Theorem 16.22.114A1DC8C4E020

Proof. Write $w=P_\theta [u]$, and let $v$ belong to the displayed family. By proposition 16.21.2, $w\leq P_\theta [v]$, and the potentials $w$, $u$, $v$ and $P_\theta [v]$ all have the same positive mass. Moreover $v\leq P_\theta [v]$, because $v\leq 0$ is a competitor for $P_\theta [v]$. Hence $w$ and $v$ both belong to $\mathcal E(X,\theta ,P_\theta [v])$, a class whose reference potential has positive mass. By proposition 16.21.3, $\mu _ w$ is carried by $\{ w=0\} $, and since $v\leq 0$, it gives no mass to $\{ w<v\} $. The domination principle, theorem 16.22.10, yields $w\geq v$. Finally, $w$ itself belongs to the family: it is nonpositive, it satisfies $u-\sup _ Xu\leq w$, and it has the same mass as $u$ by proposition 16.21.2. This proves the formula.

The potential $P_\theta [w]$ is nonpositive, satisfies $u\preceq w\preceq P_\theta [w]$, and has the same mass as $u$ by proposition 16.21.2. Hence it belongs to the family, and the formula gives $P_\theta [w]\leq w$. Conversely, $w\leq 0$ is a competitor for $P_\theta [w]$, so $P_\theta [w]=w$. Since the mass of $w$ is positive, $w$ is a model potential. $\square$