Pome

Proposition 18.4.42506B7AD6F19

Proposition 18.4.4.2506B7AD6F19 For every $u\in \mathrm{PSH}(X,\theta )$,

\[ u-\sup _ Xu\leq P_\theta [u]\leq P_\theta [u]_{\mathcal I}\leq V_\theta . \]

Moreover,

\[ \begin{gathered} P_\theta [u]\sim _{\mathcal I}u,\qquad P_\theta [P_\theta [u]]_{\mathcal I}=P_\theta [u]_{\mathcal I},\\ P_\theta [P_\theta [u]_{\mathcal I}]=P_\theta [u]_{\mathcal I}. \end{gathered} \]

Proof. Let $h\leq 0$ be a competitor for the ordinary envelope, so that $h\preceq u$. The function $\max (h,u-\sup _ Xu)$ is nonpositive, ordinarily equivalent to $u$, and dominates $h$. Since ordinarily equivalent potentials have the same multiplier ideals, it is a competitor for the ideal envelope. Taking regularized suprema gives $P_\theta [u]\leq P_\theta [u]_{\mathcal I}$; the two outer inequalities follow directly from the defining families.

Projection and monotonicity of local integrability now give, for every $\lambda >0$,

\[ \mathcal I(\lambda u) \subseteq \mathcal I(\lambda P_\theta [u]) \subseteq \mathcal I(\lambda P_\theta [u]_{\mathcal I}) =\mathcal I(\lambda u). \]

Thus the ordinary envelope preserves all multiplier ideals, and it has the same ideal envelope as $u$. For $q=P_\theta [u]_{\mathcal I}$, the inequalities already proved and idempotence give $q\leq P_\theta [q]\leq P_\theta [q]_{\mathcal I}=q$. $\square$

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