Theorem 12.35.47B32818E6010

Proof. Existence. The smooth equation and theorem 12.2.3 give a smooth $v$ with

Then $\theta +\mathrm{dd}^{\mathrm{c}}(\sigma _ D+v)=[D]+\Omega \geq 0$. Off $D$, the extended measure of $u_0=\sigma _ D+v$ is $\Omega ^ n$, and both measures give zero mass to $D$. This proves existence.

Reduction to the residual background. Let $u$ satisfy the two inequalities. Removing the divisor as in the proof of proposition 12.35.1, we obtain

where the difference is first taken in $L^1$ and then replaced by its canonical representative. On $X\setminus D$, the two currents differ only by smooth changes in the decomposition of their local potentials, so their extended measures agree there, and both measures vanish on $D$. Hence

Rigidity. Take $\mathrm{d}V=\mu $. The zero function solves the smooth equation $\Omega ^ n=c\mu $ in the background $\Omega $, so its subsolution threshold is $b_\Omega (1)=c$, by proposition 15.28.7. For $n\geq 2$, the normalized $(n-1)$st power of the Gauduchon metric conformal to $\Omega $ is a smooth strictly positive maximizer of the dual problem. By corollary 12.10.12, any two threshold subsolutions differ by a constant; comparing $w$ with zero shows that $w$ is constant. For $n=1$, the positive current $\Omega +\mathrm{dd}^{\mathrm{c}}w$ dominates its extended measure, hence dominates $c\mu =\Omega $, while both have total mass $\int _ X\Omega =c$ by Stokes' theorem. Thus $\mathrm{dd}^{\mathrm{c}}w=0$, and $w$ is again constant.

Solutions in the ordinary type. Finally, let $u$ be in the ordinary singularity type of $\sigma _ D$ with $\mathrm{MA}^{\mathrm{ext}}_\theta (u)=c'\mu $ for some $c'\geq 0$. Then $\nu (u,D_ j)=a_ j$, and the same reduction gives a bounded $w\in \mathrm{PSH}(X,\Omega )$ with $\mathrm{MA}_\Omega (w)=c'\mu $. Suppose $c'<c$. For every constant $C$, the bounded functions $w$ and $C$ satisfy $\mathrm{MA}_\Omega (w)=(c'/c)\mathrm{MA}_\Omega (C)$, so theorem 14.19.10 gives $w\geq C$, which is impossible for a bounded $w$. The case $c'>c$ is symmetric, with the roles of $w$ and the constants exchanged. Therefore $c'=c$, and the rigidity just proved gives $u=u_0+\mathrm{constant}$. $\square$