Theorem 13.21.2BE27C58EE55B

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. $\square$