Theorem 17.5.68972585EBF32

Proof. Put $\nu =g\, \mathrm dV$, a non-pluripolar measure of mass $m$. Every $\theta $-psh potential is integrable for $\mathrm dV$, hence for $\nu $, so $\mathcal D_\nu =\mathcal E^1(X,\theta ,\phi )$; we write $F=F_\nu $. By the shift formula for the energy and $\nu (X)=m$, the functional $F$ is invariant under adding constants, so we may restrict it to potentials $v$ with $\sup _ Xv=0$. For such $v$, lemma 16.22.2 gives $v\leq \phi $, hence $E(v)\leq 0$. Since $\mathrm dV\leq c_ V\, \omega ^ n$ for a constant $c_ V$, proposition 10.3.1 shows that $\int _ X|v|\, \mathrm dV$ is bounded independently of $v$, and $\phi \in L^1(\mathrm dV)$, so

with $C$ independent of $v$. Consequently $\sup F\in [0,C]$, since $F(\phi )=0$ and $F(v)\leq 0+C$.

Let $(v_ j)$ be a normalized maximizing sequence with $F(v_ j)\geq -1$. Then $E(v_ j)=F(v_ j)+\int _ X(v_ j-\phi )g\, \mathrm dV\geq -C-1$. By corollary 17.4.5 a subsequence converges in $L^1(X)$ to some $u\in \mathcal E^1(X,\theta ,\phi )$, and $\sup _ Xu=0$, because the supremum is continuous for $L^1$ convergence of potentials (proposition 10.3.1). The linear term $v\mapsto \int _ X(v-\phi )g\, \mathrm dV$ is continuous for $L^1$ convergence because $g$ is bounded, and the energy is upper semicontinuous by proposition 17.4.4. Hence $F(u)\geq \varlimsup _ jF(v_ j)=\sup F$, so $u$ is a maximizer, and theorem 17.5.5 gives $\mu _ u=\nu $. $\square$