Theorem 20.3.30.7EFA5C202804 Suppose $\Omega $ is a smooth real $(1,1)$-form on $Y$ for which $T=\mathcal T_\Omega (u;Y)$ is defined, and let $h$ be a Hermitian form on $Z$. Then
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20.3.31
\begin{equation} \mathcal T^\pi _{\pi ^*\Omega }(u;Y)=T\circ \pi , \qquad \mathcal T^\pi _{\pi ^*\Omega +\varepsilon h}(u;Y) \searrow T\circ \pi \quad (\varepsilon \downarrow 0). \end{equation}
The convergence is pointwise and in $L^1(Z)$, and the limit is independent of $h$.
Let $\delta _ m\geq 0$ tend to zero and suppose, for all sufficiently large $m$, that
\[ b_ m:=(u_ m|_ Y)\circ \pi \in \mathrm{PSH}(Z,\pi ^*\Omega +\delta _ m h). \]
Then
\[ P_{\pi ^*\Omega +\delta _ m h}[b_ m] =P_{\pi ^*\Omega +\delta _ m h}[b_ m]_{\mathcal I} \longrightarrow T\circ \pi \quad \text{in }L^1(Z). \]
If $\delta _{2^ j}$ decreases, these envelopes decrease pointwise to $T\circ \pi $ along the dyadic indices.
Proof.
For any nonempty family of trace competitors, the argument with finite maxima and semicontinuity in Step 1 of the proof of theorem 20.3.27 shows that the envelope is its largest member. By proposition 5.43.18, every $\pi ^*\Omega $-plurisubharmonic function is uniquely of the form $a\circ \pi $ with $a\in \mathrm{PSH}(Y,\Omega )$, and since $\pi $ is surjective, $a\leq 0$ if and only if $a\circ \pi \leq 0$. If $a$ satisfies all trace inequalities over $Y$, then $a\circ \pi $ satisfies those over $Z$, since every modification of $Z$, composed with $\pi $, is a modification of $Y$.
Conversely, suppose that $a\circ \pi $ satisfies the inequalities over $Z$. Let $E$ be a prime divisor on a smooth compact modification $V\to Y$. Resolving the graph of the bimeromorphic map between $V$ and $Z$ gives a smooth compact modification $W$ dominating both. The map $W\to V$ is an isomorphism near the generic point of $E$, so the strict transform $F$ of $E$ has the same ideal orders and generic Lelong numbers as $E$, and the inequality at $F$ over $Z$ is exactly the required inequality at $E$. Thus the two families of competitors correspond under pullback, and the largest member on $Z$ is $T\circ \pi $. This proves the first identity.
For $\varepsilon >0$, the pullback $T\circ \pi $ is still a competitor for $\mathcal T^\pi _{\pi ^*\Omega +\varepsilon h}(u;Y)$. These envelopes lie between $T\circ \pi $ and zero and decrease as $\varepsilon \downarrow 0$, so their limit $S$ satisfies $S\geq T\circ \pi $ and is $\pi ^*\Omega $-plurisubharmonic. The $L^1$ convergence of pullbacks and semicontinuity of Lelong numbers preserve every trace inequality, so $S$ belongs to the unperturbed family. Maximality gives $S\leq T\circ \pi $, hence equality. Since the lower bound $T\circ \pi $ is integrable on $Z$, dominated convergence gives $L^1$ convergence.
For the final assertion, denote the analytic envelope in the statement by $q_ m$. The proof of theorem 20.3.27 applies to the global coherent ideals $\mathfrak a_ m(u;Y)\mathcal O_ Z$: resolving each of them and applying lemma 5.38.11 gives $T\circ \pi \preceq b_ m$. Hence
\[ T\circ \pi \leq q_ m\leq 0,\qquad \nu (q_ m\circ \rho ,F) =\frac1m\operatorname {ord}_ F\mathfrak a_ m(u;Y) \]
for every divisor $F$ over $Z$, where ordinary and ideal envelopes agree by proposition 18.4.8. Every $L^1$ cluster point is $\pi ^*\Omega $-plurisubharmonic and satisfies all trace inequalities, so the first identity in (20.3.31) identifies it with $T\circ \pi $, and compactness gives convergence of the whole sequence. Subadditivity orders the dyadic analytic types, decreasing dyadic backgrounds then order the envelopes, and their pointwise limit is the same $L^1$ limit.
$\square$
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