Pome

Corollary 15.13.324CD508AEB8D

Corollary 15.13.3.24CD508AEB8D Let $u,v\in \mathrm{PSH}(X,\theta )$ satisfy $u,v\geq \rho -C$ for some $C\geq 0$ and $u\sim v$.

  1. If $\mathrm{MA}_\theta (u)\leq c\mathrm{MA}_\theta (v)$ for some $c\geq 0$, then $c\geq 1$.

  2. If $\mathrm{MA}_\theta (u)\leq \mathrm{e}^{\lambda (u-v)}\mathrm{MA}_\theta (v)$ for some $\lambda >0$, then $u\geq v$.

In particular, let $\mu $ be a positive measure. Among potentials bounded below by $\rho $ minus a constant, each ordinary singularity type contains at most one solution of $\mathrm{MA}_\theta (u)=\mathrm{e}^{\lambda u}\mu $, and the constant $c>0$ in $\mathrm{MA}_\theta (u)=c\mu $ is uniquely determined within each type.

Proof. If $c<1$, apply the theorem to $u$ and $v+A$ for every constant $A>0$; the constants in its hypotheses may depend on $A$. The resulting inequality $u\geq v+A$ for all $A$ is impossible. For the second assertion, fix $a>0$ and restrict the measure inequality to $\{ u<v-a\} $. There the factor is at most $\mathrm{e}^{-\lambda a}<1$, so the theorem gives $u\geq v-a$; now let $a\downarrow 0$. For two solutions of the exponential equation, apply the second assertion in both orders; for two solutions of the normalized equation, apply the first. $\square$

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