Pome011162B56B40Theorem 17.5.5.011162B56B40 Let $\nu $ be a non-pluripolar measure on $X$, and suppose that $\mathcal D_\nu =\mathcal E^1(X,\theta ,\phi )\cap L^1(\nu )$ is nonempty. Then $\phi \in L^1(\nu )$, and for the functional
a potential $u\in \mathcal D_\nu $ maximizes $F_\nu $ if and only if $\mu _ u=\nu $. In particular, $F_\nu $ has no maximizer unless $\nu (X)=m$.
Proof. Let $v\in \mathcal D_\nu $. Then $v\leq \phi +C$ for a constant $C$, and $\phi \leq 0$. Thus the negative part of $\phi $ is at most that of $v$ plus $C$, and the positive part of $\phi $ vanishes; hence $\phi \in L^1(\nu )$. If $\mu _ u=\nu $, the tangent bound of lemma 17.4.10 gives
that is, $F_\nu (v)\leq F_\nu (u)$.
Conversely, let $u$ be a maximizer and $f\in C(X,\mathbb R)$. By proposition 17.5.4, the potentials $u_ t=P_\theta (u+tf)$ have bounded difference from $u$ and finite energy, so they belong to $\mathcal D_\nu $; moreover $u_0=u$ and $u_ t\leq u+tf$ everywhere, hence $\nu $-almost everywhere. Therefore
The left side vanishes at $t=0$ and is differentiable there by proposition 17.5.4; since it attains its maximum at $t=0$, its derivative vanishes there. Thus $\int _ Xf\, \mu _ u=\int _ Xf\, \mathrm d\nu $ for every $f\in C(X,\mathbb R)$, and lemma 17.5.3(1) gives $\mu _ u=\nu $. Since $\mu _ u(X)=m$, a maximizer can exist only if $\nu (X)=m$. $\square$
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