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The preceding argument controls the extremal level sets of $u-v$ through connectedness of the set where the density is positive. When the background satisfies the Guan–Li condition below, invariance of the total mass gives the ordinary comparison principle instead. An auxiliary solution whose density is larger on a strict sublevel set of $u-v$ then forces this set to have zero mass for the common Monge–Ampère measure, and the domination principle proves uniqueness.
Set $V=\int _ X\omega ^ n$. The Guan–Li condition
0B3AF72C81CB
14.25.1
\begin{equation} \mathrm{dd}^{\mathrm{c}}\omega =0,\qquad \mathrm{dd}^{\mathrm{c}}(\omega ^2)=0 \end{equation}
is equivalent to invariance of the total mass $\int _ X\mathrm{MA}_\omega (u)$ by lemma 11.1.3, theorem 11.1.7. In dimension one it holds automatically, for degree reasons.
Theorem 14.25.2.27DED19DBC11 Assume (14.25.1), let $p>1$, and let $f\geq 0$ belong to $L^ p(X,\omega ^ n)$ with positive integral. Then there is exactly one pair $(c,u)$, with $c>0$, $u\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$ and $\sup _ Xu=0$, satisfying
\[ (\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=c f\, \omega ^ n. \]
Moreover
\[ c=\frac{\int _ X\omega ^ n}{\int f\, \omega ^ n}, \]
and $u$ is continuous. Thus two bounded solutions with the same $L^ p$ density differ by a constant, without any positive lower bound on $f$.
We first show that the ordinary comparison principle is equivalent to invariance of the mass.
Theorem 14.25.3.FE27B72E2DB5 Condition (14.25.1) holds if and only if every pair of bounded $\omega $-plurisubharmonic functions $a,b$ satisfies
A25B3D75AF64
14.25.4
\begin{equation} \int _{\{ a<b\} }\mathrm{MA}_\omega (b) \leq \int _{\{ a<b\} }\mathrm{MA}_\omega (a). \end{equation}
For the converse implication, it suffices to require this inequality for smooth potentials.
Proof.
Under (14.25.1), theorem 11.1.7 gives $\int _ X\mathrm{MA}_\omega (a)=V$ for every bounded potential $a$; for $n=1$ this is Stokes' theorem.
Set $q=\max (a,b)$. On $\{ a<b\} $, plurifine locality gives $\mathrm{MA}_\omega (q)=\mathrm{MA}_\omega (b)$. On $\{ a\geq b\} $, one has $a=q$, and $a\leq q$ everywhere, so bounded ordered contact (lemma 5.28.3) gives
\[ \mathbf1_{\{ a\geq b\} }\mathrm{MA}_\omega (a) \leq \mathbf1_{\{ a\geq b\} }\mathrm{MA}_\omega (q). \]
The total masses of $a$, $b$ and $q$ all equal $V$. Hence
\[ \int _{\{ a<b\} }\mathrm{MA}_\omega (b)=V-\int _{\{ a\geq b\} }\mathrm{MA}_\omega (q) \leq V-\int _{\{ a\geq b\} }\mathrm{MA}_\omega (a)=\int _{\{ a<b\} }\mathrm{MA}_\omega (a), \]
which is (14.25.4).
Conversely, comparison for smooth potentials implies that all smooth potentials have the same mass: given a smooth pair $a,b$, add to $b$ a constant so large that $\{ a<b\} =X$, apply (14.25.4), and then interchange the roles of $a$ and $b$. Comparison with the zero potential shows that the common mass is $V$. By theorem 11.1.7, lemma 11.1.3, this implies (14.25.1) when $n\geq 2$; for $n=1$ that condition holds for degree reasons.
$\square$
For the uniqueness argument we need, in addition to comparison, auxiliary solutions and the mixed inequality (5.28.9). The auxiliary solutions come from continuous existence: if $G\geq 0$ lies in $L^ p(X,\omega ^ n)$ and $\int _ XG\omega ^ n=V$, theorem 14.19.2 gives a continuous solution of $\mathrm{MA}_\omega (\rho )=cG\omega ^ n$ with a constant $c>0$, and since all bounded potentials have mass $V$, this constant equals one. Thus
A548048D0134
14.25.5
\begin{equation} \mathrm{MA}_\omega (\rho )=G\omega ^ n \end{equation}
has a continuous solution $\rho $ for every such $G$, in particular for the density with an indicator factor used below.
of 14.25.2.
We use the domination principle in the form of the case $c=0$ of theorem 14.19.10: if $\mathrm{MA}_\omega (a)(\{ a<b\} )=0$ for bounded $\omega $-plurisubharmonic functions $a,b$, then $a\geq b$.
We first prove that two bounded solutions $u,v$ of $\mathrm{MA}_\omega (u)=\mathrm{MA}_\omega (v)=\mu =F\omega ^ n$, with $F\in L^ p$, $p>1$, differ by a constant. The mass identity gives $\mu (X)=V$. If $u-v$ is nonconstant, add a constant to $u$ so that $u-v$ has supremum $s>0$ and infimum $-s$. Since $\{ v<u\} $ and $\{ u<v\} $ are disjoint, after interchanging $u$ and $v$ if necessary, the set $S=\{ v<u\} $ satisfies $\mu (S)\leq V/2$. Define
\[ b_0=\frac{V-2\mu (S)}V\geq 0, \qquad G=2F\mathbf1_ S+b_0. \]
This nonnegative $L^ p$ function has total integral $V$, so we can choose a continuous solution $\rho $ of (14.25.5). Take $C_0$ with $\rho -C_0\leq u$, and set
\[ q_\delta =(1-\delta )u+\delta (\rho -C_0),\qquad E_\delta =\{ v<q_\delta \} \subset S, \qquad 0<\delta <1. \]
Since $G\geq 2F$ on $S$, the mixed inequality (5.28.9) gives, on $S$,
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14.25.6
\begin{equation} \mathrm{MA}_\omega (q_\delta )\geq \bigl ((1-\delta )+\delta 2^{1/n}\bigr )^ n\mu =:A_\delta \mu , \qquad A_\delta >1. \end{equation}
Ordinary comparison (14.25.4), applied to the pair $v,q_\delta $, yields
\[ A_\delta \mu (E_\delta ) \leq \int _{E_\delta }\mathrm{MA}_\omega (q_\delta ) \leq \int _{E_\delta }\mathrm{MA}_\omega (v)=\mu (E_\delta ). \]
Thus $\mu (E_\delta )=0$, that is, $\mathrm{MA}_\omega (v)(\{ v<q_\delta \} )=0$. Bounded domination gives $v\geq q_\delta $ everywhere, and letting $\delta \downarrow 0$ gives $v\geq u$, which contradicts $\sup _ X(u-v)=s>0$. Therefore $u-v$ is constant. The argument never divides by $F$, so $F$ may vanish on an open set.
Finally, for the density $f$ in the theorem, put $I=\int _ Xf\omega ^ n>0$ and $c=V/I$. By (14.25.5) with $G=cf$, there is a continuous solution, which we normalize by subtracting its supremum. Any other bounded solution, with constant $c'$, satisfies $V=c'I$ by mass invariance, hence $c'=c$; the uniqueness just proved and the common normalization then identify the potentials. In particular, every bounded solution with this $L^ p$ density is continuous, although continuity was not assumed.
$\square$
Pome
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