Proof.
By corollary 16.9.36, we may assume that $\theta _1,\theta _2$ are both Kähler forms. We denote them by $\omega _1,\omega _2$ instead. Let $\omega =\omega _1+\omega _2$.
1+2 $\implies $ 3. It suffices to show that for each $r=0,\ldots ,n$,
\[ 2\int _ X \omega _{(\varphi _ i+\psi _ i)\lor (\varphi +\psi )}^ r\wedge \omega ^{n-r}-\int _ X \omega _{\varphi _ i+\psi _ i}^ r\wedge \omega ^{n-r}-\int _ X \omega _{\varphi +\psi }^ r\wedge \omega ^{n-r}\to 0. \]
Observe that for each $i\in I$,
\[ (\varphi _ i+\psi _ i)\lor (\varphi +\psi )\leq (\varphi _ i\lor \varphi )+(\psi _ i \lor \psi ). \]
The expression with the actual rooftop is non-negative and is bounded above by the same expression with this larger potential in place of $(\varphi _ i+\psi _ i)\lor (\varphi +\psi )$. Thus, it suffices to show that
\[ 2\int _ X \omega _{(\varphi _ i\lor \varphi )+(\psi _ i\lor \psi )}^ r\wedge \omega ^{n-r}-\int _ X \omega _{\varphi _ i+\psi _ i}^ r\wedge \omega ^{n-r}-\int _ X \omega _{\varphi +\psi }^ r\wedge \omega ^{n-r}\to 0. \]
The left-hand side is a linear combination of
\[ 2\int _ X \omega _{1,\varphi _ i\lor \varphi }^ a\wedge \omega _{2,\psi _ i\lor \psi }^{r-a}\wedge \omega ^{n-r}-\int _ X \omega _{1,\varphi _ i}^ a\wedge \omega _{2,\psi _ i}^{r-a} \wedge \omega ^{n-r}-\int _ X \omega _{1,\varphi }^ a\wedge \omega _{2,\psi }^{r-a} \wedge \omega ^{n-r} \]
with $a=0,\ldots ,r$. Observe that $\varphi _ i\lor \varphi \xrightarrow {d_ S}\varphi $ and $\psi _ i\lor \psi \xrightarrow {d_ S}\psi $ by corollary 16.9.9, each term tends to $0$ by theorem 16.9.30.
1+3 $\implies $ 2. For each $C\geq 1$, from the direction we already proved together with 16.9.41,
\[ C\varphi _ i+\psi _ i\xrightarrow {d_{S}} C\varphi +\psi . \]
By theorem 16.9.30, for each $j=0,\ldots ,n$,
\[ \begin{split} & \lim _{i\in I} \int _ X \Bigl (C\omega _1+\omega _2+\mathrm{dd}^{\mathrm{c}}(C\varphi _ i+\psi _ i)\Bigr )^ j\wedge \omega _2^{n-j}\\ = & \int _ X \Bigl (C\omega _1+\omega _2+\mathrm{dd}^{\mathrm{c}}(C\varphi +\psi )\Bigr )^ j\wedge \omega _2^{n-j}. \end{split} \]
Both sides are polynomials in $C$. Comparing the constant terms gives
FF05887CF686
16.9.46
\begin{equation} \lim _{i\in I} \int _ X \omega _{2,\psi _ i}^ j\wedge \omega _2^{n-j}=\int _ X \omega _{2,\psi }^ j\wedge \omega _2^{n-j}. \end{equation}
Therefore, 2 follows if $\psi _ i\geq \psi $ for each $i$ by 16.9.7.
Next we prove the general case. By the direction that we already proved, we know that $\varphi _ i+\psi \xrightarrow {d_{S}} \varphi +\psi $. By proposition 16.9.42, we have
\[ \varphi _ i+(\psi _ i\lor \psi )\xrightarrow {d_{S}} \varphi +\psi . \]
It follows from the special case above that $\psi _ i\lor \psi \xrightarrow {d_{S}}\psi $. It follows from (16.9.46) and corollary 16.9.33 that 2 holds.
2+3 $\implies $ 1. This is similar.
$\square$
Pome
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