15.13 Comparison above a strict singular barrier1FCAFCAE9C0E

Proof. Fix $\varepsilon >0$ and set $v_\varepsilon =\max (u,v-\varepsilon )$, which equals $u$ near the boundary. We claim that

throughout $B$. On $\{ u>v-\varepsilon \} $ this follows from locality, since $v_\varepsilon =u$ there. On $\{ u\leq v-\varepsilon \} $ it follows from ordered contact and from the assumed inequality, which applies because this set is contained in $\{ u<v\} $.

Let $q$ be the plurisubharmonic envelope on $B$ of the obstacle $u-v_\varepsilon $, in the sense of (5.27.8). This obstacle is bounded, quasi-continuous and zero near the boundary. Hence it lies between $Ah$ and zero for a sufficiently large $A$, where $h<0$ is a smooth strictly plurisubharmonic defining function for the ball. Thus $q$ tends to zero at the boundary, and (5.27.8) shows that $(\mathrm{dd}^{\mathrm{c}}q)^ n$ is carried by $D=\{ q=u-v_\varepsilon \} $.

Since $q\leq u-v_\varepsilon $ quasi-everywhere, the canonical representatives satisfy $q+v_\varepsilon \leq u$. Ordered contact, followed by the measure inequality above, yields

The positive expansion of $(\theta _{v_\varepsilon }+\mathrm{dd}^{\mathrm{c}}q)^ n$ contains both $\mathrm{MA}_\theta (v_\varepsilon )$ and $(\mathrm{dd}^{\mathrm{c}}q)^ n$. Thus $\mathbf1_ D(\mathrm{dd}^{\mathrm{c}}q)^ n=0$, and since $(\mathrm{dd}^{\mathrm{c}}q)^ n$ is carried by $D$, we get $(\mathrm{dd}^{\mathrm{c}}q)^ n=0$. The classical Bedford–Taylor comparison principle for bounded plurisubharmonic functions with zero boundary values implies $q=0$. Consequently $u\geq v_\varepsilon $, first quasi-everywhere and then everywhere. Finally let $\varepsilon \downarrow 0$. $\square$

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 $X$. $\square$

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. $\square$