Proposition 17.7.106DDFC01E2315

Proof. For each $s$ the potential $u_1-s$ belongs to $\mathcal E(X,\theta ,\phi )$, so the rooftop $P_\theta (u_0,u_1-s)$ is not identically $-\infty $, by proposition 17.4.7, and the affine curve $P_\theta (u_0,u_1-s)+ts$ is a competitor: its value at $t=0$ is at most $u_0$ and its value at $t=1$ is at most $u_1$. Hence the right side is at most $G_ t$.

Conversely, let $(v_ t)$ be a competitor and $s\in \mathbb R$. By corollary 5.12.2, applied with the affine function $\ell (t)=ts$ in charts after adding smooth local potentials of $p^*\theta $, the function $q_ s=\inf _{0<t<1}(v_ t-ts)$ is $\theta $-psh or identically $-\infty $ on each chart. By lemma 17.7.3, for almost every $x$ the function $t\mapsto v_ t(x)$ is finite and convex, hence bounded below on $(0,1)$ by an affine function (lemma 17.7.2(2)), so $q_ s$ is finite almost everywhere and therefore $\theta $-psh. Letting $t\to 0$ and $t\to 1$ in $q_ s\leq v_ t-ts$, and using the $L^1$ convergence of $v_ t$ to its endpoint potentials, which are at most $u_0$ and $u_1$, we get $q_ s\leq u_0$ and $q_ s\leq u_1-s$ almost everywhere, hence everywhere, and therefore $q_ s\leq P_\theta (u_0,u_1-s)$.

Now fix $x$ and $t_0\in (0,1)$. By lemma 17.7.3(1), the fiber $t\mapsto v_ t(x)$ is either identically $-\infty $, in which case $v_{t_0}(x)=-\infty $, or finite. In the second case lemma 17.7.2(2) gives $c\in \mathbb R$ with $v_ t(x)\geq v_{t_0}(x)+c(t-t_0)$ for all $t$, so that $q_ c(x)=\inf _ t(v_ t(x)-ct)=v_{t_0}(x)-ct_0$, and hence $v_{t_0}(x)=q_ c(x)+ct_0\leq \sup _ s(q_ s(x)+t_0s)$. In either case $v_{t_0}(x)\leq \sup _ s\{ P_\theta (u_0,u_1-s)(x)+t_0s\} $, since $q_ s\leq P_\theta (u_0,u_1-s)$. Taking the regularized supremum over all competitors and using lemma 17.7.6 proves the formula. The same lemma, applied to the affine competitors, gives the last assertion. $\square$