13.21 Monotone limitsA8EAC258D61B

Proof. Step 1: the envelopes. Translate so that $u\leq 0$, and set $p_ j=P_\theta [u_ j]$, $p=P_\theta [u]$ and $s=(\lim _ jp_ j)^*$. The $p_ j$ increase, and $u\leq s\leq p\leq V_\theta $. By theorem 13.19.5, the lower volume of $p$ is positive and $u\in \mathcal E_{\mathrm{env}}(X,\theta ,p)$. Upward stability (proposition 16.28.62) puts $s$ in the same class.

The ordinary envelope can use $V_\theta $ in place of zero as obstacle (see the remark after definition 5.40.4); that is, $p_ j=P_\theta [u_ j](V_\theta )$. Hence lemma 13.19.4 gives

The fixed bounded quasi-continuous weight $\chi =\min (1,(p-s)_+)$, defined arbitrarily on the pluripolar set where $p-s$ is undefined, vanishes on $\{ p_ j=V_\theta \} $, because $p_ j\leq s\leq p\leq V_\theta $. Since $p_ j\to s$ in capacity by lemma 5.29.19, lemma 13.14.8 gives $\int _ X\chi \, \mu _ s=0$. Thus $\mu _ s$ vanishes on $\{ s<p\} $, and theorem 13.19.1, with reference $p$ and coefficient $c=0$, yields $s=p$. This proves the assertions about the envelopes.

Step 2: the upper volumes of the envelopes. Choose any $w\in \mathrm{PSH}(X,\theta )$ with $w\sim p$, and $K\geq 0$ with $p-K\leq w\leq p+K$. Define $r_ j=P_\theta [u_ j](w)$; this is meaningful because $u_ j-K\leq p-K\leq w$, so that $P_\theta (u_ j,w)\not\equiv -\infty $. For $C\geq 0$ the competitors compare as follows:

The leftmost potential lies below $u_ j+C$ and below $p_ j-K\leq p-K\leq w$. For the upper bound, note that the middle term minus $K$ lies below $w-K\leq p\leq 0$ and is more singular than $u_ j$, so it is at most $p_ j$. Taking the increasing limit in $C$ gives

The $r_ j$ increase with $j$ and converge in capacity to $r=(\lim _ jr_ j)^*$, and $p-K\leq r\leq w\leq p+K$. Thus $r\sim p$, so $r\in \mathcal E_{\mathrm{env}}(X,\theta ,p)$. Lemma 13.19.4 gives $\mu _{r_ j}\leq \mathbf1_{\{ r_ j=w\} }\mu _ w$, and weighted lower semicontinuity with the weight $\min (1,(w-r)_+)$ shows that $\mu _ r$ vanishes on $\{ r<w\} $. Domination with reference $p$ gives $r=w$. Therefore

where $r_ j\sim p_ j$ justifies the last inequality. Taking the supremum over $w\sim p$ and using volume monotonicity proves

Step 3: the volumes of $u_ j$. Put $\delta =\underline{\operatorname {vol}}_\theta (p)>0$ and $D_ j=\overline{\operatorname {vol}}_\theta (p)-\overline{\operatorname {vol}}_\theta (p_ j)$. Since $D_ j\to 0$, we can choose $b_ j>1$ with $b_ j\to \infty $ and $b_ j^ nD_ j<\delta $ for all large $j$. With $a_ j=1-b_ j^{-1}$, lemma 13.18.1 gives $h_ j\in \mathrm{PSH}(X,\theta )$ such that

for all sufficiently large $j$. Addition, homogeneity and monotonicity of the lower volume, together with theorem 13.19.5, give

Since $a_ j\to 1$, this proves the convergence of the lower volumes, and shows that $\underline{\operatorname {vol}}_\theta (u_ j)>0$ for all large $j$. For such $j$, theorem 13.19.5 identifies the upper volumes of $u_ j$ and $p_ j$, and also those of $u$ and $p$; Step 2 therefore gives the asserted upper-volume limit. Finally, proposition 13.18.3, applied to $u_ j\preceq u$, gives the uniform convergence of Lelong numbers on each fixed modification. $\square$

Proof. Step 1: the fixed point and positivity of the limit volume. Since $q\leq q_ j\leq 0$, monotonicity of the envelope in the background and in the potential gives $q\leq P_\theta [q]\leq P_{\theta _ j}[q]\leq P_{\theta _ j}[q_ j]=q_ j$. Letting $j\to \infty $ proves $P_\theta [q]=q$. By theorem 13.17.1, the lower volumes $\underline{\operatorname {vol}}_{\theta _ j}(q_ j)$ decrease; let $\ell $ be their limit. Monotonicity also gives $\underline{\operatorname {vol}}_\theta (q)\leq \ell $, so the lower-volume assertion holds if $\ell =0$.

Suppose $\ell >0$. We show that then $\underline{\operatorname {vol}}_\theta (q)>0$; the lower-volume assertion will then follow from the volume limits proved below for limits of positive lower volume. Choose a Hermitian form $\Omega \geq \theta _1$, put $K=v_+(\Omega )$, and choose

Since $\Omega \geq \theta _ j$ and $\underline{\operatorname {vol}}_{\theta _ j}(q_ j)\geq \ell $, lemma 13.20.1, applied in background $\theta _ j$ with $r=V_{\theta _ j}$, shows that $Q_ b(q_ j,V_{\theta _ j})$ is defined for every $j$, with the same $b$. Put $h_ j=P_{\theta _ j}[Q_ b(q_ j,V_{\theta _ j})]$. For each competitor $s$ for this envelope ($s\leq 0$ and $s\preceq Q_ b(q_ j,V_{\theta _ j})$), the potential $aV_{\theta _ j}+(1-a)s$ lies below zero and is more singular than $aV_{\theta _ j}+(1-a)Q_ b(q_ j,V_{\theta _ j})\leq q_ j$. Since $q_ j=P_{\theta _ j}[q_ j]$, it lies below $q_ j$, and taking the envelope gives

The functions $h_ j$ have supremum zero and are $\Omega $-plurisubharmonic. Normalized compactness and the uniform convergence of the backgrounds give a subsequential $L^1$ limit $h\in \mathrm{PSH}(X,\theta )$. Also $V_{\theta _ j}\searrow V_\theta $, since the decreasing limit is $\theta $-plurisubharmonic and lies between $V_\theta $ and zero. Passing to the limit in the convex inequality and using lemma 5.24.3 gives

Moreover, $v_-(\Omega )=\underline{\operatorname {vol}}_\Omega (0)\geq \underline{\operatorname {vol}}_{\theta _1}(q_1)\geq \ell >0$ by volume monotonicity. By bigness, choose $\rho \leq V_\theta $ with $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \kappa \Omega $ for some $\kappa >0$. Proposition 13.17.2, applied to $0\in \mathrm{PSH}(X,\kappa \Omega )$ and $\rho \in \mathrm{PSH}(X,\theta -\kappa \Omega )$, gives $\underline{\operatorname {vol}}_\theta (\rho )\geq \kappa ^ n v_-(\Omega )$. Monotonicity and addition applied to the convex combination above now yield

It remains to prove the volume limits when $\underline{\operatorname {vol}}_\theta (q)>0$.

Step 2: a fixed background. Suppose first that $\theta _ j=\theta $ for every $j$. Put

The limit exists because the upper volumes decrease, by theorem 13.17.1, and are finite, by bounded mass; thus $e_ k\searrow 0$. After discarding finitely many indices, choose $b_ k>1$ tending to infinity with $b_ k^ n e_ k<\delta $, and set $a_ k=1-b_ k^{-1}$.

For every $j\geq k$, the upper-volume deficit $\overline{\operatorname {vol}}_\theta (q_ k)-\overline{\operatorname {vol}}_\theta (q_ j)$ is at most $e_ k$, whereas the lower volume of $q_ k$ is at least $\delta $. Thus lemma 13.18.1 gives $h_{k,j}\in \mathrm{PSH}(X,\theta )$ with

Since $q_ k\leq 0$ and $q_ j=P_\theta [q_ j]$, combining $q_ k$ with the competitors for $P_\theta [h_{k,j}]$, as in the proof of proposition 16.28.65, gives

The potentials $P_\theta [h_{k,j}]$ have supremum zero. For fixed $k$, normalized compactness gives a subsequential $L^1$ limit $h_ k$ as $j\to \infty $. Passing to the limit and applying lemma 5.24.3 gives

For either volume, homogeneity, addition and monotonicity now yield

Letting $k\to \infty $ proves both volume limits in the fixed background.

Step 3: varying backgrounds. Fix a Hermitian form $\omega $. By corollary 13.20.3, choose $w\leq q$ and $d>0$ with $\theta +\mathrm{dd}^{\mathrm{c}}w\geq d\omega $. Choose $\varepsilon _ j>0$ decreasing to zero with $\theta _ j\leq \theta +\varepsilon _ j\omega $, and put

The convex combination $f_ j$ is $\theta $-plurisubharmonic, because its background is bounded above by $\theta $:

Since $w\leq q\leq q_ k$, we have $f_ k\leq q_ k\leq q_ j$ for $k\geq j$, and upper regularization gives $F_ j\leq q_ j$. Also $f_ k\to q$ in $L^1$, so a subsequence converges almost everywhere, and $q\leq F_ j$ by lemma 5.24.3.

Background monotonicity now gives

The $r_ j$ decrease to $q$. Their lower volumes are at least $\underline{\operatorname {vol}}_\theta (q)>0$, so theorem 13.19.5 makes them ordinary models, and Step 2 applies to the sequence $r_ j$ in the fixed background $\theta $. On the other hand, addition (with $\eta =(1-c_ j)(\theta -d\omega )$) and background monotonicity give, for either volume,

Since $c_ j\to 1$, Step 2 shows that the outer terms tend to the same limit. The volumes $\operatorname {vol}_{\theta _ j}(q_ j)$ decrease by monotonicity in both background and potential, which proves the asserted decreasing convergence.

Step 4: Lelong numbers. By proposition 13.20.6, the volumes of the fixed potential $q$ in background $\theta _ j$ tend to its volumes in background $\theta $. Hence the upper-volume deficit between $q$ and $q_ j$, measured in $\theta _ j$, tends to zero, while the lower volume of $q_ j$ is at least $\underline{\operatorname {vol}}_\theta (q)>0$. Apply proposition 13.18.3 in each background $\theta _ j$. Since $\theta _ j\leq \theta _1$, its proof works with one Lelong bound $C_\pi $ for $\pi ^*\theta _1$, independent of $j$. This proves the uniform convergence on each fixed modification, and taking generic values gives the divisorial assertion. $\square$

Proof. Normalize $u_ j\leq 0$ by a common translation. Set $p_ j=P_{\theta _ j}[u_ j]$, $p=P_\theta [u]$ and $q=\lim _ jp_ j$. Since $\underline{\operatorname {vol}}_{\theta _ j}(u_ j)\geq \underline{\operatorname {vol}}_\theta (u)>0$, theorem 13.19.5 in background $\theta _ j$ shows that every $p_ j$ is an ordinary model with the same volumes as $u_ j$. The envelopes decrease, and the uniform convergence of the backgrounds gives $q\in \mathrm{PSH}(X,\theta )$. Since $u\leq p\leq q$, the limit $q$ has positive lower volume, and theorem 13.21.2 gives, for either volume,

If $q=p$, these equalities and theorem 13.19.5 show that both volumes of $u_ j$ converge to those of $u$; in particular the upper volumes do. Conversely, if the upper volumes converge, then $u$ and $q$ have the same upper volume, so $u\in \mathcal E_{\mathrm{vol}}(X,\theta ,q)$. By corollary 13.19.8, $P_\theta [u]=q$, that is, $q=p$. For the Lelong assertion, proposition 13.20.6 shows that the upper-volume deficit between $u$ and $u_ j$ in background $\theta _ j$ tends to zero. Apply proposition 13.18.3 with the common lower bound $\underline{\operatorname {vol}}_\theta (u)>0$ and the uniform Lelong bound for $\pi ^*\theta _1$. $\square$

Proof. Fix a Hermitian form $\omega $. Uniform convergence gives $\theta _ j\leq \theta +\varepsilon _ j\omega $ with $\varepsilon _ j\to 0$, so the volume limits follow from the two estimates in proposition 13.20.6; monotonicity makes them decreasing. In the positive-lower-volume case, apply proposition 13.21.3 to the constant sequence $u_ j=u$. $\square$

Proof. Put $T=\mathcal T_\Omega (u;Y)$ and $r=\dim Y$. Normalize the majorant $v$ of proposition 20.3.24 by $\sup _ Yv=0$. It belongs to $\mathrm{PSH}(Y,\theta |_ Y)\subset \mathrm{PSH}(Y,\Omega /2)$ and is a competitor for $T$, so $v\leq T$. For every $w\in \mathrm{PSH}(Y,\Omega )$ with $w\sim T$, a constant translate of $v$ lies below $w$, so $P_{\Omega /2}(w)\not\equiv -\infty $. By lemma 16.28.69, applied on $Y$ with $a=1/2$,

Since $\Omega -\theta |_ Y\geq \Omega _0$ is Hermitian and the curvature errors of the approximants tend to zero, the restrictions $u_ m|_ Y$ belong to $\mathrm{PSH}(Y,\Omega )$ for all sufficiently large $m$. By theorem 20.3.27, their dyadic ordinary envelopes $q_ j=P_\Omega [u_{2^ j}|_ Y]$ decrease to $T$. Theorem 13.21.2, applied on $Y$ in the fixed background $\Omega $, and the analytic type of $u_{2^ j}|_ Y$ give

as required. $\square$

Proof. Every ideal model is an ordinary model by corollary 18.4.4, and the retraction theorem gives $\Gamma ^{h/j}_\tau \searrow \Gamma _\tau $ in the decreasing backgrounds $\theta +h/j$. Thus theorem 13.21.2 proves the identity for the initial lower volumes, including the case where the limit is zero.

Suppose that this limit $\ell $ is positive. For $\tau \leq a$, the slice is $\Gamma _ a$. For $a<\tau <c$, choose $r\in (\tau ,c)$ and put $s=(\tau -a)/(r-a)$, so that $\tau =(1-s)a+sr$. Concavity gives

Homogeneity, proposition 13.17.2 and volume monotonicity imply

Theorem 13.21.2 now gives both volume limits for every fixed $\tau <c$.

It remains to show that the slices keep their multiplier ideals in the limit. Fix $\tau <c$ and write $q=\Gamma _\tau $, $q_ j=\Gamma ^{h/j}_\tau $ and $\theta _ j=\theta +h/j$. By proposition 13.20.6, the upper volumes of the fixed potential $q$ in the backgrounds $\theta _ j$ converge to its upper volume in $\theta $, which is the limit of the upper volumes of the $q_ j$. Consequently

After discarding finitely many indices, choose $b_ j>1$ with $b_ j\to +\infty $ and $b_ j^ nD_ j<\underline{\operatorname {vol}}_{\theta _ j}(q_ j)$. Since $q\leq q_ j$, the stretching lemma lemma 13.18.1 in background $\theta _ j$ gives $v_ j\in \mathrm{PSH}(X,\theta _ j)$ such that

Choose a Hermitian form $\Omega \geq \theta +h$, so that all $v_ j$ are $\Omega $-plurisubharmonic. By lemma 5.26.2, there is $\alpha >0$ such that $\mathcal I(\alpha v_ j)=\mathcal O_ X$ for all $j$; additive normalization does not change these ideals. Let $E$ be a local divisor over a coordinate neighborhood $U$, with generic Lelong number $\nu _ E$ and log discrepancy $A_ U(E)$ as in section 5.38. Lemma 5.38.9, with $m=\alpha $, gives $\nu _ E(v_ j)\leq A_ U(E)/\alpha $. Taking Lelong numbers in the convex inequality therefore yields

Compatibility gives all components the same multiplier ideals, so $\nu _ E(q_ j)$ is independent of $j$, and hence equal to $\nu _ E(q)$. Blowing up a point on a local modification converts its pointwise Lelong number into such a divisorial number; on a curve the point itself is a divisor. Thus all Lelong numbers of $q$ and $q_ j$ on all local modifications agree, and theorem 5.37.3 gives $q\sim _{\mathcal I}q_ j$. This holds at every parameter $\tau <c$, so proposition 16.29.11 proves recovery. $\square$