B22CE32B08DB
Let $\theta $ be a smooth closed real $(1,1)$-form on $X$ representing a big cohomology class. Fix a model potential $\phi \in \mathrm{PSH}(X,\theta )_{>0}$.
Definition 16.5.1.6444F0A28464 We define
\[ \begin{aligned} \mathrm{PSH}(X,\theta ;\phi )\coloneqq & \left\{ \eta \in \mathrm{PSH}(X,\theta ): \eta \preceq \phi \right\} ,\\ \mathcal{E}^{\infty }(X,\theta ;\phi )\coloneqq & \left\{ \eta \in \mathrm{PSH}(X,\theta ): \eta \sim \phi \right\} ,\\ \mathcal{E}(X,\theta ;\phi )\coloneqq & \left\{ \eta \in \mathrm{PSH}(X,\theta ;\phi ): \int _ X \theta _{\eta }^ n=\int _ X \theta _{\phi }^ n\right\} ,\\ \mathcal{E}^1(X,\theta ;\phi )\coloneqq & \left\{ \eta \in \mathcal{E}(X,\theta ;\phi ): \int _ X |\phi -\eta |\, \theta _{\eta }^ n<\infty \right\} . \end{aligned} \]
Potentials in the last three classes are said to have relatively minimal singularities, full mass and finite energy relative to $\phi $ respectively.
We have the following inclusions:
A8BA75F03A09
16.5.2
\begin{equation} \mathcal{E}^{\infty }(X,\theta ;\phi )\subseteq \mathcal{E}^1(X,\theta ;\phi )\subseteq \mathcal{E}(X,\theta ;\phi )\subseteq \mathrm{PSH}(X,\theta ;\phi ). \end{equation}
The only non-trivial part is the first inclusion, which follows from the monotonicity theorem theorem 13.13.11.
When $\phi =V_{\theta }$, we usually write
\[ \begin{aligned} \mathcal{E}^{\infty }(X,\theta ;V_{\theta })=& \mathcal{E}^{\infty }(X,\theta ),\\ \mathcal{E}(X,\theta ;V_{\theta })=& \mathcal{E}(X,\theta ),\\ \mathcal{E}^{1}(X,\theta ;V_{\theta })=& \mathcal{E}^{1}(X,\theta ). \end{aligned} \]
Potentials in the three classes are said to have minimal singularities, full mass and finite energy respectively. The relation (16.5.2) can be written as
\[ \mathcal{E}^{\infty }(X,\theta )\subseteq \mathcal{E}^{1}(X,\theta ) \subseteq \mathcal{E}(X,\theta ) \]
in this case.
The $P$-envelope can be used to characterize the full mass classes:
Proposition 16.5.4.557595391303 Let $\varphi \in \mathrm{PSH}(X,\theta )$. Then the following are equivalent:
88F668B67D98
$\varphi \in \mathcal{E}(X,\theta ;\phi )$;
D4B2E9D2786A
$P_{\theta }[\varphi ]=\phi $.
Proof.
2 $\implies $ 1. Assume 2. By proposition 16.3.10,
\[ \int _ X\theta _\varphi ^ n=\int _ X\theta _{P_\theta [\varphi ]}^ n=\int _ X\theta _\phi ^ n. \]
Moreover, since the mass is positive, theorem 16.3.15 implies that
\[ \varphi \preceq P_\theta [\varphi ]=\phi . \]
Thus $\varphi \in \mathcal E(X,\theta ;\phi )$.
1 $\implies $ 2. Assume 1. Then $\varphi \in \mathrm{PSH}(X,\theta )_{>0}$. Note that $\phi $ is a candidate of $P_{\theta }[\varphi ]$ as in (16.3.16). So $P_{\theta }[\varphi ]\geq \phi $. The reverse inequality is trivial.
$\square$
We have the following comparison principle.
Proposition 16.5.5 (Comparison principle).FDFD0A6F6287 Fix $j\in \{ 0,\ldots ,n\} $. Let $\theta _1,\ldots ,\theta _ j$ be closed smooth real $(1,1)$-forms on $X$ and $\psi _ i\in \mathrm{PSH}(X,\theta _ i)$ for $i=1,\ldots ,j$. Suppose that $\varphi ,\psi \in \mathcal{E}(X,\theta ;\phi )$. Then
EAAF267DE59E
16.5.6
\begin{equation} \int _{\{ \varphi <\psi \} } \theta _{\varphi }^{n-j} \wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\geq \int _{\{ \varphi <\psi \} } \theta _{\psi }^{n-j} \wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}. \end{equation}
As a consequence, we also have
D3168D446564
16.5.7
\begin{equation} \int _{\{ \varphi \leq \psi \} } \theta _{\varphi }^{n-j} \wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\geq \int _{\{ \varphi \leq \psi \} } \theta _{\psi }^{n-j} \wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}. \end{equation}
Proof.
Observe that $\varphi \lor \psi \in \mathcal{E}(X,\theta ;\phi )$, as a consequence of the monotonicity theorem theorem 13.13.11. Using the plurifine locality of the non-pluripolar product proposition 13.11.4 1, we compute
\[ \begin{aligned} & \int _ X \theta _{\varphi \lor \psi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\\ \geq & \int _{\{ \varphi >\psi \} } \theta _{\varphi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}+\int _{\{ \varphi <\psi \} } \theta _{\psi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\\ =& \int _ X \theta _{\varphi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}-\int _{\{ \varphi \leq \psi \} } \theta _{\varphi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\\ & +\int _{\{ \varphi <\psi \} } \theta _{\psi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\\ =& \int _ X \theta _{\varphi \lor \psi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}-\int _{\{ \varphi \leq \psi \} } \theta _{\varphi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\\ & +\int _{\{ \varphi <\psi \} } \theta _{\psi }^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}, \end{aligned} \]
where in the last step, we applied proposition 16.3.10 to $\varphi $ and $\varphi \lor \psi $. Therefore,
\[ \begin{aligned} \int _{\{ \varphi \leq \psi \} }& \theta _{\varphi }^{n-j} \wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}\\ & \geq \int _{\{ \varphi <\psi \} } \theta _{\psi }^{n-j} \wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}. \end{aligned} \]
Applying the preceding inequality to $\varphi +\epsilon $ in place of $\varphi $, and using $\theta _{\varphi +\epsilon }=\theta _\varphi $, we get
\[ \int _{\{ \varphi +\epsilon \leq \psi \} }\theta _\varphi ^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j} \geq \int _{\{ \varphi +\epsilon <\psi \} }\theta _\psi ^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}. \]
As $\epsilon \searrow 0$, by monotone convergence theorem, we obtain
\[ \int _{\{ \varphi <\psi \} }\theta _\varphi ^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j} \geq \int _{\{ \varphi <\psi \} }\theta _\psi ^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}, \]
which proves (16.5.6).
Finally, the (16.5.7) follows by subtracting (16.5.6) (with the role of $\varphi $ and $\psi $ switched) from the total mixed mass equality
\[ \int _ X\theta _\varphi ^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}= \int _ X\theta _\psi ^{n-j}\wedge \theta _{1,\psi _1}\wedge \cdots \wedge \theta _{j,\psi _ j}, \]
which itself follows from proposition 16.3.10 and proposition 16.5.4.
$\square$
The full mass potentials are essential in resolving the Monge–Ampère equations. We recall the following two Aubin–Yau type theorems. Recall that a measure $\mu $ on $X$ is
non-pluripolar if it is a Radon measure and $\mu (K)=0$ for each pluripolar set $K\subseteq X$.
Theorem 16.5.8.6F1A42885C28 Let $\mu $ be a non-pluripolar measure on $X$ with $\mu (X)=\int _ X \theta _{\phi }^ n$. Then there is a unique $\varphi \in \mathcal{E}(X,\theta ;\phi )$ such that
\[ \theta _{\varphi }^ n=\mu ,\quad \sup _ X \varphi =0. \]
Theorem 16.5.9.0E437FB98B97 Fix $\lambda >0$. Let $\mu $ be a non-pluripolar measure on $X$ with $\mu (X)>0$. Then there is a unique $\varphi \in \mathcal{E}(X,\theta ;\phi )$ such that
\[ \theta _{\varphi }^ n=\mathrm{e}^{\lambda \varphi }\mu . \]
Furthermore, for any $\psi \in \mathcal{E}(X,\theta ;\phi )$ satisfying
\[ \theta _{\psi }^ n\geq \mathrm{e}^{\lambda \psi }\mu , \]
we have $\varphi \geq \psi $.
The proofs are beyond the scope of this book; we refer to
[DDNL19log].
In order to handle the finite energy classes, it is convenient to introduce the following quantity:
Definition 16.5.10.8A07570966F8 We define the Monge–Ampère energy $E^{\phi }_{\theta }\colon \mathcal{E}^{\infty }(X,\theta ;\phi )\rightarrow \mathbb {R}$ as follows
AB28CEAE8B4A
16.5.11
\begin{equation} E^{\phi }_{\theta }(\varphi )\coloneqq \frac{1}{n+1}\sum _{j=0}^ n \int _ X (\varphi -\phi )\, \theta _{\varphi }^ j\wedge \theta _{\phi }^{n-j}. \end{equation}
More generally, we extend $E_{\theta }^{\phi }$ to a functional $E^{\phi }_{\theta }\colon \mathrm{PSH}(X,\theta ;\phi )\rightarrow [-\infty ,\infty )$ as follows
899E1F34D705
16.5.12
\begin{equation} E^{\phi }_{\theta }(\varphi )\coloneqq \inf \left\{ E^{\phi }_{\theta }(\psi ):\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi ),\varphi \leq \psi \right\} . \end{equation}
We write $E_{\theta }$ instead of $E_{\theta }^{\phi }$ when $\phi =V_{\theta }$.
The set in (
16.5.12) is non-empty, since $\phi +\sup _ X \varphi $ is a candidate. Therefore, $E^{\phi }_{\theta }(\varphi )<\infty $.
Note that
BC7744A40FEC
16.5.13
\begin{equation} E^{\phi }_{\theta }(\varphi +C)=E^{\phi }_{\theta }(\varphi )+C\int _ X \theta _{\phi }^ n \end{equation}
for any $\varphi \in \mathrm{PSH}(X,\theta ;\phi )$ and $C\in \mathbb {R}$.
Lemma 16.5.14.93210731F7FD The functional $E_{\theta }^{\phi }\colon \mathcal{E}^{\infty }(X,\theta ;\phi )\rightarrow \mathbb {R}$ is increasing. In particular, the extended definition (16.5.12) agrees with (16.5.11) on $\mathcal{E}^{\infty }(X,\theta ;\phi )$, and is increasing as well.
Proof.
Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$, we have
\[ \begin{aligned} & (n+1)E^{\phi }_{\theta }(\varphi )-(n+1)E^{\phi }_{\theta }(\psi )-\sum _{j=0}^ n \int _ X (\varphi -\psi )\, \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j} \\ =& \sum _{j=0}^ n \int _ X (\varphi -\phi )\, \theta _{\varphi }^ j\wedge \theta _{\phi }^{n-j}-\sum _{j=0}^ n \int _ X (\psi -\phi )\, \theta _{\psi }^ j\wedge \theta _{\phi }^{n-j}-\sum _{j=0}^ n \int _ X (\varphi -\psi )\, \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}\\ =& \sum _{j=0}^ n \int _ X (\varphi -\psi )\, \left( \theta _{\varphi }^ j\wedge \theta _{\phi }^{n-j}-\theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j} \right)\\ & +\sum _{j=0}^ n \int _ X (\psi -\phi )\, \left( \theta _{\varphi }^ j\wedge \theta _{\phi }^{n-j}-\theta _{\psi }^ j\wedge \theta _{\phi }^{n-j} \right)\\ =& \sum _{\substack {j+a+b=n-1,\\ { j,a,b\geq 0
}}} (n-j)\int _ X (\psi -\phi )\, \left( \theta _{\phi }^ a\wedge \theta _{\psi }^{b+1}\wedge \theta _{\varphi }^ j- \theta _{\phi }^ a\wedge \theta _{\psi }^{b}\wedge \theta _{\varphi }^{j+1}\right)\\ & +\sum _{\substack {j+a+b=n-1,\\ { j,a,b\geq 0
}}} (n-j)\int _ X (\psi -\phi )\, \left( \theta _{\phi }^ j\wedge \theta _{\varphi }^{a+1}\wedge \theta _{\psi }^ b- \theta _{\phi }^ j\wedge \theta _{\varphi }^{a}\wedge \theta _{\psi }^{b+1}\right)\\ =& 0, \end{aligned} \]
where the third equality follows from the integration by parts formula. See [Xia19b, Lu21] for the proof in the context of non-pluripolar products.
In other words,
F7FEEE32D362
16.5.15
\begin{equation} E^{\phi }_{\theta }(\varphi )-E^{\phi }_{\theta }(\psi )=\frac{1}{n+1}\sum _{j=0}^ n \int _ X(\varphi -\psi )\, \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}. \end{equation}
The monotonicity follows.
$\square$
Lemma 16.5.16.8D9EA7C5CC7C Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Then
6E4685F8B030
16.5.17
\begin{equation} \int _ X (\varphi -\psi )\, \theta _{\psi }^ n \geq E_{\theta }^{\phi }(\varphi )-E_{\theta }^{\phi }(\psi )\geq \int _ X (\varphi -\psi )\, \theta _{\varphi }^ n. \end{equation}
Proof.
Thanks to (16.5.13), we may assume that $\varphi \geq \psi $.
As we have seen in (16.5.15), the middle term can be written as
\[ \frac{1}{n+1}\sum _{j=0}^ n \int _ X(\varphi -\psi )\, \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}. \]
We claim that for each individual $j$, the inequality (16.5.17) holds:
\[ \int _ X (\varphi -\psi )\, \theta _{\psi }^ n \geq \int _ X(\varphi -\psi )\, \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j} \geq \int _ X (\varphi -\psi )\, \theta _{\varphi }^ n. \]
Thanks to proposition 16.5.5, we have
\[ \begin{aligned} \int _ X(\varphi -\psi )\, \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}=& \int _0^{\infty } \int _{\{ \varphi >\psi +t\} } \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}\, \mathrm{d}t\\ \geq & \int _0^{\infty } \int _{\{ \varphi >\psi +t\} } \theta _{\varphi }^ n\, \mathrm{d}t \\ =& \int _ X (\varphi -\psi )\, \theta _{\varphi }^ n. \end{aligned} \]
The other inequality is similar.
$\square$
Lemma 16.5.18.B42B448077CF The function $E^{\phi }_{\theta }\colon \mathcal{E}^{\infty }(X,\theta ;\phi )\rightarrow \mathbb {R}$ is concave.
Proof.
Let $\varphi _0,\varphi _1\in \mathcal{E}^{\infty }(X,\theta ;\phi )$. We write $\varphi _ t=t\varphi _1+(1-t)\varphi _0$ for each $t\in [0,1]$.
We claim that for each $t\in [0,1]$, we have
806CE06D337B
16.5.19
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}E^{\phi }_{\theta }(\varphi _ t) =\int _ X \left(\varphi _1-\varphi _0\right)\, \theta _{\varphi _ t}^ n. \end{equation}
At the boundary $t=0,1$, the derivative is understood as the one-sided derivative.
Since $\varphi _0,\varphi _1\in \mathcal{E}^{\infty }(X,\theta ;\phi )$, the difference $\varphi _1-\varphi _0$ is quasi-everywhere bounded. Hence all mixed integrals below are finite. We compute the right derivative first. Fix $t\in [0,1)$ and take $s>0$ small enough so that $t+s\leq 1$. By the cocycle formula (16.5.15),
\[ \begin{aligned} \frac{E^{\phi }_{\theta }(\varphi _{t+s})-E^{\phi }_{\theta }(\varphi _ t)}{s} =& \frac{1}{(n+1)s}\sum _{j=0}^ n \int _ X(\varphi _{t+s}-\varphi _ t)\, \theta _{\varphi _{t+s}}^ j\wedge \theta _{\varphi _ t}^{n-j}\\ =& \frac{1}{n+1}\sum _{j=0}^ n \int _ X(\varphi _1-\varphi _0)\, \theta _{\varphi _{t+s}}^ j\wedge \theta _{\varphi _ t}^{n-j}. \end{aligned} \]
For each fixed $j$, the last integral is a polynomial in $s$ by the multi-linearity of the non-pluripolar product proposition 13.11.4 6. Letting $s\to 0+$, it converges to
\[ \int _ X(\varphi _1-\varphi _0)\, \theta _{\varphi _ t}^ n. \]
Thus the right derivative is given by the right-hand side of (16.5.19). The proof for the left derivative is identical, using $s<0$ with $t+s\in [0,1]$. Hence the left and right derivatives agree for $t\in (0,1)$, and the endpoint derivatives are the corresponding one-sided derivatives.
Now it suffices to show that the derivative (16.5.19) is decreasing in $t$. Take $0\leq s<t \leq 1$. We wish to show that
\[ \int _ X (\varphi _1-\varphi _0)\, \theta _{\varphi _ t}^ n\leq \int _ X (\varphi _1-\varphi _0)\, \theta _{\varphi _ s}^ n. \]
This can be equivalently written as
\[ \int _ X (\varphi _ t-\varphi _ s)\, \theta _{\varphi _ t}^ n\leq \int _ X (\varphi _ t-\varphi _ s)\, \theta _{\varphi _ s}^ n. \]
In particular, replacing $\varphi _0$ by $\varphi _ s$ and $\varphi _1$ by $\varphi _ t$, we are reduced to proving the following
\[ \int _ X (\varphi _1-\varphi _0)\, \theta _{\varphi _1}^ n\leq \int _ X (\varphi _1-\varphi _0)\, \theta _{\varphi _0}^ n. \]
But this is just a consequence of 16.5.16.
$\square$
Proposition 16.5.20.37860FDCEAB5 Let $(\varphi _ i)_{i\in I}$ be a decreasing net in $\mathrm{PSH}(X,\theta ;\phi )$ with
\[ \varphi =\inf _{i\in I}\varphi _ i\not\equiv -\infty . \]
Then
BBEAC6CFC4B5
16.5.21
\begin{equation} \lim _{i\in I} E_{\theta }^{\phi }(\varphi _ i)=E_{\theta }^{\phi }(\varphi ). \end{equation}
Proof.
We first handle the case where $I=\mathbb {Z}_{>0}$. Namely, $(\varphi _ i)_ i$ is a decreasing sequence. Thanks to 16.5.14, we know that $E_{\theta }^{\phi }(\varphi _ i)$ is decreasing in $i$. So the limit in (16.5.21) exists and the $\geq $ inequality holds. Conversely, let $\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ and $\varphi \leq \psi $. We need to show that
\[ E_{\theta }^{\phi }(\psi )\geq \lim _{i\to \infty } E_{\theta }^{\phi }\left(\varphi _ i\lor \psi \right). \]
Replacing the sequence $(\varphi _ i)_ i$ by $(\varphi _ i\lor \psi )_ i$, we are reduced to proving the $\leq $ direction of (16.5.21) when $\varphi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
In this case, thanks to 16.5.16, we have
\[ E_{\theta }^{\phi }(\varphi _ i)-E_{\theta }^{\phi }(\varphi )\leq \int _ X (\varphi _ i-\varphi )\, \theta _{\varphi }^ n. \]
Our assertion follows from the monotone convergence theorem.
We now treat the general net case. By 16.5.14, the net $E_{\theta }^{\phi }(\varphi _ i)$ is decreasing and bounded below by $E_{\theta }^{\phi }(\varphi )$. Thus
\[ L\coloneqq \lim _{i\in I}E_{\theta }^{\phi }(\varphi _ i) \]
exists and satisfies $L\geq E_{\theta }^{\phi }(\varphi )$. It remains to prove the reverse inequality.
Since $\varphi \leq \varphi _ i$ for every $i\in I$, the hypotheses of corollary 5.17.5 are satisfied, and hence $\varphi _ i\xrightarrow {L^1}\varphi $. For each $k\geq 1$, choose $a_ k\in I$ such that
\[ \| \varphi _ i-\varphi \| _{L^1}<k^{-1},\quad i\geq a_ k. \]
We can recursively choose an increasing sequence $(i_ k)_{k\geq 1}$ in $I$ such that $i_ k\geq a_ k$ for all $k$. Then $(\varphi _{i_ k})_ k$ is a decreasing sequence and $\varphi _{i_ k}\to \varphi $ in $L^1$. Its pointwise decreasing limit is therefore equal to $\varphi $ almost everywhere, hence everywhere by proposition 5.14.6. Applying the sequence case to $(\varphi _{i_ k})_ k$, we get
\[ \lim _{k\to \infty }E_{\theta }^{\phi }(\varphi _{i_ k})=E_{\theta }^{\phi }(\varphi ). \]
Since $L\leq E_{\theta }^{\phi }(\varphi _{i_ k})$ for every $k$, we conclude $L\leq E_{\theta }^{\phi }(\varphi )$.
$\square$
Proposition 16.5.22.E2221B6939BE Let $\varphi \in \mathcal{E}(X,\theta ;\phi )$. The following are equivalent:
0DF993971FDE
$\varphi \in \mathcal{E}^1(X,\theta ;\phi )$;
4FACCBDBBADB
$E^{\phi }_{\theta }(\varphi )>-\infty $.
Proof.
Fix $\varphi \in \mathcal{E}(X,\theta ;\phi )$. Without loss of generality, we may assume that $\varphi \leq 0$. In particular, $\varphi \leq P_{\theta }[\varphi ]\leq P_{\theta }[\phi ]=\phi $.
2 $\implies $ 1. Assume 2. Using the plurifine locality of the non-pluripolar product proposition 13.11.4 1, we have that $\mathds {1}_{\{ \varphi >\phi -C\} } \theta _{\varphi \lor (\phi -C)}^ n=\mathds {1}_{\{ \varphi >\phi -C\} } \theta _{\varphi }^ n$ converges strongly to $\theta _{\varphi }^ n$ as $C\to \infty $ by the dominated convergence theorem. Therefore, for a fixed $D>0$,
\[ \begin{aligned} \int _ X \Bigl ( \varphi \lor (\phi -D)-\phi \Bigr )\, \theta _{\varphi }^ n=& \lim _{C\to \infty } \int _{\{ \varphi >\phi -C\} } \Bigl ( \varphi \lor (\phi -D)-\phi \Bigr )\, \theta _{\varphi \lor (\phi -C)}^ n\\ \geq & \varlimsup _{C\to \infty } \int _{\{ \varphi >\phi -C\} } \Bigl ( \varphi \lor (\phi -C)-\phi \Bigr )\, \theta _{\varphi \lor (\phi -C)}^ n\\ \geq & \varlimsup _{C\to \infty } \int _ X \Bigl ( \varphi \lor (\phi -C)-\phi \Bigr )\, \theta _{\varphi \lor (\phi -C)}^ n\\ \geq & \varlimsup _{C\to \infty } (n+1)E_{\theta }^{\phi }\Bigl (\varphi \lor (\phi -C) \Bigr )\\ \geq & (n+1)E_{\theta }^{\phi }(\varphi ). \end{aligned} \]
Here the last but one inequality follows from (16.5.11) and the fact that $\varphi \lor (\phi -C)\leq \phi $. Letting $D\to \infty $, we conclude 1.
1 $\implies $ 2. Assume 1. Thanks to 16.5.16, we know that for each $C>0$,
\[ \begin{aligned} E_{\theta }^{\phi }\Bigl (\varphi \lor (\phi -C)\Bigr )\geq & \int _ X \Bigl (\varphi \lor (\phi -C)-\phi \Bigr )\, \theta _{\varphi \lor (\phi -C)}^ n\\ =& \int _{\left\{ \varphi >\phi -C \right\} } \Bigl (\varphi -\phi \Bigr )\, \theta _{\varphi \lor (\phi -C)}^ n+ \int _{\left\{ \varphi \leq \phi -C \right\} } (-C)\, \theta _{\varphi \lor (\phi -C)}^ n \\ =& \int _{\{ \varphi +C>\phi \} } \left(\varphi +C-\phi \right)\, \theta _{\varphi }^ n-C\int _ X \theta _{\phi }^ n\\ \geq & \int _ X \left(\varphi +C-\phi \right)\, \theta _{\varphi }^ n-C\int _ X \theta _{\phi }^ n\\ =& \int _ X (\varphi -\phi )\, \theta _{\varphi }^ n. \end{aligned} \]
Due to proposition 16.5.20, we have
\[ \lim _{C\to \infty }E_{\theta }^{\phi }\Bigl (\varphi \lor (\phi -C)\Bigr )=E_{\theta }^{\phi }(\varphi ), \]
so 2 holds.
$\square$
Corollary 16.5.23.555B1CEACB0F The function $E^{\phi }_{\theta }\colon \mathrm{PSH}(X,\theta ;\phi )\rightarrow [-\infty ,\infty )$ is concave.
Proof.
Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta ;\phi )$ and $t\in (0,1)$. We need to show that
3BD672AA942B
16.5.24
\begin{equation} E^{\phi }_{\theta }\Bigl ( t\varphi +(1-t)\psi \Bigr )\geq tE^{\phi }_{\theta }(\varphi )+(1-t)E^{\phi }_{\theta }(\psi ). \end{equation}
For this purpose, observe that when $C\to \infty $, we have
\[ t\Bigl (\varphi \lor (\phi -C)\Bigr )+(1-t)\Bigl (\psi \lor (\phi -C)\Bigr )\searrow t\varphi +(1-t)\psi . \]
Observe that
\[ \varphi \lor (\phi -C),\psi \lor (\phi -C)\in \mathcal{E}^{\infty }(X,\theta ;\phi ). \]
By 16.5.18, for each $C>0$, we have
\[ E_{\theta }^{\phi }\Biggl ( t\Bigl (\varphi \lor (\phi -C)\Bigr )+(1-t)\Bigl (\psi \lor (\phi -C)\Bigr ) \Biggr )\geq tE_{\theta }^{\phi }\Bigl (\varphi \lor (\phi -C)\Bigr )+(1-t)E_{\theta }^{\phi }\Bigl (\psi \lor (\phi -C)\Bigr ). \]
Letting $C\to \infty $, we conclude (16.5.24) using proposition 16.5.20.
$\square$
Corollary 16.5.25.B5D23DBE0536 The sets $\mathrm{PSH}(X,\theta ;\phi )$, $\mathcal{E}(X,\theta ;\phi )$, $\mathcal{E}^1(X,\theta ;\phi )$, $\mathcal{E}^{\infty }(X,\theta ;\phi )$ are all convex.
Proof.
The sets $\mathrm{PSH}(X,\theta ;\phi )$ and $\mathcal{E}^{\infty }(X,\theta ;\phi )$ are trivially convex.
We next handle $\mathcal{E}(X,\theta ;\phi )$. In this case, suppose that $\varphi ,\psi \in \mathcal{E}(X,\theta ;\phi )$ and $t\in (0,1)$. Then we compute
\[ \begin{split} \int _ X \theta _{t\varphi +(1-t)\psi }^ n=\sum _{j=0}^ n \binom {n}{j}t^ j (1-t)^{n-j} \int _ X \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}\\ =\sum _{j=0}^ n \binom {n}{j}t^ j (1-t)^{n-j} \int _ X \theta _{\phi }^ n=\int _ X \theta _{\phi }^ n, \end{split} \]
where we applied proposition 16.3.10 in the second equality. Therefore, $t\varphi +(1-t)\psi \in \mathcal{E}(X,\theta ;\phi )$.
Finally, let us consider $\mathcal{E}^1(X,\theta ;\phi )$. In this case, suppose that $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$ and $t\in (0,1)$. We have already shown that $t\varphi +(1-t)\psi \in \mathcal{E}(X,\theta ;\phi )$. Furthermore, the energy of this potential is finite due to corollary 16.5.23. Therefore, $t\varphi +(1-t)\psi \in \mathcal{E}^1(X,\theta ;\phi )$ in view of proposition 16.5.22.
$\square$
Lemma 16.5.26.8C85369857E4 Let $\varphi ,\psi ,\gamma \in \mathcal{E}(X,\theta ;\phi )$. Assume that $\gamma \geq \varphi \lor \psi $. Then
\[ \int _ X (\gamma -\varphi )\, \theta _{\psi }^ n\leq 2\int _ X (\gamma -\varphi )\, \theta _{\varphi }^ n+2\int _ X (\gamma -\psi )\, \theta _{\psi }^ n. \]
Proof.
Observe that
6786C4A2C71B
16.5.27
\begin{equation} \{ \gamma >\varphi +2t\} \subseteq \{ \gamma >\psi +t \} \cup \{ \psi >\varphi +t\} \, . \end{equation}
So
\[ \begin{aligned} \int _ X (\gamma -\varphi )\, \theta _{\psi }^ n =& 2\int _0^{\infty } \int _{\{ \gamma >\varphi +2t\} }\, \theta _{\psi }^ n\, \mathrm{d}t \\ \leq & 2\int _0^{\infty } \int _{\{ \gamma >\psi +t\} }\, \theta _{\psi }^ n\, \mathrm{d}t+ 2\int _0^{\infty } \int _{\{ \psi >\varphi +t\} }\, \theta _{\psi }^ n\, \mathrm{d}t\\ \leq & 2\int _ X (\gamma -\psi ) \, \theta _{\psi }^ n+2\int _0^{\infty } \int _{\{ \psi >\varphi +t\} }\, \theta _{\varphi }^ n\, \mathrm{d}t\\ \leq & 2\int _ X (\gamma -\psi ) \, \theta _{\psi }^ n+2\int _0^{\infty } \int _{\{ \gamma >\varphi +t\} }\, \theta _{\varphi }^ n\, \mathrm{d}t\\ =& 2\int _ X (\gamma -\psi ) \, \theta _{\psi }^ n+2\int _ X (\gamma -\varphi ) \, \theta _{\varphi }^ n, \end{aligned} \]
where the second line follows from (16.5.27), while the third line follows from proposition 16.5.5.
$\square$
Corollary 16.5.28.61DB6FF8D92F Let $\varphi ,\psi ,\gamma _1,\ldots ,\gamma _ n\in \mathcal{E}^1(X,\theta ;\phi )$. Then
987A6BCB813E
16.5.29
\begin{equation} \int _ X (\psi -\varphi )\, \theta _{\gamma _1}\wedge \cdots \wedge \theta _{\gamma _ n} \end{equation}
is well-defined and is in $\mathbb {R}$.
Proof.
Step 1. We first consider the case $\psi =\phi $.
We can find $C>0$ so that $\varphi \leq \phi +C$. Therefore, $\phi -\varphi \geq -C$ almost everywhere with respect to $\theta _{\gamma _1}\wedge \cdots \wedge \theta _{\gamma _ n}$. So the integral (16.5.29) is well-defined and takes value in $\mathbb {R}\cup \{ \infty \} $. We show that the integral is finite. For this purpose, we may replace $\varphi $ by $\varphi -C$ and assume that $\varphi \leq \phi $. Put
\[ \eta =n^{-1}(\gamma _1+\cdots +\gamma _ n). \]
Then by the multi-linearity of the non-pluripolar product proposition 13.11.4 6 and positivity, we have
\[ \int _ X (\phi -\varphi )\, \theta _{\gamma _1}\wedge \cdots \wedge \theta _{\gamma _ n}\leq n^ n\int _ X (\phi -\varphi )\, \theta _{\eta }^ n. \]
Thanks to corollary 16.5.25, $\eta \in \mathcal{E}^1(X,\theta ;\phi )$. Choose $A>0$ so that $\eta \leq \phi +A$. Applying 16.5.26 to $\varphi ,\eta $ and $\phi +A$, we get
\[ \int _ X(\phi +A-\varphi )\, \theta _{\eta }^ n \leq 2\int _ X(\phi +A-\varphi )\, \theta _{\varphi }^ n +2\int _ X(\phi +A-\eta )\, \theta _{\eta }^ n<\infty . \]
Hence $\int _ X(\phi -\varphi )\, \theta _{\eta }^ n<\infty $, and Step 1 follows.
Step 2. We handle the general case. Take $C>0$ so that
\[ \psi \leq \phi +C,\quad \varphi \leq \phi +C. \]
Then
\[ (\psi -C)-\phi \leq \psi -\varphi \leq \phi -(\varphi -C). \]
Step 1 applied to $\varphi -C$ and to $\psi -C$ shows that $\phi -(\varphi -C)$ and $\phi -(\psi -C)$ are integrable with respect to $\theta _{\gamma _1}\wedge \cdots \wedge \theta _{\gamma _ n}$. Hence the positive and negative parts of $\psi -\varphi $ are both integrable, so (16.5.29) is well-defined and finite.
$\square$
Proposition 16.5.30.4280CB108A7E Given $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$, we have the following cocycle equality
F8225060CBAC
16.5.31
\begin{equation} E^{\phi }_{\theta }(\psi )-E^{\phi }_{\theta }(\varphi )=\frac{1}{n+1}\sum _{j=0}^ n \int _ X (\psi -\varphi )\, \theta _{\psi }^ j\wedge \theta _{\varphi }^{n-j}. \end{equation}
As a consequence of (
16.5.31), for $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$, (
16.5.11) continues to hold. Note that each term on the right-hand side of (
16.5.31) is finite due to corollary
16.5.28.
Proof.
Since the conclusion is known when $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ (see (16.5.15)), it suffices to prove that for each $k=0,\ldots ,n$, we have
9D55DDF004D1
16.5.32
\begin{equation} \lim _{C\to \infty } \int _ X \Bigl ( \varphi \lor (\phi -C)-\psi \lor (\phi -C) \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=\int _ X (\varphi -\psi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}. \end{equation}
For this purpose, we may assume that $\varphi ,\psi \leq \phi $ and it suffices to establish the following:
989753C108DD
16.5.33
\begin{equation} \lim _{C\to \infty } \int _ X \Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=\int _ X (\varphi -\phi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}. \end{equation}
Fix $C>0$. We compute the difference as follows:
\[ \begin{aligned} & \int _ X \Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}-\int _ X (\varphi -\phi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}\\ =& \int _{\{ \min \{ \varphi ,\psi \} >\phi -C\} }\left( \varphi -\phi \right) \, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}-\int _ X (\varphi -\phi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}\\ & +\int _{\{ \min \{ \varphi ,\psi \} \leq \phi -C\} }\Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}\\ =& -\int _{\{ \min \{ \varphi ,\psi \} \leq \phi -C\} }\left( \varphi -\phi \right) \, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}+\\ & +\int _{\{ \min \{ \varphi ,\psi \} \leq \phi -C\} }\Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k} \end{aligned} \]
We first handle the first term. In order to show that it converges to $0$ when $C\to \infty $, it suffices to show that
\[ \int _ X\left( \phi -\varphi \right) \, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}<\infty . \]
For this purpose, it suffices to prove a stronger inequality:
0BA9923E2163
16.5.34
\begin{equation} \int _ X\left( \phi -\varphi \right) \, \theta _{(\varphi +\psi )/2}^ n<\infty . \end{equation}
From corollary 16.5.25, we know that $(\varphi +\psi )/2\in \mathcal{E}^1(X,\theta ;\phi )$. Therefore, (16.5.34) follows from 16.5.26.
It remains to handle the second term. It then suffices to prove the following:
F913F2DBE96C
16.5.35
\begin{equation} \begin{split} \lim _{C\to \infty } C\int _{\{ \varphi \leq \phi -C\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=& 0,\\ \lim _{C\to \infty } C\int _{\{ \psi \leq \phi -C\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=& 0. \end{split} \end{equation}
By symmetry, it suffices to handle the first. Observe the following inclusions:
\[ \{ \varphi \leq \phi -C\} \subseteq \left\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\Bigl ( \psi \lor (\phi -C)+\phi -C \Bigr ) \right\} \subseteq \{ \varphi \leq \phi -C/2\} . \]
By proposition 16.5.5, we find
\[ \begin{aligned} & C\int _{\{ \varphi \leq \phi -C\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}\\ \leq & C\int _{\left\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )\right\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}\\ \leq & 2^{n-k}C\int _{\left\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )\right\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )}^{n-k}\\ \leq & 2^{n-k}C\int _{\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )\} } \theta _{\varphi \lor (\phi -C)}^ n\\ \leq & 2^{n-k}C\int _{\{ \varphi \leq \phi -C/2\} } \theta _{\varphi \lor (\phi -C)}^ n\\ = & 2^{n-k}C\int _{\{ \varphi \leq \phi -C/2\} } \theta _{\varphi }^ n. \end{aligned} \]
Here on the final step, we have applied the following argument:
\[ \begin{aligned} & \int _{\left\{ \varphi \leq \phi -C/2 \right\} } \theta _{\varphi \lor (\phi -C)}^ n & \\ = & \int _ X \theta _{\varphi \lor (\phi -C)}^ n-\int _{\left\{ \varphi > \phi -C/2 \right\} } \theta _{\varphi \lor (\phi -C)}^ n & \\ = & \int _ X \theta _{\varphi }^ n-\int _{\left\{ \varphi > \phi -C/2 \right\} } \theta _{\varphi }^ n & \textup{by proposition \href{../../tag/9E81D36B1290/}{13.11.4} \href{../../tag/9E81D36B1290/#sgpt-1C036AD6}{1}}\\ =& \int _{\left\{ \varphi \leq \phi -C/2 \right\} } \theta _{\varphi }^ n. & \end{aligned} \]
On the other hand,
\[ \varlimsup _{C\to \infty } C\int _{\{ \varphi \leq \phi -C/2 \} } \theta _{\varphi }^ n\leq 2 \varlimsup _{C\to \infty } \int _{\{ \varphi \leq \phi -C \} }(\phi -\varphi )\, \theta _{\varphi }^ n=0, \]
since $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$. We conclude (16.5.35).
$\square$
Proposition 16.5.36.51CFFEE97C34 Let $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$. Then
3.132
\begin{equation} \int _ X \left(\psi -\varphi \right)\theta ^ n_{\psi }\leq E_{\theta }^{\phi }(\psi )-E_{\theta }^{\phi }(\varphi )\leq \int _ X \left(\psi -\varphi \right)\theta ^ n_{\varphi }. \end{equation}
Proof.
Thanks to (16.5.32), this can be reduced to the case where $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$, which is established in 16.5.16.
$\square$
Proposition 16.5.37.3DA562555236 Let $(\varphi _ i)_{i\in I}$ be a net in $\mathcal{E}^1(X,\theta ;\phi )$. Assume that $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$ and either one of the following conditions holds:
ACF55428594E
$(\varphi _ i)_ i$ is a decreasing net with $\varphi =\inf _{i\in I}\varphi _ i$;
CB79DD733120
$(\varphi _ i)_ i$ is an increasing net with almost everywhere limit $\varphi $.
Then
60942A58F028
16.5.38
\begin{equation} \lim _{i\in I} E_{\theta }^{\phi }(\varphi _ i)=E_{\theta }^{\phi }(\varphi ). \end{equation}
Proof.
The decreasing case is already proved in proposition 16.5.20, so let us focus on the case where $(\varphi _ i)_ i$ is increasing.
We first handle the case where $I=\mathbb {Z}_{>0}$. Namely, $(\varphi _ i)_ i$ is an increasing sequence.
Step 1. We first prove the result when $\varphi _ j,\varphi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ for each $j\geq 1$.
Due to corollary 13.13.12, we have
\[ \lim _{j\to \infty } \int _ X \theta _{\varphi }^ k\wedge \theta _{\varphi _ j}^{n-k}=\int _ X \theta _{\varphi }^ n \]
for each $k=0,\ldots ,n$. By (16.5.15), we have
\[ \begin{aligned} 0\leq & E_{\theta }^{\phi }(\varphi )-E_{\theta }^{\phi }(\varphi _ j)\\ =& \frac{1}{n+1}\sum _{k=0}^ n \int _ X(\varphi -\varphi _ j)\, \theta _{\varphi }^ k\wedge \theta _{\varphi _ j}^{n-k}. \end{aligned} \]
We then apply theorem 13.13.9 to conclude that the right-hand side converges to $0$.
Step 2. We prove the general case. The $\leq $ direction in (16.5.38) follows from the monotonicity of $E_{\theta }^{\phi }$, as proved in 16.5.14. It remains to prove the $\geq $ direction.
After subtracting a constant from all $\varphi _ j$ and from $\varphi $, which shifts all the energies by the same constant in view of (16.5.13), we may assume that $\varphi \leq \phi $. For each $C>0$ and $j\geq 1$, we let
\[ \varphi _ C\coloneqq \varphi \lor (\phi -C),\quad \varphi _{j,C}\coloneqq \varphi _ j\lor (\phi -C). \]
By Step 1, for each $C>0$, we have
\[ \lim _{j\to \infty } E_{\theta }^{\phi }(\varphi _{j,C})= E_{\theta }^{\phi }(\varphi _{C})\geq E_{\theta }^{\phi }(\varphi ). \]
It suffices therefore to show that
7AAAE969D14B
16.5.39
\begin{equation} \lim _{C\to \infty } E_{\theta }^{\phi }(\varphi _{j,C})-E_{\theta }^{\phi }(\varphi _{j})=0\quad \textup{uniformly for $j$ in a tail}. \end{equation}
Fix $i\geq 1$. For $j\geq i$, we compute
\[ \begin{aligned} & E_{\theta }^{\phi }(\varphi _{j,C})-E_{\theta }^{\phi }(\varphi _{j})\\ \leq & \int _ X \left( \varphi _{j,C}-\varphi _{j}\right)\, \theta _{\varphi _{j}}^ n & \textup{by proposition \href{../../tag/51CFFEE97C34/}{16.5.36}}\\ \leq & \int _{\{ \varphi _ j\leq \phi -C\} } (\phi -C-\varphi _ j)\, \theta _{\varphi _ j}^ n\\ =& \int _ C^{\infty }\int _{\{ \varphi _ j\leq \phi -t\} } \theta _{\varphi _ j}^ n\, \mathrm{d}t\\ \leq & \int _ C^{\infty }\int _{\{ \varphi _ i\leq (\varphi _ j+\phi -t)/2\} } \theta _{\varphi _ j}^ n\, \mathrm{d}t\\ \leq & 2^ n\int _ C^{\infty }\int _{\{ \varphi _ i\leq (\varphi _ j+\phi -t)/2\} } \theta _{(\varphi _ j+\phi -t)/2}^ n\, \mathrm{d}t\\ \leq & 2^ n\int _ C^{\infty }\int _{\{ \varphi _ i\leq (\varphi _ j+\phi -t)/2\} } \theta _{\varphi _ i}^ n\, \mathrm{d}t & \textup{by proposition \href{../../tag/FDFD0A6F6287/}{16.5.5}}\\ \leq & 2^ n\int _ C^{\infty }\int _{\{ \varphi _ i\leq \phi -t/2\} } \theta _{\varphi _ i}^ n\, \mathrm{d}t\\ = & 2^{n+1}\int _{\{ \varphi _ i< \phi -C/2\} } (\phi -\varphi _ i-C/2)\, \theta _{\varphi _ i}^ n\\ = & 2^{n+1}\int _{\{ \varphi _ i< \phi -C/2\} } (\phi -\varphi _ i)\, \theta _{\varphi _ i}^ n-C 2^ n\int _{\{ \varphi _ i< \phi -C/2\} }\theta _{\varphi _ i}^ n. \end{aligned} \]
As $C\to \infty $, the first term converges to $0$ since $\varphi _ i\in \mathcal{E}^1(X,\theta ;\phi )$. As for the second term,
\[ \lim _{C\to \infty }C\int _{\{ \varphi _ i< \phi -C/2 \} } \theta _{\varphi _ i}^ n\leq 2 \lim _{C\to \infty }\int _{\{ \varphi _ i< \phi -C/2 \} }(\phi -\varphi _ i)\, \theta _{\varphi _ i}^ n = 0. \]
We conclude (16.5.39).
We now pass from sequences to nets. By 16.5.14, the net $(E_{\theta }^{\phi }(\varphi _ i))_{i\in I}$ is increasing and bounded above by $E_{\theta }^{\phi }(\varphi )$. Let
\[ L\coloneqq \lim _{i\in I}E_{\theta }^{\phi }(\varphi _ i). \]
Then $L\leq E_{\theta }^{\phi }(\varphi )$.
By proposition 5.14.6, we have
\[ \varphi =\operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _ i. \]
Moreover, $\varphi _ i\leq \varphi $ for all $i\in I$, so the family is locally uniformly bounded above. Applying Choquet's lemma proposition 5.14.3, we can find a countable subset $J\subseteq I$ such that
\[ \varphi =\operatorname *{sup}_{j\in J}{}\! {}^{*}\varphi _ j. \]
Enumerate $J=\{ j_1,j_2,\ldots \} $, repeating elements if necessary. Since $I$ is directed, we can choose an increasing sequence $(i_ k)_{k\geq 1}$ in $I$ such that $i_ k$ dominates $j_1,\ldots ,j_ k$. It follows that $(\varphi _{i_ k})_ k$ increases almost everywhere to $\varphi $. Applying the sequence case to this sequence gives
\[ \lim _{k\to \infty }E_{\theta }^{\phi }(\varphi _{i_ k})=E_{\theta }^{\phi }(\varphi ). \]
Since $E_{\theta }^{\phi }(\varphi _{i_ k})\leq L$ for every $k$, we conclude $E_{\theta }^{\phi }(\varphi )\leq L$.
$\square$
Next we want to prove that $\mathcal{E}^1(X,\theta ;\phi )$ is closed under the rooftop operator. For this purpose, we shall need a few preliminary results.
We shall approximate the rooftop operator by solutions of certain Monge–Ampère equations.
Lemma 16.5.40.4C2A27A8F97A Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Then there is $\gamma \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ such that
4E22C768C562
16.5.41
\begin{equation} \theta _{\gamma }^ n=\mathrm{e}^{\gamma -\varphi }\, \theta _{\varphi }^ n+\mathrm{e}^{\gamma -\psi }\, \theta _{\psi }^ n. \end{equation}
It is not clear if $\int _ X\mathrm{e}^{-\varphi }\theta _{\varphi }^ n<\infty $, hence we cannot say that $\mathrm{e}^{-\varphi }\theta _{\varphi }^ n+\mathrm{e}^{-\psi }\theta _{\psi }^ n$ is a Radon measure, so theorem
16.5.9 is not directly applicable.
Proof.
For each $j\geq 1$, let $\varphi _ j\coloneqq \varphi \lor (-j)$, $\psi _ j\coloneqq \psi \lor (-j)$. These functions are not necessarily $\theta $-psh. Let
\[ \mu _ j=\mathrm{e}^{-\varphi _ j}\theta _{\varphi }^ n+\mathrm{e}^{-\psi _ j}\theta _{\psi }^ n. \]
By theorem 16.5.9, we can find $\gamma _ j\in \mathcal{E}(X,\theta ;\phi )$ such that
5B35DF529607
16.5.42
\begin{equation} \theta _{\gamma _ j}^ n=\mathrm{e}^{\gamma _ j}\mu _ j=\mathrm{e}^{\gamma _ j-\varphi _ j}\theta _{\varphi }^ n+\mathrm{e}^{\gamma _ j-\psi _ j}\theta _{\psi }^ n. \end{equation}
Take a constant $C>0$ so that $\psi -2C\leq \varphi \leq \psi +2C$. Let
\[ \eta \coloneqq \frac{\varphi +\psi }{2}-C-n\log 2. \]
Then $\eta \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ and we have the following inequalities:
\[ \begin{aligned} \theta _{\eta }^ n =& \theta _{(\varphi +\psi )/2}^ n\\ \geq & 2^{-n}\theta _{\varphi }^ n+2^{-n}\theta _{\psi }^ n\\ \geq & 2^{-n}\mathrm{e}^{(\varphi +\psi )/2}\mathrm{e}^{-C}\left( \mathrm{e}^{-\varphi }\theta _{\varphi }^ n+\mathrm{e}^{-\psi }\theta _{\psi }^ n \right)\\ \geq & 2^{-n}\mathrm{e}^{(\varphi +\psi )/2}\mathrm{e}^{-C}\left( \mathrm{e}^{-\varphi _ j}\theta _{\varphi }^ n+\mathrm{e}^{-\psi _ j}\theta _{\psi }^ n \right)\\ = & \mathrm{e}^{\eta }\mu _ j. \end{aligned} \]
Hence, $\gamma _ j\geq \eta $ by theorem 16.5.9. By theorem 16.5.9 again, $\gamma _ j$ is decreasing in $j$; let $\gamma =\inf _{j>0}\gamma _ j$. Then $\gamma \geq \eta $, hence $\gamma \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
Now observe that
\[ \theta _{\gamma _ j}^ n\rightharpoonup \theta _{\gamma }^ n \]
as $j\to \infty $, by theorem 13.13.9. Finally observe that there is a constant $C'>0$ so that
\[ \gamma _1\leq \varphi +C',\quad \gamma _1\leq \psi +C', \]
since $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Therefore, for each $j>0$,
\[ \gamma _ j\leq \varphi _ j+C',\quad \gamma _ j\leq \psi _ j+C'. \]
Hence, (16.5.41) follows from (16.5.42) and the dominated convergence theorem by letting $j\to \infty $.
$\square$
We also need a few integral estimates.
Lemma 16.5.43.BAAC43C43BD5 Let $\varphi ,\psi ,\gamma \in \mathcal{E}(X,\theta ;\phi )$. Assume that $\varphi \leq \psi \leq \gamma $. Then
\[ \int _ X (\gamma -\psi )\, \theta _{\psi }^ n\leq 2^{n+1}\int _ X (\gamma -\varphi )\, \theta _{\varphi }^ n. \]
Proof.
Observe that for any $t\geq 0$,
7058C51CA443
16.5.44
\begin{equation} \begin{split} \{ \gamma >\psi +2t\} \setminus \{ \psi =-\infty \} \subseteq \left\{ (\gamma +\psi )/2>\psi +t\right\} \\ \subseteq \left\{ (\gamma +\psi )/2>\varphi +t\right\} \subseteq \{ \gamma >\varphi +t\} . \end{split} \end{equation}
So
\[ \begin{aligned} \int _ X (\gamma -\psi )\, \theta _{\psi }^ n =& 2\int _0^{\infty } \int _{\{ \gamma -\psi >2t\} }\, \theta _{\psi }^ n\, \mathrm{d}t & \\ \leq & 2\int _0^{\infty } \int _{\left\{ (\gamma +\psi )/2>\psi +t\right\} }\, \theta _{\psi }^ n\, \mathrm{d}t & \textup{by (\href{../../tag/7058C51CA443/}{16.5.44})}\\ \leq & 2^{n+1}\int _0^{\infty } \int _{\left\{ (\gamma +\psi )/2>\varphi +t\right\} }\, \theta _{(\gamma +\psi )/2}^ n\, \mathrm{d}t & \\ \leq & 2^{n+1}\int _0^{\infty } \int _{\left\{ (\gamma +\psi )/2>\varphi +t\right\} }\, \theta _{\varphi }^ n\, \mathrm{d}t & \textup{by proposition \href{../../tag/FDFD0A6F6287/}{16.5.5}}\\ \leq & 2^{n+1}\int _0^{\infty } \int _{\left\{ \gamma >\varphi +t\right\} }\, \theta _{\varphi }^ n\, \mathrm{d}t & \textup{by (\href{../../tag/7058C51CA443/}{16.5.44})}\\ =& 2^{n+1} \int _ X (\gamma -\varphi )\, \theta _{\varphi }^ n & . \end{aligned} \]
This completes the proof.
$\square$
Lemma 16.5.45.7BA1823201EA Let $\varphi _ j\in \mathcal{E}(X,\theta ;\phi )$ ($j\in \mathbb {Z}_{>0}$) and $\gamma \in \mathcal{E}(X,\theta ;\phi )$. Assume that $\varphi _ j\leq \gamma $ for each $j$ and that $\varphi _ j$ converges to $\varphi \in \mathrm{PSH}(X,\theta )$ with respect to the $L^1$-topology. Assume that there is a constant $A>0$ such that for any $j>0$,
\[ \int _ X (\varphi _ j-\gamma )\, \theta _{\varphi _ j}^ n\geq -A. \]
Then $\varphi \in \mathcal{E}(X,\theta ;\phi )$ and
2E33CEABAB85
16.5.46
\begin{equation} \int _ X (\varphi -\gamma )\, \theta _{\varphi }^ n\geq - 2^{n+3}A. \end{equation}
Proof.
Replacing simultaneously $\gamma ,\varphi _ j,\varphi $ by $\gamma -B,\varphi _ j-B,\varphi -B$ for $B\gg 1$, we may assume that $\gamma \leq 0$. The assumptions and the conclusion are invariant under this translation. Since $\gamma \in \mathcal E(X,\theta ;\phi )$ and $P_\theta [\phi ]=\phi $, we then have $\gamma \leq \phi $.
Step 1. Assume that $(\varphi _ j)_ j$ is decreasing. In this case, we prove
B09C34234351
16.5.47
\begin{equation} \int _ X (\varphi -\gamma )\, \theta _{\varphi }^ n\geq -4A. \end{equation}
By 16.5.26, for any $j, k>0$,
\[ \int _ X (\varphi _ j-\gamma )\, \theta _{\varphi _ k}^ n\geq -4A. \]
For any $C>0$,
\[ \int _ X \Bigl (\varphi _ j \lor (\gamma -C)-\gamma \Bigr )\, \theta _{\varphi _ k}^ n\geq \int _ X (\varphi _ j-\gamma ) \, \theta _{\varphi _ k}^ n\geq -4A. \]
Letting $k\to \infty $, by theorem 13.13.9, we find
\[ \int _ X \Bigl (\varphi _ j \lor (\gamma -C)-\gamma \Bigr )\, \theta _{\varphi }^ n\geq -4A. \]
Letting $j\to \infty $, by the monotone convergence theorem, we get
\[ \int _ X \Bigl (\varphi \lor (\gamma -C)-\gamma \Bigr )\, \theta _{\varphi }^ n\geq -4A. \]
Then we let $C\to \infty $, again by the monotone convergence theorem, we conclude (16.5.47).
Step 2. In general, let
\[ \psi _ j=\operatorname *{sup}_{k\geq j}{}\! {}^{*} \varphi _ k. \]
Then $\varphi =\inf _{j>0}\psi _ j$ due to corollary 5.14.8.
For each $C>0$, let
\[ \psi _{j,C}=\psi _ j \lor (\gamma -C),\quad \varphi _ C=\varphi \lor (\gamma -C). \]
Then $\psi _{j,C},\varphi _ C\in \mathcal{E}(X,\theta ;\phi )$ since $\gamma -C\in \mathcal{E}(X,\theta ;\phi )$ thanks to the monotonicity theorem theorem 13.13.11. Observe that $\psi _{j,C}$ decreases to $\varphi _ C$ as $j\to \infty $. Moreover,
\[ \gamma \geq \psi _{j,C} \geq \psi _ j\geq \varphi _ j. \]
By 16.5.43,
\[ \int _ X (\psi _{j,C}-\gamma )\, \theta _{\psi _{j,C}}^ n\geq -2^{n+1}A. \]
By Step 1,
053AD38249AF
16.5.48
\begin{equation} \int _ X(\varphi _ C-\gamma )\, \theta _{\varphi _ C}^ n\geq -2^{n+3}A. \end{equation}
In particular, by the plurifine locality of the non-pluripolar product proposition 13.11.4 1,
\[ \int _{\{ \varphi >\gamma -C\} }(\varphi -\gamma )\, \theta _{\varphi }^ n\geq -2^{n+3}A. \]
Letting $C\to \infty $, we conclude (16.5.46). As a consequence,
\[ \int _ X (\gamma -\varphi )\, \theta _{\varphi }^ n<\infty . \]
We still have to prove that $\varphi \in \mathcal{E}(X,\theta ;\phi )$. In fact, by (16.5.48), for any $C>0$,
\[ \int _{\{ \varphi \leq \gamma -C\} }\, \theta _{\varphi _ C}^ n\leq \frac{1}{C}\int _ X (\gamma -\varphi _ C)\, \theta _{\varphi _ C}^ n\leq 2^{n+3}C^{-1}A. \]
Using the monotonicity theorem theorem 13.13.11 and the plurifine locality of the non-pluripolar product proposition 13.11.4 1, we find
\[ \int _ X \theta _{\phi }^ n=\int _ X \theta _{\varphi _ C}^ n=\int _{\{ \varphi \leq \gamma -C\} }\, \theta _{\varphi _ C}^ n+\int _{\{ \varphi >\gamma -C\} }\, \theta _{\varphi }^ n. \]
Letting $C\to \infty $, we conclude that
\[ \int _ X \theta _{\varphi }^ n=\int _ X \theta _{\phi }^ n. \]
This completes the proof.
$\square$
Proposition 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 $.
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.
$\square$
Proposition 16.5.52.420BBF2E1362 Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$ be potentials such that $\varphi \preceq \psi \preceq \phi $. Assume that $\varphi \in \mathcal{E}(X,\theta ;\phi )$ (resp. $\mathcal{E}^1(X,\theta ;\phi )$, $\mathcal{E}^{\infty }(X,\theta ;\phi )$), then so is $\psi $.
Proof.
We may assume that $\varphi \leq \psi $.
The case $\mathcal{E}^{\infty }(X,\theta ;\phi )$ is trivial. The case $\mathcal{E}(X,\theta ;\phi )$ follows from theorem 13.13.11. The case $\mathcal{E}^1(X,\theta ;\phi )$ follows from the characterization of $\mathcal{E}^1(X,\theta ;\phi )$ in proposition 16.5.22 and the monotonicity of $E_{\theta }^{\phi }$ proved in 16.5.14.
$\square$
Proposition 16.5.53.322796EE46DE Let $(\varphi _ i)_{i\in I}$ be a uniformly bounded from above non-empty family in $\mathcal{E}(X,\theta ;\phi )$ (resp. $\mathcal{E}^1(X,\theta ;\phi )$, $\mathcal{E}^{\infty }(X,\theta ;\phi )$), then so is $\operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _ i$.
Proof.
Thanks to proposition 16.5.52, it suffices to show that
\[ \operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _ i\preceq \phi . \]
Since $\phi $ is model and $\varphi _ i\preceq \phi $, we know that
\[ \varphi _ i-\sup _ X \varphi _ i \leq \phi \]
for any $i\in I$. Since $(\varphi _ i)_{i\in I}$ is uniformly bounded from above by assumption, our assertion follows.
$\square$
Proposition 16.5.54.75FA2659491D Let $\varphi ,\psi \in \mathcal{E}(X,\theta ;\phi )$. Then
\[ P_{\theta }[\varphi ](\psi )=\operatorname *{sup}_{C > 0}{}\! {}^{*} (\varphi +C)\land \psi =\psi . \]
Proof.
The first equality follows from the definition of $P_{\theta }[\varphi ](\psi )$. We focus on the second equality.
Since for each $C\geq 0$,
\[ (\varphi \land \psi +C)\land \psi \leq (\varphi +C)\land \psi \leq \psi , \]
we may replace $\varphi $ by $\varphi \land \psi $ (cf. proposition 16.5.49) and assume that $\varphi \leq \psi $.
Let
\[ \gamma \coloneqq \operatorname *{sup}_{C > 0}{}\! {}^{*} (\varphi +C)\land \psi . \]
Observe that $\gamma \leq \psi $, and $\gamma \in \mathcal{E}(X,\theta ;\phi )$ due to proposition 16.5.49 and proposition 16.5.53. Then corollary 16.3.14 guarantees that
\[ \int _{\{ \gamma <\psi \} } \theta _{\gamma }^ n=0. \]
Therefore, we can apply the domination principle theorem 13.13.43 to conclude that $\gamma =\psi $.
$\square$
Lemma 16.5.55.88E2DBCC3C35 Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Define
\[ \eta _ t\coloneqq \Bigl ( (1-t)\varphi +t\psi \Bigr )\land \psi ,\quad t\in [0,1]. \]
Then $E_{\theta }^{\phi }(\eta _ t)$ is differentiable for $t\in [0,1]$ and
F8B858957CF5
16.5.56
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}E_{\theta }^{\phi }(\eta _ t)=\int _{X} \Bigl ( \psi -\min \{ \varphi ,\psi \} \Bigr )\, \theta _{\eta _ t}^ n. \end{equation}
The derivatives at $t=0$ or $1$ are understood as one-sided derivatives.
Proof.
Let us prove (16.5.56) with right-derivative instead of derivative, and $t\in [0,1)$. The left-derivative case is completely parallel.
For each $t\in [0,1]$, we let
\[ f_ t=\min \{ (1-t)\varphi +t\psi , \psi \} . \]
Fix $t\in [0,1)$ and $s>0$ small enough so that $t+s<1$. Thanks to proposition 16.5.36, we have
\[ \begin{aligned} E_{\theta }^{\phi }(\eta _{t+s})-E_{\theta }^{\phi }(\eta _{t})\leq & \int _ X \left(\eta _{t+s}-\eta _ t\right)\, \theta _{\eta _ t}^ n \\ =& \int _ X \left(\eta _{t+s}-f_ t\right)\, \theta _{\eta _ t}^ n & \textup{by \href{../../tag/4CF7246E6E5A/}{16.3.3}}\\ \leq & \int _ X \left(f_{t+s}-f_ t \right)\, \theta _{\eta _ t}^ n\\ =& s\int _ X \Bigl (\psi -\min \{ \varphi ,\psi \} \Bigr )\, \theta _{\eta _ t}^ n. \end{aligned} \]
Similarly, we get
\[ E_{\theta }^{\phi }(\eta _{t+s})-E_{\theta }^{\phi }(\eta _{t})\geq s\int _ X \Bigl (\psi -\min \{ \varphi ,\psi \} \Bigr )\, \theta _{\eta _{t+s}}^ n. \]
Observe that $\eta _{t+s}$ converges in capacity (in fact, even uniformly) to $\eta _ t$ as $s\to 0+$. The desired result (16.5.56) is then just a consequence of theorem 13.12.8.
$\square$
Lemma 16.5.57.8B9E394D5AE5 Let $\varphi ,\psi \in \mathcal{E}^{1}(X,\theta ;\phi )$, and set
\[ \varphi _ t\coloneqq (1-t)\varphi +t\psi ,\qquad t\in [0,1]. \]
Then the function
\[ t\mapsto E_{\theta }^{\phi }(\varphi _ t) \]
is differentiable on $[0,1]$, where the derivative at the endpoints is understood as the one-sided derivative. Moreover,
B09DAFFD8EA2
16.5.58
\begin{equation} \frac{\mathrm d}{\mathrm dt}E_{\theta }^{\phi }(\varphi _ t) = \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n. \end{equation}
Proof.
By corollary 16.5.25, $\varphi _ t\in \mathcal E^1(X,\theta ;\phi )$ for all $t\in [0,1]$. For $a=0,\ldots ,n$, put
\[ M_ a=\int _ X(\psi -\varphi )\, \theta _{\psi }^ a\wedge \theta _{\varphi }^{n-a}. \]
Note that each $M_ a$ is finite by corollary 16.5.28. By the multi-linearity of the non-pluripolar product proposition 13.11.4 6, the quantity
\[ \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n \]
is therefore a polynomial in $t$ with coefficients given by the $M_ a$'s. In particular, the right-hand side of (16.5.58) is finite and continuous in $t$.
We prove the formula for the right derivative at a point $t\in [0,1)$. Let $s>0$ be small enough so that $t+s\leq 1$. By the cocycle formula proposition 16.5.30, we have
\[ \begin{aligned} \frac{E_{\theta }^{\phi }(\varphi _{t+s})-E_{\theta }^{\phi }(\varphi _ t)}{s} =& \frac{1}{(n+1)s}\sum _{j=0}^ n \int _ X(\varphi _{t+s}-\varphi _ t)\, \theta _{\varphi _{t+s}}^ j\wedge \theta _{\varphi _ t}^{n-j} \\ =& \frac{1}{n+1}\sum _{j=0}^ n \int _ X(\psi -\varphi )\, \theta _{\varphi _{t+s}}^ j\wedge \theta _{\varphi _ t}^{n-j}. \end{aligned} \]
Again by proposition 13.11.4 6, for each fixed $j$ the last integral is a polynomial in $s$, expressed as a linear combination of the $M_ a$'s with coefficients depending on $t$. Letting $s\to 0+$, it converges to
\[ \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n. \]
Therefore,
\[ \lim _{s\to 0+} \frac{E_{\theta }^{\phi }(\varphi _{t+s})-E_{\theta }^{\phi }(\varphi _ t)}{s} = \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n. \]
The proof for the left derivative is identical, using $s<0$ with $t+s\in [0,1]$. Thus the left and right derivatives agree for $t\in (0,1)$, and the endpoint derivatives are given by the same formula.
$\square$
In particular, we have a diamond-like inequality.
Corollary 16.5.59.4A849C02F386 Let $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$. Then
B424028202D9
16.5.60
\begin{equation} E_{\theta }^{\phi }(\varphi \lor \psi )+E_{\theta }^{\phi }(\varphi \land \psi )\geq E_{\theta }^{\phi }(\psi )+E_{\theta }^{\phi }(\varphi ). \end{equation}
Proof.
Step 1. We first reduce to the case where $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
Assume that (16.5.60) is known in that case. For each $C>0$, we let
\[ \varphi _ C=\varphi \lor (\phi -C),\quad \psi _ C=\psi \lor (\phi -C). \]
Then
\[ E_{\theta }^{\phi }(\varphi _ C\lor \psi _ C)+E_{\theta }^{\phi }(\varphi _ C\land \psi _ C)\geq E_{\theta }^{\phi }(\psi _ C)+E_{\theta }^{\phi }(\varphi _ C). \]
As $C\to \infty $, we have
\[ \varphi _ C\searrow \varphi ,\quad \psi _ C\searrow \psi ,\quad \varphi _ C\lor \psi _ C\searrow \varphi \lor \psi , \]
and, by the monotonicity and maximality of the rooftop,
\[ \varphi _ C\land \psi _ C\searrow \varphi \land \psi . \]
Applying proposition 16.5.20 to these four decreasing sequences gives (16.5.60).
Step 2. We assume that $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
Thanks to 16.5.57 and the plurifine locality of the non-pluripolar product proposition 13.11.4 1, we have
\[ \begin{aligned} E_{\theta }^{\phi }(\varphi \lor \psi )-E_{\theta }^{\phi }(\varphi )=& \int _0^1\int _ X (\varphi \lor \psi -\varphi )\, \theta _{(1-t)\varphi +t(\varphi \lor \psi )}^ n\, \mathrm{d}t\\ =& \int _0^1 \int _{\{ \psi >\varphi \} } (\psi -\varphi )\, \theta _{(1-t)\varphi +t\psi }^ n\, \mathrm{d}t. \end{aligned} \]
On the other hand,
\[ \begin{aligned} & E_{\theta }^{\phi }(\psi )-E_{\theta }^{\phi }(\varphi \land \psi )\\ =& \int _0^1 \int _ X (\psi -\min \{ \varphi ,\psi \} )\, \theta _{((1-t)\varphi +t\psi )\land \psi }^ n\, \mathrm{d}t\\ \leq & \int _0^1 \int _{\{ \varphi <\psi \} } (\psi -\varphi )\, \theta _{(1-t)\varphi +t\psi }^ n\, \mathrm{d}t& \textup{by \href{../../tag/4CF7246E6E5A/}{16.3.3}}. \end{aligned} \]
Adding these equations together, we conclude (16.5.60).
$\square$
Pome
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