Pome

Proposition 16.4.2CF5FECD1B547

Proposition 16.4.2.CF5FECD1B547

  1. C852A00815E6

    Suppose that $\varphi \in \mathrm{PSH}(X,\theta )$ and $\lambda \in \mathbb {R}_{>0}$. Then

    \[ P_{\lambda \theta }[\lambda \varphi ]=\lambda P_{\theta }[\varphi ]. \]
  2. 054C3F14E799

    Suppose that $\varphi _1\in \mathrm{PSH}(X,\theta _1)$ and $\varphi _2\in \mathrm{PSH}(X,\theta _2)$. Then

    \[ P_{\theta _1+\theta _2}[\varphi _1+\varphi _2]\geq P_{\theta _1}[\varphi _1]+P_{\theta _2}[\varphi _2]. \]
  3. 7E5E2644837B

    Suppose that $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$ and $\varphi \preceq \psi $. Then

    \[ P_{\theta }[\varphi ]\leq P_{\theta }[\psi ]. \]

Proof. 1 This follows by rescaling the candidates in the definition of the $P$-envelope.

2 Suppose that $\psi _1\in \mathrm{PSH}(X,\theta _1)$ and $\psi _2\in \mathrm{PSH}(X,\theta _2)$ satisfy

\[ \psi _ i\leq 0,\quad \psi _ i\preceq \varphi _ i \]

for $i=1,2$. Then

\[ \psi _1+\psi _2\leq 0,\quad \psi _1+\psi _2\preceq \varphi _1+\varphi _2. \]

It follows from (16.3.6) that

\[ \psi _1+\psi _2\leq P_{\theta _1+\theta _2}[\varphi _1+\varphi _2]. \]

Since $\psi _1$ and $\psi _2$ are arbitrary, we conclude.

3 This follows by inclusion of the defining candidate sets. $\square$

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