Theorem 18.27.40E6CABEB995A

Proof. Suppose first that $v\sim \sigma _\Delta +q$ and $\theta +\mathrm{dd}^{\mathrm{c}}v\geq \epsilon h$ with $\epsilon >0$. By lemma 18.27.2, $\nu (v,D_ j)=a_ j$. Apply lemma 5.38.1 with background $\theta -\epsilon h$ to obtain

For $\Delta =0$ this is the original strict inequality. The $L^1$ difference $v-\sigma _\Delta $ consequently has a global quasi-plurisubharmonic representative $q'$ with $B+\mathrm{dd}^{\mathrm{c}}q'\geq \epsilon h$. Off the divisor it satisfies $q'\sim q$; these inequalities extend everywhere by lemma 5.24.3. Thus $q'$ has zero point Lelong numbers, and lemma 18.27.3 gives $B+\mathrm{dd}^{\mathrm{c}}f>0$ for a smooth $f$.

Conversely, compactness gives $B+\mathrm{dd}^{\mathrm{c}}f\geq \epsilon h$ for some $\epsilon >0$. Choose $q_\delta $ in (18.27.1) with $\delta <\epsilon /2$. Then $v=\sigma _\Delta +q_\delta +f$ is in the specified ordinary type, and Poincaré–Lelong gives