Theorem 13.24.3.E9FC137BBE4E Let $u\in \mathrm{PSH}(X,\theta )$, put $q=P_\theta [u]_{\mathcal I}$, and suppose $\underline{\operatorname {vol}}_\theta (q)>0$. There are analytic $b_ j\in \mathrm{PSH}(X,\theta )$ with $b_ j\to u$ in $L^1$ such that the ordinary models $r_ j=P_\theta [b_ j]$ satisfy
34D208A53BBD
13.24.4
\begin{equation} d_\theta (r_ j,q)\longrightarrow 0,\qquad d_\theta (r_ j,u)\longrightarrow \mathcal V^+_{\mathcal I,\theta }(u) -\overline{\operatorname {vol}}_\theta (u). \end{equation}
Their lower volumes converge to $\underline{\operatorname {vol}}_\theta (q)$, and $r_ j\to q$ in $L^1$ and capacity. In particular $d_\theta (r_ j,u)\to 0$ if and only if $P_\theta [u]=P_\theta [u]_{\mathcal I}$.
Proof.
Step 1: the approximants. Fix a Hermitian form $h$. Bigness and theorem 5.36.1 give an analytic $\rho \in \mathrm{PSH}(X,\theta )$ with $\rho \leq 0$ and a rational $\kappa >0$ such that $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \kappa h$. Choose the dyadic regularizations $a_ j$ as in (18.7.5), enlarging their errors to positive rational numbers $\epsilon _ j\downarrow 0$. Put
\[ c_ j=\frac{\kappa }{\kappa +\epsilon _ j},\qquad b_ j=c_ ja_ j+(1-c_ j)\rho . \]
Then $\theta +\mathrm{dd}^{\mathrm{c}}b_ j\geq ((1-c_ j)\kappa -c_ j\epsilon _ j)h=0$. The coefficients are rational, so $b_ j$ has analytic singularities by proposition 5.34.1. Since $a_ j\to u$ in $L^1$ and $c_ j\to 1$, also $b_ j\to u$ in $L^1$.
Step 2: two-sided bounds. Choose $A>0$ with $\theta \leq Ah$, and set
\[ \widehat\theta _ j=\theta +(1+A/\kappa )\epsilon _ jh, \qquad t_ j=P_{\widehat\theta _ j}[a_ j]. \]
These backgrounds decrease to $\theta $ and dominate $\theta +\epsilon _ jh$, so the $a_ j$ also satisfy (18.7.5) with the larger errors $(1+A/\kappa )\epsilon _ j$. Theorem 18.7.6 therefore gives $t_ j\searrow q$. Moreover
\[ c_ j\widehat\theta _ j-\theta =(1-c_ j)(Ah-\theta )+c_ j\epsilon _ jh\geq 0. \]
By proposition 18.4.8, $r_ j\sim b_ j$. Hence $r_ j/c_ j$ is a nonpositive $\widehat\theta _ j$-plurisubharmonic function that is more singular than $a_ j$, because $b_ j\leq c_ ja_ j$; it is therefore a competitor for $t_ j$, and $r_ j\leq c_ jt_ j$. Also $q\leq t_ j\leq c_ jt_ j$.
For a lower bound, recall that $q\sim _{\mathcal I}u$, and $u\preceq _{\mathcal I}a_ j$ because $a_ j\geq u$. Addition and scaling preserve this order by theorem 5.37.3, lemma 5.37.4. Thus $c_ jq+(1-c_ j)\rho \preceq _{\mathcal I}b_ j$. This potential lies below zero, so envelope order and proposition 18.4.8 give
\[ c_ jq+(1-c_ j)\rho \leq P_\theta [b_ j]_{\mathcal I}=r_ j \leq c_ jt_ j. \]
Step 3: volumes. For either singularity volume, homogeneity, addition and monotonicity now yield
\[ c_ j^ n\operatorname {vol}_\theta (q) \leq \operatorname {vol}_\theta (r_ j) \leq \operatorname {vol}_\theta (\max (r_ j,q)) \leq c_ j^ n\operatorname {vol}_{\widehat\theta _ j}(t_ j). \]
The last comparison uses both $\theta \leq c_ j\widehat\theta _ j$ and $\max (r_ j,q)\leq c_ jt_ j$. By theorem 13.21.2, the outer terms tend to $\operatorname {vol}_\theta (q)$. This proves the convergence of both volumes of $r_ j$ and the first limit in (13.24.4). In particular, after discarding finitely many indices, the ordinary models $r_ j$ and $q$ have a common positive lower-volume bound, so $r_ j\to q$ in $L^1$ and in capacity by proposition 13.22.4, theorem 13.23.3.
For the second limit, translate $u$ so that $u\leq q$; this does not change $d_\theta (r_ j,u)$. Then
\[ \overline{\operatorname {vol}}_\theta (r_ j) \leq \overline{\operatorname {vol}}_\theta (\max (r_ j,u)) \leq \overline{\operatorname {vol}}_\theta (\max (r_ j,q)). \]
Both bounds tend to $\overline{\operatorname {vol}}_\theta (q)$, so $d_\theta (r_ j,u)=2U(\max (r_ j,u))-U(r_ j)-U(u)$ tends to $U(q)-U(u)$, as claimed. The last equivalence follows from proposition 13.24.2.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.