Corollary 18.26.24.9829FC8151FB Assume $0<v_-(\omega )\leq v_+(\omega )<+\infty $, as in proposition 18.26.23, and fix $0<t_0<1/C$. Set $p_ t=P_\omega [tH]$ for $0<t\leq t_0$, define $p_0=0$ separately, and write
\[ L(t)=\underline{\operatorname {vol}}_\omega (p_ t),\qquad U(t)=\overline{\operatorname {vol}}_\omega (p_ t). \]
For $0\leq t\leq t_0$,
\[ (1-Ct)^ n v_-(\omega )\leq L(t)\leq U(t) \leq v_+(\omega )-Mt^ n. \]
The models are pairwise distinct, and $p_ s\geq p_ t$ for $s\leq t$. For either $V=L$ or $V=U$, and $0\leq s<t\leq t_0$,
\[ \left(\frac{1-Ct}{1-Cs}\right)^ n V(s) \leq V(t)\leq V(s). \]
Consequently
\[ d_\omega (p_ s,p_ t)=|U(s)-U(t)| \leq \frac{nC v_+(\omega )}{1-Ct_0}|s-t|, \qquad \delta _\omega ^{\mathrm{pol}}(p_ s,p_ t)=0. \]
With the restriction of $d_\omega $, the family is a complete ordered interval, isometric to $[0,v_+(\omega )-U(t_0)]$. Every model with positive parameter is not an ideal model and has trivial multiplier ideals. For any $0<\delta <v_-(\omega )$, $t_0$ can be chosen so that the whole interval lies in $\mathcal M_\delta (\omega )$.
Proof.
Put $u_ t=tH$ for $t>0$ and $u_0=0$. For $0<t<1/C$, the potential $u_ t$ is $Ct\omega $-plurisubharmonic. Adding it to the zero potential in the background $(1-Ct)\omega $, proposition 13.17.2 and homogeneity show that either singularity volume of $u_ t$ is at least $(1-Ct)^ n$ times the corresponding volume of zero. In particular, the lower volume of $u_ t$ is positive on the whole range $0<t<1/C$, and ordinary projection preserves both volumes by theorem 13.19.5. The upper bound follows from proposition 18.26.21, and the zero endpoint has $L(0)=v_-(\omega )$ and $U(0)=v_+(\omega )$.
Near $S$, one has $u_ t=th$. After a fixed smooth local correction, invariance of residual measures and lemma 18.26.15 show that $u_ t$ has residual measure of mass $Mt^ n$, and by theorem 13.19.6 so does $p_ t$. Zero has residual measure zero, since its corrected top-degree measure is smooth. Thus two models of the family coincide only if their parameters are equal.
For $s<t$, set $c=(1-Ct)/(1-Cs)$. Then $0<c<1$ and
\[ u_ t=cu_ s+(t-cs)H,\qquad C(t-cs)=1-c. \]
The second summand is $(1-c)\omega $-plurisubharmonic. Addition and homogeneity give $V(t)\geq c^ nV(s)$, whereas monotonicity gives $V(t)\leq V(s)$. The same computation holds for $s=0$, with the separately defined zero potential. Monotonicity of the envelope gives $p_ s\geq p_ t$, so (13.22.2) gives $d_\omega (p_ s,p_ t)=U(s)-U(t)$. Since $U(s)\leq v_+(\omega )$, the Lipschitz bound follows from
\[ 1-c^ n\leq n(1-c) =\frac{nC(t-s)}{1-Cs} \leq \frac{nC(t-s)}{1-Ct_0}. \]
By this bound, $U$ is continuous. Distinct ordinary models of positive lower volume are separated by proposition 13.22.4, so $U$ is strictly decreasing. Thus $t\mapsto v_+(\omega )-U(t)$ identifies the family isometrically with the closed interval in the statement, which proves completeness.
The support and Lelong-number argument in the proof of proposition 18.26.17 applies on the whole range $0<t<1/C$: all first polar currents vanish and all multiplier ideals are trivial. The deficit $Mt^ n$ of the upper volume separates every positive parameter from zero, which proves that the models with positive parameter are not ideal models and that their polar comparisons vanish. Finally, choose $t_0<1/C$ with $(1-Ct_0)^ nv_-(\omega )>\delta $ to place the whole interval in $\mathcal M_\delta (\omega )$.
$\square$
Pome
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