PomeCorollary 11.45.13.86F268014BE0 For the fixed data, $c_0$ is the largest coefficient $c>0$ for which $u\in \mathrm{PSH}(X,\theta )$ with $u\preceq \phi _{a,d}$ satisfies $\mathrm{MA}^{\mathrm{ext}}_\theta (u)\geq c f\, \mathrm{d}W$. Among separated potentials $u=\psi +r(x)\preceq \phi _{a,d}$, the critical subsolutions are exactly the translates of $\psi +w_0$. Thus $v_0$ is their unique supremum-zero member, and $s_0$ is their greatest member with $\int _ Xu\, \mathrm{d}\nu _0\leq 0$.
Corollary 11.45.1386F268014BE0
Proof. For a separated critical subsolution, (11.45.6) and the product expansion give $\mathrm{MA}^{\mathrm{ext}}_{a\omega _1}(r)\geq A_0\omega _1$. Closed curve truncations have mass $a$, so the non-pluripolar measure has mass at most $a$. Since $\int A_0\omega _1=a$, it must equal $A_0\omega _1$. The positive closed current $a\omega _1+\mathrm{dd}^{\mathrm{c}}r$ also has mass $a$, so it has no remaining measure on the pole set. Consequently $\mathrm{dd}^{\mathrm{c}}r=(A_0-a)\omega _1$ as distributions, and $r=w_0+C$ because harmonic distributions on $\mathbb P^1$ are constant. Supremum zero gives $C=-S_0$, while the mean condition gives $C\leq -J$.
For a general $u\preceq \phi _{a,d}$ with coefficient $c>0$, write $u=\psi +r$ by (11.45.6) and put $B_ r=\sup _ Xr$. Since $\psi \leq 0$, the potential $u+\sigma ^{-1}\log c-B_ r$ is an exponential subsolution for every $\sigma >0$. Greatestness gives
\[ u+\sigma ^{-1}\log c-B_ r\leq s_\sigma . \]
At a point off $D$ where $r$ is finite, subtract $\sigma ^{-1}\log c_0$, multiply by $\sigma $ and use the fixed-data case of theorem 11.45.9. This yields $\log c\leq \log c_0$. The potential $v_0$ attains $c_0$. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.