Proposition 18.4.6.EAFF378DAF48 For $u,v\in \mathrm{PSH}(X,\theta )$,
\[ u\preceq _{\mathcal I}v\quad \Longrightarrow \quad P_\theta [u]_{\mathcal I}\leq P_\theta [v]_{\mathcal I}. \]
For $a>0$ one has $P_{a\theta }[a u]_{\mathcal I}=aP_\theta [u]_{\mathcal I}$. For smooth real $(1,1)$-forms $\theta _ i$ and potentials $u_ i\in \mathrm{PSH}(X,\theta _ i)$, $i=1,2$,
\[ P_{\theta _1}[u_1]_{\mathcal I}+P_{\theta _2}[u_2]_{\mathcal I} \leq P_{\theta _1+\theta _2}[u_1+u_2]_{\mathcal I}. \]
Proof.
Let $h$ be a competitor for $P_\theta [u]_{\mathcal I}$. Since $h\sim _{\mathcal I}u\preceq _{\mathcal I}v$, the Lelong-number criterion in theorem 5.37.3 shows that $\max (h,v-\sup _ Xv)$ is multiplier-ideal equivalent to $v$. This function is nonpositive and dominates $h$, so it is a competitor for $P_\theta [v]_{\mathcal I}$; taking the supremum over $h$ proves that the envelope preserves the order. Multiplication by $a$ is a bijection between the defining families in the scaling identity. For the sum, $P_{\theta _1}[u_1]_{\mathcal I}+P_{\theta _2}[u_2]_{\mathcal I}$ is nonpositive and $(\theta _1+\theta _2)$-plurisubharmonic, and it is multiplier-ideal equivalent to $u_1+u_2$ by projection and lemma 5.37.4. It is therefore a competitor for the envelope on the right.
$\square$
Pome
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