Theorem 11.41.6C87E852ED386

Proof. The correspondence lemma and theorem 15.28.4 for the Hermitian form $\beta $ prove the first assertion, including attainment. This assertion allows additional poles and excess or infinite mass. For smooth positive $f$, theorem 14.2.1 gives a smooth $w$ with $\beta +\mathrm{dd}^{\mathrm{c}}w>0$ and $(\beta +\mathrm{dd}^{\mathrm{c}}w)^ n=c_\beta (f)f\, \mathrm dV$. The potential $\varphi +w$ belongs to the ordinary type and hence to all three classes. The class containments and the first assertion give equality of their coefficients; bounded-data normalization identifies $b_\beta (f)=c_\beta (f)$. The exact measure is finite.

It remains to prove strictness. Let $v$ be the smooth solution for $\theta $, and put $\sigma =\theta +\mathrm{dd}^{\mathrm{c}}v$, $\eta =\beta +\mathrm{dd}^{\mathrm{c}}w$. The one-sided subsolution gives $c_\beta (f)\le b_\theta (f)=c_\theta (f)$. If equality held, then on $X\setminus S$ the smooth function $a=\varphi +w-v$ would satisfy

The form $Q$ is smooth and strictly positive, so it defines a real second-order operator $L$ with positive definite principal matrix and no zeroth-order term. Also $a\to -\infty $ near $S$, so its maximum is attained inside a component $U$ of $X\setminus S$. The strong maximum principle makes $a$ constant on $U$. To recall its local argument, put $z=\max _ Ua-a\ge 0$. If a positive region of $z$ had a boundary zero inside $U$, choose a ball in that region touching the zero set, with closure in a coordinate chart. On its outer half-annulus, the barrier $e^{-A|x-y|^2}-e^{-AR^2}$ has strictly positive $L$ for large $A$. A sufficiently small multiple lies below $z$ on the two boundary spheres. A negative interior minimum of the difference would have nonnegative $L$, whereas its equation gives strictly negative $L$. Thus the comparison holds throughout the annulus and gives a positive inward derivative at the touching zero, contradicting $\nabla z=0$. Connectedness propagates the zero set to all of $U$. The boundary of $U$ is nonempty and lies in $S$: otherwise $U$ would be both open and closed in connected $X$, although $S\ne \varnothing $. Constancy then contradicts $a\to -\infty $ at that boundary. This proves strictness; no semipositivity of the divisor's smooth curvature representative is required. $\square$