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 $f$.
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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