Proposition 13.29.5D21EE85B9F1F

Proof. The form $\theta _1-\lambda \theta _0$ has integral zero, so the scalar Poisson equation gives $h$, as in lemma 5.29.22. Write $p=p_{\theta _0}(\mu )$. Scaling the competitors in the ordinary envelope shows that $\lambda p$ is an ordinary model for $\lambda \theta _0$. The gauge construction in the proof of proposition 13.29.1 therefore makes $P_{\theta _1}[\lambda p-h]$ an ordinary model in background $\theta _1$. The curvature current of $\lambda p-h$ in this background is $\lambda (\theta _0+\mathrm{dd}^{\mathrm{c}}p)$, so its polar measure is $\lambda \mu $. The construction preserves this measure, which gives the first identity in (13.29.6).

The function $\lambda p-h+\min _ Yh$ is a competitor for $q$, which gives the lower bound. If $v\leq 0$ is any competitor for $q$, then $v+h-\max _ Yh$ is $\lambda \theta _0$-plurisubharmonic, nonpositive and more singular than $\lambda p$, so it is at most $\lambda p$. Taking the upper envelope gives the upper bound. These inequalities hold also at the poles, and off the common pole set they imply $|q-\lambda p|\leq \operatorname {osc}_ Yh$.

At zero mass, the polar measures exhaust the curvature currents of both $p$ and $q$. Hence $\mathrm{dd}^{\mathrm{c}}(q-\lambda p+h)=0$, and the uniqueness of harmonic functions together with the normalization $\sup _ Yq=0$ gives the last formula. $\square$