PomeProposition 16.5.49.AAD590A053BB Assume that $\varphi ,\psi \in \mathcal{E}(X,\theta ;\phi )$ (resp. $\mathcal{E}^1(X,\theta ;\phi )$, $\mathcal{E}^{\infty }(X,\theta ;\phi )$), then so is $\varphi \land \psi $.
Proposition 16.5.49AAD590A053BB
Proof. The case of $\mathcal{E}^{\infty }(X,\theta ;\phi )$ is trivial.
We consider the case $\mathcal{E}(X,\theta ;\phi )$. It follows from proposition 16.3.25 that $\varphi \land \psi \in \mathrm{PSH}(X,\theta )$. By theorem 16.3.35, we have
\[ \int _ X \theta _{\varphi \land \psi }^ n\geq \int _ X \theta _{\phi }^ n. \]
By the monotonicity theorem theorem 13.13.11, equality holds. By theorem 16.3.15, we conclude that
\[ P_{\theta }[\varphi \land \psi ]=\phi . \]
Finally, assume that $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$. We may assume that $\varphi ,\psi \leq \phi $. For each $j\geq 1$, consider the approximations:
\[ \varphi _ j\coloneqq \varphi \lor (\phi -j)\, ,\quad \psi _ j\coloneqq \psi \lor (\phi -j). \]
By 16.5.40, we can take $\gamma _ j\in \mathcal{E}^{\infty }(X,\theta ;\phi )$ solving the following equation:
\[ \theta _{\gamma _ j}^ n=\mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n+\mathrm{e}^{\gamma _ j-\psi _ j}\, \theta _{\psi _ j}^ n. \]
In particular,
\[ \theta _{\gamma _ j}^ n\geq \mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n. \]
We claim that this implies $\gamma _ j\leq \varphi _ j$. Indeed, by proposition 16.5.5,
\[ \int _{\{ \varphi _ j<\gamma _ j\} }\theta _{\varphi _ j}^ n \geq \int _{\{ \varphi _ j<\gamma _ j\} }\theta _{\gamma _ j}^ n \geq \int _{\{ \varphi _ j<\gamma _ j\} }\mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n. \]
Hence
\[ \int _{\{ \varphi _ j<\gamma _ j\} }\left(\mathrm{e}^{\gamma _ j-\varphi _ j}-1\right)\theta _{\varphi _ j}^ n=0. \]
It follows that $\int _{\{ \varphi _ j<\gamma _ j\} }\theta _{\varphi _ j}^ n=0$, and the domination principle theorem 13.13.43 gives $\gamma _ j\leq \varphi _ j$. Similarly, $\gamma _ j\leq \psi _ j$. In particular, $\gamma _ j\leq \varphi _ j\land \psi _ j$. We claim that
081622986279
16.5.50
\begin{equation} \int _ X (\gamma _ j-\phi )\, \theta _{\gamma _ j}^ n >-C \end{equation}
for some $C$ independent of $j$.
Assume the claim is true for now. We get immediately that
\[ \sup _{X}\gamma _ j=\sup _{X\setminus \{ \phi =-\infty \} }(\gamma _ j-\phi )\geq -\frac{C}{\int _ X \theta _{\phi }^ n}, \]
where the first equality follows from proposition 16.3.21. By proposition 5.17.3, after possibly extracting a subsequence, we may assume that $\gamma _ j\to \gamma \in \mathrm{PSH}(X,\theta )$ in $L^1$-topology. Then $\gamma \in \mathcal{E}^1(X,\theta ;\phi )$ by 16.5.45 and proposition 16.5.22. Since $\gamma \leq \varphi \land \psi \leq \phi $, the monotonicity theorem theorem 13.13.11 gives
\[ \int _ X\theta _{\varphi \land \psi }^ n=\int _ X\theta _\phi ^ n. \]
Moreover, by the monotonicity of $E^\phi _\theta $,
\[ E^\phi _\theta (\varphi \land \psi )\geq E^\phi _\theta (\gamma )>-\infty . \]
Thus $\varphi \land \psi \in \mathcal E^1(X,\theta ;\phi )$ by proposition 16.5.22.
Now we prove the claim (16.5.50). By symmetry, it suffices to prove
\[ \int _ X (\phi -\gamma _ j) \mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n\leq C. \]
But note that
\[ \int _ X (\phi -\gamma _ j) \mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n=\int _ X (\phi -\varphi _ j) \mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n+\int _ X (\varphi _ j-\gamma _ j) \mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n. \]
But $x \mathrm{e}^{-x}\leq \mathrm{e}^{-1}$ for $x\geq 0$ (with maximum at $x=1$), so the second term is bounded; it remains to prove
\[ \int _ X (\phi -\varphi _ j) \mathrm{e}^{\gamma _ j-\varphi _ j}\, \theta _{\varphi _ j}^ n\leq C. \]
As $\gamma _ j\leq \varphi _ j$, it suffices to prove
B0A9B146545E
16.5.51
\begin{equation} \int _ X (\phi -\varphi _ j)\, \theta _{\varphi _ j}^ n\leq C. \end{equation}
We compute, using (16.5.11) and the fact that the $k<n$ terms $\int _ X(\varphi _ j-\phi )\theta _{\varphi _ j}^ k\wedge \theta _\phi ^{n-k}$ are all non-positive (since $\varphi _ j\leq \phi $),
\[ \int _ X (\varphi _ j-\phi )\, \theta _{\varphi _ j}^ n\geq (n+1)E_{\theta }^{\phi }(\varphi _ j). \]
Since $\varphi _ j\searrow \varphi $ and $E_\theta ^{\phi }$ is continuous along decreasing sequences (proposition 16.5.20), $E_\theta ^{\phi }(\varphi _ j)$ is bounded below uniformly in $j$. Thus, (16.5.51) follows. $\square$
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