15.37 Relative equations under positive lower mass31A655BFE398

Proof. It suffices to prove $u\geq w$ for $w=\max (u,v)$. Upward stability (proposition 16.28.62) gives $w\in \mathcal E_\varphi $, and locality gives $\mu _ u\leq c\mu _ w$ on $\{ u<w\} $. For $b>1$ put $q_ b=Q_ b(u,w)$, which belongs to $\mathcal E_\varphi $ by lemma 16.28.63. It satisfies $q_ b\leq u$, and by contact support its measure is carried by

On $D_ b$ the potential $b^{-1}q_ b+(1-b^{-1})w$ touches $u$ from below, so positivity of mixed products and ordered contact give

Choose $b$ so large that $(1-b^{-1})^ n>c$. Comparing this inequality with $\mu _ u\leq c\mu _ w$ on $D_ b\cap \{ u<w\} $ forces both $\mu _ w$ and $\mu _{q_ b}$ to vanish there. Since $\mu _ w(X)$ may be infinite, this subtraction is first carried out on common finite truncation levels of the potentials, where all measures are finite; exhausting the complements of the pole sets then gives the assertion for the extended measures.

It follows that $\mu _{q_ b}$ is carried by $D_ b\cap \{ u=w\} \subseteq \{ q_ b=u\} $. Since $q_ b\leq u$, ordered contact applied once more gives

with coefficient one, independent of $b$.

As $b\to \infty $ through the integers, the potentials $q_ b$ decrease, because the obstacles $bu-(b-1)w=u-(b-1)(w-u)$ do. Suppose that they decreased to $-\infty $ identically. Then theorem 5.25.2 would give $M_ b:=\sup _ Xq_ b\to -\infty $, and (15.37.7) would imply

because $\mu _ u$ is finite and does not charge the pole set of $u$. This is impossible, so the decreasing limit $q$ is not identically $-\infty $. Now fix $a>0$. On $\{ u<w-a\} $, the defining inequality of $q_ b$ gives $q_ b\leq u-(b-1)a$, so $q=-\infty $ at every point of this set where $u$ is finite. Since neither $q$ nor $u$ is identically $-\infty $, their pole sets have zero smooth volume, and hence so does $\{ u<w-a\} $. Thus $u\geq w$ almost everywhere, and then everywhere by lemma 5.24.3. In particular $u\geq v$. $\square$

Proof. Translate both potentials by a common constant so that $u\leq v\leq \varphi \leq 0$. Fix $\delta $ with $\varepsilon <\delta <1$, put $\beta =1-\delta $, and set

This envelope is not identically $-\infty $ by theorem 16.28.70, applied to $v$. By lemma 16.28.68,

The stretching lemma lemma 16.28.63, applied with reference zero, shows that $z=P_\theta (w/\beta )$ belongs to $\mathcal E_0$. Scaling the background and each defining competitor gives

Indeed, if $a\in \mathrm{PSH}(X,\theta )$ lies below $t z$, then $\beta a\in \mathrm{PSH}(X,\beta \theta )$ lies below $t\beta z$, for every $t\geq 1$.

Since $q\leq v\leq \varphi $, we have

off the pole sets. Thus the envelope

is not identically $-\infty $. Here the difference obstacle is taken off the pluripolar pole sets; it is quasi-continuous and finite almost everywhere. The general envelope theorem gives a finite upper bound for $k$ and the inequality $k+q\leq u$. Since $\beta z\leq k$, upward stability gives $k\in \mathcal E_{\mathrm{env}}(X,\beta \theta ,0)$.

Write $\nu _ k=\mathrm{MA}^{\mathrm{ext}}_{\beta \theta }(k)$. By contact support, $\nu _ k$ is carried by

outside the pole sets. Positivity of mixed products and ordered contact give

These inequalities are first proved on common finite truncation levels and then pass to the limit by exhaustion. On $D\cap \{ k<0\} $ we have $u=k+q<q\leq v$, so the hypothesis gives $\nu _ k(\{ k<0\} )=0$.

This vanishing implies that $\nu _ k$ is finite. Indeed, $\widehat k=\max (k,-1)$ is a bounded $\beta \theta $-plurisubharmonic function, and locality gives

We now apply theorem 15.37.6 in the background $\beta \theta $, with reference zero, to the potentials $k$ and $0$ and the coefficient zero. Its hypotheses hold: $\beta \theta $ is semipositive and big, the reference zero satisfies the model condition, $\nu _ k$ is finite and vanishes on $\{ k<0\} $, and

The scaling identity follows from the bijection $a\mapsto \beta a$ between bounded potentials in the two backgrounds. The theorem yields $k\geq 0$, and hence $u\geq q$.

It remains to let $\delta \uparrow 1$. Fix $\delta _0\in (\varepsilon ,1)$ and put $r=P_{\delta _0\theta }(v)$. For $\delta _0<\delta <1$, set $t=(\delta -\delta _0)/(1-\delta _0)$. The potential $t v+(1-t)r$ is $\delta \theta $-plurisubharmonic and lies below $v$. Therefore

Letting $t\uparrow 1$ at points where $v$ and $r$ are finite gives $u\geq v$ almost everywhere, and then everywhere by lemma 5.24.3. Together with $u\leq v$, this proves equality. $\square$

Proof. Put $w=\max (u,v)\in \mathcal E_\varphi $ and choose $b>1$ with $(1-b^{-1})^ n>c$. The stretching lemma gives $p=Q_ b(u,w)\in \mathcal E_\varphi $ and $p\leq u$. By contact support, its measure is carried by

On $D$, positivity and ordered contact imply

On $\{ u<w\} $, locality gives $\mu _ u\leq c\mu _ w$. We restrict both inequalities to $D\cap \{ u<w\} $ and to common finite truncation levels, where all measures are finite and may be subtracted. Since $(1-b^{-1})^ n>c$, it follows that $\mu _ p$ vanishes there. Exhausting the finite levels, we conclude that $\mu _ p$ is carried by $D\cap \{ u=w\} \subseteq \{ p=u\} $; that is, $\mu _ p(\{ p<u\} )=0$.

Now lemma 15.37.9, applied to $p\leq u$, both in $\mathcal E_\varphi $, gives $p=u$. Substituting this into the defining inequality $b^{-1}p+(1-b^{-1})w\leq u$ yields $w\leq u$ off the poles, hence everywhere by lemma 5.24.3. Thus $v\leq w\leq u$. $\square$

Proof. Fix $a>0$. On $\{ u<v-a\} $ the exponential inequality gives $\mu _ u\leq \mathrm{e}^{-\lambda a}\mu _ v$; here all weights are evaluated off the pluripolar pole sets. The theorem, applied to $v-a$, gives $u\geq v-a$, and we let $a\downarrow 0$. For the second assertion, suppose $c<1$. Then the theorem applies to $v+A$ for every real $A$, since $\mu _{v+A}=\mu _ v$, and gives $u\geq v+A$. This is impossible at a point where $u$ and $v$ are both finite. $\square$

Proof. We first bound stretched rooftops of the tail of the sequence from below and then add the finitely many remaining terms.

The suprema of an $L^1$-convergent sequence are bounded, so (15.37.2) gives a constant $C$ with $u_ j\leq \varphi +C$ for all $j$. After a common translation we may assume $u_ j\leq \varphi \leq 0$, and then $u\leq \varphi $. Since $\mu _{u_ j}\leq \nu $, lemma 5.44.2 gives

We extract, once and for all, a subsequence with $I_ j\leq 2^{-j}$ along which $u_ j\to u$ almost everywhere for smooth volume.

For $N\geq j$ put $p_{j,N}=P_\theta (u_ j,\ldots ,u_ N)$. These rooftops belong to $\mathcal E_\varphi $ by proposition 16.28.64. Partition their contact sets according to the first index at which the minimum is attained. Ordered contact on these disjoint pieces gives both

The first bound has no factor $N$, because each piece is controlled by the restriction of the same measure $\nu $ to that piece.

Fix a real number $B\geq 1$ and set $r_{j,N}=Q_ B(p_{j,N},\varphi )$. It belongs to $\mathcal E_\varphi $ by the stretching lemma, so $\mu _{r_{j,N}}(X)\geq \kappa $. With $D_{j,N}=\{ B^{-1}r_{j,N}+(1-B^{-1})\varphi =p_{j,N}\} $, contact support and positivity give

On this contact set, $\varphi \leq 0$ gives $\mathrm{e}^{p_{j,N}}\leq \mathrm{e}^{r_{j,N}/B}$. Consequently

On the other hand, choose $K>0$ with $B^ n\nu (\{ u\leq -K\} )<\kappa /2$; this is possible because $u\not\equiv -\infty $ and $\nu $ does not charge its pole set. The measure domination then gives $\mu _{r_{j,N}}(\{ u\leq -K\} )<\kappa /2$, and the lower mass bound yields

Choose $j$ so large that $B^ n2^{1-j}<\mathrm{e}^{-K}\kappa /4$. Then (15.37.15) bounds $\sup _ Xr_{j,N}$ from below, independently of $N$. Since the sequence decreases in $N$ and is bounded above by $\varphi $, its limit $R$ is not identically $-\infty $. It satisfies, for every $\ell \geq j$,

The second inequality follows already from $r_{j,N}\leq p_{j,N}\leq \varphi $.

It remains to include the finitely many earlier terms and to verify every defining condition of the relative class. Fix $b\geq 1$, apply the preceding construction with $B=2b$, and let $j$ and $R$ be as above. If $j>1$, let $p_0=P_\theta (u_1,\ldots ,u_{j-1})$ and $S=Q_{2b}(p_0,\varphi )$. By the finite-rooftop and stretching lemmas, neither is identically $-\infty $, and $S\leq \varphi $. If $j=1$, set $S=\varphi $. The potential $H=(R+S)/2$ is $\theta $-plurisubharmonic. For a tail index $\ell \geq j$ we use (15.37.16) and $S\leq \varphi $; for an earlier index we use the analogous inequality for $S$ and $R\leq \varphi $. In either case,

For $b=1$ this shows that $h=P_\theta (\inf _\ell u_\ell )$ is not identically $-\infty $. For general $b$ the left side is a competitor for $h$, so the same inequality holds with $h$ in place of every $u_\ell $; this is the defining condition of the relative class for $b$. Also $h\leq \varphi $. Thus $h\in \mathcal E_\varphi $.

Finally $h\leq u_ j$ for all $j$, so the almost-everywhere convergence and lemma 5.24.3 give $h\leq u\leq \varphi $. Upward stability proves $u\in \mathcal E_\varphi $. Undoing the common translation proves the lemma as stated. $\square$

Proof. The bounds pass to $u$ almost everywhere along a subsequence, and then everywhere by lemma 5.24.3; upward stability gives $u\in \mathcal E_\varphi $. After translating the entire family and its bounds by a common constant, we may assume that all potentials are nonpositive.

It suffices to prove convergence in capacity along a further subsequence of every subsequence. Extract so that $I_ j\leq 2^{-j}$ and $u_ j\to u$ almost everywhere for smooth volume, and form

All these envelopes lie above $h$ and below $\varphi +C$; hence none of them is identically $-\infty $, and all belong to $\mathcal E_\varphi $. Partition the contact set of $p_{j,N}$ according to the first index $\ell $ with $p_{j,N}=u_\ell $. Ordered contact on each part gives

Only integrals of nonnegative functions occur here, so the estimate holds even if a rooftop has infinite mass.

As $N\to \infty $, $p_{j,N}\downarrow p_ j$; as $j\to \infty $, the $p_ j$ increase, and their regularized limit $p$ is not identically $-\infty $. Both monotone limits hold in capacity by lemma 5.29.19. Moreover $h\leq p\leq u\leq \varphi +C$, where $p\leq u$ follows from the almost-everywhere convergence of the extracted $u_ j$. Thus $p\in \mathcal E_\varphi $.

Apply lemma 13.14.8, first as $N\to \infty $ and then as $j\to \infty $, with the bounded quasi-continuous weights in the preceding estimate. These weights converge in capacity because the exponential is Lipschitz on a common upper half-line. We obtain

The second equality follows off the poles, which the extended measures ignore. The zero-domination lemma lemma 15.37.9 now gives $p=u$; no finiteness of $\mu _ p$ is needed.

Finally, the upper envelopes $U_ j=(\sup _{\ell \geq j}u_\ell )^*$ decrease to $u$, by theorem 5.25.2 and the almost-everywhere convergence. Applying lemma 5.29.19 to $p_ j\uparrow u$ and $U_ j\downarrow u$, the inequalities $p_ j\leq u_ j\leq U_ j$ give convergence in capacity along this subsequence, and hence for the original sequence. $\square$

Proof. Start with an arbitrary subsequence. By lemma 15.37.13, a further subsequence has a common lower envelope $h\in \mathcal E_\varphi $, and $u\in \mathcal E_\varphi $. Inequality (15.37.2) and Hartogs' lemma give a common upper bound $u_ j\leq \varphi +C$.

After a common translation, assume $u_ j,u\leq 0$. Since $\mu _{u_ j}\leq \nu $, lemma 5.44.2 gives

Thus proposition 15.37.17 gives convergence in capacity along this subsequence. For the measures themselves, the common lower bound gives

since $\nu $ is finite and does not charge the pole set of $h\not\equiv -\infty $. The tail criterion for smooth backgrounds, theorem 13.15.23, therefore gives $\mu _{u_ j}\rightharpoonup \mu _ u$, and weak limits preserve domination by the fixed finite measure $\nu $. Every subsequence of the original sequence has a further subsequence with these conclusions and the same limit $u$, so both convergences hold for the original sequence. $\square$

Proof. Translate all $u_ j$ by one constant so that $u_ j\leq \varphi \leq 0$; this changes $A$ by a fixed factor. Translate each $\psi _ j$ so that it takes values in $[-2L,0]$; this does not change its measure. Set $q=P_{\varepsilon \theta }(\varphi )$ and choose $\rho \in \mathrm{PSH}(X,\theta )$ and $a>0$ such that

Here $q\not\equiv -\infty $ by the model condition, and $\rho $ exists because $\theta $ is big. Fix $0<\delta <(1-\varepsilon )a$.

For a fixed real $b\geq 1$, put $r_{j,b}=Q_ b(u_ j,\varphi )$. The stretching lemma puts it in $\mathcal E_\varphi $, and $r_{j,b}\leq u_ j\leq 0$. Contact support and positivity imply

On this contact set, $\varphi \leq 0$ gives $u_ j\leq r_{j,b}/b$. Thus (15.37.22) yields

where the weights are evaluated off the pluripolar pole sets.

Choose $C_ b\geq 0$ so large that

and define

This is a $\theta $-plurisubharmonic function, not identically $-\infty $, since

Consequently $\mu _{s_{j,b}}\geq \delta ^ n\omega _{\psi _ j}^ n$: this inequality holds away from the poles by the local positive product inequality, and then globally because the bounded reference measure does not charge pluripolar sets. Also $s_{j,b}\leq -C_ b$ and $s_{j,b}\preceq \varphi $. Together with (15.37.23), we conclude

If $v_-(\theta )>0$, apply corollary 15.37.11 with first potential $r_{j,b}$ and second potential $s_{j,b}$. It gives $r_{j,b}\geq s_{j,b}$.

In the second alternative, $r_{j,b}\sim \varphi $: if $u_ j\geq \varphi -K_ j$, the potential $\varphi -bK_ j$ is a competitor for $r_{j,b}$. Hence the smaller-background envelope $P_{\varepsilon \theta }(r_{j,b})$ is not identically $-\infty $, and $s_{j,b}\preceq r_{j,b}$. Therefore corollary 15.2.3 also gives $r_{j,b}\geq s_{j,b}$, without a lower-mass hypothesis.

Since $\psi _ j\geq -2L$, we now have a lower bound independent of $j$:

The potential $H_ b$ is $\theta $-plurisubharmonic and not identically $-\infty $. By the definition of $r_{j,b}$, therefore,

For $b=1$ this shows that the infimum envelope $h$ is not identically $-\infty $. For general $b$, the left side is a competitor for $h$, so it lies below $h$; it is thus a competitor in the defining condition of the relative class for $b$. Since $h\leq \varphi $, we obtain $h\in \mathcal E_\varphi $. Undoing the common translation finishes the proof. $\square$

Proof. The common envelope gives $h\leq u_ j\leq \varphi +C$. Along an almost-everywhere convergent subsequence, lemma 5.24.3 gives $h\leq u\leq \varphi +C$, and upward stability gives the conclusion. $\square$

Proof. Translate so that $u_ j,u\leq 0$. The functions $\mathrm{e}^{u_ j},\mathrm{e}^ u$ are uniformly bounded quasi-plurisubharmonic functions with a common lower curvature bound. They converge in $L^1$ for smooth volume, since the exponential is Lipschitz on $(-\infty ,0]$. Hence lemma 5.44.7, lemma 5.44.2 give

Thus the integration error in proposition 15.37.17 tends to zero, and that proposition gives convergence in capacity and $u\in \mathcal E_\varphi $.

By lemma 5.29.16, the bounded reference measures $\omega _{\psi _ j}^ n$ have uniformly bounded mass and satisfy a common capacity estimate. Hence, for a constant $B$ independent of $j$ and $T$,

uniformly in $j$. The first term tends to zero because $h$ is a quasi-plurisubharmonic function not identically $-\infty $, and the second because $\nu $ is finite and does not charge the pole set of $h$. The total masses $\mu _{u_ j}(X)$ are uniformly bounded by the assumed domination. The tail criterion for smooth backgrounds, theorem 13.15.23, therefore gives finiteness of $\mu _ u$ and the asserted weak convergence. $\square$

Proof. For the reference zero, $P_{\epsilon \theta }(0)=0$ for every $\epsilon >0$. Letting $\epsilon \downarrow 0$ in theorem 16.28.70 therefore gives

Positive exponent. By lemma 15.24.4, write $\mu =f\nu $ with $\nu =\mathrm{MA}_\theta (\psi )$, where $\psi \in \mathrm{PSH}(X,\theta )$ is bounded and $0\leq f\in L^1(\nu )$. For positive integers $j$, let $f_ j=\min (f,j)$. These densities are bounded, and the measures $f_ j\nu $ are nonzero. By theorem 15.10.1 there are bounded $u_ j\in \mathrm{PSH}(X,\theta )$ with

These bounded potentials belong to $\mathcal E_0$ by proposition 16.28.62. Since $f_ j$ increases with $j$, exponential order (corollary 15.37.11) shows that $u_ j$ decreases; in particular $u_ j\leq u_1$. On the other hand, since $f_ j\leq f$,

so the suprema of the $u_ j$ have a common lower bound. Hence the decreasing limit $u$ is a $\theta $-plurisubharmonic function, not identically $-\infty $, and $u_ j\to u$ in $L^1(X)$. Moreover

Applying theorem 15.37.19 with reference zero, we obtain $u\in \mathcal E_0$ and weak convergence of the Monge–Ampère measures to $\mathrm{MA}^{\mathrm{ext}}_\theta (u)$. The right-hand sides converge in total variation to $\mathrm{e}^{\lambda u}f\nu $, since the functions $\mathrm{e}^{\lambda u_ j}f_ j$ converge $\nu $-almost everywhere and are bounded by $\mathrm{e}^{\lambda \sup _ Xu_1}f\in L^1(\nu )$. This proves the equation, and corollary 15.37.11 gives uniqueness.

Zero exponent. For each positive integer $j$, let $u_ j\in \mathcal E_0$ be the solution just constructed with exponent $1/j$. To bound the normalization, we compare with the solution $s\in \mathcal E_0$ for exponent one, so that $\mathrm{MA}^{\mathrm{ext}}_\theta (s)=\mathrm{e}^ s\mu $. Set $H=\max (0,\sup _ Xs)$ and $w_ j=s+(j-1)H$. Then $w_ j/j\geq s$, and

Exponential order, now with exponent $1/j$, gives $u_ j\leq w_ j$. Put

Then $\sup _ Xv_ j=0$ and $0<c_ j\leq \mathrm{e}^ H$, because $M_ j\leq \sup _ Xw_ j\leq jH$. By (15.37.29),

so $c_ j\geq v_-(\theta )/\mu (X)>0$. After extraction, theorem 5.25.2 gives $v_ j\to v$ in $L^1(X)$, where $v\not\equiv -\infty $ and $\sup _ Xv=0$, and $c_ j\to c>0$.

By lemma 5.44.5, $a_ j=\mathrm{e}^{v_ j/j}\to 1$ in $L^1(\mu )$. Consequently

Applying theorem 15.37.19 again, we obtain $v\in \mathcal E_0$ and identify the limit as $\mathrm{MA}^{\mathrm{ext}}_\theta (v)$; thus $\mathrm{MA}^{\mathrm{ext}}_\theta (v)=c\mu $. The scalar assertion of corollary 15.37.11, applied in both directions to two solutions, proves that $c$ is unique. $\square$

of theorem 15.37.4. The proof has four steps: continuous positive densities, bounded densities, integrable densities, and finally the zero exponent.

Let $\mu $ be the given nonzero finite non-pluripolar measure. Apply lemma 15.24.4 to the Hermitian, hence semipositive and big, background $\omega $ to write

Step 1: positive continuous density. Fix $\lambda >0$ and suppose first that $f$ is continuous and strictly positive. By theorem 5.32.1, choose smooth strictly $\omega $-plurisubharmonic $\psi _ j\downarrow \psi $. These potentials are uniformly bounded; write $\| \psi _ j\| _\infty \leq L$ and $\nu _ j=\omega _{\psi _ j}^ n$. Bounded Bedford–Taylor continuity gives $\nu _ j\rightharpoonup \nu $. The measure $f\nu _ j$ has a continuous positive density with respect to smooth volume, so it belongs to every finite $L^ p$ space. Hence the prescribed ordinary-type theorem theorem 15.12.1 gives $u_ j\sim \varphi $ with

We first bound the $u_ j$ from above, uniformly in $j$. Choose $K$ with $\mathrm{e}^{-\lambda (\psi _ j+K)}\leq \min _ Xf$ for all $j$. Then

where the last inequality follows from $\omega \geq \theta $ and positivity of extended mixed products. Apply corollary 15.2.3 in the background $\omega $, with first potential $\psi _ j+K$ and second potential $u_ j$. Its hypotheses hold: the first potential is bounded, so it admits a smaller-background envelope, and $u_ j\preceq \psi _ j+K$. Thus $u_ j\leq \psi _ j+K\leq L+K=:C$.

By (15.37.2), $u_ j\leq \varphi +C$. Since

proposition 15.37.21 gives a single $h\in \mathcal E_\varphi $ below all $u_ j$. The bounds $h\leq u_ j\leq \varphi +C$ and theorem 5.25.2 give, after extraction, an $L^1$ limit $u\not\equiv -\infty $. The moving-measure theorem theorem 15.37.26 now applies, because $\psi _ j\to \psi $ in capacity and $\mu _{u_ j}\leq \mathrm{e}^{\lambda C}\| f\| _\infty \nu _ j$. It gives $u\in \mathcal E_\varphi $, convergence in capacity and $\mu _{u_ j}\rightharpoonup \mu _ u$.

The right-hand sides converge weakly to $\mathrm{e}^{\lambda u}f\nu $. Indeed, $\mathrm{e}^{\lambda u_ j}-\mathrm{e}^{\lambda u}\to 0$ in capacity with a common bound, so its integral against $\nu _ j$ tends to zero by the uniform capacity estimate for these measures. The fixed function $\mathrm{e}^{\lambda u}$ is bounded and quasi-continuous. Approximating it by a continuous function outside an open set of arbitrarily small capacity, bounded product continuity and the same capacity estimate allow this fixed weight in $\nu _ j\rightharpoonup \nu $. Multiplication by the continuous function $f$ and by a continuous test function preserves both estimates. Hence $\mu _ u=\mathrm{e}^{\lambda u}f\nu $.

Step 2: bounded density. Suppose next that $0\leq f\in L^\infty (\nu )$ and $\int _ Xf\, \mathrm{d}\nu >0$. Choose positive continuous functions $f_ j$, uniformly bounded by a constant $F$, with $f_ j\to f$ almost everywhere for $\nu $. Such a sequence is obtained from the density of continuous functions in $L^1(\nu )$ by clipping to a common interval, adding positive constants tending to zero, and extracting a summably convergent subsequence.

There is a bounded nonnegative function $g$ with $g\nu \ne 0$ and $g\leq f_ j$ for all $j$. Indeed, choose $a>0$ with $\nu (\{ f\geq 2a\} )>0$. By Egorov's theorem, there is a measurable subset $E\subseteq \{ f\geq 2a\} $ of positive measure on which the convergence is uniform; hence $f_ j\geq a$ on $E$ for all $j\geq N$, for some $N$. The finitely many remaining functions $f_1,\ldots ,f_{N-1}$ are positive and continuous, so they have positive minima on $X$. If $b>0$ is a lower bound for these minima and for $a$, then $g=b\mathbf1_ E$ works.

By Step 1, let $u_ j\in \mathcal E_\varphi $ solve $\mu _{u_ j}=\mathrm{e}^{\lambda u_ j}f_ j\nu $. The theorem for the reference zero, theorem 15.37.28, applied to the finite nonzero measure $g\nu $, gives $s\in \mathcal E_0$ satisfying

Apply corollary 15.37.11 with reference zero, first potential $s\in \mathcal E_0$ and second potential $u_ j\preceq 0$. Since $g\nu \leq f_ j\nu $, it gives $u_ j\leq s$. With $C=\sup _ Xs$ we have

where the second inequality uses $\mu _{u_ j}(X)\geq \kappa $. These two bounds on the suprema give a subsequence converging in $L^1$ to some $u\not\equiv -\infty $. The fixed-measure continuity theorem theorem 15.37.19 puts $u$ in $\mathcal E_\varphi $ and identifies the weak limit of $\mu _{u_ j}$ as $\mu _ u$.

By lemma 5.44.2, $\mathrm{e}^{\lambda u_ j}\to \mathrm{e}^{\lambda u}$ in $L^1(\nu )$, and $f_ j\to f$ in $L^1(\nu )$ by dominated convergence. Since both factors are uniformly bounded,

Thus the right-hand sides converge in total variation, which proves the equation for $f$.

Step 3: integrable density. For general $f\in L^1(\nu )$, apply Step 2 to $f_ j=\min (f,j)$ and obtain $u_ j\in \mathcal E_\varphi $ with $\mu _{u_ j}=\mathrm{e}^{\lambda u_ j}f_ j\nu $. Since $f_ j\leq f_{j+1}$, exponential order gives $u_{j+1}\leq u_ j$. Furthermore

Hence the decreasing limit $u$ is not identically $-\infty $, and the fixed-measure continuity theorem gives $u\in \mathcal E_\varphi $ and $\mu _{u_ j}\rightharpoonup \mu _ u$. Dominated convergence gives $\mathrm{e}^{\lambda u_ j}f_ j\to \mathrm{e}^{\lambda u}f$ in $L^1(\nu )$, so $\mu _ u=\mathrm{e}^{\lambda u}\mu $. Uniqueness follows from corollary 15.37.11, applied in both directions.

Step 4: zero exponent. For each positive integer $j$, Step 3 with exponent $1/j$ gives $u_ j\in \mathcal E_\varphi $ with $\mu _{u_ j}=\mathrm{e}^{u_ j/j}\mu $. By theorem 15.37.28, choose $s\in \mathcal E_0$ solving

Put $H=\max (0,\sup _ Xs)$ and $w_ j=s+(j-1)H$. Then $w_ j/j\geq s$, whence

As in Step 2, exponential order applies with reference zero, first potential $w_ j\in \mathcal E_0$ and second potential $u_ j\preceq 0$. It gives $u_ j\leq w_ j\leq jH$. Only an upper bound for $s$ is used here; the mass of $s$ is finite by its equation.

Set $M_ j=\sup _ Xu_ j$, $v_ j=u_ j-M_ j$ and $c_ j=\mathrm{e}^{M_ j/j}$. Then

The lower bound for $c_ j$ follows from $\mu _{v_ j}(X)\geq \kappa $ and $v_ j\leq 0$. Extract so that $v_ j\to v$ in $L^1(X)$ and $c_ j\to c>0$; the limit $v$ is not identically $-\infty $ and satisfies $\sup _ Xv=0$.

By lemma 5.44.5, $a_ j=\mathrm{e}^{v_ j/j}\to 1$ in $L^1(\mu )$. Consequently $\mu _{v_ j}=c_ ja_ j\mu \to c\mu $ in total variation. The fixed-measure continuity theorem gives $v\in \mathcal E_\varphi $ and $\mu _ v=c\mu $. Finally, the scalar assertion of corollary 15.37.11, applied in both directions, shows that $c$ is unique. $\square$

Proof. By theorem 15.37.28, applied with $\theta =h$, there are $c>0$ and $w\in \mathcal E_{\mathrm{env}}(X,h,0)$ with $\mathrm{MA}^{\mathrm{ext}}_ h(w)=cf\, \mathrm{d}V$. If $s\in \mathrm{PSH}(X,h)$ satisfies $\mathrm{MA}^{\mathrm{ext}}_ h(s)\geq af\, \mathrm{d}V$ with $a>0$, then

Every $h$-plurisubharmonic $s$ satisfies $s\preceq 0$, so the scalar assertion of corollary 15.37.11, with reference zero, gives $c/a\geq 1$. Taking the maximum over all subsolution coefficients $a$, theorem 15.28.4 yields $b_ h(f)\leq c$. The reverse inequality holds because $w$ itself is a subsolution with coefficient $c$. The same argument identifies the scalar of every solution in this envelope class with $b_ h(f)$.

For $\lambda >0$, theorem 15.37.28 also gives a solution $s_\lambda $ of the exponential equation in the envelope class. Since $s_\lambda $ is in particular a subsolution, the greatest-solution property gives $s_\lambda \leq u_\lambda $. Upward stability (proposition 16.28.62) places $u_\lambda $ and its translate $v_\lambda $ in that class.

Write $c_\lambda =\exp (\lambda \sup _ Xu_\lambda )$. By corollary 15.35.3, $c_\lambda \to b_ h(f)>0$ and

Since $v_\lambda \leq 0$ and $c_\lambda $ stays bounded for small $\lambda $, these measures are dominated by one finite non-pluripolar measure $C f\, \mathrm{d}V$. Applying theorem 15.37.19 to the given $L^1$-convergent sequence $v_{\lambda _ j}$, we obtain class membership, convergence in capacity and the equation for its limit. $\square$

Proof. Use the form and the two potentials of 14.29.1. Since the manifold is projective, lemma 19.9.12 gives bounded mass. Fix a Kähler form $\Omega $ on $X$ and choose $\varepsilon >0$ with $\varepsilon \Omega \leq \omega $. Lower-volume monotonicity, theorem 13.17.1, applied to the zero potential in both backgrounds, gives

Here the potentials of the same ordinary type as zero are exactly the bounded ones, so these lower volumes are the infima defining $v_-$.

The two smooth potentials belong to $\mathcal E_{\mathrm{env}}(X,\omega ,0)$ by proposition 16.28.62, and their extended measures are the ordinary top powers. Their common measure $\nu $ is finite and does not charge pluripolar sets. The construction gives $c=1$, and corollary 15.37.11 shows that no other positive scalar is possible. The density and its zero set were computed in 14.29.1. $\square$