PomeCorollary 15.36.11.E75339160354 Every separated threshold subsolution $r=r(y)$ with $r\in \mathrm{PSH}(Y,\kappa )$ and $\mathrm{MA}^{\mathrm{ext}}_ H(r)\geq bq_\alpha \mu $ equals $u_\alpha +d$ for some constant $d$. In particular $u_\alpha $ is its unique supremum-zero member. If $\alpha <k/(k+1)$, then $u_\alpha -M_\alpha $ is the greatest such subsolution whose $q_\alpha $-mean on $Y$ is nonpositive.
Corollary 15.36.11E75339160354
Proof. The product formula reduces the inequality to $\mathrm{MA}^{\mathrm{ext}}_\kappa (r)\geq q_\alpha \kappa ^ k$. Closed Stokes gives each bounded truncation mass one, so the extended measure of $r$ has mass at most one. The inequality therefore gives equality and full mass. Closed-class uniqueness yields $r=u_\alpha +d$. In the finite-moment range its mean is $M_\alpha +d$, which is nonpositive exactly when $d\leq -M_\alpha $. $\square$
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