15.12 Equations in a prescribed ordinary typeC6737BA51F0E

Proof. Since $u\sim \varphi $, the smaller-background envelope $P_{\varepsilon \theta }(u)$ is not identically $-\infty $. If $F=0$, then theorem 15.2.2, applied to $u$ and $u+1$ with coefficient zero, would give $u\geq u+1$; hence $N=\| F\| _ q>0$.

Apply lemma 15.3.1 in the semipositive big background $(1-\varepsilon )\theta $ to $F/N$, whose $L^ q$ norm is one. It gives a bounded $s$ with $-1\leq s\leq 0$ and a constant $b_ q>0$, depending only on this fixed background and on $q$, such that

Put $h=P_{\varepsilon \theta }(\varphi )+s$. This is a $\theta $-plurisubharmonic function with $h\preceq u$. Since $\theta _ h$ is the sum of the positive currents $\varepsilon \theta +\mathrm{dd}^{\mathrm{c}}P_{\varepsilon \theta }(\varphi )$ and $(1-\varepsilon )\theta +\mathrm{dd}^{\mathrm{c}}s$, the same density bound holds for $\mathrm{MA}^{\mathrm{ext}}_\theta (h)$. To justify this, we truncate $P_{\varepsilon \theta }(\varphi )$ at finite levels, use positivity of bounded mixed products, and exhaust the set where $P_{\varepsilon \theta }(\varphi )$ is finite; its pole set has zero volume.

If $N<b_ q$, we would have

Since $N/b_ q<1$, theorem 15.2.2 would then imply $u\geq h+t$ for every $t$, which is impossible at a point where both potentials are finite. Thus $N\geq b_ q$. The construction of $s$ used no property of $\varphi $, which proves the last assertion. $\square$

Proof. The upper bound follows directly from the fixed-point property: since $u\leq 0$ and $u\preceq \varphi $, the function $u$ is a competitor for $P_\theta [\varphi ]=\varphi $.

For the lower bound we construct a subsolution $\psi $ of the type of $\varphi $ whose measure dominates a multiple of $\mathrm{e}^{-\gamma u}\mathrm{MA}_\theta (u)$, and then apply singular domination. Put $\varphi _\varepsilon =P_{\varepsilon \theta }(\varphi )$, so that $\varphi _\varepsilon \leq \varphi \leq 0$. Uniform exponential integrability (lemma 5.26.2) for the normalized potentials $u$ and exponential integrability for the fixed function $\varphi _\varepsilon $ allow us to choose $\gamma >0$ and $1<q<p$ such that

has a uniform $L^ q$ bound. Explicitly, we first choose $q<p$ and apply Hölder's inequality with exponents $p/q$ and $p/(p-q)$; then we choose $\gamma $ so small that the exponential factor is controlled by the Cauchy–Schwarz inequality and the two exponential integrability estimates.

Theorem 15.3.5, applied in the semipositive big background $(1-\varepsilon )\theta $ to the density $1+g$, gives a bounded $v$ with $\sup _ Xv=0$ and a constant $c>0$ such that

Both $\| v\| _\infty $ and $c^{-1}$ are uniformly bounded. Adding the constant one to $g$ makes the root integral of the density at least one, so no lower bound for the integral of $f$ is needed.

Put $h=\varphi _\varepsilon +v\leq \varphi $, which is $\theta $-plurisubharmonic, and define

where the formula for $\psi $ is evaluated off the pole sets. Then $0\leq \chi \leq 1$, $0<\chi '\leq 1$, and $\chi ''>0$. For bounded potentials with $h\leq \varphi $, lemma 5.28.13, applied with $G(x,y)=x+\chi (y-x)$, gives

For unbounded potentials we apply this bounded assertion to $\varphi _ k=\max (\varphi ,-k)$ and $h_ k=\max (h,-k)\leq \varphi _ k$. Since $G$ is nondecreasing in both variables, $G(\varphi _ k,h_ k)$ decreases as $k$ increases. Its limit is $\theta $-plurisubharmonic and lies between $\varphi $ and $\varphi +1$; hence it is not identically $-\infty $, and it equals $\psi $. Locality on common finite levels and positivity of bounded products give

where the second inequality uses $\mathrm{MA}^{\mathrm{ext}}_\theta (h)\geq ((1-\varepsilon )\theta +\mathrm{dd}^{\mathrm{c}}v)^ n\geq cg\, \mathrm{d}V$. These inequalities are first obtained off the pole sets; since pole sets have zero volume, they extend by exhaustion to all Borel sets.

Write $T=-\varphi _\varepsilon \geq 0$. If $C_1\geq 1+\| v\| _\infty $, then $\varphi -h+1\leq T+C_1$. Since

we obtain, with a uniform $d>0$,

Choose $K=2+\gamma ^{-1}|\log d|$. On $\{ u<\psi -K\} $ we have $u<1-K$, because $\psi \leq \varphi +1\leq 1$; hence

The potentials $u$ and $\psi -K$ have the same ordinary singularity type. Moreover, $u\geq \varphi -C_ u$ for some constant $C_ u$, so $P_{\varepsilon \theta }(u)\geq \varphi _\varepsilon -C_ u$ is not identically $-\infty $. Theorem 15.2.2 therefore yields $u\geq \psi -K\geq \varphi -K$. All constants used to choose $K$ are uniform over the family in the theorem; the possibly nonuniform constant $C_ u$ serves only to show that the smaller-background envelope of $u$ is not identically $-\infty $, and it enters no estimate. $\square$

Proof. Let $v$ be the bounded solution of $\mathrm{MA}_\theta (v)=\mathrm{e}^{\lambda v}f\, \mathrm{d}V$ given by theorem 15.3.3, and write $M=\sup _ Xv$ and $a=v-M\leq 0$. The rooftop envelope $p=P_\theta (a,\varphi )$ satisfies $\varphi -\| a\| _\infty \leq p\leq \varphi $, so $p\sim \varphi $. By the rooftop inequality of proposition 16.28.6,

The second term vanishes: by proposition 16.28.7, the measure of $\varphi $ is carried by $\{ \varphi =0\} $, whereas $\varphi <a\leq 0$ on the set in question. On the first contact set, $a=p$ and $\mathrm{MA}_\theta (a)=\mathrm{e}^{\lambda (a+M)}f\, \mathrm{d}V$. Thus $\mathrm{MA}^{\mathrm{ext}}_\theta (p)\leq \mathrm{e}^{\lambda (p+M)}f\, \mathrm{d}V$, and $w=p+M$ has the required properties. $\square$

of theorem 15.12.1, part (1). Step 1: a maximizing sequence. Fix $\lambda >0$. By lemma 15.12.4, there is $w\sim \varphi $ such that

Consider all potentials $z\sim \varphi $ for which

and let $B$ be the supremum of their weighted masses $\nu _ z(X)$. This family is nonempty, since it contains $w$, and $B\leq \mu (X)<+\infty $. Choose $u_ j$ in the family with $\nu _{u_ j}(X)\to B$. Since $\nu _0\leq \nu _{u_ j}$, corollary 15.2.3 gives $u_ j\leq w$.

Step 2: uniform bounds. We need a lower bound for the suprema of the $u_ j$. Put $\varphi _\varepsilon =P_{\varepsilon \theta }(\varphi )\leq 0$, and let $\psi $ be the bounded solution of

given by theorem 15.3.3. For $h=\psi +\varphi _\varepsilon $, positivity of the splitting $\theta _ h=(\varepsilon \theta +\mathrm{dd}^{\mathrm{c}}\varphi _\varepsilon )+((1-\varepsilon )\theta +\mathrm{dd}^{\mathrm{c}}\psi )$ gives

Here $h\preceq u_ j$, because $h\preceq \varphi $ and $u_ j\sim \varphi $, and each $u_ j$ admits a smaller-background envelope because $u_ j\sim \varphi $. Corollary 15.2.3 therefore gives $u_ j\geq h$. Consequently

and the lower bound is finite.

We now apply theorem 15.12.3 to $v_ j=u_ j-M_ j$. Their measures satisfy

so $\varphi -C\leq v_ j\leq \varphi $. Together with the bounds for $M_ j$, this shows that all $u_ j$ lie within a fixed bounded distance of $\varphi $.

Step 3: the limit. Extract an $L^1$ limit $u$. By corollary 15.9.4, applied with the dominating measure $\mathrm{e}^{\lambda \sup _ Xw}\mu $, the convergence holds in capacity, $u\sim \varphi $, and

After a further extraction we may also assume $\nu _{u_ j}\rightharpoonup \nu $. Since $\nu _0\leq \nu _{u_ j}\leq \mu $, the limit satisfies the same inequalities. We claim that

Indeed, $\mathrm{e}^{\lambda u_ j}\to \mathrm{e}^{\lambda u}$ in $L^1(\mu )$ by lemma 5.44.2, and all the measures $\nu _{u_ j}$ are at most $\mu $; this controls the error caused by changing the weight. The fixed bounded weight $\mathrm{e}^{\lambda u}$ can be approximated in $L^1(\mu )$ by continuous functions, and since both $\nu _{u_ j}$ and $\nu $ are dominated by $\mu $, this approximation is uniform in the integrals. The weak convergence $\nu _{u_ j}\rightharpoonup \nu $ then gives the claim, and the same argument applies after multiplication by any continuous test function.

Since $\mathrm{e}^{\lambda u_ j}\nu _{u_ j}=\mathrm{MA}^{\mathrm{ext}}_\theta (u_ j)$, it follows that

The second identity holds off the pole set of $u$, which has zero $\mu $-measure. Hence $u$ belongs to the family and attains the supremum $B$. Write $\nu =g\, \mathrm{d}V$, with $0\leq g\leq f$.

Step 4: no deficit. Suppose that $g\neq f$ in $L^1(\mathrm{d}V)$. Applying lemma 15.12.4 to $f-g$, we find $v\sim \varphi $ with

This measure is nonzero. In fact, no potential $v$ admitting a smaller-background envelope has zero Monge–Ampère measure, since theorem 15.2.2, applied with coefficient zero to $v$ and $v+1$, would give $v\geq v+1$. Thus $\nu _ v(X)>0$.

The sum $\nu +\nu _ v$ is finite, so theorem 15.11.1 gives $a\sim \varphi $ with

This measure still lies between $\nu _0$ and $\mu $, but its mass is $B+\nu _ v(X)>B$, a contradiction. Therefore $g=f$, and $u$ solves the required equation. Uniqueness in the prescribed ordinary type follows from corollary 15.2.3. $\square$

of theorem 15.12.1, part (2). For each $j\geq 1$, part (1) gives $u_ j\sim \varphi $ with

Write $M_ j=\sup _ Xu_ j$, $v_ j=u_ j-M_ j$ and $c_ j=\mathrm{e}^{M_ j/j}$. Let $h$ be a bounded normalized solution of $\mathrm{MA}_\theta (h)=\alpha f\, \mathrm{d}V$, with $\alpha >0$, given by theorem 15.3.5.

For $H_ j=h+\| h\| _\infty +j\log \alpha $ we have

The bounded function $H_ j$ admits a smaller-background envelope, and $u_ j\preceq H_ j$, so corollary 15.2.3 gives $u_ j\leq H_ j$. In particular,

The lower density bound of lemma 15.12.2, applied to $v_ j$ with $q=p$, gives

so the $c_ j$ are also bounded away from zero.

Since $\sup _ Xv_ j=0$ and $\mathrm{MA}^{\mathrm{ext}}_\theta (v_ j)=c_ j\mathrm{e}^{v_ j/j}f\, \mathrm{d}V\leq C f\, \mathrm{d}V$, theorem 15.12.3 gives $\varphi -C'\leq v_ j\leq \varphi $. Extract a subsequence with $v_ j\to v$ in $L^1(X)$ and $c_ j\to c>0$. By corollary 15.9.4, the convergence holds in capacity, $v\sim \varphi $, and the Monge–Ampère measures converge weakly. Meanwhile, lemma 5.44.5 gives

Thus the right-hand sides converge in total variation to $cf\, \mathrm{d}V$, and $\mathrm{MA}^{\mathrm{ext}}_\theta (v)=cf\, \mathrm{d}V$.

For the uniqueness of the constant, suppose that $a,b\sim \varphi $ solve the normalized equation with constants $c_ a<c_ b$. For every real $t$,

so theorem 15.2.2 gives $a\geq b+t$. This is impossible as $t\to +\infty $, at any point where both potentials are finite. Interchanging the roles of the two constants proves that $c$ is unique. $\square$