PomeLemma 16.9.29.2E78B4FEFFF9 Let $\varphi \in \mathrm{PSH}(X,\theta )$. For each $t\in (0,1)$, let $\varphi _ t=(1-t)\varphi +t V_{\theta }$. Then
\[ \varphi _ t\xrightarrow {d_ S}\varphi \]
as $t\to 0+$.
Lemma 16.9.292E78B4FEFFF9
Proof. Since $\varphi \preceq \varphi _ t$, by 16.9.7, we need to show that for each $j=1,\dots ,n$, we have
\[ \lim _{t\to 0+} \int _ X \theta _{\varphi _ t}^ j\wedge \theta _{V_{\theta }}^{n-j}=\int _ X \theta _{\varphi }^ j\wedge \theta _{V_{\theta }}^{n-j}. \]
For this purpose, we compute
\[ \begin{aligned} & \int _ X \theta _{\varphi _ t}^ j\wedge \theta _{V_{\theta }}^{n-j}- \int _ X \theta _{\varphi }^ j\wedge \theta _{V_{\theta }}^{n-j}\\ =& \left((1-t)^ j-1\right)\int _ X \theta _{\varphi }^ j\wedge \theta _{V_{\theta }}^{n-j}\\ & +\sum _{i=0}^{j-1} \binom {j}{i}(1-t)^ i t^{j-i}\int _ X\theta _{\varphi }^ i\wedge \theta _{V_{\theta }}^{n-i}. \end{aligned} \]
As $t\to 0+$, the right-hand side clearly tends to $0$. $\square$
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