13.25 Model comparison under background changesB2557A0A9699

Proof. The map $u\mapsto u-f$ is a bijection from $\mathrm{PSH}(X,\theta )$ to $\mathrm{PSH}(X,\theta ')$ that preserves curvature currents, singularity types and hence both singularity volumes. Also $\max (u-f,v-f)=\max (u,v)-f$, so

By proposition 18.5.1, $G_ f(p)+f\sim p$, that is, $G_ f(p)\sim p-f$. Thus $G_ f(p)$ has the same volumes as $p$, and it is an ordinary model by theorem 13.19.5; in particular $G_ f(p)\in \mathcal M_\delta (\theta ')$. Since the volume comparison depends only on ordinary types, the displayed identity follows. Taking the ordinary envelope of $G_ f(p)+f\sim p$ gives $P_\theta [G_ f(p)+f]=P_\theta [p]=p$. The same argument with the two backgrounds interchanged proves that the map is bijective with the stated inverse.

Subtraction of a smooth function and bounded changes preserve analytic singularities. If $p=P_\theta [p]_{\mathcal I}$, proposition 18.5.1 also gives

The ideal envelope is an ordinary model by corollary 18.4.4, and ordinary models of the same ordinary type coincide; hence $P_{\theta '}[G_ f(p)]_{\mathcal I}=P_{\theta '}[p-f]_{\mathcal I}=G_ f(p)$. The inverse map gives the converse. $\square$

Proof. Normalize $u\leq 0$, so that $u\leq V_\beta $, and note that $\rho \leq V_\beta $. Since $\Omega \geq \beta $ and $\underline{\operatorname {vol}}_\beta (u)\geq \delta $, lemma 13.20.1, applied in the big background $\beta $, shows that $H=Q_{b_\delta }(u,V_\beta )$ is defined. The potential

satisfies $w\leq u$ and $\beta +\mathrm{dd}^{\mathrm{c}}w\geq (1-b_\delta ^{-1})(\theta +\mathrm{dd}^{\mathrm{c}}\rho )\geq \kappa _\delta h$. The addition argument in the proof of proposition 13.20.6, with $h$, $\kappa _\delta $ and $t=\varepsilon /\kappa _\delta $ in place of $\omega $, $\delta $ and $t$, therefore gives the upper bound, with constants independent of $\beta $ and $u$. The lower bound is volume monotonicity. $\square$

Proof. The preceding proposition, with $\beta =\theta $ and $\gamma =\eta $, applies to $u$, $v$ and $m=\max (u,v)$, and their upper volumes in background $\theta $ are at most $v_+(\Omega )=K$. Thus, for $x=u,v,m$, the errors

lie between zero and $K((1+\varepsilon /\kappa _\delta )^ n-1)$. The difference $d_\eta (u,v)-d_\theta (u,v)$ equals $2e_ m-e_ u-e_ v$, whose absolute value is at most twice this bound. Finally, the projection invariance in proposition 13.22.4, applied in background $\eta $, gives $d_\eta (u,v)=d_\eta (P_\eta [u],P_\eta [v])$, which is the comparison appearing in (13.25.4). $\square$

Proof. Suppose first that $p=P_\theta [p]_{\mathcal I}$. By theorem 13.24.3, there are analytic ordinary models $r_ j$ in background $\theta $ with $d_\theta (r_ j,p)\to 0$ and a common positive lower-volume bound. Corollary 13.25.3 and projection invariance in background $\eta _ j$ give $d_{\eta _ j}(r_ j,p)\to 0$, so $a_ j=r_ j$ works. One can also use the ordinary models $a_ j=P_{\eta _ j}[r_ j]$, since $P_{\eta _ j}[r_ j]\sim r_ j$ by proposition 18.4.8.

Conversely, suppose that such $a_ j$ exist. Normalize them below zero, and put $m_ j=\max (a_ j,p)$ and $\delta =\underline{\operatorname {vol}}_\theta (p)>0$. Inequality (13.22.2) and proposition 13.25.2 give

Indeed, the upper volume of $p$ in background $\eta _ j$ tends to its upper volume in background $\theta $, and the differences of $\overline{\operatorname {vol}}_{\eta _ j}(a_ j)$ and $\overline{\operatorname {vol}}_{\eta _ j}(m_ j)$ from $\overline{\operatorname {vol}}_{\eta _ j}(p)$ are bounded in absolute value by $d_{\eta _ j}(a_ j,p)$.

Enlarge the errors $\varepsilon _ j$ to positive rational numbers tending to zero, and put

As in the proof of theorem 13.24.3, $\theta +\mathrm{dd}^{\mathrm{c}}b_ j\geq 0$ and $b_ j$ has analytic singularities. Choose $A>0$ with $\theta \leq Ah$, and set

Then $\widehat\theta _ j\geq \eta _ j$ and $c_ j\widehat\theta _ j\geq \theta $. Since $r_ j\sim b_ j$ and $\rho \leq 0$, the potential $r_ j/c_ j$ is a nonpositive $\widehat\theta _ j$-plurisubharmonic function more singular than $a_ j$, hence more singular than $m_ j$. Therefore $r_ j\leq c_ j\widehat m_ j$. Also $p\leq \widehat m_ j\leq c_ j\widehat m_ j$.

The lower volume of $m_ j$ in background $\eta _ j$ is at least $\delta $. For large $j$, proposition 13.25.2 applies with $\beta =\eta _ j$, $\gamma =\widehat\theta _ j$ and $\varepsilon =(1+A/\kappa )\varepsilon _ j$. Consequently

By theorem 13.19.5, $\widehat m_ j$ has the same upper volume in background $\widehat\theta _ j$ as $m_ j$. On the other hand,

Addition, applied to $c_ ja_ j\in \mathrm{PSH}(X,c_ j\eta _ j)$ and $(1-c_ j)\rho \in \mathrm{PSH}(X,(1-c_ j)(\theta -\kappa h))$, and background monotonicity therefore bound the upper volume of $b_ j$ below by $c_ j^ n\overline{\operatorname {vol}}_{\eta _ j}(a_ j)$. Since $r_ j\sim b_ j$, the two comparisons yield

Here the last inequality uses $\theta \leq c_ j\widehat\theta _ j$ and $\max (r_ j,p)\leq c_ j\widehat m_ j$. Both outer terms tend to $\overline{\operatorname {vol}}_\theta (p)$, which proves $d_\theta (r_ j,p)\to 0$. By proposition 18.4.8, $r_ j=P_\theta [b_ j]_{\mathcal I}$ is an ideal model, so theorem 13.24.5 gives $p=P_\theta [p]_{\mathcal I}$.

For large $j$, the upper-volume deficit from $r_ j$ to $\max (r_ j,p)$ is small, and lemma 13.18.1 with $b=2$ gives a convex lower bound for $r_ j$ whose first term is $\frac12\max (r_ j,p)$. Volume addition gives $\underline{\operatorname {vol}}_\theta (r_ j)\geq 2^{-n}\delta $. Thus proposition 13.22.4, theorem 13.23.3 give the asserted $L^1$ and capacity convergence of the models $r_ j$. $\square$