ScalingStacks

Proof. Let $\chi$ be a compactly supported nonnegative smooth function on the square $\{|x_{i}|<1\}\subset\mathbb{R}^{n}$, equal to one on $\{|x_{i}|\leq 1/2\}$. We identify $\chi$ with $\chi\circ s^{-1}\text{Log}$, and denote $\omega_{std}=\sqrt{-1}\sum d\log z_{i}\wedge d\overline{\log z_{i}}$. Then

$$-Cs^{-2}\omega_{std}\leq\sqrt{-1}\partial\bar{\partial}\chi\leq Cs^{-2}\omega_{std}.$$

The basic obervation is that if $T$ is a positive current of bidegree $(n-1,n-1)$, then by integration by part,

$$\begin{split}&\int_{E}\sqrt{-1}\partial\bar{\partial}u\wedge T\leq\int_{\text{supp}(\chi)}\chi\sqrt{-1}\partial\bar{\partial}u\wedge T\\ &=\int_{\text{supp}(\chi)}u\sqrt{-1}\partial\bar{\partial}\chi\wedge T\leq Cs^{-2}\int_{\text{supp}(\chi)}\omega_{std}\wedge T.\end{split}$$

Iterating this argument to lower the power of $\sqrt{-1}\partial\bar{\partial}u$,

$$\int_{E}(\sqrt{-1}\partial\bar{\partial}u)^{n}\leq Cs^{-2n}\int_{U}\omega_{std}^{n}\leq Cs^{-n}.$$

The second statement is proved similarly by removing $\sqrt{-1}\partial\bar{\partial}v$ factors iteratively. ∎