PomeLemma 16.9.50.CC25BA40D9FE Fix a Kähler form $\omega $ on $X$. As $\epsilon \to 0$ (from both sides), we have
11D8415D35F8
16.9.51
\begin{equation} V_{\theta +\epsilon \omega }\xlongrightarrow {d_ S}V_{\theta }. \end{equation}
Lemma 16.9.50CC25BA40D9FE
Proof. There are two assertions to prove, as detailed in the two steps.
Step 1. We first handle the case where $\epsilon \to 0+$.
In this case (16.9.51) means
\[ V_{\theta +\epsilon \omega }\xlongrightarrow {d_{S,\theta +\omega }}V_{\theta } \]
as $\epsilon \to 0+$. So thanks to corollary 16.9.14 and proposition 16.4.4, it suffices to prove the following:
E3D4ADBA54B9
16.9.52
\begin{equation} \inf _{\epsilon >0} P_{\theta +\omega }\left[ V_{\theta +\epsilon \omega } \right]=P_{\theta +\omega }\left[ V_{\theta } \right]. \end{equation}
First observe that
\[ V_{\theta }=\inf _{\epsilon >0} V_{\theta +\epsilon \omega }. \]
In fact, the $\leq $ direction is trivial. As for the reverse inequality, it suffices to observe that the right-hand side lies in $\mathrm{PSH}(X,\theta )$. Therefore, due to proposition 16.4.4,
\[ \lim _{\epsilon \to 0+} \int _ X \left( \theta +\epsilon \omega +\mathrm{dd}^{\mathrm{c}}V_{\theta +\epsilon \omega } \right)^ n=\int _ X \theta _{V_{\theta }}^ n. \]
Therefore, for all $\epsilon >0$ small enough, we can find $\eta _{\epsilon }\in \mathrm{PSH}(X,\theta +\epsilon \omega )$ and $a_{\epsilon }>0$ decreasing to $0$ so that
\[ (1-a_\epsilon )V_{\theta +\epsilon \omega } + a_\epsilon \eta _\epsilon \leq V_{\theta }. \]
Therefore, thanks to proposition 16.4.2,
\[ (1-a_\epsilon )P_{\theta +\omega }\left[ V_{\theta +\epsilon \omega } \right] + a_\epsilon P_{\theta +\omega }\left[\eta _\epsilon \right] \leq P_{\theta +\omega }\left[V_{\theta }\right]. \]
Since the envelopes $P_{\theta +\omega }[\eta _\epsilon ]$ have supremum $0$, they are uniformly bounded in $L^1$ by proposition 5.17.3; hence $a_\epsilon P_{\theta +\omega }[\eta _\epsilon ]\to 0$ in $L^1$. Letting $\epsilon \to 0$ and using proposition 5.14.6, we conclude (16.9.52).
Step 2. We then handle the case where $\epsilon \to 0-$.
In this case, (16.9.51) simply means
\[ V_{\theta -\epsilon \omega }\xrightarrow {d_{S,\theta }}V_{\theta } \]
as $\epsilon \to 0+$. But this follows from corollary 16.9.12 if we can prove
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16.9.53
\begin{equation} \operatorname *{sup}_{\epsilon >0}{}\! {}^{*} V_{\theta -\epsilon \omega }=V_{\theta }, \end{equation}
where we understand that $\epsilon $ is small enough so that $\theta -\epsilon \omega $ represents a big cohomology class. Since $\{ \theta \} $ is big, we can find $\varphi \in \mathrm{PSH}(X,\theta )$ so that
\[ \varphi \leq 0,\quad \theta _{\varphi }\geq \delta \omega \]
for some $\delta >0$. For a small $\epsilon >0$, we then have
\[ \theta +\mathrm{dd}^{\mathrm{c}}\left( \left(1-\frac{\epsilon }{\delta }\right)V_{\theta }+ \frac{\epsilon }{\delta }\varphi \right)\geq \epsilon \omega . \]
Therefore,
\[ \left(1-\frac{\epsilon }{\delta }\right)V_{\theta }+ \frac{\epsilon }{\delta }\varphi \leq V_{\theta -\epsilon \omega }. \]
Letting $\epsilon \to 0+$, we then find
\[ V_{\theta }\leq \operatorname *{sup}_{\epsilon >0}{}\! {}^{*} V_{\theta -\epsilon \omega }. \]
The reverse inequality is trivial, hence (16.9.53) is established. $\square$
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