16.22 Extrapolation and model singularities472B5CD7929C

Proof. Fix a smooth probability volume form $\mathrm dV$ on $X$, put $m=\int _ X\langle \theta _\psi ^ n\rangle $, and consider

The set $\mathcal S$ contains $\psi $, and $H$ takes values in $[H(\psi ),0]$ on it. We construct an increasing sequence in $\mathcal S$ that almost maximizes $H$ at every stage. Put $v_0=\psi $. Given $v_ k\in \mathcal S$, let

and choose $v_{k+1}\in \mathcal S$ with $v_{k+1}\geq v_ k$ and $H(v_{k+1})\geq s_ k-2^{-k}$. The regularized limit $\phi =(\sup _ kv_ k)^*$ is a potential with $\psi \leq \phi \leq 0$, and $\int _ X\langle \theta _\phi ^ n\rangle =m$ by corollary 13.8.9; thus $\phi \in \mathcal S$. Since $v_ k$ increases to $\phi $ almost everywhere, between the integrable bounds $\psi $ and $0$, dominated convergence gives $H(v_ k)\to H(\phi )$.

Let $w\in \mathcal S$ with $w\geq \phi $. Then $w\geq v_ k$ for every $k$, so

and letting $k\to \infty $ gives $H(w)\leq H(\phi )$. Since $w\geq \phi $, it follows that $w=\phi $ almost everywhere, hence everywhere. Now $P_\theta [\phi ]$ lies between $\phi $ and $0$, because $\phi \leq 0$ is a competitor for it, and it has mass $m$ by proposition 16.21.2. Thus $P_\theta [\phi ]\in \mathcal S$ and $P_\theta [\phi ]\geq \phi $, so $P_\theta [\phi ]=\phi $. Since $m>0$, $\phi $ is a model potential. $\square$

Proof. The potential $\tau -\sup _ X\tau $ is nonpositive and more singular than $\phi $, so it is a competitor for $P_\theta [\phi ]=\phi $. Hence $\tau -\phi \leq \sup _ X\tau $ wherever $\phi $ is finite. Conversely, $\phi \leq 0$, so $\tau -\phi \geq \tau $ wherever $\phi $ is finite. The pole set of $\phi $ is pluripolar, and removing a pluripolar set does not change the supremum of the quasi-plurisubharmonic function $\tau $: an upper bound for $\tau $ off a Lebesgue null set holds everywhere. This proves the reverse inequality. $\square$

Proof. (1) Off the pole set of $u$, the functions $u$ and $v$ are finite and $f=u+(a-1)(u-v)\leq u$, because $u\leq v$; on the pole set, $f=-\infty $. Thus $f$ is a Borel function with $f\leq u$. Let $K$ be a compact subset of a chart. Since $u$ and $v$ are quasi-continuous (lemma 16.11.2(1)), there are open sets of arbitrarily small capacity outside which both are continuous on $K$, as functions with values in $[-\infty ,\infty )$. On the complement in $K$ of such a set, $f$ is continuous at every point where $u$ is finite, because $v\geq u$ is finite near such a point, and at a pole of $u$ it is continuous with value $-\infty $, because $f\leq u$. Hence $f$ is quasi-continuous. Off the pole set of $u$, the inequalities $\eta \leq f$ and $a^{-1}\eta +(1-a^{-1})v\leq u$ are equivalent. Since the pole set is pluripolar, and an inequality between quasi-plurisubharmonic functions that holds quasi-everywhere holds everywhere, this proves the equivalence.

(2) By (1), the potentials satisfying the inequality are exactly the competitors for $f$; since $f\leq u$, they lie below $u$ quasi-everywhere, hence everywhere. Thus $\mathcal A_\theta (f)$ is nonempty, by assumption, and uniformly bounded above, and proposition 16.11.4 shows that $\eta $ is its largest member; in particular $\eta \leq u$ and $\gamma \leq u$. By theorem 16.13.4, $\mu _\eta $ is carried by $\{ \eta =f\} $. At a point of this set at which $u$ is finite, $\eta =au+(1-a)v$ means $\gamma =u$; since $\mu _\eta $ does not charge the pole set of $u$, it is carried by $D$. Moreover $\mu _\gamma \geq a^{-n}\mu _\eta $, and lemma 16.13.1, applied to $\gamma \leq u$, gives $\mathbf1_ D\mu _\gamma \leq \mathbf1_ D\mu _ u$. Hence

Proof. We first construct one potential satisfying (16.22.6) as the limit of a decreasing sequence of envelopes, and then show that $E_ a(u,v)$ is the largest solution and has positive mass.

Reduction. Since $u\preceq v$, there are constants $A,B$ with $u-A\leq v-B\leq 0$. If $\eta '$ satisfies (16.22.6) for the pair $u-A$, $v-B$, then $\eta '+aA+(1-a)B$ satisfies it for $u$, $v$; the envelopes of the two pairs differ by the same constant, and masses do not change. We may therefore assume that $u\leq v\leq 0$. Then $m=\int _ X\langle \theta _ v^ n\rangle \geq \int _ X\langle \theta _ u^ n\rangle >0$, and lemma 16.22.1 provides a model potential $\phi \geq v$ with $\int _ X\langle \theta _\phi ^ n\rangle =m$. Since $1-a<0$, we have $au+(1-a)\phi \leq au+(1-a)v$; we first construct a potential below $au+(1-a)\phi $.

A decreasing sequence of envelopes. For $j\geq 1$ put $u_ j=\max \{ u,\phi -j\} $, so that $\phi -j\leq u_ j\leq \phi $; in particular, $u_ j$ and $\phi $ have the same pole set. Put

where $f_ j=-\infty $ on the pole set of $\phi $, as in lemma 16.22.3 for the pair $u_ j\leq \phi $. Outside this pole set, $f_ j=\phi +a(u_ j-\phi )$ lies between $\phi -aj$ and $\phi $. Hence the potential $\phi -aj$ is a competitor, and every competitor is at most $\phi $, so $\eta _ j$ is a potential with

where the masses agree because $\eta _ j\sim \phi \sim u_ j$. Since $u_ j$ decreases with $j$, so do $f_ j$ and $\eta _ j$.

A contact bound. Put $\gamma _ j=a^{-1}\eta _ j+(1-a^{-1})\phi $ and $D_ j=\{ \gamma _ j=u_ j>-\infty \} $. By lemma 16.22.3(2), applied to the pair $u_ j\leq \phi $, we have $\gamma _ j\leq u_ j$ and

A tail estimate. Fix $k>0$ and let $j>k$. On $D_ j$ we have $\eta _ j=\phi +a(u_ j-\phi )$, so there $\eta _ j\leq \phi -ak$ if and only if $u_ j\leq \phi -k$, that is, if and only if $u\leq \phi -k$. The plurifine open set $\{ u>\phi -k\} $ is contained in $\{ u>\phi -j\} $, on which $u_ j=u$; hence $\mu _{u_ j}=\mu _ u$ on $\{ u>\phi -k\} $, and $\mu _{u_ j}(\{ u\leq \phi -k\} )=m-\mu _ u(\{ u>\phi -k\} )$, pole sets having measure zero. Together with (16.22.7) this gives

The envelopes do not collapse. Suppose that $\eta _ j$ decreases to $-\infty $ identically. Then $\sup _ X\eta _ j\to -\infty $ by corollary 10.3.6. Since $\eta _ j\preceq \phi $ and $\phi $ is a model potential, lemma 16.22.2 gives $\eta _ j-\phi \leq \sup _ X\eta _ j$ off the pole set of $\phi $. Hence, for each $k$, $\eta _ j\leq \phi -ak$ everywhere once $j$ is large, and the left side of (16.22.8) equals $\int _ X\langle \theta _{\eta _ j}^ n\rangle =m$. Since $\phi $ is finite wherever $u$ is, $\mu _ u(\{ u>\phi -k\} )\to \int _ X\langle \theta _ u^ n\rangle $ as $k\to \infty $, and (16.22.8) yields $m\leq a^ n\bigl (m-\int _ X\langle \theta _ u^ n\rangle \bigr )$, which contradicts (16.22.5). Consequently $\eta _ j$ decreases to a potential $\eta $. Letting $j\to \infty $ in $\gamma _ j\leq u_ j$ gives $a^{-1}\eta +(1-a^{-1})\phi \leq u$, and since $v\leq \phi $, also (16.22.6).

The largest solution. By lemma 16.22.3(1), applied to the pair $u\leq v$, the potentials satisfying (16.22.6) are exactly the competitors for the obstacle $f=au+(1-a)v$. The potential $\eta $ constructed above is one of them, so part (2) of the lemma shows that $E_ a=E_ a(u,v)=P_\theta (f)$ is the largest potential satisfying (16.22.6), and that $E_ a\leq u$.

Positive mass. Choose $b>a$ that still satisfies (16.22.5). The preceding steps, applied to $b$, give $E_ b=E_ b(u,v)$. The potential

is a competitor for the obstacle $f$ of $E_ a$: quasi-everywhere, $E_ b\leq bu+(1-b)v$, and therefore

Hence $q\leq E_ a$, and the monotonicity of mass together with $\mu _ q\geq (1-a/b)^ n\mu _ u$ yields

If $\int _ X\langle \theta _ u^ n\rangle =\int _ X\langle \theta _ v^ n\rangle $, then (16.22.5) holds for every $a>1$, so $b$ can be taken arbitrarily large, and we obtain $\int _ X\langle \theta _{E_ a}^ n\rangle \geq \int _ X\langle \theta _ u^ n\rangle $. The reverse inequality follows from $E_ a\leq u$. $\square$

Proof. Put $m=\int _ X\langle \theta _{V_\theta }^ n\rangle $. Since $u\preceq V_\theta $ and $\int _ X\langle \theta _ u^ n\rangle >0$, we can choose $a>1$ so close to one that $a^ n\bigl (m-\int _ X\langle \theta _ u^ n\rangle \bigr )<m$. Theorem 16.22.4 gives $\eta \in \mathrm{PSH}(X,\theta )$ with $a^{-1}\eta +(1-a^{-1})V_\theta \leq u$. Since the class is big, it contains a closed current $T\geq \delta _0\omega $ for some $\delta _0>0$, and by theorem 10.2.15 we can write $T=\theta +\mathrm{dd}^{\mathrm{c}}\psi $ with $\psi $ quasi-plurisubharmonic; after subtracting a constant, $\psi \in \mathrm{PSH}(X,\theta )$ satisfies $\psi \leq 0$ and $\theta _\psi \geq \delta _0\omega $, and $\psi \leq V_\theta $ because $\psi $ is a competitor for $V_\theta $. Therefore

Proof. After reducing to the case $u\leq v$, we show that the extrapolation envelopes $E_ a(u,v)$ agree with $u$ almost everywhere for $\mu _ u$. Hence they do not collapse as $a\to \infty $, while their limit equals $-\infty $ quasi-everywhere on $\{ u<v\} $, so that this set is pluripolar.

The potential $\max \{ u,v\} $ is more singular than $\chi $ and less singular than $u$, so

thus $\max \{ u,v\} \in \mathcal E(X,\theta ,\chi )$. Since $\{ u<\max \{ u,v\} \} =\{ u<v\} $, we may replace $v$ by $\max \{ u,v\} $ and assume that $u\leq v$. Subtracting the same constant from $u$ and $v$, we may also assume $v\leq 0$. Let $m>0$ be the common mass of $u$, $v$ and $\chi $.

For $a>1$ let $\eta _ a=E_ a(u,v)$, $\gamma _ a=a^{-1}\eta _ a+(1-a^{-1})v$ and $D_ a=\{ \gamma _ a=u>-\infty \} $. By theorem 16.22.4, $\eta _ a$ is a potential of mass $m$; in particular its obstacle $au+(1-a)v$ has a competitor, and lemma 16.22.3(2), applied to the pair $u\leq v$, gives $\eta _ a\leq u$, $\gamma _ a\leq u$ and

Since $u\leq v$, the hypothesis means that $\mu _ u$ is carried by $\{ u=v\} $, so $\mu _{\eta _ a}$ is carried by $D_ a\cap \{ u=v\} $. At a point of this set, $a^{-1}\eta _ a+(1-a^{-1})u=u$, that is, $\eta _ a=u$. Hence $\mu _{\eta _ a}$ is carried by $\{ \eta _ a=u\} $, and lemma 16.13.1, applied to $\eta _ a\leq u$, gives $\mu _{\eta _ a}\leq \mathbf1_{\{ \eta _ a=u\} }\mu _ u\leq \mu _ u$. Both measures have mass $m$, so they are equal, and $\eta _ a=u$ holds $\mu _ u$-almost everywhere. In particular

where the last integral is positive because $m>0$ and $\mu _ u$ does not charge the pole set of $u$.

As $a$ increases, the obstacle $v+a(u-v)$ decreases, so $\eta _ a$ decreases. If $\eta _ a$ decreased to $-\infty $ identically as $a\to \infty $ through the integers, dominated convergence, with $0\leq \mathrm e^{\eta _ a}\leq 1$, would make the left side of the last display tend to zero, a contradiction. Hence the limit $\eta $ along the integers is a potential. Each $\eta _ a$ satisfies $\eta _ a\leq v+a(u-v)$ outside a pluripolar set; let $N$ be the union of these sets over the integers $a\geq 2$. At a point outside $N$ where $u$ is finite and $u<v$, we get $\eta \leq v+a(u-v)$ for every integer $a$, hence $\eta =-\infty $. Thus $\{ u<v\} $ is contained in the union of $N$ and the pole sets of $u$ and $\eta $, which is pluripolar. Therefore $u\geq v$ quasi-everywhere, and hence everywhere. $\square$

Proof. Write $w=P_\theta [u]$, and let $v$ belong to the displayed family. By proposition 16.21.2, $w\leq P_\theta [v]$, and the potentials $w$, $u$, $v$ and $P_\theta [v]$ all have the same positive mass. Moreover $v\leq P_\theta [v]$, because $v\leq 0$ is a competitor for $P_\theta [v]$. Hence $w$ and $v$ both belong to $\mathcal E(X,\theta ,P_\theta [v])$, a class whose reference potential has positive mass. By proposition 16.21.3, $\mu _ w$ is carried by $\{ w=0\} $, and since $v\leq 0$, it gives no mass to $\{ w<v\} $. The domination principle, theorem 16.22.10, yields $w\geq v$. Finally, $w$ itself belongs to the family: it is nonpositive, it satisfies $u-\sup _ Xu\leq w$, and it has the same mass as $u$ by proposition 16.21.2. This proves the formula.

The potential $P_\theta [w]$ is nonpositive, satisfies $u\preceq w\preceq P_\theta [w]$, and has the same mass as $u$ by proposition 16.21.2. Hence it belongs to the family, and the formula gives $P_\theta [w]\leq w$. Conversely, $w\leq 0$ is a competitor for $P_\theta [w]$, so $P_\theta [w]=w$. Since the mass of $w$ is positive, $w$ is a model potential. $\square$

Proof. Write $w_ C=P_\theta (u+C,\phi )$. Each $w_ C$ with $C$ large has the singularity type of $u$, hence mass $\int _ X\langle \theta _ u^ n\rangle $, and $w_ C$ increases almost everywhere to $w$; by corollary 13.8.9, $\int _ X\langle \theta _ w^ n\rangle =\int _ X\langle \theta _ u^ n\rangle $. By corollary 16.13.5, together with $w_ C\leq \phi $ and $w_ C\leq w\leq \phi $,

As $C\to \infty $, the sets $\{ u+C\leq \phi \} $ decrease to the pole set of $u$, which is $\mu _ u$-null, because $\phi $ is bounded above; so the first term tends to zero in total variation by dominated convergence. Moreover $\mu _{w_ C}\to \mu _ w$ weakly by corollary 13.8.9. Integrating continuous functions $\chi \geq 0$ and applying lemma 16.12.1, we obtain the asserted measure bound.

If (1) holds, then $w$ and $\phi $ have the same positive mass and both belong to $\mathcal E(X,\theta ,\phi )$. By the measure bound, $\mu _ w$ gives no mass to $\{ w<\phi \} $, and the domination principle, theorem 16.22.10, gives $w\geq \phi $; hence (2). Conversely, if (2) holds, then $u$ has the mass of $w=\phi $, which is (1).

To prove that (2) implies (3), normalize $\phi \leq 0$. This is harmless: replacing $\phi $ by $\phi +c$ replaces $w_ C$ by $P_\theta (u+C-c,\phi )+c$, hence $w$ by $w+c$, so (2) is preserved, and $P_\theta [u]$ and $P_\theta [\phi ]$ do not change. Every $w_ C$ is then a nonpositive potential with the singularity type of $u$, hence a competitor for $P_\theta [u]$, so $\phi =w\leq P_\theta [u]$. Since (2) implies (1), $u$ has the positive mass of $\phi $, so $P_\theta [u]$ is a model potential by theorem 16.22.11. Monotonicity and $u\preceq \phi $ now give

Finally, if (3) holds, then by proposition 16.21.2 the potential $u$ has the mass of $P_\theta [u]=P_\theta [\phi ]$, which is the mass of $\phi $; this is (1). $\square$