16.9.26 Convergence theorems0BF3C2E4260B

Proof. Fix $t\in (0,1]$. We write

for any $i\in I$.

By corollary 16.9.9, it suffices to show that for each $j=0,\ldots ,n$,

Observe that

So after binomial expansion, (16.9.28) follows from corollary 16.9.9. $\square$

Proof. Since $\varphi \preceq \varphi _ t$, by 16.9.7, we need to show that for each $j=1,\dots ,n$, we have

For this purpose, we compute

As $t\to 0+$, the right-hand side clearly tends to $0$. $\square$

Proof. Since $d_ S$ is a pseudometric, in order to establish the continuity of mixed masses, it suffices to consider sequences instead of nets. So we may assume that $I=\mathbb {Z}_{>0}$ as ordered sets.

Step 1. We reduce to the case where $\varphi _ j^ k,\varphi _ j$ all have positive masses and there is a constant $\delta >0$ such that for all $j=1,\ldots ,n$ and $k\in I$,

Assuming in this case, the theorem has been proved, let us handle the general case. Take $t\in (0,1)$. By 16.9.27, we have

as $k\to \infty $ for each $j=1,\ldots ,n$. Note that for all $k\in I$,

Assume that we have proved the special case of the theorem, we have

More explicitly, for fixed $k$ the left-hand side is the value at $t$ of a polynomial $Q_ k(t)$ of degree at most $n$, and the right-hand side is the value of a polynomial $Q(t)$ of degree at most $n$. Choose $n+1$ distinct numbers $t_0,\ldots ,t_ n\in (0,1)$. Lagrange interpolation expresses $Q_ k(0)-Q(0)$ as a fixed linear combination of the numbers $Q_ k(t_ a)-Q(t_ a)$, which tend to $0$. Hence the same formula holds at $t=0$. From this, (16.9.31) follows.

Step 2. We now assume that the assumptions mentioned at the beginning of Step 1 are met. Note that we may assume that $\varphi _ j^ k$, $\varphi _ j$ are model potentials for all $j=1,\ldots ,n$, $k>0$ by proposition 16.9.6 and corollary 16.3.20.

We first handle the following monotone situation. Assume that, for each fixed $i$, the sequence $(\varphi _ i^ k)_ k$ is either increasing or decreasing. Let $\tilde{\varphi }_ i$ denote its pointwise limit, namely

in the decreasing case and

in the increasing case. We assume moreover that $\tilde{\varphi }_ i\sim _ P\varphi _ i$ for every $i$. Since mixed non-pluripolar masses are invariant under replacing each entry by a $P$-equivalent potential, it suffices in this monotone situation to prove convergence to the mixed mass of the $\tilde{\varphi }_ i$'s.

After reordering the indices, we may assume that there is $i_0\in \{ 0,\ldots ,n\} $ such that $(\varphi _ i^ k)_ k$ is decreasing for $i\leq i_0$ and increasing for $i>i_0$. Therefore, for each $k>0$, using theorem 13.13.11, we have

Using corollary 13.13.12, we therefore conclude that

It remains to prove

By theorem 13.13.11, for each $k>0$, we have

When proving (16.9.32), we may replace $\varphi _ j^ k$ by $\tilde{\varphi }_ j$ whenever $j>i_0$, $k>0$. Thus, we are reduced to the case where for all $i$, $(\varphi _ i^ k)_ k$ is decreasing.

By proposition 16.4.4, the masses $\int _ X\theta _{i,\varphi _ i^ k}^ n$ decrease to $\int _ X\theta _{i,\tilde{\varphi }_ i}^ n$. Hence, thanks to 13.13.30, for each $i=1,\ldots ,n$, we may take an increasing sequence $(b^ k_ i)_ k$ tending to $\infty $ satisfying

where the upper endpoint is understood as $\infty $ if the denominator is $0$. For each such $b_ i^ k$, 13.13.30 gives $\rho _ i^ k\in \mathrm{PSH}(X,\theta _ i)$ with

Then by theorem 13.13.11 again,

Letting $k\to \infty $, we conclude (16.9.32). This proves the monotone situation.

We now return to the general case. It suffices to prove that any subsequence of

has a further subsequence converging to the desired limit. Starting with an arbitrary subsequence, we pass to a further subsequence, still indexed by $k$, so that the construction in proposition 16.9.17 applies simultaneously to the finitely many sequences $(\varphi _ i^ k)_ k$, $i=1,\ldots ,n$.

Thus, after discarding finitely many terms and relabeling, for each $i$ there are model potentials $\eta _ i^ k,\psi _ i^ k\in \mathrm{PSH}(X,\theta _ i)$ such that $(\eta _ i^ k)_ k$ is increasing, $(\psi _ i^ k)_ k$ is decreasing, and

For the upper sequence, we take $\psi _ i^ k=P_{\theta _ i}[\operatorname *{sup}_{\ell \geq k}{}\! {}^{*}\varphi _ i^\ell ]$; as in (16.9.21), this gives $\inf _ k\psi _ i^ k=\varphi _ i$. For the lower sequence, the construction in proposition 16.9.17 together with proposition 16.4.3 shows that the $\eta _ i^ k$'s are model after discarding finitely many terms, and the proof of proposition 16.9.17 shows that their increasing limit is $P$-equivalent to $\varphi _ i$.

By theorem 13.13.11, for every $k$,

The monotone situation just proved applies to both the lower and upper sequences, and both limiting mixed masses are equal to

The squeeze theorem therefore gives the desired convergence along this further subsequence. This concludes the proof. $\square$

Proof. The equivalence between 2 and 3 follows directly from 16.9.7.

1 $\implies $ 2. That $\varphi _ i\lor \varphi \xrightarrow {d_ S}\varphi $ follows from corollary 16.9.9. While (16.9.34) follows from theorem 16.9.30.

2 $\implies $ 1. By (16.9.10), we need to show that for each $j=0,\ldots ,n$, we have

This follows from theorem 16.9.30 and (16.9.34). $\square$

Proof. 1 $\implies $ 2. It suffices to show that for each $j=0,\ldots ,n$, we have

Note that this quantity is a linear combination of terms of the following form:

where $r=0,\ldots ,j$. By theorem 16.9.30, it suffices to show that $\varphi \lor \varphi _ i \xrightarrow {d_ S}\varphi $. But this follows from corollary 16.9.33.

2 $\implies $ 1. From the direction we already proved, for each $C\geq 1$, we have

By theorem 16.9.30, it follows that

for all $j=0,\ldots ,n$. It follows that

By corollary 16.9.33, it remains to show that $\varphi _ i\lor \varphi \xrightarrow {d_{S,\theta }}\varphi $. By corollary 16.9.33 again, we know that $\varphi _ i\lor \varphi \xrightarrow {d_{S,\theta +\omega }}\varphi $. So it suffices to apply (16.9.37) to $\varphi _ i\lor \varphi $ instead of $\varphi _ i$, and we conclude by 16.9.7. $\square$

Proof. First note that for each $j=0,\ldots ,n$,

So thanks to corollary 16.9.33, it remains to argue that for all $j=0,\ldots ,n$,

Observe that for $t\in (0,1)$, we have

so the desired limit follows by the same binomial expansion. $\square$

Proof. By corollary 16.9.36, we may replace $\theta $ by $\theta +\omega $, where $\omega $ is a Kähler form on $X$. In particular, we may assume that $\varphi _ i,\psi _ i,\eta _ i\in \mathrm{PSH}(X,\theta )_{>0}$ for all $i\in I$. By proposition 16.9.6, we may assume that $\varphi _ i,\psi _ i,\eta _ i$ are model potentials for all $i\in I$ and hence $\psi _ i \leq \varphi _ i\leq \eta _ i$ for all $i\in I$.

It follows from theorem 13.13.11 that for each $k=0,\ldots ,n$, we have

for all $i\in I$. By theorem 16.9.30, the limits, with respect to $i\in I$, of the two outer terms are $\int _ X \theta _{\varphi }^ k\wedge \theta _{V_{\theta }}^{n-k}$. It follows that

By corollary 16.9.33, it remains to prove that $\varphi _ i \lor \varphi \xrightarrow {d_ S}\varphi $. By corollary 16.9.33 and proposition 16.8.28, up to replacing $\psi _ i$ (resp. $\varphi _ i$, $\eta _ i$) by $\psi _ i\lor \varphi $ (resp. $\varphi _ i\lor \varphi $, $\eta _ i\lor \varphi $), we may assume from the beginning that $\psi _ i,\varphi _ i,\eta _ i\geq \varphi $. Now $\varphi _ i\xrightarrow {d_ S}\varphi $ by (16.9.40) and 16.9.7. $\square$

Proof. We have $V_{\lambda \theta }=\lambda V_{\theta }$. The middle expression in (16.9.10) for the pair $(\lambda \varphi _ i,\lambda \varphi )$ in the class $\lambda \theta $ is therefore $\lambda ^ n$ times the corresponding expression for $(\varphi _ i,\varphi )$ in the class $\theta $. The equivalence follows from corollary 16.9.9. $\square$

Proof. Since $d_ S$ is a pseudometric, we may assume that both nets are actually sequences and $I=\mathbb {Z}_{>0}$. By corollary 16.9.36, we may assume that the masses $\int _ X \theta _{\varphi }^ n>0$, $\int _ X \theta _{\psi }^ n>0$.

By proposition 16.9.6 and proposition 16.8.28, replacing all potentials by their $P$-envelopes changes neither the hypotheses nor the conclusion. Thus we may assume that the $\varphi _ j$'s, the $\psi _ j$'s, $\varphi $ and $\psi $ are all model. After discarding finitely many terms, the masses of the two sequences are bounded below by a positive constant. Applying proposition 16.9.17 to both sequences and using the sandwich criterion corollary 16.9.39, it is enough to treat the case where each of $(\varphi _ j)_ j$ and $(\psi _ j)_ j$ is monotone. In this reduced situation, $(\varphi _ j)_ j$ (resp. $(\psi _ j)_ j$) converges to $\varphi $ (resp. $\psi $) almost everywhere due to proposition 16.9.47, corollary 16.9.12 and proposition 16.4.9.

We handle three cases separately.

Step 1. Assume that both sequences are increasing.

In this case, we have $\varphi _ j\lor \psi _ j\nearrow \varphi \lor \psi $ almost everywhere. Therefore, $\varphi _ j\lor \psi _ j\xrightarrow {d_ S} \varphi \lor \psi $ by corollary 16.9.12.

Step 2. Assume that one sequence, say $(\varphi _ j)_ j$ is increasing while the other is decreasing. Then we have

Thanks to corollary 16.9.39, it suffices to show that both sides converge to $\varphi \lor \psi $ with respect to $d_ S$. So we reduce to the case where both sequences are decreasing.

Step 3. Assume that both sequences are decreasing.

In this case, due to corollary 16.9.14, it suffices to show that

The $\geq $ direction follows from theorem 13.13.11; it remains to argue the $\leq $ direction.

Thanks to 13.13.30, we may find a sequence $(\epsilon _ j)_ j$ in $(0,1)$ with limit $0$ and a sequence $(\eta _ j)_ j$ in $\mathrm{PSH}(X,\theta )_{>0}$ such that

It follows that for each $j\geq 1$, we have

Therefore by theorem 13.13.11,

Letting $j\to \infty $, we find that

Therefore, in order to prove (16.9.43), we may assume that one of the sequences is constant; let us say $\psi _ j=\psi $ for all $j$. Repeating the same argument as before and constructing $(\epsilon _ j)_ j$, $(\eta _ j)_ j$ as above, we get

Letting $j\to \infty $, we conclude (16.9.43). $\square$

Proof. It follows from proposition 16.9.42 that

By 16.8.13, we have $\varphi _ i\lor \psi _ i\sim _ P \psi _ i$ for all $i\in I$. In particular, by proposition 16.9.6,

By proposition 16.9.6 again, $\varphi \lor \psi \sim _ P\psi $ and hence $\varphi \preceq _ P \psi $ by 16.8.13. $\square$

Proof. By corollary 16.9.36, we may assume that $\theta _1,\theta _2$ are both Kähler forms. We denote them by $\omega _1,\omega _2$ instead. Let $\omega =\omega _1+\omega _2$.

1+2 $\implies $ 3. It suffices to show that for each $r=0,\ldots ,n$,

Observe that for each $i\in I$,

The expression with the actual rooftop is non-negative and is bounded above by the same expression with this larger potential in place of $(\varphi _ i+\psi _ i)\lor (\varphi +\psi )$. Thus, it suffices to show that

The left-hand side is a linear combination of

with $a=0,\ldots ,r$. Observe that $\varphi _ i\lor \varphi \xrightarrow {d_ S}\varphi $ and $\psi _ i\lor \psi \xrightarrow {d_ S}\psi $ by corollary 16.9.9, each term tends to $0$ by theorem 16.9.30.

1+3 $\implies $ 2. For each $C\geq 1$, from the direction we already proved together with 16.9.41,

By theorem 16.9.30, for each $j=0,\ldots ,n$,

Both sides are polynomials in $C$. Comparing the constant terms gives

Therefore, 2 follows if $\psi _ i\geq \psi $ for each $i$ by 16.9.7.

Next we prove the general case. By the direction that we already proved, we know that $\varphi _ i+\psi \xrightarrow {d_{S}} \varphi +\psi $. By proposition 16.9.42, we have

It follows from the special case above that $\psi _ i\lor \psi \xrightarrow {d_{S}}\psi $. It follows from (16.9.46) and corollary 16.9.33 that 2 holds.

2+3 $\implies $ 1. This is similar. $\square$

Proof. First observe that by our assumption, $(P_{\theta }[\varphi _ i])_{i\in I}$ is a decreasing net. By the continuity of non-pluripolar masses corollary 16.9.13, we have

Replacing $I$ by a tail, we may therefore assume that $\varphi _ i\in \mathrm{PSH}(X,\theta )_{>0}$ for each $i\in I$. Since $\varphi _ i\sim _ P P_\theta [\varphi _ i]$, replacing $\varphi _ i$ by $P_\theta [\varphi _ i]$ does not change its $d_ S$-class by proposition 16.9.6. We may assume that each $\varphi _ i$ is model and $(\varphi _ i)_{i\in I}$ is decreasing. In this case, by proposition 16.4.4, the potential

is model in $\mathrm{PSH}(X,\theta )_{>0}$ with

Therefore, by corollary 16.9.14, we find $\varphi _ i\xrightarrow {d_ S}\psi $. In particular, $\varphi \sim _ P \psi $ by proposition 16.9.6. Therefore,

Therefore, (16.9.48) follows. $\square$

Proof. Let $(\varphi _ i)_{i\in \mathbb {Z}_{>0}}$ be a sequence in $\mathrm{PSH}(X,\theta )_{>0}$ such that $\varphi _ i\xrightarrow {d_ S}\varphi \in \mathrm{PSH}(X,\theta )_{>0}$. We want to show that

We may assume that the $\varphi _ i$'s and $\varphi $ are all model potentials by proposition 16.9.6.

By proposition 16.9.17 and corollary 16.9.39, we may assume that $(\varphi _ i)_ i$ is either increasing or decreasing. In the increasing case, we apply proposition 16.6.30 and corollary 16.9.12, while in the decreasing case, we apply proposition 16.6.29, proposition 16.4.4 and corollary 16.9.14. $\square$

Proof. There are two assertions to prove, as detailed in the two steps.

Step 1. We first handle the case where $\epsilon \to 0+$.

In this case (16.9.51) means

as $\epsilon \to 0+$. So thanks to corollary 16.9.14 and proposition 16.4.4, it suffices to prove the following:

First observe that

In fact, the $\leq $ direction is trivial. As for the reverse inequality, it suffices to observe that the right-hand side lies in $\mathrm{PSH}(X,\theta )$. Therefore, due to proposition 16.4.4,

Therefore, for all $\epsilon >0$ small enough, we can find $\eta _{\epsilon }\in \mathrm{PSH}(X,\theta +\epsilon \omega )$ and $a_{\epsilon }>0$ decreasing to $0$ so that

Therefore, thanks to proposition 16.4.2,

Since the envelopes $P_{\theta +\omega }[\eta _\epsilon ]$ have supremum $0$, they are uniformly bounded in $L^1$ by proposition 5.17.3; hence $a_\epsilon P_{\theta +\omega }[\eta _\epsilon ]\to 0$ in $L^1$. Letting $\epsilon \to 0$ and using proposition 5.14.6, we conclude (16.9.52).

Step 2. We then handle the case where $\epsilon \to 0-$.

In this case, (16.9.51) simply means

as $\epsilon \to 0+$. But this follows from corollary 16.9.12 if we can prove

where we understand that $\epsilon $ is small enough so that $\theta -\epsilon \omega $ represents a big cohomology class. Since $\{ \theta \} $ is big, we can find $\varphi \in \mathrm{PSH}(X,\theta )$ so that

for some $\delta >0$. For a small $\epsilon >0$, we then have

Therefore,

Letting $\epsilon \to 0+$, we then find

The reverse inequality is trivial, hence (16.9.53) is established. $\square$