Proposition 14.19.34.880520C4A2B1 Suppose $u\in \mathrm{PSH}(X,\omega )\cap C(X)$ and $c>0$ satisfy $\omega _ u^ n=cf\, \mathrm{d}V$. Set $m=\inf _ Xu$ and $M=\sup _ Xu$. Then, for every $\lambda >0$, the solution of (14.19.4) satisfies
3B8781BD8BF5
14.19.35
\begin{equation} u+\frac{\log c}{\lambda }-M \leq u_\lambda \leq u+\frac{\log c}{\lambda }-m. \end{equation}
In particular,
\[ \| \lambda u_\lambda -\log c\| _\infty \leq \lambda (M-m), \qquad \mathrm{e}^{\lambda u_\lambda }\longrightarrow c \quad \hbox{uniformly on }X\hbox{ as }\lambda \downarrow 0. \]
Consequently, $c$ is the only constant for which the equation $\omega _ v^ n=cf\, \mathrm{d}V$ has a solution $v\in \mathrm{PSH}(X,\omega )\cap C(X)$.
Proof.
Set $a=u+\lambda ^{-1}\log c-M$ and $b=u+\lambda ^{-1}\log c-m$. These translates of $u$ have the same Monge–Ampère measure as $u$, and since $m\leq u\leq M$,
\[ \mathrm{e}^{\lambda a}f\, \mathrm{d}V \leq cf\, \mathrm{d}V=\omega _ a^ n=\omega _ b^ n \leq \mathrm{e}^{\lambda b}f\, \mathrm{d}V. \]
Thus $\mathrm{e}^{-\lambda a}\omega _ a^ n\geq f\, \mathrm{d}V=\mathrm{e}^{-\lambda u_\lambda }\omega _{u_\lambda }^ n\geq \mathrm{e}^{-\lambda b}\omega _ b^ n$. Applying corollary 14.19.12 first to the pair $(u_\lambda ,a)$ and then to the pair $(b,u_\lambda )$ gives $a\leq u_\lambda \leq b$, which is (14.19.35). Multiplying these inequalities by $\lambda $ and using $m\leq u\leq M$ gives the norm bound, and the uniform convergence follows.
If $(v,c')$ is another such pair, the same argument applied to $v$ gives $\lambda u_\lambda \to \log c'$, because $u_\lambda $ depends only on $f$ and $\lambda $. Hence $c'=c$.
$\square$
Pome
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