Theorem 16.3.15.7ADF30509E4E Assume that $\varphi \in \mathrm{PSH}(X,\theta )_{>0}$. Then
2B585EFD6781
16.3.16
\begin{equation} P_{\theta }[\varphi ]=\sup \left\{ \psi \in \mathrm{PSH}(X,\theta ): \psi \leq 0,\varphi \preceq \psi , \int _ X \theta _{\varphi }^ n=\int _ X \theta _{\psi }^ n\right\} . \end{equation}
In particular, in this case,
A43EDA3894B0
16.3.17
\begin{equation} P_{\theta }\Bigl [P_{\theta }[\varphi ]\Bigr ]=P_{\theta }[\varphi ]. \end{equation}
Proof.
Let $\psi $ be a candidate of the right-hand side of (16.3.16). It follows from theorem 16.3.12 that
\[ \int _{\{ P_{\theta }[\varphi ]<\psi \} } \theta _{P_{\theta }[\varphi ]}^ n\leq \int _{\{ P_{\theta }[\varphi ]<\psi \} \cap \{ P_{\theta }[\varphi ]=0\} } \theta ^ n=0. \]
On the other hand, we have
\[ P_{\theta }[\varphi ]\preceq P_{\theta }[\psi ],\quad \psi \preceq P_{\theta }[\psi ] \]
and all these potentials have the same mass due to proposition 16.3.10. So the domination principle theorem 13.13.43 is applicable and gives
\[ P_{\theta }[\varphi ]\geq \psi . \]
Hence we get the $\geq $ direction in (16.3.16).
On the other hand, $P_{\theta }[\varphi ]$ itself is a candidate of the right-hand side of (16.3.16), as a consequence of proposition 16.3.10. Therefore, (16.3.16) follows.
As for (16.3.17), the $\geq $ direction is trivial. For the other direction, let $\psi \in \mathrm{PSH}(X,\theta )$ be a potential satisfying
\[ \psi \leq 0,\quad P_{\theta }[\varphi ]\preceq \psi ,\quad \int _ X \theta _{P_{\theta }[\varphi ]}^ n=\int _ X \theta _{\psi }^ n. \]
Then it follows from proposition 16.3.10 that
\[ \psi \leq 0,\quad \varphi \preceq \psi ,\quad \int _ X \theta _{\varphi }^ n=\int _ X \theta _{\psi }^ n. \]
In view of (16.3.16), we conclude the $\leq $ direction of (16.3.17).
$\square$
Pome
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