Theorem 15.13.2F71978C42841

Proof. We first modify the barrier. Let $r=a\rho -A$, where $a>1$ is sufficiently close to one and $A$ is sufficiently large. Since

we can arrange $\theta _ r\geq \omega /2$ and $r-v\leq -1$, and moreover $u-r\to +\infty $ at $Z$, because $u-r\geq (a-1)(-\rho )+A-C_0$. When $Z$ is empty, the last property is not needed.

Choose $0<\delta <1$ with $(1-\delta )^ n>c$, and put $v_\delta =(1-\delta )v+\delta r$. Then

Suppose that the infimum $m$ of $u-v_\delta $ on $\Omega $ were negative. Since $u-v_\delta \to +\infty $ at $Z$, a sequence $x_ j$ along which $u-v_\delta $ tends to $m$ has a subsequence converging to a point $x_0\in \Omega $. Choose a coordinate ball $B\Subset \Omega $ centered at $x_0$, identify it with $\{ |z|<1\} $, and take

The function $w=v_\delta +\delta \eta +m$ is $\theta $-plurisubharmonic on $B$, and $\liminf _{\partial B}(u-w)\geq 0$ because $\eta $ vanishes on $\partial B$. Also

Thus $\{ u<w\} \subset \{ u<v\} $, and positivity of mixed products gives on this set

By lemma 15.13.1, $u\geq w$ on $B$. Evaluating at $x_ j$ gives $u(x_ j)-v_\delta (x_ j)\geq m+\delta \eta (x_ j)$, which contradicts $\eta (x_0)>0$ in the limit. Hence $u\geq v_\delta $ on $\Omega $. Letting $\delta \downarrow 0$ along values with $(1-\delta )^ n>c$, we obtain $u\geq v$ on $\Omega $, and lemma 5.24.3 extends this inequality to $X$. $\square$