ScalingStacks

Proof. The $L^{1}$-estimate follows from the sup estimate as above, so the real problem is to transfer bounds between different $E_{i}^{0}$. This is nontrivial because the necks connecting $E_{i}^{0}$ with each other are highly degenerate.

Given one divisor $E_{0}$ such that $\int_{E_{0}^{0}}u\omega_{\mathcal{X}}|_{X_{t}}^{n}\geq-C,$ we produce a good test function $v$ by Lemma 2.1. Integrating by parts,

$$\int_{X_{t}}v\sqrt{-1}\partial\bar{\partial}u\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}=\int_{X_{t}}u\sqrt{-1}\partial\bar{\partial}v\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}.$$

The LHS is the difference of $\int_{X_{t}}v(\omega_{t}+\sqrt{-1}\partial\bar{\partial}u)\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}$ and $\int_{X_{t}}v\omega_{t}\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}$, and since $-\log|t|\gtrsim v\geq 0$ both terms are bounded between $0$ and $C$. Thus

$$|\int_{X_{t}}u\sqrt{-1}\partial\bar{\partial}v\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}|\leq C.$$

Now the form $\sqrt{-1}\partial\bar{\partial}v\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}$ can only be negative on $\{h\sim|t|^{2}\}=E_{0}^{0}$, and is bounded below by $-C\omega_{\mathcal{X}}|_{X_{t}}^{n}$. Thus the positive part of the signed measure $u\sqrt{-1}\partial\bar{\partial}v\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}$ has total mass controlled by $\int_{E_{0}^{0}}|u|\omega_{\mathcal{X}}|_{X_{t}}^{n}\leq C$. Consequently, the negative part of the signed measure must also have total mass $\leq C$.

By construction, for any divisor $E_{j}$ intersecting $E_{0}$ there is a nontrivial amount of $\sqrt{-1}\partial\bar{\partial}v\wedge\omega_{\mathcal{X}}|_{X_{t}}^{n-1}$-measure inside $E_{j}^{0}$. This forces $\sup_{E_{j}^{0}}u\geq-C$. To summarize, we have transferred the sup bound from $E_{0}^{0}$ to any $E_{j}^{0}$ with $E_{j}\cap E_{0}\neq\emptyset$. Since the central fibre $X_{0}$ is connected, in at most $|I|$ steps this sup bound is transferred to all $E_{i}^{0}$ with $i\in I$. ∎