Proposition 16.3.25.F7C0BACE770D Assume that $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$ and
E6B68D406648
16.3.26
Then $\varphi \land \psi \in \mathrm{PSH}(X,\theta )$.
F7C0BACE770DProposition 16.3.25.F7C0BACE770D Assume that $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$ and
E6B68D406648
16.3.26
Then $\varphi \land \psi \in \mathrm{PSH}(X,\theta )$.
Proof. Without loss of generality, we may assume that $\varphi ,\psi \leq 0$. For simplicity, we write
Note that due to our assumption (16.3.26), $\varphi ,\psi \in \mathrm{PSH}(X,\theta )_{>0}$, and hence so is $\varphi \lor \psi $ thanks to the monotonicity theorem theorem 13.13.11. By corollary 16.3.20, $\eta $ is a model potential, and $\int _ X\theta _\eta ^ n=\int _ X\theta _{\varphi \lor \psi }^ n>0$.
Take $C>0$ large enough, so that
E47535691487
16.3.27
This is possible thanks to proposition 13.11.4 4. Fix $C'>C$. Write
Then observe that $\gamma _{C'}\sim \eta $, and
Suppose, for contradiction, that $\varphi \land \psi \equiv -\infty $. Then
Thanks to proposition 16.3.21, for each $C'>C$,
since $\eta $ is a model potential. It follows that
74C044AB05CB
16.3.28
In particular, we can take $C'$ large enough so that
For such $C'$, we have
where the second line follows from 16.3.3, the fourth line follows from (16.3.27). This contradicts the fact that $\gamma _{C'}\sim \eta $ in view of theorem 13.13.11. $\square$
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