Lemma 15.19.34A0FCD6D9FCD

Proof. Write $\mu =f\, \mathrm{d}V$, where $f$ is smooth and positive. The strict barrier makes $\theta $ big. By lemma 15.3.1, there is a bounded $s\in \mathrm{PSH}(X,\theta )$ with $-1\leq s\leq 0$ and $\mathrm{MA}_\theta (s)\geq b\mu $ for some fixed $b>0$. If $c_\varepsilon <b$, positivity gives

for every constant $A$. Strict bounded domination would imply $u_\varepsilon \geq s+A$ for every $A$, which is absurd. Hence $c_\varepsilon \geq b$.

For an upper bound, choose a fixed Gauduchon form $\gamma $. The arithmetic–geometric mean inequality and Stokes give

The first integral is fixed and positive, giving $c_\varepsilon \leq B$.

For the lower bound on $u_\varepsilon $, note that all $u_\varepsilon $ are normalized $(\theta +\eta )$-plurisubharmonic functions. Uniform exponential integrability, lemma 5.26.2, gives $a>0$ such that $g_\varepsilon =\mathrm{e}^{-a u_\varepsilon }f$ has a common $L^2$ bound. Also $g_\varepsilon \geq f$. By theorem 15.3.3, the solutions

are uniformly bounded. Put $B_0=\max \{ 1,B\} $ and $z_\varepsilon =w_\varepsilon -a^{-1}\log B_0$. Then

The comparison principle for the exponential equation in the background $\theta +\varepsilon \eta $ gives $z_\varepsilon \leq u_\varepsilon \leq 0$, and $z_\varepsilon $ is bounded below uniformly in $\varepsilon $. $\square$