18.7 Limits of analytic envelopesE69A0F80536C

Proof. Set $p=P_\theta [u]_{\mathcal I}$ and $q_ j=P_{\theta +\epsilon _ jh}[a_ j]_{\mathcal I}$. Projection and the ideal order give

The first inequality holds because $p$ is a competitor in the larger background, and the second follows from proposition 18.4.6, since $p\sim _{\mathcal I}u\preceq _{\mathcal I}a_ j$. All $q_ j$ have supremum zero and belong to $\mathrm{PSH}(X,\theta +Ch)$ for a fixed $C\geq \sup _ j\epsilon _ j$. By normalized compactness, every subsequence therefore has a further subsequence converging in $L^1$. Let $q$ be such a limit. Passing to the limit in the current inequalities gives $q\in \mathrm{PSH}(X,\theta )$, and lemma 5.24.3 gives $p\leq q\leq 0$.

Fix a local modification $\pi :Z\to U$ and a prime divisor $E\subset Z$. By lemma 5.43.19, the chosen subsequence still converges in $L^1_{\mathrm{loc}}$ after pullback. Since $q_ j\sim _{\mathcal I}a_ j$, the two functions have the same divisorial Lelong numbers, and the pointwise semicontinuity in lemma 5.38.7 yields

Indeed, at every point of $E$ the pullback of each $q_ j$ has Lelong number at least $\nu _ E(q_ j)$; it remains to apply the pointwise inequality at a point of $E$ where the pullback of $q$ has Lelong number $\nu _ E(q)$. On the other hand, $p\leq q$ gives $\nu _ E(q)\leq \nu _ E(p)=\nu _ E(u)$. Thus $q$ and $u$ have the same Lelong numbers along all local divisors. Blowing up a point of any local modification recovers the pointwise Lelong number there as a generic divisorial number; on a curve the point is already a divisor. By theorem 5.37.3, $q\sim _{\mathcal I}u$.

Consequently $q$ is a competitor for $p$, so $q\leq p$. Every cluster value is therefore $p$, which proves convergence of the whole sequence. For analytic $a_ j$, proposition 18.4.8 identifies the ordinary and ideal envelopes, and the last assertion follows from uniqueness of $L^1$ limits and lemma 5.24.3. $\square$

Proof. Fix a local divisor $E$ over a coordinate neighborhood $U$, and write $A=A_ U(E)$ for its log discrepancy. By lemma 5.38.9, the assumed inclusion implies

for all large $j$. Taking the lower limit in $j$ first, and then letting $m\to \infty $ and $\delta \downarrow 0$, we obtain $\liminf _ j\nu _ E(a_ j)\geq \nu _ E(u)$. The opposite inequality $\nu _ E(a_ j)\leq \nu _ E(u)$ holds for every $j$ because $u\preceq _{\mathcal I}a_ j$. Hence proposition 18.7.1 applies. $\square$

Proof. Choose $u$ smooth, nonconstant and normalized, with $\theta +\mathrm{dd}^{\mathrm{c}}u>0$, and apply the construction in theorem 18.6.4 to the smooth probability measure $\nu =\theta +\mathrm{dd}^{\mathrm{c}}u$. Since $\nu $ has no atoms, the largest mass of a cell tends to zero along partitions of vanishing diameter. Choose the rational approximations of the weights so that their total absolute error tends to zero. The resulting atomic probability measures $\nu _ j$ converge weakly to $\nu $, and their largest atoms tend to zero.

By the same theorem, their normalized Green potentials $a_ j$ have analytic singularities, satisfy $P_\theta [a_ j]=P_\theta [a_ j]_{\mathcal I}=a_ j$, and converge to $u$ in $L^1$. At every point $y\in Y$,

Since local modifications of a smooth curve are isomorphisms, this gives all the required divisorial limits. As $u$ is bounded and $\theta >0$, both of its envelopes are zero, and since $u$ is not constant, the envelopes do not converge to them. Each $a_ j$ has a logarithmic pole, so $u\preceq _{\mathcal I}a_ j$ fails. $\square$

Proof. By proposition 18.4.8, $q_ j=P_{\theta +\epsilon _ jh}[a_ j]_{\mathcal I}\sim a_ j$. Since the backgrounds decrease and $a_{j+1}\preceq a_ j$, the function $q_{j+1}$ is a competitor for $P_{\theta +\epsilon _ jh}[a_ j]$. Hence $q_{j+1}\leq q_ j$.

The inequality $a_ j\geq u$ gives $u\preceq _{\mathcal I}a_ j$. For every local divisor $E$, lemma 5.38.9 gives

By proposition 18.7.1, $q_ j\to p:=P_\theta [u]_{\mathcal I}$ in $L^1$, and $p\leq q_ j\leq 0$. The decreasing limit of the $q_ j$ is therefore bounded below by $p$; it is a $\theta $-plurisubharmonic function that agrees with $p$ almost everywhere, hence everywhere. This identifies the limit in terms of $\theta $ and the multiplier-ideal class of $u$ alone, which proves the asserted independence. $\square$

Proof. The theorem identifies both limits of the $q_ j$ as $P_\theta [u]_{\mathcal I}$ for every such choice, and its $L^1$ convergence gives the integral formula. By proposition 18.4.4, the limiting difference is nonnegative. Its integral therefore vanishes exactly when the two envelopes agree almost everywhere, hence everywhere by lemma 5.24.3. This proves all the equivalences. $\square$

Proof. Use $h=\omega $ in (18.7.5). Enlarge the error bounds to positive rational numbers $\epsilon _ j\downarrow 0$, and set

Then $c_ j\uparrow 1$ is rational and

The analytic coefficient of $b_ j$ is $c_ j/m_ j$. Moreover $b_ j\to u$ in $L^1$, since $a_ j\to u$ there and the sequence has bounded $L^1$ norm. Because $a_{j+1}\preceq a_ j$, $c_{j+1}\geq c_ j$, and each $a_ j$ is bounded above, one has $b_{j+1}\preceq b_ j$.

Put $q_ j=P_{(1+\epsilon _ j)\omega }[a_ j]=P_{(1+\epsilon _ j)\omega }[a_ j]_{\mathcal I}$. The preceding theorem gives $q_ j\searrow P_\omega [u]_{\mathcal I}$ pointwise and in $L^1$. Since $b_ j$ has analytic singularities, proposition 18.4.8 and the homogeneity in proposition 18.4.6 give

Here $q_ j\leq 0$ decreases and $c_ j\uparrow 1$, so $r_ j$ also decreases, with pointwise limit $P_\omega [u]_{\mathcal I}$. Moreover

Since the lower bound is integrable, dominated convergence gives convergence in $L^1$.

The distance in proposition 18.6.1 from $[b_ j]_{\mathcal I}$ to $[u]_{\mathcal I}$ is exactly the $L^1$ distance from $r_ j$ to $P_\omega [u]_{\mathcal I}$, so it tends to zero. The last assertion follows because the limit of the $r_ j$ has been identified. $\square$