Proof.
Without loss of generality, we may assume that $\varphi \leq \psi \leq 0$.
Step 1. We show the existence of $\eta \in \mathrm{PSH}(X,\theta )$ satisfying (13.13.32).
Step 1.1. We first consider the case where $\psi \sim \varphi $. In this case, we can take $\eta =\varphi -C$ enough $C>0$. But since by (13.13.34), we also have $P_{\theta }(a\varphi +(1-a)\psi )\leq \psi $, we find that
\[ P_{\theta }\Bigl (a\varphi +(1-a)\psi \Bigr )\sim \varphi \sim \psi \]
in this case.
Step 1.2. We make a first reduction of the problem.
Now for each $\gamma \in \mathrm{PSH}(X,\theta )$, we let
\[ P_{\theta }[\gamma ]\coloneqq \operatorname *{sup}_{}{}\! {}^{*}\Bigl \{ \eta \in \mathrm{PSH}(X,\theta ): \eta \leq \min \{ \gamma +C,0\} \textup{ for some }C>0 \Bigr \} {}^{\href{#sgpt-a0000015147}{6}}. \]
Observe that due to corollary 13.13.12 and theorem 13.13.11, we have
00E5A7157825
13.13.35
\begin{equation} \int _ X \theta _{P_{\theta }[\gamma ]}^ n=\int _ X \theta _{\gamma }^ n. \end{equation}
Further, $0\geq P_{\theta }[\gamma ]\geq \gamma $ if $\gamma \leq 0$.
We now define an increasing transfinite sequence $(\psi _{\alpha })_{\alpha }$ consisting of non-positive functions in $\mathrm{PSH}(X,\theta )$ as follows:
1A558B3DDD3E
$\psi _0=\psi $;
F7EF40CB8904
$\psi _{\alpha +1}=P_{\theta }[\psi _{\alpha }]$;
04DDC5115F70
if $\alpha $ is a limit ordinal, we define
\[ \psi _{\alpha }=\operatorname *{sup}_{\beta <\alpha }{}\! {}^{*} \psi _{\beta }. \]
Observe that
\[ \int _ X \theta _{\psi _{\alpha }}^ n=\int _ X \theta _{\psi }^ n \]
by corollary 13.13.12 and (13.13.35). Since $\mathrm{PSH}(X,\theta )$ is indeed a set, the transfinite sequence must stabilize, say at some $\phi \in \mathrm{PSH}(X,\theta )$. Note that $0 \geq \phi \geq \psi $, and $P_{\theta }[\phi ]=\phi $.
Note that the interval (13.13.31) remains the same if we replace $\psi $ by $\phi $. Take $a$ in this interval, it suffices to prove the existence of $\eta \in \mathrm{PSH}(X,\theta )$ so that
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13.13.36
\begin{equation} a^{-1}\eta +\left(1-a^{-1}\right)\phi \leq \varphi . \end{equation}
Let us record the following observation for later use. Suppose that $\tau \in \mathrm{PSH}(X,\theta )$ and $\tau \preceq \phi $. Then
CEDD9EE9573E
13.13.37
\begin{equation} \sup _{\{ \phi \neq -\infty \} } (\tau -\phi )=\sup _ X \tau . \end{equation}
Observe that $\phi \leq 0$, so on the set $\{ \phi \neq -\infty \} $, we have
\[ \tau -\phi \geq \tau . \]
So the $\geq $ direction in (13.13.37) follows from corollary 5.15.21. Conversely, by assumption, we can find a constant $C>0$ so that
\[ \tau -\sup _ X \tau \leq \min \{ \phi +C,0 \} . \]
It follows that $\tau -\sup _ X \tau \leq P_{\theta }[\phi ]=\phi $. Therefore, the reverse inequality follows, and (13.13.37) follows.
For each $k>0$, we introduce
\[ \varphi _ k=\varphi \lor \left(\phi -k\right),\quad \eta _ k\coloneqq P_{\theta }\Bigl ( a \varphi _ k+(1-a)\phi \Bigr ). \]
Since $\varphi _ k\sim \phi $, using Step 1.1, we have $\eta _ k\in \mathrm{PSH}(X,\theta )$ and $\eta _ k\sim \phi $ as well.
Note that $(\eta _ k)_ k$ is decreasing in $k$. Let
\[ \eta \coloneqq \inf _{k\in \mathbb {N}} \eta _ k. \]
Note that $\eta $ automatically satisfies (13.13.36). It remains to show that $\eta \not\equiv -\infty $.
Step 1.3. We prove that $\eta \not\equiv -\infty $.
Suppose, for contradiction, that $\sup _ X \eta _ k\to -\infty $. For each $k>0$, let
\[ \gamma _ k\coloneqq a^{-1}\eta _ k+\left(1-a^{-1}\right)\phi ,\quad D_ k\coloneqq \left\{ \gamma _ k=\varphi _ k\right\} . \]
We claim that
87D0AE27AE70
13.13.38
\begin{equation} \theta _{\eta _ k}^ n\leq a^ n\mathds {1}_{D_ k}\, \theta _{\varphi _ k}^ n. \end{equation}
Since $\gamma _ k\leq \varphi _ k$ with equality on $D_ k$, 13.13.14 gives
\[ \mathds {1}_{D_ k}\, \theta _{\gamma _ k}^ n\leq \mathds {1}_{D_ k}\, \theta _{\varphi _ k}^ n. \]
On the other hand,
\[ \theta _{\gamma _ k}=a^{-1}\theta _{\eta _ k}+\left(1-a^{-1}\right)\theta _{\phi }, \]
so positivity and multilinearity of the non-pluripolar product imply
\[ a^{-n}\theta _{\eta _ k}^ n\leq \theta _{\gamma _ k}^ n. \]
Finally, it follows from theorem 13.13.24 that
\[ \mathds {1}_{D_ k}\, \theta _{\eta _ k}^ n=\theta _{\eta _ k}^ n. \]
Putting these results together, we conclude (13.13.38).
Fix $j>k>0$. Note that
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13.13.39
\begin{equation} \int _{\{ \varphi \leq \phi -k \} } \theta _{\varphi _ j}^ n=\int _ X \theta _{\varphi _ j}^ n-\int _{\{ \varphi >\phi -k \} } \theta _{\varphi _ j}^ n=\int _ X \theta _{\phi }^ n-\int _{\{ \varphi >\phi -k \} } \theta _{\varphi }^ n, \end{equation}
where we applied theorem 13.13.11 and proposition 13.11.4 1 in the second equality.
Next, we compute
\[ \begin{aligned} \int _{\left\{ \eta _ j\leq \phi -ak \right\} } \theta _{\eta _ j}^ n\leq & a^ n \int _{\left\{ \eta _ j\leq \phi -ak \right\} } \mathds {1}_{D_ j}\, \theta _{\varphi _ j}^ n & \textup{by (\href{../../tag/87D0AE27AE70/}{13.13.38})}\\ \leq & a^ n \int _{\left\{ a\varphi _ j+(1-a)\phi \leq \phi -ak \right\} \cap \left\{ \phi \neq -\infty \right\} } \theta _{\varphi _ j}^ n\\ = & a^ n \int _{\left\{ \varphi _ j\leq \phi -k \right\} } \theta _{\varphi _ j}^ n\\ = & a^ n \int _{\left\{ \varphi \leq \phi -k \right\} } \theta _{\varphi _ j}^ n\\ \leq & a^ n\left( \int _ X \theta _{\phi }^ n- \int _{\left\{ \varphi >\phi -k \right\} } \theta _{\varphi }^ n \right)& \textup{ by (\href{../../tag/2263D1F9CA2F/}{13.13.39})}. \end{aligned} \]
Next thanks to (13.13.37),
\[ \sup _{ \left\{ \phi \neq -\infty \right\} } \left(\eta _ j-\phi \right)=\sup _ X \eta _ j\to -\infty . \]
In particular, for a fixed $k$, if $j$ is large enough, we have
\[ \left\{ \eta _ j\leq \phi -ak \right\} =X. \]
Therefore, for a fixed $k$, for any large enough $j$,
\[ \int _ X \theta _{\phi }^ n=\int _ X \theta _{\eta _ j}^ n\leq a^ n\left( \int _ X \theta _{\phi }^ n- \int _{\{ \varphi >\phi -k\} } \theta _{\varphi }^ n\right). \]
Letting $k\to \infty $, we find
\[ \int _ X \theta _{\phi }^ n\leq a^ n \left(\int _ X \theta _{\phi }^ n-\int _ X \theta _{\varphi }^ n \right). \]
This is a contradiction with our choice of $a$.
Step 2. Next we argue that $P_{\theta }(a\varphi +(1-a)\psi )\in \mathrm{PSH}(X,\theta )_{>0}$. Choose
\[ a'\in \left( a, \left(\frac{\int _ X \theta _{\psi }^ n}{\int _ X \theta _{\psi }^ n-\int _ X \theta _{\varphi }^ n}\right)^{1/n} \right). \]
It follows from (13.13.33) that
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13.13.40
\begin{equation} P_{\theta }\Bigl (a\varphi +(1-a)\psi \Bigr )\geq \frac{a}{a'} P_{\theta }\Bigl (a'\varphi +(1-a')\psi \Bigr )+\frac{a'-a}{a'}\varphi . \end{equation}
Therefore, by theorem 13.13.11, we have
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13.13.41
\begin{equation} \int _ X \theta _{P_{\theta }\left(a\varphi +(1-a)\psi \right)}^ n\geq \frac{(a'-a)^ n}{a'^ n}\int _ X \theta _{\varphi }^ n>0. \end{equation}
We conclude the proof.
$\square$
Pome
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