Proof.
Translate all $u_ j$ by one constant so that $u_ j\leq \varphi \leq 0$; this changes $A$ by a fixed factor. Translate each $\psi _ j$ so that it takes values in $[-2L,0]$; this does not change its measure. Set $q=P_{\varepsilon \theta }(\varphi )$ and choose $\rho \in \mathrm{PSH}(X,\theta )$ and $a>0$ such that
\[ q\leq \varphi , \qquad \rho \leq 0,\qquad \theta +\mathrm{dd}^{\mathrm{c}}\rho \geq a\omega . \]
Here $q\not\equiv -\infty $ by the model condition, and $\rho $ exists because $\theta $ is big. Fix $0<\delta <(1-\varepsilon )a$.
For a fixed real $b\geq 1$, put $r_{j,b}=Q_ b(u_ j,\varphi )$. The stretching lemma puts it in $\mathcal E_\varphi $, and $r_{j,b}\leq u_ j\leq 0$. Contact support and positivity imply
\[ \mu _{r_{j,b}} \leq b^ n\mathbf1_{D_{j,b}}\mu _{u_ j}, \qquad D_{j,b}=\{ b^{-1}r_{j,b}+(1-b^{-1})\varphi =u_ j\} . \]
On this contact set, $\varphi \leq 0$ gives $u_ j\leq r_{j,b}/b$. Thus (15.37.22) yields
53CEBC73788B
15.37.23
\begin{equation} \mathrm{e}^{-\lambda r_{j,b}/b}\mu _{r_{j,b}} \leq b^ n A\omega _{\psi _ j}^ n, \end{equation}
where the weights are evaluated off the pluripolar pole sets.
Choose $C_ b\geq 0$ so large that
\[ \delta ^ n\exp (\lambda C_ b/b)\geq b^ n A, \]
and define
\[ s_{j,b}=q+(1-\varepsilon )\rho +\delta \psi _ j-C_ b. \]
This is a $\theta $-plurisubharmonic function, not identically $-\infty $, since
\[ \begin{split} \theta +\mathrm{dd}^{\mathrm{c}}s_{j,b} & = (\varepsilon \theta +\mathrm{dd}^{\mathrm{c}}q) +(1-\varepsilon )(\theta +\mathrm{dd}^{\mathrm{c}}\rho ) +\delta \mathrm{dd}^{\mathrm{c}}\psi _ j\\ & \geq \delta \omega _{\psi _ j}. \end{split} \]
Consequently $\mu _{s_{j,b}}\geq \delta ^ n\omega _{\psi _ j}^ n$: this inequality holds away from the poles by the local positive product inequality, and then globally because the bounded reference measure does not charge pluripolar sets. Also $s_{j,b}\leq -C_ b$ and $s_{j,b}\preceq \varphi $. Together with (15.37.23), we conclude
\[ \mathrm{e}^{-\lambda s_{j,b}/b}\mu _{s_{j,b}} \geq b^ n A\omega _{\psi _ j}^ n \geq \mathrm{e}^{-\lambda r_{j,b}/b}\mu _{r_{j,b}}. \]
If $v_-(\theta )>0$, apply corollary 15.37.11 with first potential $r_{j,b}$ and second potential $s_{j,b}$. It gives $r_{j,b}\geq s_{j,b}$.
In the second alternative, $r_{j,b}\sim \varphi $: if $u_ j\geq \varphi -K_ j$, the potential $\varphi -bK_ j$ is a competitor for $r_{j,b}$. Hence the smaller-background envelope $P_{\varepsilon \theta }(r_{j,b})$ is not identically $-\infty $, and $s_{j,b}\preceq r_{j,b}$. Therefore corollary 15.2.3 also gives $r_{j,b}\geq s_{j,b}$, without a lower-mass hypothesis.
Since $\psi _ j\geq -2L$, we now have a lower bound independent of $j$:
\[ H_ b=q+(1-\varepsilon )\rho -2\delta L-C_ b \leq r_{j,b}. \]
The potential $H_ b$ is $\theta $-plurisubharmonic and not identically $-\infty $. By the definition of $r_{j,b}$, therefore,
\[ b^{-1}H_ b+(1-b^{-1})\varphi \leq u_ j \quad \hbox{for every }j. \]
For $b=1$ this shows that the infimum envelope $h$ is not identically $-\infty $. For general $b$, the left side is a competitor for $h$, so it lies below $h$; it is thus a competitor in the defining condition of the relative class for $b$. Since $h\leq \varphi $, we obtain $h\in \mathcal E_\varphi $. Undoing the common translation finishes the proof.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.