Theorem 13.26.3218C2CE831D1

Proof. Write $c=c_\varepsilon $, $s=1-c$, $v=cp+s\rho $ and $T(p)=T^\rho _{\eta ,\varepsilon }(p)$. Since $s\rho \in \mathrm{PSH}(X,s(\theta -\kappa h))$, addition gives $\operatorname {vol}_{\theta -s\kappa h}(v)\geq c^ n\operatorname {vol}_\theta (p)$, and since $\theta -s\kappa h=\theta -\varepsilon h\leq \eta $, monotonicity gives the lower bound in (13.26.4) for $v$. In particular $v$ has positive lower volume in background $\eta $, so theorem 13.19.5 shows that $T(p)=P_\eta [v]$ is an ordinary model with the same two volumes as $v$.

Since $\varepsilon <\kappa /2$, we have $c\geq 1/2$, and the choice of $B$ gives

Since $\eta \leq \theta +\varepsilon h\leq c\widehat\theta _\varepsilon $, every competitor $w$ for $T(p)$, that is, every nonpositive $\eta $-plurisubharmonic $w\preceq v$, satisfies $w/c\in \mathrm{PSH}(X,\widehat\theta _\varepsilon )$ and $w/c\preceq p$, because $\rho \leq 0$. Taking envelopes, and using that $v\leq 0$ is itself a competitor, yields

Volume monotonicity, together with theorem 13.19.5 in the larger background $\widehat\theta _\varepsilon $, proves the upper bound in (13.26.4).

Now let $q\in \mathcal M_\delta (\theta )$ and $m=\max (p,q)$. The same pointwise bounds give

Consequently the upper volumes of $T(p)$, $T(q)$ and of their maximum lie between $c^ n$ times the respective upper volumes of $p$, $q$, $m$ in the backgrounds $\theta $ and $\widehat\theta _\varepsilon $. The lower volumes of $p$, $q$ and $m$ are at least $\delta $, and their upper volumes are at most $K$. By proposition 13.25.2 with $\beta =\theta $ and $\gamma =\widehat\theta _\varepsilon $, each of the three errors $\overline{\operatorname {vol}}_\eta (T(x))-c^ n\overline{\operatorname {vol}}_\theta (x)$, $x=p,q,m$ (with $T(m)$ read as $\max (T(p),T(q))$), is nonnegative and at most $c^ nK((1+B\varepsilon /\kappa _\delta )^ n-1)$. As in the proof of corollary 13.25.3, this proves (13.26.5).

For fixed $p$, corollary 13.21.4, lemma 5.29.19 give $P_{\widehat\theta _\varepsilon }[p]\to p$ in $L^1$ and in capacity along every decreasing sequence $\varepsilon \downarrow 0$. An arbitrary sequence $\varepsilon _ j\to 0$ is handled by comparing with the decreasing sequence $\sup _{k\geq j}\varepsilon _ k$, since the envelopes increase with the background and lie above $p$. Quasi-continuity and integrability give $cp+s\rho \to p$ in both senses. The bounds (13.26.6) prove the asserted limits.

To compare two choices along a sequence $\eta _ j$, denote their coefficients by $c_ j^{(i)}$ and their upper backgrounds by $\theta +B_ i\varepsilon _ j^{(i)}h$, $i=0,1$. Put $c_ j=\min _ i c_ j^{(i)}$ and $a_ j=\max _ i B_ i\varepsilon _ j^{(i)}$. Since the envelopes are nonpositive, both transported models lie below $c_ jP_{\theta +a_ jh}[p]$. Moreover $\eta _ j\leq c_ j(\theta +a_ jh)$, by the form inequality above for an index attaining the smaller coefficient. The upper volume of the maximum of the two transported models is therefore at most $c_ j^ n\overline{\operatorname {vol}}_{\theta +a_ jh}(p)$, which tends to $\overline{\operatorname {vol}}_\theta (p)$. Each of their individual upper volumes is at least $(c_ j^{(i)})^ n\overline{\operatorname {vol}}_\theta (p)$ and at most the upper volume of the maximum, so it has the same limit. Hence their $d_{\eta _ j}$-comparison tends to zero. $\square$