Proposition 20.2.26.DB90B2D3ADEB Let $\varphi \in \mathrm{QPSH}(X)$ with $\nu (\varphi ,Y)=0$. Assume that $Y$ is smooth. Then for any $\lambda >0$, we have
E9AC29FB30B6
20.2.27
DB90B2D3ADEBProposition 20.2.26.DB90B2D3ADEB Let $\varphi \in \mathrm{QPSH}(X)$ with $\nu (\varphi ,Y)=0$. Assume that $Y$ is smooth. Then for any $\lambda >0$, we have
E9AC29FB30B6
20.2.27
Proof. Take a Kähler form $\omega $ on $X$ such that $\omega _{\varphi }$ is a Kähler current.
Let $(\varphi _ j)_ j$ be a quasi-equisingular approximation of $\varphi $ in $\mathrm{PSH}(X,\omega )$.
For each $j\geq 1$, we have
Indeed, $\operatorname{Tr}_ Y(\varphi )$ is $P$-equivalent to the decreasing $d_ S$-limit of the restrictions $\varphi _ j|_ Y$, hence it is more singular than each $\varphi _ j|_ Y$.
For any $\lambda '>\lambda >0$, we can find $j>0$ so that
By corollary 16.8.14 and theorem 5.16.26, we have
Thanks to the strong openness theorem theorem 5.16.21, we conclude (20.2.27). $\square$
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