Pome

Proposition 18.4.8EB95710FF4D7

Proposition 18.4.8.EB95710FF4D7 Let $a\in \mathrm{PSH}(X,\theta )$ have analytic singularities. Then

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

In particular the common envelope has analytic singularities with the same coefficient and ideal as $a$, up to a bounded remainder.

Proof. Set $q=P_\theta [a]_{\mathcal I}$. Projection gives $q\sim _{\mathcal I}a$, so lemma 5.38.11 gives $q\preceq a$. Since $q\leq 0$, it is a competitor for $P_\theta [a]$. Together with proposition 18.4.4, this yields

\[ a-\sup _ Xa\leq P_\theta [a]\leq q\leq P_\theta [a]. \]

Hence the two envelopes are equal and ordinarily equivalent to $a$. Since they differ from $a$ by a bounded function, they inherit its analytic presentation. $\square$

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