16.25 Singularity constraints and relative capacity2B1B1CB179EE

Proof. We begin with the final identities. A competitor for $P_\theta [u](0)$ is a nonpositive potential more singular than $u$, exactly as in definition 16.21.1; and every potential is more singular than $V_\theta $, by minimality, so the families defining $P_\theta [V_\theta ](f)$ and $P_\theta (f)$ coincide.

(1) The maximum of two members of the family in definition 16.25.1 is again a member. By theorem 5.10.2, applied on finitely many coordinate balls on which $\theta $ has smooth local potentials, there are members $v_1,v_2,\ldots $ with $(\sup _ kv_ k)^*=P_\theta [u](f)$, and the maxima $w_ k=\max \{ v_1,\ldots ,v_ k\} $ form an increasing sequence of members with the same regularized supremum. By corollary 5.47.11, $\sup _ kw_ k=P_\theta [u](f)$ outside a pluripolar set, and each $w_ k$ lies below $f$ outside a pluripolar set; since countable unions of pluripolar sets are pluripolar (theorem 5.47.8), $P_\theta [u](f)\leq f$ quasi-everywhere. (The envelope itself need not be more singular than $u$; see example 16.22.12.) The other assertions follow from the corresponding properties of the families of competitors.

(2) A competitor $h$ for the obstacle $\min \{ u+k,f\} $ satisfies $h\leq u+k$ quasi-everywhere, hence everywhere, and $h\leq f$ quasi-everywhere; so it belongs to the family of definition 16.25.1. In particular these competitors are uniformly bounded above, $P_\theta (u+k,f)$ is defined, and when it is a potential, it belongs to the family. Conversely, every member $v$ of the family satisfies $v\leq u+k$ for all sufficiently large integers $k$, hence $v\leq \min \{ u+k,f\} $ quasi-everywhere and $v\leq P_\theta (u+k,f)$. Thus a fixed member shows that the rooftop envelopes, which increase with $k$, are potentials for all large $k$, and the family and the rooftop envelopes have the same supremum. This proves the representation.

(3) Normalize $u\leq 0$. If $|f|\leq C$ quasi-everywhere, then $u-C$ is a competitor, and for large $k$ each rooftop envelope $P_\theta (u+k,f)$ lies between $u-C$ and $u+k$, so it has the singularity type of $u$ and mass $\int _ X\langle \theta _ u^ n\rangle $. By (2) and corollary 13.8.9, $P_\theta [u](f)$ also has this mass. Comparing $f$ with the constants $-C$ and $C$, and using (1) together with $P_\theta [u](0)=P_\theta [u]$, gives the two bounds. $\square$

Proof. Write $p=P_\theta (f)$ and choose $C$ with $|f|\leq C$. Then $V_\theta -C$ is a competitor for $f$, and every competitor $v$ satisfies $v-C\leq 0$, hence $v\leq V_\theta +C$; thus $V_\theta -C\leq p\leq V_\theta +C$. Consequently $p$ has minimal singularities, and its mass is that of $V_\theta $, namely $\operatorname {vol}(\{ \theta \} )>0$, by definition 13.9.1, proposition 13.9.2. A potential lies below $f$ quasi-everywhere if and only if it lies below $p$: one direction is the maximality of $p$, the other holds because $p\leq f$ quasi-everywhere. Hence the two constrained families coincide, and $w=P_\theta [u](p)$. Since $u\preceq p$, this is the relative envelope of definition 16.22.13, by proposition 16.25.2. The measure bound in proposition 16.22.14, which applies because $p$ has positive mass, gives the first inequality. By theorem 16.13.4, $\mu _ p$ is carried by $\{ p=f\} $, so $\mu _ w$ is carried by $\{ w=p\} \cap \{ p=f\} \subset \{ w=f\} $. Finally, $f-w$ is nonnegative quasi-everywhere and vanishes $\mu _ w$-almost everywhere, which gives the integral identity. $\square$

Proof. For $\varepsilon >0$ put $q=\max \{ u,v-\varepsilon \} $. Write $T_ i=\theta _ i'+\mathrm{dd}^{\mathrm{c}}\varphi _ i$ with smooth closed real $(1,1)$-forms $\theta _ i'$ and $\varphi _ i\in \mathrm{PSH}(X,\theta _ i')$ (theorem 10.2.15). Since $q\preceq \phi $, the monotonicity of mixed masses in corollary 13.8.8, applied with $n$ factors to the potentials $q,\ldots ,q,\varphi _1,\ldots ,\varphi _{n-p}$ and $\phi ,\ldots ,\phi ,\varphi _1,\ldots ,\varphi _{n-p}$, gives $\lambda _ q(X)\leq \lambda _\phi (X)$. On the disjoint plurifine open sets $\{ u<v-\varepsilon \} $ and $\{ u>v-\varepsilon \} $, $q$ equals $v-\varepsilon $ and $u$ respectively, so plurifine locality gives

Writing $\lambda _ u(\{ u>v-\varepsilon \} )=\lambda _ u(X)-\lambda _ u(\{ u\leq v-\varepsilon \} )$, enlarging $\{ u\leq v-\varepsilon \} $ to $\{ u<v\} $ up to the common pole set of $u$ and $v$, which carries no mass, and letting $\varepsilon \downarrow 0$, we obtain the first inequality.

Now let $\int _ X\langle \theta _\phi ^ n\rangle >0$ and $u,v\in \mathcal E(X,\theta ,\phi )$. By proposition 16.22.14, $P_\theta [u]=P_\theta [v]=P_\theta [\phi ]$. With the potentials $\varphi _ i$ of the currents $T_ i$ chosen above, the preservation of mixed masses in proposition 16.21.2, applied to the $n$ potentials $u,\ldots ,u,\varphi _1,\ldots ,\varphi _{n-p}$ and to the corresponding tuples for $v$ and $\phi $, gives $\lambda _ u(X)=\lambda _ v(X)=\lambda _\phi (X)$. Thus the last two terms in the inequality cancel. Applying the inequality with $u$ and $v$ interchanged, we get $\lambda _ u(\{ v<u\} )\leq \lambda _ v(\{ v<u\} )$, and subtracting this from $\lambda _ v(X)=\lambda _ u(X)$ gives $\lambda _ v(\{ u\leq v\} )\leq \lambda _ u(\{ u\leq v\} )$. $\square$

Proof. Every competitor has the singularity type of $\phi $ and hence mass $\int _ X\langle \theta _\phi ^ n\rangle $; since $\phi $ itself is a competitor, the capacity is finite and $\operatorname {Cap}_{\theta ,\phi }(X)=\int _ X\langle \theta _\phi ^ n\rangle $. Monotonicity, countable subadditivity and continuity along increasing sequences hold for each measure $\mu _ v$ and pass to the supremum; for the last property, interchange the supremum over the sequence with the supremum over competitors. Inner regularity follows in the same way from the inner regularity of each finite measure $\mu _ v$ (lemma 16.12.1): $\operatorname {Cap}_{\theta ,\phi }(A)=\sup _ v\sup _ K\mu _ v(K)=\sup _ K\operatorname {Cap}_{\theta ,\phi }(K)$, the inner suprema being taken over the compact sets $K\subset A$. Each $\mu _ v$ gives no mass to pluripolar sets. Adding a constant $c$ to $\phi $ corresponds to the bijection $v\mapsto v+c$ of competitors, which does not change $\mu _ v$.

For the comparison, note first that $\phi \leq \psi +C$ and $\psi \leq \phi +C$ hold almost everywhere, hence everywhere; in particular $\phi $ and $\psi $ have the same pole set. Put $t=(2C+1)^{-1}$. If $\phi -1\leq v\leq \phi $, the potential $q=(1-t)\psi +tv-tC$ satisfies

so it is a competitor for $\operatorname {Cap}_{\theta ,\psi }$, and $\mu _ q\geq t^ n\mu _ v$. This proves the upper comparison, and exchanging $\phi $ and $\psi $ proves the lower one. The last assertion follows from the bijection $v\mapsto v-g$ between the two families of competitors, which preserves the current $\theta +\mathrm{dd}^{\mathrm{c}}v$. $\square$

Proof. Let $v$ be a competitor for $\operatorname {Cap}_{\theta ,\phi }$, and put $q=(1-t)\phi +tv-s$. Then $\phi -s-t\leq q\leq \phi -s$, so

Moreover $q\sim \phi $ and $\mu _ q\geq t^ n\mu _ v$. Proposition 16.25.4, with $p=n$, applied to $u$ and $q$, gives

and taking the supremum over $v$ proves the inequality. If $u\in \mathcal E(X,\theta ,\phi )$, then $\int _ X\langle \theta _ u^ n\rangle =\int _ X\langle \theta _\phi ^ n\rangle $. With $t=1$, the right side tends to zero as $s\to \infty $ by dominated convergence, because the sets $\{ u<\phi -s\} $ decrease to the pole set of $u$, which is $\mu _ u$-null. $\square$