Theorem 20.3.275FE15106F340

Proof. Step 1: $T$ is the largest trace competitor. The argument of proposition 20.3.24, applied with the assumed curvature bounds, gives a normalized $\Omega $-plurisubharmonic limit $v$ of a subsequence of the $a_ m$, whose divisorial Lelong numbers dominate $\tau _{Y,u}$. Hence the family defining $T$ is nonempty, and its supremum lies between $v$ and zero. The family is stable under finite maxima, since the Lelong number of a maximum is the minimum of the two Lelong numbers, also after pullback. By Choquet's lemma and finite maxima, there is an increasing sequence of trace competitors whose limit has upper semicontinuous regularization $T$. Convergence of the pullbacks and semicontinuity of Lelong numbers show that $T$ satisfies every defining inequality. Thus $T$ is itself a trace competitor, and it is the largest one.

Step 2: a common lower barrier. Fix $m$, and let $\pi :Z\to Y$ be a log resolution of $\mathfrak a_ m(u;Y)$. For each prime component $E$ of the resulting divisor, (20.3.3) gives

The criterion of lemma 5.38.11, which involves only these finitely many divisors, gives $T\preceq a_ m$. Since $T\leq 0$ and $\delta _ m\geq 0$, the function $T$ is a competitor for $q_ m$. Moreover, proposition 18.4.8 identifies the two envelopes in (20.3.28) and gives $q_ m\sim a_ m$. Therefore

for every divisor over $Y$.

Step 3: convergence. All $q_ m$ have supremum zero and a common smooth lower bound for their curvature. Let $q$ be any $L^1$ cluster point. It belongs to $\mathrm{PSH}(Y,\Omega )$, lies between $T$ and zero, and satisfies

along the convergent subsequence, by the pullback and semicontinuity argument of proposition 20.3.24 on each fixed compact modification. Thus $q$ is a trace competitor, so $q\leq T$, and hence $q=T$. Normalized compactness now gives $L^1$ convergence of the whole sequence.

Step 4: monotonicity and the fixed-point identities. Subadditivity gives $\mathfrak a_{2m}(u;Y)\subseteq \mathfrak a_ m(u;Y)^2$, hence $a_{2m}\preceq a_ m$. When the backgrounds $\Omega +\delta _{2^ j}h$ decrease, this implies $q_{2^{j+1}}\leq q_{2^ j}$. The lower bound by $T$ shows that the pointwise limit is not identically $-\infty $, and since it agrees with $T$ almost everywhere by $L^1$ convergence, it equals $T$ everywhere. The family defining $T$ depends only on $\Omega $ and on the multiplier ideals of $u$, which proves the asserted independence.

Finally, $P_\Omega [T]_{\mathcal I}$ has the same multiplier ideals as $T$, hence the same divisorial Lelong numbers, so it is a trace competitor. Maximality and the inequality $P_\Omega [T]_{\mathcal I}\geq T$ give $P_\Omega [T]_{\mathcal I}=T$, and the identity $P_\Omega [T]=T$ follows from corollary 18.4.4. $\square$