13.19 Domination and ordinary envelopes832E0568D7C3

Proof. Put $\ell =\underline{\operatorname {vol}}_\theta (\varphi )>0$ and $\mathcal E_\varphi =\mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. We may replace $v$ by $\max (u,v)$: by locality, the hypothesis is unchanged on $\{ u<v\} $, and by upward stability (proposition 16.28.62) the new $v$ lies in $\mathcal E_\varphi $. Thus we may assume $u\leq v$. Translate $u,v$ by the same constant so that $v\leq \varphi $, and then translate all three potentials so that $\varphi \leq 0$. The idea is to stretch $u$ away from $v$; the stretched potentials keep lower volume $\ell $, and this is incompatible with a strict inequality $u<v$ on a set of positive volume. Set

Lemma 16.28.63 gives $q_ b\in \mathcal E_\varphi $ and $q_ b\leq u$. Since $b^{-1}q_ b+(1-b^{-1})v\leq u$ with contact set $D_ b$, contact support and positivity of the mixed products give

Here $\mu _{q_ b}$ is carried by $D_ b$. Choose $b$ so large that $(1-b^{-1})^ n>c$. On $D_ b\cap \{ u<v\} $ the hypothesis bounds the right-hand side by $c\mu _ v$, and subtracting finite measures shows that $\mu _{q_ b}$ vanishes there. Thus $\mu _{q_ b}$ is carried by $D_ b\cap \{ u=v\} \subset \{ q_ b=u\} $. Since $q_ b\leq u$, ordered contact for this pair and theorem 13.18.5, which gives $\underline{\operatorname {vol}}_\theta (q_ b)=\ell $, now yield

Suppose now that $u<v$ somewhere. Then there is $\delta >0$ such that $\{ u<v-\delta \} $ has positive Lebesgue measure; otherwise $u\geq v$ almost everywhere, and hence everywhere by lemma 5.24.3. On this set the convex inequality gives $q_ b\leq u-(b-1)\delta $. The family $q_ b$ decreases in $b$ and is bounded above by $u$. Its limit is $-\infty $ on a set of positive measure, so it cannot be $\theta $-plurisubharmonic and is therefore identically $-\infty $. Hence theorem 5.25.2 gives $\sup _ Xq_ b\to -\infty $. The sets $\{ q_ b=u\} \subset \{ u\leq \sup _ Xq_ b\} $ then have $\mu _ u$-mass tending to zero, because $\mu _ u$ is finite and puts no mass on the pole set of $u$. This contradicts (13.19.2). $\square$

Proof. Fix $\delta >0$. On $\{ u<v-\delta \} $ the weighted inequality gives $\mu _ u\leq \mathrm{e}^{-\lambda \delta }\mu _ v=\mathrm{e}^{-\lambda \delta }\mu _{v-\delta }$. Theorem 13.19.1, applied to $v-\delta $ with $c=\mathrm{e}^{-\lambda \delta }<1$, gives $u\geq v-\delta $; now let $\delta \downarrow 0$. If the second assertion failed, the theorem would apply with the same $c<1$ to every translate $v+A$, $A\in \mathbb R$, and would give $u\geq v+A$ for all $A$. This is impossible at a point where $u$ and $v$ are both finite. $\square$

Proof. Set $p_ C=P_\theta (u+C,w)$ for $C\geq 0$ and $p=P_\theta [u](w)$. The envelopes $p_ C$ increase with $C$, lie between $p_0$ and $w$, and converge to $p$ in capacity by lemma 5.29.19. By the contact theorem, $\mu _{p_ C}$ is carried by $\{ p_ C=\min (u+C,w)\} $. Assigning the points where $u+C=w$ to the second term, ordered contact gives

The first measure has total mass at most

because $\mu _ u$ is finite and non-pluripolar. Testing against nonnegative continuous functions and applying the lower semicontinuity of lemma 13.14.8 gives $\mu _ p\leq \mu _ w$.

The bounded quasi-continuous weight $\chi =\min (1,(w-p)_+)$, defined arbitrarily on the pluripolar set where $w-p$ is undefined, vanishes on $\{ p_ C=w\} $, since $p_ C\leq p\leq w$. Thus $\int _ X\chi \, \mu _{p_ C}\leq \varepsilon _ C$. Weighted lower semicontinuity gives $\int _ X\chi \, \mu _ p=0$, so $\mu _ p$ is carried by $\{ p=w\} $. Together with $\mu _ p\leq \mu _ w$, this proves the lemma. $\square$

Proof. Since $u\preceq p$, monotonicity gives $\underline{\operatorname {vol}}_\theta (u)\leq \underline{\operatorname {vol}}_\theta (p)$ and $\overline{\operatorname {vol}}_\theta (u)\leq \overline{\operatorname {vol}}_\theta (p)$. In particular, the lower volumes agree if $\underline{\operatorname {vol}}_\theta (p)=0$. Suppose henceforth that $\underline{\operatorname {vol}}_\theta (p)>0$. Normalize $u\leq 0$, so that $u\leq p$, and choose any $w\in \mathrm{PSH}(X,\theta )$ with $w\sim p$. Fix $K\geq 0$ with $p-K\leq w\leq p+K$. The rooftop envelope $p_ C=P_\theta (u+C,w)$ satisfies $u-K\leq p_ C\leq u+C$ for each $C\geq 0$, so it is $\theta $-plurisubharmonic and equivalent to $u$. Its regularized increasing limit $r=P_\theta [u](w)$ satisfies

Indeed, the leftmost potential lies below $u+C$, and below $p-K\leq w$ because $P_\theta (u+C+K,0)\leq P_\theta [u]=p$; it is therefore a competitor for the rooftop. Letting $C\to \infty $ gives the lower bound for $r$. Thus $r\sim p$, and proposition 16.28.62 puts $r$ in $\mathcal E_{\mathrm{env}}(X,\theta ,p)$.

By lemma 13.19.4, $\mu _ r$ vanishes on $\{ r<w\} $. Theorem 13.19.1, with coefficient $c=0$ and reference $p$, gives $r\geq w$; hence $r=w$. Since $p_ C\to r$ in capacity, lower semicontinuity of the total mass (lemma 13.14.8) and $p_ C\sim u$ yield

Taking the supremum over $w\sim p$ gives $\overline{\operatorname {vol}}_\theta (p)\leq \overline{\operatorname {vol}}_\theta (u)$, and hence equality of the upper volumes. Theorem 13.18.5 then gives $u\in \mathcal E_{\mathrm{env}}(X,\theta ,p)$ and equality of the lower volumes. Finally, proposition 16.28.65 gives $P_\theta [p]=P_\theta [u]=p$. $\square$

Proof. Step 1: the rooftops. Adding a constant to $u$ changes neither $p$ nor the family $(p_ m)_{m>0}$, apart from a shift of the parameter $m$. We may therefore assume $u\leq 0$. Then

where the increasing limit is understood after upper semicontinuous regularization. Thus $p_ m\sim u$, and $p_ m$ and $p$ are locally bounded off $S$. By theorem 13.19.5, $p$ is an ordinary model with the same volumes as $u$. By lemma 5.29.19, $p_ m\to p$ in capacity. On compact subsets of $X\setminus S$ the potentials have a common bounded range, so all bounded local products converge there, by the bounded product theory of section 5.27.

Step 2: full measures. Let $v\in \mathrm{PSH}(X,\omega )$ be locally bounded off $S$, and put $v_ k=\max (v,-k)$. On each chart choose a smooth $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq \omega $, so that the functions $v_ k+q$ are plurisubharmonic and decrease to $v+q$. Theorem 5.27.18 then gives full weak limits in every degree,

Indeed, $\mathrm{dd}^{\mathrm{c}}v_ k=\mathrm{dd}^{\mathrm{c}}(v_ k+q)-\mathrm{dd}^{\mathrm{c}}q$, and the powers can be expanded in the closed positive powers of $\mathrm{dd}^{\mathrm{c}}(v_ k+q)$ and smooth forms. The local limits patch together because the approximating bounded currents are global. The limits are currents of order zero with finite mass on the compact manifold $X$, and $\Phi _ v$ is positive. Stokes' theorem for the bounded potentials $v_ k$ gives $\int _ XW_ n(v)=0$. Consequently

The measure $\Phi _ v$ equals $\mathrm{MA}_\omega (v)$ off $S$. Every closed positive power of degree less than $n$ in this expansion has coefficient measures vanishing on $S$, by corollary 5.34.3. Thus only the top closed power contributes to $\mathbf1_ S\Phi _ v$. This identifies $\mathbf1_ S\Phi _ v$ with $\mathbf1_ S(\mathrm{dd}^{\mathrm{c}}(v+q))^ n_{\mathrm{full}}$ and proves independence of $q$. These assertions are local on each chart and do not require a single chart containing $S$. Finally, $J(v)\leq v_+(\omega )$, since every truncation $v_ k$ has mass at most $v_+(\omega )$.

Step 3: convergence of the lower-degree powers. We need local trace bounds for the closed powers of $p_ m$ that are uniform in $m$. Fix a relatively compact coordinate neighborhood and write $\kappa =\mathrm{dd}^{\mathrm{c}}|z|^2$. Choose $B>0$ with $B\kappa \geq \omega $ there, and $A>1$ with $A\omega -B\kappa \geq \kappa $. For $p_{m,k}=\max (p_ m,-k)$, the closed current $T_{m,k}=B\kappa +\mathrm{dd}^{\mathrm{c}}p_{m,k}$ is positive and

Expanding $Q_{m,k}^ n$ with the smooth positive difference $Q_{m,k}-T_{m,k}\geq \kappa $ gives, for $1\leq j\leq n$,

The last bound holds because $Q_{m,k}=A(\omega +\mathrm{dd}^{\mathrm{c}}(p_{m,k}/A))$ and $p_{m,k}/A$ is a bounded $\omega $-plurisubharmonic function. Letting $k\to \infty $ gives trace bounds, independent of $m$, for every full closed power of $\mathrm{dd}^{\mathrm{c}}(p_ m+B|z|^2)$.

For $j<n$, every subsequential weak limit of these closed positive $(j,j)$-currents puts no mass on $S$, again by corollary 5.34.3. Off $S$ the bounded capacity convergence identifies the limit with the corresponding full power of $\mathrm{dd}^{\mathrm{c}}(p+B|z|^2)$, which also puts no mass on $S$. Hence the limit is unique. Expanding back in the smooth correction and using a finite coordinate cover yields

Hence (13.19.7) gives $J(p_ m)\to J(p)$.

Step 4: the residual measures. Let $R=\mathbf1_ S\Phi _ u$. Since $p_ m-u$ is bounded, the equality case of lemma 5.27.19, applied on each chart after a common smooth correction, gives $\mathbf1_ S\Phi _{p_ m}=R$ for every $m$. The total masses $J(p_ m)$ are bounded by $v_+(\omega )$, so along any sequence $m_ i\to \infty $ a subsequence of $\Phi _{p_{m_ i}}$ converges weakly to a measure $\eta $. Since $\Phi _{p_ m}-R$ is positive, so is $\eta -R$. Off $S$, $\eta $ equals $\mathrm{MA}_\omega (p)$, by the bounded convergence of Step 1, and its total mass is $\lim _ mJ(p_ m)=J(p)$ by Step 3. The measure $\Phi _ p$ has the same total mass and the same restriction to $X\setminus S$, so

On the other hand, $u\leq p$, so the inequality in lemma 5.27.19 gives $\mathbf1_ S\Phi _ p\leq R$ on each chart. Together with the previous inequality, this forces $\Phi _ p(S)=R(S)$, hence $\mathbf1_ S\Phi _ p=R$, which is the asserted equality of residual measures. It also gives $\mathbf1_ S\eta =R$, since $\eta -R\geq 0$ and $\eta (S)=R(S)$. As $\eta $ and $\Phi _ p$ also agree off $S$, we conclude that $\eta =\Phi _ p$. Every weak subsequential limit of $\Phi _{p_ m}$ is therefore $\Phi _ p$, which proves the convergence $\Phi _{p_ m}\to \Phi _ p$. $\square$

Proof. The first equality is theorem 13.18.5. Membership in $\mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$ implies $P_\theta [u]=P_\theta [\varphi ]=\varphi $ by proposition 16.28.65. Conversely, if $P_\theta [u]=\varphi $, then theorem 13.19.5 puts $u$ in $\mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. $\square$