Proposition 18.26.236136A0B71CE1

Proof. Taking the infimum and the supremum in proposition 18.26.21 gives the bounds for the singularity volumes of $tH$. The lower volume is positive, so by theorem 13.19.5 ordinary projection preserves both volumes. Since $tH\leq p_ t\leq 0$ and $\sup _ XtH=0$, the model is normalized.

The type of zero consists of the bounded potentials, so its upper volume is $v_+(\omega )$. Since $p_ t\leq 0$, (13.22.2) gives $d_\omega (p_ t,0)=v_+(\omega )-\overline{\operatorname {vol}}_\omega (p_ t)\geq Mt^ n$. In particular $p_ t\ne 0$.

The part of the proof of proposition 18.26.17 that uses supports and Lelong numbers applies to this model: its pole set lies in $S$, its polar restriction is closed and positive, and every exceptional divisor whose image lies in $S$ maps to a point. Since the curve potential $g$ has zero Lelong numbers, every divisorial coefficient vanishes. The lower bound $p_ t\geq tH$ and Skoda's integrability theorem give trivial multiplier ideals at all weights. Hence the ideal envelope of $p_ t$ is the zero model, while $p_ t\ne 0$. All first polar comparisons with zero vanish, which proves that $p_ t$ is not an ideal model and lies in the closure of the analytic models.

Given $0<\delta <v_-(\omega )$, choose $t_0<1/C$ with $Mt_0^ n<v_-(\omega )-\delta $. The lower bound for the volume then puts zero and every $p_ t$ with $0<t\leq t_0$ in $\mathcal M_\delta (\omega )$.

On a compact complex surface, a Gauduchon form has finite upper mass by theorem 11.4.2, theorem 11.1.7. By the metric independence in proposition 13.14.15, every Hermitian form then has finite upper mass, and corollary 12.19.15 gives positive lower mass. $\square$