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Theorem 14.19.10C0DEE578BC5D

Theorem 14.19.10.C0DEE578BC5D Let $u,v\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$, and let $0\leq c<1$. If

\[ \mathbf1_{\{ u<v\} }\omega _ u^ n \leq c\, \mathbf1_{\{ u<v\} }\omega _ v^ n, \]

then $u\geq v$. In particular, $\omega _ u^ n(\{ u<v\} )=0$ implies $u\geq v$.

Proof. Suppose $m=\inf _ X(u-v)<0$. Choose $\varepsilon >0$ so small that

\[ (1-\varepsilon )^ n>c, \qquad 2\varepsilon \| v\| _\infty <-m/4, \]

omitting the second restriction if $v=0$. Let $m_\varepsilon $ and $U_{\varepsilon ,s}$ be defined as in theorem 14.19.6 for this pair $u,v$. Since $|m_\varepsilon -m|\leq \varepsilon \| v\| _\infty $, for all sufficiently small $s>0$ the set $U=U_{\varepsilon ,s}$ is contained in $\{ u<v\} $: there

\[ u-v<m_\varepsilon +s-\varepsilon v \leq m+2\varepsilon \| v\| _\infty +s<0. \]

The set $U$ is nonempty by the definition of the infimum and has positive volume by lemma 5.24.4.

Decrease $s$ further so that theorem 14.19.6 applies and $(1-4B_0n^2s\varepsilon ^{-3})^ n(1-\varepsilon )^ n>c$. Write the first factor as $q^ n$. Positivity of every mixed term gives

\[ \omega _{(1-\varepsilon )v}^ n \geq (1-\varepsilon )^ n\omega _ v^ n+\varepsilon ^ n\omega ^ n. \]

Modified comparison and the assumption would therefore imply

\[ q^ n(1-\varepsilon )^ n\int _ U\omega _ v^ n +q^ n\varepsilon ^ n\int _ U\omega ^ n \leq \int _ U\omega _ u^ n\leq c\int _ U\omega _ v^ n. \]

All integrals are finite, so the choice $q^ n(1-\varepsilon )^ n>c$ and the positive volume of $U$ make this impossible. Hence $m\geq 0$. $\square$

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