ScalingStacks

Proof. Let the global maximum of $u$ be achieved at $q_{0}\in E_{J}^{0}$, and denote the local potential of $u$ as $u_{\beta}$. Without loss of generality $u_{\beta}\leq 0$. We have $u_{\beta}(q_{0})\geq-C$ since $|u-u_{\beta}|\leq C$. Applying the mean value inequality around $q_{0}$, we find that the local average function $\bar{u}_{\beta}$ produced in Lemma 2.2 satisfies $\sup\bar{u}_{\beta}\geq-C$ for another uniform constant $C$. By the convexity of $\bar{u}_{\beta}$ its sup is almost achieved at the boundary of the chart, which is contained in a union of less deep strata $E_{J^{\prime}}^{0}$ with $J^{\prime}\subsetneq J$. Thus we can find a point $q^{\prime}$ with $u(q^{\prime})\geq-C$ that belongs to a less deep stratum; an induction shows that there is some $i\in I$, such that $\sup_{E_{i}^{0}}u\geq-C$.

For the $L^{1}$-bound we recall the following Harnack inequality argument. Suppose a coordinate ball $B(q,3R)$ is contained in a local chart in a small neighbourhood of $E_{i}^{0}$. Applying the mean value inequality to the local psh function associated to $u$, we see for $y\in B(q,R)$ that

hence the Harnack inequality

Applying this to a chain of balls connecting any two points in $E_{i}^{0}$ gives the $L^{1}$-bound $\int_{E_{i}^{0}}u\omega_{\mathcal{X}}|_{X_{t}}^{n}\geq-C^{\prime}$; the bound is uniform because the number of balls involved in the chain can be controlled independent of $t$. ∎

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,

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

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 $i\in I$. ∎