Pome

Proposition 16.6.11363E28FF4A61

Proposition 16.6.11.363E28FF4A61 Let $\varphi \in \mathrm{PSH}(X,\theta )$. Then

\[ \varphi \sim _{\mathcal{I}} P_{\theta }[\varphi ]_{\mathcal{I}}. \]

In particular,

\[ P_{\theta }\Bigl [ P_{\theta }[\varphi ]_{\mathcal{I}} \Bigr ]_{\mathcal{I}}=P_{\theta }[\varphi ]_{\mathcal{I}} \]

and the upper semicontinuous regularization in (16.6.8) is not necessary.

Proof. In view of proposition 16.6.2, it suffices to show that for $k \in \mathbb {Z}_{>0}$, we have

5888DEC3C930 16.6.12
\begin{equation} \mathcal{I}(k\varphi )=\mathcal{I}(kP_{\theta }[\varphi ]_{\mathcal{I}}). \end{equation}

By proposition 5.14.3, we can find $\psi _ i\in \mathrm{PSH}(X,\theta )$ ($i\in \mathbb {Z}_{>0}$) such that $\psi _ i\leq 0$, $\psi _ i\sim _{\mathcal{I}}\varphi $ for all $i\geq 1$ and

\[ \operatorname *{sup}_{i>0}{}\! {}^{*}\psi _ i=P_{\theta }[\varphi ]_{\mathcal{I}}. \]

By proposition 16.6.5, we may replace $\psi _ i$ by $\psi _1\lor \cdots \lor \psi _ i$ and assume that the sequence $\psi _ i$ is increasing. In this case, it follows from the strong openness theorem theorem 5.16.21 that for each $k\in \mathbb {Z}_{>0}$, we have

\[ \mathcal{I}(k\varphi )=\mathcal{I}(k\psi _ j)=\mathcal{I} (kP_{\theta }[\varphi ]_{\mathcal{I}}) \]

for $j$ large enough. $\square$

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.