Theorem 12.14.6ECAB6BFB33FD

Proof. The coefficient $a$ is nonnegative, and $\theta +\eta h>0$ for every $\eta >0$; this proves semipositivity and nefness. Since $a$ is nonconstant, $\mathrm{d}\theta =\mathrm{d}a\wedge w\ne 0$.

Suppose that $\theta +\mathrm{dd}^{\mathrm{c}}v\geq \delta h$ for a quasi-plurisubharmonic function $v$ and some $\delta >0$. Choose a nonzero smooth function $f\geq 0$ on $B$ supported in $\{ a<\delta \} $. Its pullback $S=f\omega ^ m$ is a smooth semipositive $(m,m)$-form, closed because it has top degree on the base. Positivity and distributional Stokes would give

This contradiction shows that $\theta $ is not big, since every Hermitian form on $X$ dominates a positive multiple of $h$.

For the lower bound on the volume, pull back the bounded potential $\varphi $ of lemma 12.14.2. The positive current $\Omega =A+a w=\theta +\mathrm{dd}^{\mathrm{c}}\varphi $ is a smooth strictly positive form off $Z=\{ p\} \times \mathbb P^1$, where

Let $\gamma $ be a Gauduchon metric. Since $\varphi $ is bounded, distributional Stokes lets us replace $\theta $ by $\Omega $ in the numerator of its quotient. The set $Z$ has zero smooth volume and $\Omega \wedge \gamma ^{n-1}$ is a positive measure, so the pointwise arithmetic–geometric mean inequality off $Z$ gives

This lower bound holds for every test in definition 12.2.1; no power of the current $\Omega $ along $Z$ is needed.

For the upper bound, put $c_\epsilon =n/H_\epsilon $ and $\Omega _\epsilon =A_\epsilon +(a+\epsilon )w=\theta _\epsilon +\mathrm{dd}^{\mathrm{c}}\varphi _\epsilon $. Then $\Omega _\epsilon ^ n=c_\epsilon \mu $. The metrics in (12.14.8) satisfy

The first term has top degree on the base and the second is a product of closed forms, so $T_\epsilon $ is closed; that is, $\gamma _\epsilon $ is balanced, and in particular Gauduchon. Smooth Stokes and (12.14.5) give

so the quotient of $\gamma _\epsilon $ at $\theta _\epsilon $ equals $c_\epsilon ^{1/n}$. The same arithmetic–geometric mean argument, applied to the smooth representative $\Omega _\epsilon $, bounds every Gauduchon quotient at $\theta _\epsilon $ below by $c_\epsilon ^{1/n}$. This proves that $\gamma _\epsilon $ minimizes, and gives the formula for $\mathcal V_\mu (\theta _\epsilon )$. Since $\theta \leq \theta _\epsilon $, the metrics $\gamma _\epsilon $, used as tests for $\theta $, also give $\lambda _\mu (\theta )\leq c_\epsilon ^{1/n}$. Letting $\epsilon \downarrow 0$ proves (12.14.7). By theorem 12.11.1, this value is also the limit of the volumes of the Hermitian perturbations $\theta +\epsilon h$. $\square$