _ X\mathrm e^{\lambda u}\, \mathrm d\nu $ by corollary 17.5.12, so $F_\lambda (u)\geq \varlimsup _ jF_\lambda (v_ j)=\sup F_\lambda $, and $u$ is a maximizer. By lemma 17.5.9, $\mu _ u=\mathrm e^{\lambda u}\nu $. Uniqueness in $\mathcal E(X,\theta ,\phi )$ is proposition 17.5.7. $\square$

Proof. Put $g_ j=\inf _{k\geq j}f_ k$, so that $g_ j\uparrow f$ $\nu $-almost everywhere. We first prove that $\mu _{a_ j}\geq g_ j\nu $ for the tail envelopes $a_ j=(\sup _{k\geq j}q_ k)^*$, and then let $j\to \infty $. For $l\geq j$, the finite maximum $\max \{ q_ j,\ldots ,q_ l\} $ has measure at least $g_ j\nu $. Indeed, let $r_1,r_2\in \mathrm{PSH}(X,\theta )$ satisfy $\mu _{r_1},\mu _{r_2}\geq g_ j\nu $, and put $r=\max \{ r_1,r_2\} $. The set $\{ r_1\geq r_2\} $ is the contact set $\{ r_1=r\} $, so lemma 16.13.1, applied to $r_1\leq r$, gives $\mu _ r\geq \mu _{r_1}\geq g_ j\nu $ on $\{ r_1\geq r_2\} $; in the same way $\mu _ r\geq g_ j\nu $ on $\{ r_2>r_1\} $. Hence $\mu _ r\geq g_ j\nu $. Since $\mu _{q_ k}\geq f_ k\nu \geq g_ j\nu $ for $k\geq j$, induction on $l$ proves the claim.

The suprema $\sup _ Xq_ k$ are bounded, by proposition 10.3.1, so the regularized upper envelope $a_ j$ is $\theta $-psh, and it equals $\sup _{k\geq j}q_ k$ almost everywhere (theorem 5.10.2, applied in charts after adding smooth local potentials of $\theta $). The finite maxima $\max \{ q_ j,\ldots ,q_ l\} $ therefore increase almost everywhere to $a_ j$ as $l\to \infty $, and their measures converge weakly to $\mu _{a_ j}$ by corollary 13.8.9. The fixed measure $g_ j\nu \leq \mu _{q_ j}$ is finite, so the inequality passes to the weak limit, by lemma 17.5.3(2):

By lemma 17.4.3, $a_ j$ decreases pointwise to $q$.

To let $j\to \infty $ we truncate. Fix $C>0$ and an index $j_0$. The truncations $a_ j^ C=\max \{ a_ j,V_\theta -C\} $ have uniformly bounded difference from $V_\theta $: they are at least $V_\theta -C$, and since the suprema of the $a_ j$ are bounded, minimality of $V_\theta $ bounds them above by $V_\theta +C'$. They decrease to $q^ C=\max \{ q,V_\theta -C\} $, so they converge to $q^ C$ in extended capacity, by lemma 13.7.3 applied in charts after adding smooth local potentials of $\theta $. Since $q^ C$ has bounded difference from $V_\theta $, it has the mass of $V_\theta $ (corollary 13.8.8), that is, $q^ C\in \mathcal E(X,\theta ,V_\theta )$. Corollary 13.9.4, with $V_\theta $ in place of $\phi $, therefore gives $\mu _{a_ j^ C}\to \mu _{q^ C}$ weakly.

On the set $A_ C=\{ q>V_\theta -C\} $ we have $a_ j\geq q>V_\theta -C$, so $a_ j^ C=a_ j$ on the plurifine open set $\{ a_ j>V_\theta -C\} \supset A_ C$, and locality gives $\mu _{a_ j^ C}\geq \mathbf1_{A_ C}g_{j_0}\nu $ for $j\geq j_0$. Since $\mathbf1_{A_ C}g_{j_0}\nu $ is a fixed finite measure, the inequality passes to the weak limit, by lemma 17.5.3(2), and gives $\mu _{q^ C}\geq \mathbf1_{A_ C}g_{j_0}\nu $. On the plurifine open set $A_ C$ we have $q^ C=q$, so locality gives $\mu _ q\geq \mathbf1_{A_ C}g_{j_0}\nu $. Thus $\mu _ q(E)\geq \int _{E\cap A_ C}g_{j_0}\, \mathrm d\nu $ for every Borel set $E$. Let $j_0\to \infty $ and then $C\to \infty $. The sets $A_ C$ increase to the set where $q$ and $V_\theta $ are finite, whose complement is pluripolar and hence $\nu $-null, so monotone convergence gives $\mu _ q(E)\geq \int _ Ef\, \mathrm d\nu $, that is, $\mu _ q\geq f\nu $. $\square$

Proof. Let $\lambda _ j\downarrow 0$, let $u_ j$ be the solution of $\mu _{u_ j}=\mathrm e^{\lambda _ ju_ j}\nu $ given by theorem 17.5.13, and write $u_ j=w_ j+c_ j$ with $c_ j=\sup _ Xu_ j$. Since $u_ j\preceq \phi $ and $\sup _ Xw_ j=0$, lemma 16.22.2 gives $w_ j\leq \phi $, so $w_ j$ belongs to the set $K_\phi $ of corollary 17.4.5, which is compact in $L^1(X)$. On $K_\phi $ the functions

are continuous, by corollary 17.5.12, and they increase pointwise to the constant $m$, by monotone convergence, since $w\leq 0$ and the pole set of $w$ is $\nu $-null. The convergence is uniform on $K_\phi $ (Dini's theorem): for $\varepsilon >0$ the compact sets $\{ w\in K_\phi :m-H_ j(w)\geq \varepsilon \} $ decrease and have empty intersection, so they are empty for large $j$.

Comparing total masses in the equation gives $m=\mathrm e^{\lambda _ jc_ j}H_ j(w_ j)$, so

Also $\int _ X|1-\mathrm e^{\lambda _ jw_ j}|\, \mathrm d\nu =m-H_ j(w_ j)\to 0$. Hence $r_ j=\mathrm e^{\lambda _ ju_ j}=\mathrm e^{\lambda _ jc_ j}\mathrm e^{\lambda _ jw_ j}\to 1$ in $L^1(\nu )$, and $\mu _{w_ j}=\mu _{u_ j}=r_ j\nu $. Pass to a subsequence along which $r_ j\to 1$ $\nu $-almost everywhere and, by compactness of $K_\phi $, $w_ j\to u$ in $L^1(X)$ for some $u\in K_\phi $. Lemma 17.5.14 gives $\mu _ u\geq \nu $. Since $u\leq \phi $, mass monotonicity gives $\mu _ u(X)\leq m=\nu (X)$. Hence $\mu _ u=\nu $, $u$ has full mass relative to $\phi $, and $\sup _ Xu=0$ because $u\in K_\phi $. $\square$

Proof. Put $\mu =\mu _ u=\mu _ v$. Since $u\in \mathcal E(X,\theta ,\phi )$, this is a finite measure of mass $\int _ X\langle \theta _ u^ n\rangle =\int _ X\langle \theta _\phi ^ n\rangle =m$, and it is non-pluripolar, because non-pluripolar products give no mass to pluripolar sets (lemma 13.2.2). Hence the pole sets of $u$ and $v$ are $\mu $-null, and $u-v$ is defined and finite $\mu $-almost everywhere. We first show that $u-v$ is constant $\mu $-almost everywhere.

Suppose not. Let $\lambda $ be the image of $\mu $ under $u-v$, a finite Borel measure on $\mathbb R$ of total mass $m>0$, and put

with $\sup \varnothing =-\infty $ and $\inf \varnothing =+\infty $. The sets $(-\infty ,c)$ increase to $\mathbb R$ as $c\to +\infty $, so $\lambda ((-\infty ,c))\to m>0$; hence $c_-<+\infty $, and similarly $c_+>-\infty $. If $c_->-\infty $, the $\lambda $-null sets $(-\infty ,c)$ with $c<c_-$ increase to $(-\infty ,c_-)$, so $\lambda ((-\infty ,c_-))=0$; similarly $\lambda ((c_+,\infty ))=0$ if $c_+<+\infty $. Hence $c_+<c_-$ is impossible, since $\mathbb R=(-\infty ,c_-)\cup (c_+,\infty )$ would then be $\lambda $-null. If $c_-=c_+$, then $\lambda $ is carried by the point $c_-$, that is, $u-v$ is constant $\mu $-almost everywhere, contrary to our assumption. Thus $c_-<c_+$, and every $c\in (c_-,c_+)$ satisfies $\lambda ((-\infty ,c))>0$ and $\lambda ((c,\infty ))>0$, that is,

The singletons $\{ c\} $ with $\lambda (\{ c\} )>0$ are pairwise disjoint sets of positive measure, so there are only countably many of them by lemma 2.2.5. Since the interval $(c_-,c_+)$ is uncountable, we may choose $c$ in it with $\mu (\{ u-v=c\} )=0$. Replacing $v$ by $v+c$ changes neither the class $\mathcal E(X,\theta ,\phi )$ nor the measure $\mu _ v$, so we may assume $c=0$. Then $\mu (\{ u<v\} )$ and $\mu (\{ u>v\} )$ are positive, and the set $U=\{ u<v\} $ satisfies

Set $q=(m/\mu (U))^{1/n}>1$. The measure $q^ n\mathbf1_ U\mu $ is non-pluripolar and has mass $m$, so theorem 17.5.15 gives $h\in \mathcal E(X,\theta ,\phi )$ with

For $0<t<1$ put $w_ t=(1-t)v+th$. By corollary 17.3.17, $w_ t\in \mathcal E(X,\theta ,\phi )$. Since $\theta _{w_ t}=(1-t)\theta _ v+t\theta _ h$, corollary 13.5.11, applied to the currents $\theta _ v,\theta _ h$, the common measure $\mu $ and the densities $1$ and $q^ n\mathbf1_ U$, gives

Let $U_ t=\{ u<w_ t\} $, and put

The comparison principle for relative full mass, proposition 16.25.4, applied to $u$ and $w_ t$, and the preceding lower bound give

Since $1-d_ t>0$, it follows that

We let $t\downarrow 0$. At every point where $u,v,h$ are finite and $u\neq v$, we have $w_ t=v+t(h-v)$, so $\mathbf1_{U_ t}\to \mathbf1_ U$ there. These points have full $\mu $-measure, because the three pole sets and the level set $\{ u=v\} $ are $\mu $-null. By dominated convergence,

Moreover $b_ t-d_ t=ntq+O(t^2)$ and $1-d_ t=nt+O(t^2)$, so $(b_ t-d_ t)/(1-d_ t)\to q$. Passing to the limit in (17.5.18) gives $q\mu (U)\leq \mu (U)$, which contradicts $q>1$ and $\mu (U)>0$.

Hence, for the original pair, there is a finite constant $C$ such that $u=v+C$ $\mu $-almost everywhere. Both $u$ and $v+C$ belong to $\mathcal E(X,\theta ,\phi )$, and their common measure $\mu $ gives zero mass to $\{ u<v+C\} $ and to $\{ v+C<u\} $. The domination principle (theorem 16.22.10), applied in both directions, gives $u=v+C$ everywhere, including at the common poles. This proves the first assertion.

For the second, theorem 17.5.15 provides a solution with supremum zero. If $u$ and $v$ are two such solutions, then $u=v+C$, and comparing suprema gives $C=0$. $\square$

Proof. We derive a recursive inequality for the capacities of the sublevel sets of $u-\phi $ and iterate it along levels that converge to a finite depth. For $s>0$ put

The sets $E_ s$ decrease as $s$ increases, so $g$ is nonincreasing.

We first compare the measures on $E_ s$. Both $u$ and $\phi -s$ belong to $\mathcal E(X,\theta ,\phi )$, so the comparison principle of proposition 16.25.4, applied to $u$ and $\phi -s$, gives $\mu _\phi (E_ s)\leq \mu _ u(E_ s)$. By (17.5.21),

and hence $(1-a)\mu _ u(E_ s)\leq \nu (E_ s)$. Since $u$ has full mass relative to $\phi $, proposition 16.25.7 gives $t^ n\operatorname {Cap}_{\theta ,\phi }(E_{s+t})\leq \mu _ u(E_ s)$ for $0<t\leq 1$. Combining this with the last inequality and the capacity bound in (17.5.21), we obtain $t^ ng(s+t)^ n\leq B^ ng(s)^{n(1+\delta )}$, that is,

By the second assertion of proposition 16.25.7, $g(s)\to 0$ as $s\to \infty $. Choose $s_0>0$ with $2Bg(s_0)^\delta \leq 1$, and put

We claim that $g(C)=0$. If $g(s_0)=0$, then $C=s_0$ and there is nothing to prove. Otherwise, as long as $g(s_ j)>0$, define inductively

Since $g$ is nonincreasing, $t_ j\leq 2Bg(s_0)^\delta \leq 1$, and (17.5.23) with $s=s_ j$ and $t=t_ j$ gives

Hence $g(s_ j)\leq 2^{-j}g(s_0)$ and $t_ j\leq 2^{-j\delta }\cdot 2Bg(s_0)^\delta $, so that

If the construction stops because $g(s_ j)=0$ for some $j$, then $g(C)\leq g(s_ j)=0$. Otherwise $g(C)\leq g(s_ j)\leq 2^{-j}g(s_0)$ for every $j$, and again $g(C)=0$. This proves the claim.

Thus $\operatorname {Cap}_{\theta ,\phi }(E_ C)=0$. The capacity bound in (17.5.21) gives $\nu (E_ C)=0$. Since $\phi $ itself is a competitor for $\operatorname {Cap}_{\theta ,\phi }$ (definition 16.25.5), we also have $\mu _\phi (E_ C)\leq \operatorname {Cap}_{\theta ,\phi }(E_ C)=0$. The measure inequality in (17.5.21) therefore gives $\mu _ u(\{ u<\phi -C\} )=0$. As $\phi -C\in \mathcal E(X,\theta ,\phi )$, the domination principle, theorem 16.22.10, yields $u\geq \phi -C$ everywhere.

Now suppose that $I<\infty $. On $E_ s$ the weight $\phi -u$ exceeds $s$, so Markov's inequality gives $\nu (E_ s)\leq I/s$. Taking $t=1$ in the sublevel estimate and using the measure inequality on $E_ s$, we find

Choose $s=1+I(2B)^{n/\delta }/(1-a)$ and $s_0=s+1$. Since $(1-a)s>I(2B)^{n/\delta }$, we get $g(s_0)^ n\leq (2B)^{-n/\delta }$, that is, $2Bg(s_0)^\delta \leq 1$. Thus $s_0$ satisfies the condition imposed on it above, and the constant in (17.5.24) is at most $s_0+(1-2^{-\delta })^{-1}$. Since $(2B)^{n/\delta }=\bigl (2(A/(1-a))^{1/n}\bigr )^{n/\delta }$, this is the right side of (17.5.22). The lower bound $\phi -C\leq u$ remains true when $C$ is increased, which proves (17.5.22). $\square$

Proof. Apply theorem 17.5.20 with $a=0$ and $\nu =\mu _ u$. By the first inequality in (17.3.15), $I=\int _ X(\phi -u)\, \mu _ u\leq (n+1)\bigl (-E_\theta ^\phi (u)\bigr )\leq (n+1)L$, and (17.5.22) with $a=0$ gives the stated bound. $\square$

Proof. Since $\sup _ Xu=0$, lemma 16.22.2 gives $u\leq \phi $; recall also that $\sup _ X\phi =0$, by definition 16.21.1. Let $q=p/(p-1)$ be the conjugate exponent. By corollary 10.3.2, the $L^ q(X,\omega ^ n)$ norms of $u$ and $\phi $ are at most a constant $M_ q$ depending only on $X,\theta ,\omega ,q$. Hölder's inequality gives

By corollary 16.27.10 with $r=2$, the measure $\nu =f\omega ^ n$ satisfies the capacity bound in (17.5.21) with $\delta =1$ and $A=B_2\| f\| _{L^ p}/m^2$, where $B_2$ depends only on $X,\theta ,\omega ,p$. (If $f=0$ almost everywhere, take $A=1$ instead.) With $\delta =1$, the constant (17.5.22) equals $4+2^ nAI/(1-a)^2$. Inserting the bounds for $A$ and $I$, and recalling that the lower bound $\phi -C\leq u$ survives an increase of $C$, we see that theorem 17.5.20 gives the assertion with

which has the stated dependence.

Finally, the measure $f\omega ^ n$ is non-pluripolar, as noted after definition 17.5.2. If its mass is $m$, theorem 17.5.15 gives a solution $u\in \mathcal E(X,\theta ,\phi )$ of $\mu _ u=f\omega ^ n$ with $\sup _ Xu=0$, and the bound just proved, with $a=0$, shows that $u$ has bounded difference from $\phi $. $\square$

Proof. Necessity. If $u\in \mathrm{PSH}(X,\theta )$, then $T=\theta +\mathrm{dd}^{\mathrm{c}}u\geq 0$. For $\tau $ as in the statement, the definition of $\mathrm{dd}^{\mathrm{c}}u$ as a current gives $\int _ X\mathrm{dd}^{\mathrm{c}}u\wedge \tau =\int _ Xu\, \mathrm{dd}^{\mathrm{c}}\tau =0$, hence $\int _ X\theta \wedge \tau =\int _ XT\wedge \tau $, which is nonnegative by the definition of a positive current, since $\tau $ is positive, hence strongly positive.

Sufficiency. Conversely, assume (17.5.29) for all forms $\tau $ as in the statement. We extend the functional $\tau \mapsto \int _ X\theta \wedge \tau $ from $\mathrm{dd}^{\mathrm{c}}$-closed forms to a positive linear functional on all smooth real $(n-1,n-1)$-forms, regard the extension as a positive $(1,1)$-current $T$, and show that $T-\theta $ is $\mathrm{dd}^{\mathrm{c}}$ of a distribution.

A positive extension. Let $F$ be the real vector space of smooth real $(n-1,n-1)$-forms on $X$, let $S=\{ \tau \in F:\mathrm{dd}^{\mathrm{c}}\tau =0\} $, and put $e=\omega ^{n-1}$, with $e=1$ if $n=1$. Since $\omega $ is closed, $e\in S$, and $e$ is strictly positive. Define $\ell _0(\tau )=\int _ X\theta \wedge \tau $ for $\tau \in S$. By hypothesis $\ell _0(\tau )\geq 0$ when $\tau \in S$ is strictly positive. If $\tau \in S$ is merely positive, then $\tau +te\in S$ is strictly positive for every $t>0$. Indeed, at each point $x$ the cone of positive forms contains a ball about $e(x)$, for some norm; since the cone is stable under sums and positive multiples, it also contains the ball of $t$ times that radius about $\tau (x)+te(x)$. Hence $\ell _0(\tau )+t\ell _0(e)\geq 0$, and letting $t\downarrow 0$ gives $\ell _0(\tau )\geq 0$.

We next compare the forms in $F$ with $e$. Fix finitely many holomorphic charts $U_1,\ldots ,U_ N$ and compact sets $K_ a\subset U_ a$ with $\bigcup _ aK_ a=X$. In $U_ a$ write $\eta \in F$ as $\eta =\sum _{I,J}\eta ^ a_{IJ}\, \mathrm dz_ I\wedge \mathrm d\bar z_ J$, the sum running over increasing multi-indices of length $n-1$, and put

We claim that there is a constant $M>0$ such that

Let $Z$ be the finite-dimensional real vector space of real $(n-1,n-1)$-forms with constant coefficients on $\mathbb C^ n$, normed by the maximum modulus of the coefficients, and let $P\subset Z$ be the closed convex cone of positive forms (proposition 5.3.2). For $x\in U_ a$, the differential of the chart is a complex-linear isomorphism from the tangent space at $x$ onto $\mathbb C^ n$, under which $\mathrm dz_ j$ corresponds to the $j$th coordinate form. Since complex-linear isomorphisms preserve positivity, a real $(n-1,n-1)$-form $\eta $ is positive at $x$ if and only if its coefficient form $\eta ^ a(x)=\sum _{I,J}\eta ^ a_{IJ}(x)\, \mathrm dz_ I\wedge \mathrm d\bar z_ J\in Z$ lies in $P$.

Because $e$ is strictly positive, the continuous map $x\mapsto e^ a(x)$ sends the compact set $K_ a$ into the interior of $P$; the complement of this interior is closed and disjoint from the compact image, so it has positive distance $r_ a$ from it. Put $r=\frac12\min _ ar_ a$. If $\eta \in F$ satisfies $\| \eta \| \leq r$ and $x\in K_ a$, then $e^ a(x)\pm \eta ^ a(x)$ lies within distance $r<r_ a$ of $e^ a(x)$, hence in $P$; thus $e\pm \eta $ is positive at $x$. Since the sets $K_ a$ cover $X$, the forms $e\pm \eta $ are positive. Applying this to $r\eta /\| \eta \| $ for $\eta \neq 0$ proves (17.5.31) with $M=1/r$.

For $\eta \in F$, a form $\tau \in S$ with $\tau \geq \eta $ is called an $S$-majorant of $\eta $, and we put

Let $\eta \in F$ and $C=M\| \eta \| $, so that $-Ce\leq \eta \leq Ce$ by (17.5.31). The form $Ce$ is an $S$-majorant of $\eta $, so $p(\eta )\leq C\ell _0(e)$. If $\tau $ is an $S$-majorant of $\eta $, then $\tau +Ce\geq \eta +Ce\geq 0$, so $\tau +Ce\in S$ is positive and $\ell _0(\tau )\geq -C\ell _0(e)$. Thus $p$ is finite. It is sublinear: the sum of $S$-majorants of $\eta _1$ and $\eta _2$ is an $S$-majorant of $\eta _1+\eta _2$, and for $\lambda >0$ the $S$-majorants of $\lambda \eta $ are $\lambda $ times those of $\eta $. If $\eta \in S$, then $\eta $ is its own $S$-majorant, and every $S$-majorant $\tau $ of $\eta $ satisfies $\ell _0(\tau )-\ell _0(\eta )=\ell _0(\tau -\eta )\geq 0$ because $\tau -\eta \in S$ is positive. Hence $p=\ell _0$ on $S$.

Next we extend $\ell _0$ to a linear functional on $F$ dominated by $p$; this is the Hahn–Banach theorem, whose proof we recall. Let $\ell $ be a linear functional on a subspace $W\supset S$ which extends $\ell _0$ and satisfies $\ell \leq p$ on $W$, and let $\eta _0\in F\setminus W$. For $\eta _1,\eta _2\in W$,

so

Taking $\eta _1=\eta _2=0$ shows that the supremum is greater than $-\infty $ and the infimum is less than $+\infty $; hence both are finite, and we may choose a real number $c$ between them. Define $\ell (\eta +t\eta _0)=\ell (\eta )+tc$ for $\eta \in W$ and $t\in \mathbb R$. For $t>0$, the upper bound for $c$ with $\eta _2=\eta /t$, multiplied by $t$, gives $\ell (\eta )+tc\leq p(\eta +t\eta _0)$; for $t<0$, the lower bound for $c$ with $\eta _1=\eta /|t|$, multiplied by $|t|$, gives the same inequality. Thus the extension to $W+\mathbb R\eta _0$ is still dominated by $p$. The union of a chain of such extensions is again such an extension, so Zorn's lemma gives a maximal one, and by the step just proved its domain is $F$. We denote it again by $\ell $.

The functional $\ell $ is positive: if $\eta \in F$ is positive, then $0$ is an $S$-majorant of $-\eta $, so $\ell (-\eta )\leq p(-\eta )\leq 0$. It is also bounded: by (17.5.31), the form $M\| \eta \| e$ is an $S$-majorant of both $\eta $ and $-\eta $, so

We now turn $\ell $ into a current. Every smooth form on the compact manifold $X$ is a test form. For a complex form $\varphi $ of degree $2n-2$, let $\varphi '$ be its component of bidegree $(n-1,n-1)$, and put $T(\varphi )=\ell (\operatorname{Re}\varphi ')+\mathrm i\, \ell (\operatorname{Im}\varphi ')$, where $\operatorname{Re}\varphi '=(\varphi '+\overline{\varphi '})/2$ and $\operatorname{Im}\varphi '=(\varphi '-\overline{\varphi '})/(2\mathrm i)$ are real forms in $F$. This is a complex-linear functional: replacing $\varphi $ by $\mathrm i\varphi $ replaces $(\operatorname{Re}\varphi ',\operatorname{Im}\varphi ')$ by $(-\operatorname{Im}\varphi ',\operatorname{Re}\varphi ')$. Moreover, $T$ has the continuity property of a current of degree $2$. Indeed, let $Q$ be a compact subset of a chart $U$, and let $\varphi $ be supported in $Q$, with coefficients $\varphi _ J$ in real coordinates on $U$. On the compact set $K_ a\cap Q$, the coefficients of $\operatorname{Re}\varphi '$ and $\operatorname{Im}\varphi '$ in the chart $U_ a$ are linear combinations of the functions $\varphi _ J$ and $\overline{\varphi _ J}$ with coefficients that are smooth on $U\cap U_ a$, and they vanish on $K_ a\setminus Q$. Hence $\| \operatorname{Re}\varphi '\| +\| \operatorname{Im}\varphi '\| \leq C_ Q\sum _ J\sup _ Q|\varphi _ J|$ with a constant $C_ Q$ independent of $\varphi $, and (17.5.32) gives $|T(\varphi )|\leq MC_ Q\ell _0(e)\sum _ J\sup _ Q|\varphi _ J|$.

By construction $T$ vanishes on forms of bidegree other than $(n-1,n-1)$, so it is a current of bidegree $(1,1)$. It is real, because the $(n-1,n-1)$-component of $\overline\varphi $ is $\overline{\varphi '}=\operatorname{Re}\varphi '-\mathrm i\operatorname{Im}\varphi '$, so that $T(\overline\varphi )=\overline{T(\varphi )}$. It is positive, because a strongly positive test form $\varphi $ of bidegree $(n-1,n-1)$ is real and positive, so that $T(\varphi )=\ell (\varphi )\geq 0$. Since $\ell $ extends $\ell _0$, the real $(1,1)$-current $R=T-\theta $ annihilates $S$.

The current $R$ is $\mathrm{dd}^{\mathrm{c}}$-exact. We first show that $R$ is closed. If $\xi $ is a smooth form of bidegree $(n-2,n-1)$, then $\mathrm{dd}^{\mathrm{c}}\partial \xi =\frac{\mathrm i}{2\pi }\partial \bar\partial \partial \xi =-\frac{\mathrm i}{2\pi }\partial \partial \bar\partial \xi =0$, and $\mathrm{dd}^{\mathrm{c}}\bar\partial \xi '=\frac{\mathrm i}{2\pi }\partial \bar\partial \bar\partial \xi '=0$ for $\xi '$ of bidegree $(n-1,n-2)$. Since $\mathrm{dd}^{\mathrm{c}}$ is a real operator, the conjugate of a $\mathrm{dd}^{\mathrm{c}}$-closed complex $(n-1,n-1)$-form $\sigma $ is $\mathrm{dd}^{\mathrm{c}}$-closed, so the real forms $\operatorname{Re}\sigma $ and $\operatorname{Im}\sigma $ lie in $S$; as $R$ is complex-linear and annihilates $S$, it annihilates $\sigma $. In particular $R(\partial \xi )=R(\bar\partial \xi ')=0$. For a smooth $(2n-3)$-form $\gamma $, the $(n-1,n-1)$-component of $\mathrm d\gamma $ is $\partial \gamma ^{n-2,n-1}+\bar\partial \gamma ^{n-1,n-2}$; since $R$ has bidegree $(1,1)$, it follows that $R(\mathrm d\gamma )=0$. Therefore $\mathrm dR=0$.

Since $e\in S$, we also have $\int _ XR\wedge \omega ^{n-1}=0$. Let $\Lambda $ denote contraction with $\omega $, as in lemma 10.2.12; the identity $R\wedge \omega ^{n-1}=(\Lambda R)\omega ^ n/n$ shows that the trace $\Lambda R$ has mean zero. By lemma 10.2.5(2), applied to the Laplacian $L=\mathrm d^*\mathrm d$ of section 10.2, there is a real distribution $f$ with $Lf=-4\pi \Lambda R$. By lemma 10.2.12, $\Lambda \mathrm{dd}^{\mathrm{c}}f=\Lambda R$, so $R'=R-\mathrm{dd}^{\mathrm{c}}f$ is a closed real $(1,1)$-current with zero trace. By the last assertion of lemma 10.2.12, $R'$ is harmonic, and by lemma 10.2.5(1) it is smooth. If $n=1$, a $(1,1)$-form with zero trace vanishes, so $R'=0$. If $n\geq 2$, (10.2.13) gives $*R'=-R'\wedge \omega ^{n-2}/(n-2)!$, a smooth closed $(n-1,n-1)$-form; hence $*R'\in S$. Since $R$ annihilates $S$ and $\mathrm{dd}^{\mathrm{c}}{*R'}=0$,

Thus $R'=0$ and $R=\mathrm{dd}^{\mathrm{c}}f$ in every dimension.

Conclusion. The current $T=\theta +\mathrm{dd}^{\mathrm{c}}f$ is closed and positive, and it lies in the class $\{ \theta \} $, since it differs from $\theta $ by $\mathrm{dd}^{\mathrm{c}}$ of a distribution. By corollary 10.2.19, there is $u\in \mathrm{PSH}(X,\theta )$ with $\theta +\mathrm{dd}^{\mathrm{c}}u=T$. This proves sufficiency.

For the last assertion, $\{ \theta \} $ is big exactly when $\mathrm{PSH}(X,\theta -\delta \omega )\neq \varnothing $ for some $\delta >0$. Indeed, if $T\geq \delta \omega $ is a Kähler current in $\{ \theta \} $ (definition 10.4.1), then $T-\delta \omega $ is a closed positive current in the class $\{ \theta -\delta \omega \} $, so $T-\delta \omega =\theta -\delta \omega +\mathrm{dd}^{\mathrm{c}}u$ with $u\in \mathrm{PSH}(X,\theta -\delta \omega )$, by corollary 10.2.19. Conversely, if $u\in \mathrm{PSH}(X,\theta -\delta \omega )$, then $\theta +\mathrm{dd}^{\mathrm{c}}u\geq \delta \omega $ is a Kähler current in $\{ \theta \} $. Applying the first part to $\theta -\delta \omega $ shows that this holds if and only if (17.5.30) holds for that $\delta $. $\square$

Proof. We prove a pointwise inequality for smooth forms, apply it to regularizations of the potentials of $S$ on small balls in $X\setminus A$, and globalize with a partition of unity.

Step 1: a pointwise inequality. Let $\sigma $ be a smooth positive definite real $(1,1)$-form. At a given point, choose linear coordinates on the tangent space that diagonalize the Hermitian matrices of $\omega $ and $\sigma $ simultaneously, so that $\omega =\sum _ j\gamma _ j$ and $\sigma =\sum _ j\lambda _ j\gamma _ j$ with $\lambda _ j>0$, where $\gamma _ j=\mathrm i\, \mathrm dz_ j\wedge \mathrm d\bar z_ j$, and put $\mathrm dV=\gamma _1\wedge \cdots \wedge \gamma _ n$. Since $\tau $ is strictly positive (section 5.3), the form $\tau -c\, \omega ^{n-1}$ is positive at this point for some $c>0$. The product of the strongly positive form $\gamma _ j$ with this positive form is a nonnegative multiple of $\mathrm dV$ (proposition 5.3.2), and $\gamma _ j\wedge \omega ^{n-1}=(n-1)!\, \mathrm dV$, so $\gamma _ j\wedge \tau \geq c(n-1)!\, \mathrm dV$; thus $\gamma _ j\wedge \tau =t_ j\, \mathrm dV$ with $t_ j>0$. Expanding the products, we obtain

Since $\lambda _ j\sum _ k\lambda _ k^{-1}\geq 1$ for every $j$, the product of these two ratios is at least $1/n$. If $\sigma $ is only semipositive, we apply this to $\sigma +s\omega $ and let $s\downarrow 0$. Thus, for every smooth semipositive real $(1,1)$-form $\sigma $, as densities with respect to any fixed smooth volume form,

Step 2: regularization on a ball. Since the local potentials of $S$ are locally bounded on the open set $X\setminus A$, every point of $X\setminus A$ is the center of a coordinate ball $B\Subset X\setminus A$ on which $S=\mathrm{dd}^{\mathrm{c}}v$ with $v$ plurisubharmonic and bounded. Fix such a ball, put $\eta =\mathrm{dd}^{\mathrm{c}}|z|^2$ as in section 7.8, and write $\nu =g\eta ^ n$ with $g$ smooth and positive on $B$. Since $v$ is bounded on $B$, the non-pluripolar powers of $S$ on $B$ are the Bedford–Taylor powers of $v$ (see the remark following definition 13.2.4); in particular $(\mathrm{dd}^{\mathrm{c}}v)^ n=mg\eta ^ n$.

Let $v_\varepsilon =v*\rho _\varepsilon $ be the regularizations of $v$ by the radial mollifiers of section 2.5. On each ball $B'$ concentric with $B$ and of smaller radius, the functions $v_\varepsilon $ are defined for $\varepsilon $ smaller than the difference of the radii, and by proposition 5.9.2(2) they are smooth and plurisubharmonic on $B'$ and decrease to $v$ as $\varepsilon \downarrow 0$. As recalled at the beginning of section 7.8, $(\mathrm{dd}^{\mathrm{c}}v_\varepsilon )^ n=\det \mathcal H(v_\varepsilon )\, \eta ^ n$. The last assertion of proposition 7.8.3, applied on $B$ to $v$ and the continuous density $mg$, therefore gives

The right side converges to $mg$ uniformly on compact subsets of $B$, by lemma 2.5.1(1) applied to the continuous function $(mg)^{1/n}$. Hence $\sqrt{F_\varepsilon g}$ is bounded below by functions converging uniformly to $\sqrt mg$ on compact sets, and for every nonnegative $\chi \in C^\infty _ c(B)$,

Moreover, for every $\chi \in C^\infty _ c(B)$, as $\varepsilon \downarrow 0$ along any sequence,

For the first convergence, note that $\int \chi \, \mathrm{dd}^{\mathrm{c}}v_\varepsilon \wedge \tau =\int v_\varepsilon \, \mathrm{dd}^{\mathrm{c}}(\chi \tau )$, by the definition of $\mathrm{dd}^{\mathrm{c}}$ on currents (section 5.3); since $v_\varepsilon \to v$ in $L^1$ on $\operatorname{Supp}\chi $ (lemma 2.5.1(2)), this tends to $\int v\, \mathrm{dd}^{\mathrm{c}}(\chi \tau )=\int \chi \, S\wedge \tau $.

For the second convergence, choose a ball $B'$ as above containing $\operatorname{Supp}\chi $, and a sequence $\varepsilon _ j\downarrow 0$. By theorem 7.3.7, applied on $B'$ to the closed positive current $\omega $ of bidimension $(n-1,n-1)$ and to $n-1$ copies of the decreasing sequence $(v_{\varepsilon _ j})$, the currents $(\mathrm{dd}^{\mathrm{c}}v_{\varepsilon _ j})^{n-1}\wedge \omega $ converge weakly to the Bedford–Taylor product $(\mathrm{dd}^{\mathrm{c}}v)^{n-1}\wedge \omega $. By proposition 7.2.2(1) and (4), these products are the exterior products with $\omega $ of the smooth forms $(\mathrm{dd}^{\mathrm{c}}v_{\varepsilon _ j})^{n-1}$ and of the current $(\mathrm{dd}^{\mathrm{c}}v)^{n-1}$, which equals $\langle S^{n-1}\rangle $ on $B$. (For $n=1$ both sides of the second convergence equal $\int \chi \, \omega $.)

Step 3: globalization. Let $\chi $ be a smooth function with compact support in $X\setminus A$ and $0\leq \chi \leq 1$. Cover $\operatorname{Supp}\chi $ by finitely many balls $B_ l$ as in Step 2, and write $\chi =\sum _ l\chi _ l$ with nonnegative $\chi _ l\in C^\infty _ c(B_ l)$; for instance, multiply $\chi $ by a partition of unity with compact supports subordinate to the cover of $X$ by the balls $B_ l$ and $X\setminus \operatorname{Supp}\chi $ (proposition 5.2.6). In each $B_ l$ we use the notation of Step 2 with an index $l$ and put

Let $\alpha _{l,\varepsilon }$ and $\beta _{l,\varepsilon }$ be the densities of $\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon }\wedge \tau $ and $(\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon })^{n-1}\wedge \omega $ with respect to $\eta _ l^ n$. By (17.5.35) with $\sigma =\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon }$, $\alpha _{l,\varepsilon }\beta _{l,\varepsilon }\geq F_{l,\varepsilon }g_ l/n$, so the Cauchy–Schwarz inequality gives

Summing over $l$ and applying the Cauchy–Schwarz inequality to the finite sum $\sum _ l\sqrt{a_{l,\varepsilon }b_{l,\varepsilon }}$, we obtain

Now let $\varepsilon \downarrow 0$ along a sequence. By Step 2, the two sums on the left converge to $\int _ X\chi \, S\wedge \tau $ and $\int _ X\chi \langle S^{n-1}\rangle \wedge \omega $. These are at most the corresponding integrals over $X$ without $\chi $, because $S\wedge \tau $ and $\langle S^{n-1}\rangle \wedge \omega $ are positive measures and $0\leq \chi \leq 1$. By (17.5.36) and $g_ l\eta _ l^ n=\nu $, the lower limit of the right side is at least $(m/n)(\int _ X\chi \nu )^2$. Therefore

Finally, we remove the cutoff. Fix a metric $d$ inducing the topology of $X$ (lemma 5.2.7), and for $k\geq 1$ let $K_ k=\{ x\in X:d(x,A)\geq 1/k\} $, with $K_ k=X$ if $A=\varnothing $. These compact subsets of $X\setminus A$ increase to $X\setminus A$, because $A$ is closed. By proposition 5.2.6 there are smooth functions $\chi _ k\colon X\to [0,1]$ with compact support in $X\setminus A$ and $\chi _ k=1$ on $K_ k$. Since $\nu $ has a smooth density and $A$ has Lebesgue measure zero, $\nu (K_ k)\to \nu (X\setminus A)=\int _ X\nu $, so $\int _ X\chi _ k\nu \to \int _ X\nu $. Applying the last inequality to $\chi =\chi _ k$ and letting $k\to \infty $ proves (17.5.34). Both factors on its left are finite: the first is the pairing of a current with a smooth form on the compact manifold $X$, and the second is finite by theorem 13.5.4. $\square$

Proof. By proposition 13.9.2, big classes have positive volume, and every closed positive $(1,1)$-current $T$ satisfies $\int _ X\langle T^ n\rangle \leq \operatorname {vol}(\{ T\} )$. It therefore suffices to show that a pseudo-effective class $\alpha $ with $m=\operatorname {vol}(\alpha )>0$ is big, and we do so by verifying Lamari's criterion (lemma 17.5.28). Write $\alpha =\{ \theta \} $ with $\theta $ smooth and closed, choose $C>0$ such that $\theta +C\omega $ is positive definite, and put

the integrand is a positive multiple of the volume form at every point, by the positivity of mixed discriminants of positive definite matrices (lemma 5.3.3(2),(3)).

For $0<\varepsilon \leq 1$ let $\theta _\varepsilon =\theta +\varepsilon \omega $. Since $\alpha $ is pseudo-effective, there is $q\in \mathrm{PSH}(X,\theta )$ (corollary 10.2.19), and $\theta _\varepsilon +\mathrm{dd}^{\mathrm{c}}q\geq \varepsilon \omega $; thus $\{ \theta _\varepsilon \} =\alpha +\varepsilon \{ \omega \} $ is big. By proposition 16.21.2(4), $P_{\theta _\varepsilon }[V_{\theta _\varepsilon }]=V_{\theta _\varepsilon }$, and the mass of $V_{\theta _\varepsilon }$ is

the inequality holds because $\operatorname {vol}(\alpha +\varepsilon \{ \omega \} )$ decreases to $\operatorname {vol}(\alpha )$ as $\varepsilon \downarrow 0$, by definition 13.9.1, proposition 13.9.2. Hence $V_{\theta _\varepsilon }$ is a model potential of mass $m_\varepsilon $.

Fix a smooth strictly positive real $(n-1,n-1)$-form $\tau $ with $\mathrm{dd}^{\mathrm{c}}\tau =0$, normalized by $\int _ X\omega \wedge \tau =1$. The measure $m_\varepsilon \omega \wedge \tau $ has a smooth density with respect to $\omega ^ n$ and total mass $m_\varepsilon $. By corollary 17.5.26, applied in the big class $\{ \theta _\varepsilon \} $ with the reference $V_{\theta _\varepsilon }$, there are $u_\varepsilon \in \mathrm{PSH}(X,\theta _\varepsilon )$ and finite constants $B_\varepsilon $ such that

The constants $B_\varepsilon $ need not be bounded uniformly in $\varepsilon $; we only use that $u_\varepsilon $ is locally bounded outside a closed set of measure zero. By lemma 16.13.2, applied to the big class $\{ \theta _\varepsilon \} $, there is a closed pluripolar set $A_\varepsilon \subset X$ such that $V_{\theta _\varepsilon }$ is locally bounded on $X\setminus A_\varepsilon $; being pluripolar, $A_\varepsilon $ has Lebesgue measure zero in every chart (see the remarks after definition 5.47.2). Hence $u_\varepsilon $, which lies between $V_{\theta _\varepsilon }-B_\varepsilon $ and $V_{\theta _\varepsilon }$, is locally bounded on $X\setminus A_\varepsilon $, and so are the local potentials of $S_\varepsilon $, which differ from $u_\varepsilon $ by smooth local potentials of $\theta _\varepsilon $ (lemma 5.4.2).

We may thus apply lemma 17.5.33 to $S_\varepsilon $, with $A_\varepsilon $, $\nu =\omega \wedge \tau $ and $m_\varepsilon $ in place of $A$, $\nu $ and $m$:

We bound the two factors on the left. Since $\mathrm{dd}^{\mathrm{c}}\tau =0$, the first factor equals the cohomological pairing $\int _ X\theta _\varepsilon \wedge \tau =\int _ X\theta \wedge \tau +\varepsilon $. If $n\geq 2$, the form $\theta _\varepsilon +C\omega $ is positive definite, so theorem 13.5.4 gives

the last inequality holds because $\theta +(C+1)\omega =(\theta +(C+\varepsilon )\omega )+(1-\varepsilon )\omega $ and products of positive $(1,1)$-forms are positive. If $n=1$, the second factor equals $\int _ X\omega =C_0$. As the first factor is nonnegative, in both cases

Letting $\varepsilon \downarrow 0$ gives $\int _ X\theta \wedge \tau \geq m/(nC_0)$. Removing the normalization of $\tau $, we conclude that every smooth strictly positive $\mathrm{dd}^{\mathrm{c}}$-closed $(n-1,n-1)$-form $\tau $ satisfies (17.5.30) with $\delta =m/(nC_0)>0$. By lemma 17.5.28, the class $\alpha $ is big. $\square$

Proof. By proposition 16.11.4(1), $u\in \mathcal A_\theta (f)$; that is, $u\in \mathrm{PSH}(X,\theta )$ and $u\leq f$ quasi-everywhere. If $\{ \theta \} $ is not big, then $\mu _ u(X)=0$ by theorem 17.5.37, since otherwise $\theta +\mathrm{dd}^{\mathrm{c}}u$ would be a closed positive current of positive non-pluripolar mass in $\{ \theta \} $. Hence $\mu _ u=0$, and in particular the first assertion holds.

If $\{ \theta \} $ is big, the first assertion is theorem 16.13.4.

For the rooftop, the function $\min _ ju_ j$ is quasi-continuous, as a finite minimum of potentials (see the beginning of section 16.11), and it is bounded above; its envelope is $w$. Since $w$ is not identically $-\infty $, the family $\mathcal A_\theta (\min _ ju_ j)$ is nonempty, and it is uniformly bounded above because the obstacle is. By the first assertion, $\mu _ w$ is carried by $\{ w=\min _ ju_ j\} \subset \bigcup _ j\{ w=u_ j\} $. The obstacle is upper semicontinuous, so $w\leq u_ j$ everywhere by proposition 16.11.4(1),(2), and lemma 16.13.1 gives $\mathbf1_{\{ w=u_ j\} }\mu _ w\leq \mathbf1_{\{ w=u_ j\} }\mu _{u_ j}$ for every $j$. Summing over $j$ proves the rooftop inequality. $\square$

Proof. The potential $V_\theta +\min _ Xf$ lies below $f$, so $\mathcal A_\theta (f)$ is nonempty, and it is bounded above by $\max _ Xf$. Put $u=P_\theta (f)$. Since $f$ is continuous, $u\leq f$ everywhere by proposition 16.11.4(1),(2). The scalar case of theorem 16.20.2, with all $\theta _ j=\theta $, $u_ j=u$ and $f_ j=f$, gives

and corollary 17.5.38 shows that $\langle \theta _ u^ n\rangle $ is carried by $\{ u=f\} $. This proves the first identity. For $f=0$ the envelope $P_\theta (0)$ is $V_\theta $, by definition 10.3.7, which gives the second. If $\{ \theta \} $ is not big, the left sides vanish by corollary 17.5.38, and hence so do the right sides. $\square$

Proof. All the masses involved are finite by theorem 13.5.4. If one of the masses $\int _ X\langle T_ i^ n\rangle $ vanishes, the right side of (17.5.42) is zero and there is nothing to prove. Otherwise put $m_ i=\int _ X\langle T_ i^ n\rangle >0$. Both sides of (17.5.42) are positively homogeneous of degree one in each $T_ i$, by the multilinearity of non-pluripolar products (proposition 13.2.6), so after replacing $T_ i$ by $m_ i^{-1/n}T_ i$ we may assume $m_ i=1$ for all $i$.

By theorem 17.5.37, the class of each $T_ i$ is big. Let $\theta _ i$ be a smooth closed real $(1,1)$-form in the class of $T_ i$ (definition 10.2.17). By corollary 10.2.19 we can write $T_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}u_ i$ with $u_ i\in \mathrm{PSH}(X,\theta _ i)$. Set $\phi _ i=P_{\theta _ i}[u_ i]$. By the mass preservation in proposition 16.21.2(3) and by theorem 16.22.11, $\phi _ i$ is a model potential of mass one.

Let $\nu =\omega ^ n/\int _ X\omega ^ n$. This smooth probability measure is non-pluripolar, as noted after definition 17.5.2. In each big class $\{ \theta _ i\} $, theorem 17.5.15 gives

By proposition 16.22.14, $P_{\theta _ i}[w_ i]=P_{\theta _ i}[\phi _ i]=\phi _ i$. Applying the mixed mass preservation of proposition 16.21.2(3) first to the tuple $(w_1,\ldots ,w_ n)$ and then to the tuple $(u_1,\ldots ,u_ n)$, we obtain

On the other hand, proposition 13.5.8, with all densities equal to one, gives $\langle S_1\wedge \cdots \wedge S_ n\rangle \geq \nu $. Hence the common value of these mixed masses is at least $\nu (X)=1$, which is the normalized inequality. Undoing the scalings proves (17.5.42).

For the concavity, let $T,S$ be closed positive $(1,1)$-currents and $0<t<1$; for $t=0,1$ there is nothing to prove. Put $a=\int _ X\langle T^ n\rangle $ and $b=\int _ X\langle S^ n\rangle $. By multilinearity,

For $1\leq k\leq n-1$, (17.5.42) with $k$ factors $T$ and $n-k$ factors $S$ gives $\int _ X\langle T^ k\wedge S^{n-k}\rangle \geq a^{k/n}b^{(n-k)/n}$; the terms with $k=n$ and $k=0$ are $a$ and $b$, also when one of them vanishes. The binomial formula therefore yields

Taking $n$th roots proves that the function $T\mapsto (\int _ X\langle T^ n\rangle )^{1/n}$ is concave. If $a$ and $b$ are both positive, then, since the logarithm is increasing and concave,

If one of them is zero, the right side is $-\infty $ for $0<t<1$, so there is nothing to prove. $\square$