Proposition 13.22.438359A611D4C

Proof. Translate the inputs below zero, and put $p=P_\theta [u]$, $q=P_\theta [v]$, $m=\max (u,v)$ and $r=P_\theta [m]$. Since $u\leq p$, $v\leq q$ and the envelope is monotone,

All these potentials have positive lower volume. Theorem 13.19.5 and monotonicity therefore give $U(m)=U(\max (p,q))=U(r)$, while $U(u)=U(p)$ and $U(v)=U(q)$. This proves the first identity. If $d_\theta (u,v)=0$, then $U(u)=U(v)=U(m)$, since both are at most $U(m)$. Thus $u,v\in \mathcal E_{\mathrm{vol}}(X,\theta ,m)$, and theorem 13.18.5, proposition 16.28.65 give $P_\theta [u]=P_\theta [m]=P_\theta [v]$. Conversely, if $p=q$, the first identity gives $d_\theta (u,v)=d_\theta (p,p)=0$.

Now let $u,v\in \mathcal M_\delta (\theta )$ and set $s=d_\theta (u,v)$. By theorem 5.25.2, there is $C_\theta $ bounding the $L^1$ norm of every $\theta $-plurisubharmonic potential with supremum zero. Since ordinary models have supremum zero, the estimate holds if $s\geq \delta /2$. Suppose $0<s<\delta /2$ and put $t=(2s/\delta )^{1/n}$. The upper-volume deficit from $u$ to $m$ is at most $s$, and the lower volume of $m$ is at least $\delta $. Thus lemma 13.18.1, with $b=t^{-1}$, gives $h\in \mathrm{PSH}(X,\theta )$ such that

For each competitor $g$ for $P_\theta [h]$ (so $g\leq 0$ and $g\preceq h$), the potential $(1-t)m+tg$ lies below zero and is more singular than $u$. Since $u=P_\theta [u]$, it lies below $u$, and taking the envelope gives

Hence $0\leq m-u\leq t(m-P_\theta [h])\leq -tP_\theta [h]$, and $\int _ X(m-u)\, \mathrm{d}V\leq tC_\theta $ because $P_\theta [h]$ has supremum zero. The same argument applies to $v$, and adding the two estimates gives the assertion, since $|u-v|\leq (m-u)+(m-v)$. If $s=0$, the argument works for every $t\in (0,1)$; let $t\downarrow 0$. $\square$