PomeA3C93A5CDBF4Lemma 16.8.4.A3C93A5CDBF4 Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )_{>0}$. For any Kähler form $\omega $ on $X$, the following are equivalent:
ED4E919B13E0$P_{\theta }[\varphi ]\leq P_{\theta }[\psi ]$;
E3965942668C$P_{\theta +\omega }[\varphi ]\leq P_{\theta +\omega }[\psi ]$.
Proof. 1 $\implies $ 2. Assume 1. Observe that
Taking the $P_{\theta +\omega }$-envelope, we find
where the equality follows from theorem 16.3.15. In particular,
7E95C2DD4B7D
16.8.5
A similar formula holds for $\psi $. So we see that 2 holds.
2 $\implies $ 1. Assume 2. By (16.8.5), we may assume that $\varphi $ and $\psi $ are both model potentials in $\mathrm{PSH}(X,\theta )_{>0}$.
Observe that $\varphi \lor \psi \preceq P_{\theta +\omega }[\psi ]$. It follows that $P_{\theta +\omega }[\varphi \lor \psi ]\leq P_{\theta +\omega }[\psi ]$. The reverse inequality is trivial, so
From the direction we have proved, for any $C\geq 1$,
So by proposition 16.3.10,
Since both sides are polynomials in $C$, the equality extends to $C=0$, namely,
In particular, $\varphi \lor \psi \leq P_{\theta }[\psi ]=\psi $ by (16.3.16). So 1 follows. $\square$
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