Lemma 5.24.1.729BA86BCC7C If $u,v\in \mathrm{PSH}(X,\theta )$, then $\max (u,v)\in \mathrm{PSH}(X,\theta )$. Consequently, if $\theta \geq 0$ and $u\in \mathrm{PSH}(X,\theta )$, then $\max (u,-t)$ is $\theta $-plurisubharmonic for every real $t$.
Lemma 5.24.1729BA86BCC7C
Proof. Adding a common local smooth correction to $u$ and $v$ shows that $\max (u,v)$ is quasi-plurisubharmonic. To obtain the sharp lower bound for its Hessian, fix a positive constant $(1,1)$-form $\kappa $ on a coordinate ball. For each $\varepsilon >0$, every point of the ball has a neighborhood on which $\theta $ is close to a constant form $\theta '$, in the sense that $-\varepsilon \kappa \leq \theta -\theta '\leq \varepsilon \kappa $. Write $\theta '+\varepsilon \kappa =\mathrm{dd}^{\mathrm{c}}r$ with a quadratic polynomial $r$. Then $\mathrm{dd}^{\mathrm{c}}(u+r)\geq \theta +\mathrm{dd}^{\mathrm{c}}u\geq 0$, so $u+r$ and, likewise, $v+r$ are plurisubharmonic; hence so is their maximum. It follows that
on each such neighborhood, and letting $\varepsilon \to 0$ gives $\theta +\mathrm{dd}^{\mathrm{c}}\max (u,v)\geq 0$. The last assertion follows because constants belong to $\mathrm{PSH}(X,\theta )$ when $\theta \geq 0$. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.