Theorem 12.34.1FC99385C18B8

Proof. If $\theta +\mathrm{dd}^{\mathrm{c}}u\geq ah$ with $a>0$ and $u$ smooth, then integration by parts gives $\int _ X\theta \wedge T\geq a$ for every $T\in \mathscr G_ h$. Since a smooth positive representative satisfies such a bound for some $a>0$ by compactness, this proves one implication.

For the converse, work in the real Fréchet space of smooth real $(1,1)$-forms. Let $\mathcal K$ be its open convex cone of strictly positive forms, and let $\mathcal L=\mathrm{dd}^{\mathrm{c}}C^\infty (X,\mathbb R)$. If the affine subspace $\theta +\mathcal L$ is disjoint from $\mathcal K$, the Hahn–Banach separation theorem for an open convex set and a disjoint affine subspace gives a nonzero continuous real linear functional $T$ with

Replacing $\ell $ by arbitrary real multiples shows that $T(\mathcal L)=0$. Scaling a fixed $\eta \in \mathcal K$ to zero shows that $T(\theta )\leq 0$, while scaling it to infinity shows that $T(\eta )\geq 0$. Since every smooth semipositive form is a limit of strictly positive forms, $T$ is a positive current, and positivity makes it a current of order zero. The equality $T(\mathcal L)=0$ says exactly that $\mathrm{dd}^{\mathrm{c}}T=0$. Since $T\ne 0$ and $h>0$, the trace $T(h)$ is positive, and normalizing $T$ gives an element of $\mathscr G_ h$ with nonpositive pairing against $\theta $. This contradicts $m_ h(\theta )>0$ and proves the converse.

Applying the equivalence just proved to $\theta -ah$, and using $m_ h(\theta -ah)=m_ h(\theta )-a$, shows that the set in the supremum is exactly $(-\infty ,m_ h(\theta ))$. Finally, $\theta $ is nef if and only if $\theta +\varepsilon h$ has a smooth positive representative for every $\varepsilon >0$, that is, if and only if $m_ h(\theta )+\varepsilon >0$ for all such $\varepsilon $, or equivalently $m_ h(\theta )\geq 0$. $\square$