For a Hermitian background, the total Monge–Ampère mass varies with the potential, and the ordinary comparison principle fails in general (see 14.25.3). A weaker form survives. Replacing the comparison potential $v$ by $(1-\varepsilon )v$ leaves the strictly positive form $\varepsilon \omega $ in reserve, and this reserve absorbs the torsion terms. On a sublevel set of height $s$, integration by parts makes the error proportional to $s$, and a finite induction controls the intermediate mixed products. The resulting multiplicative correction tends to one as $s$ tends to zero with $\varepsilon $ fixed.
Since $X$ is compact and $\omega $ is positive, we can choose $B_0\geq 1$ such that
\[ -B_0\omega ^2\leq \mathrm{dd}^{\mathrm{c}}\omega \leq B_0\omega ^2, \qquad -B_0\omega ^3\leq \mathrm{d}\omega \wedge \mathrm d^{\mathrm c}\omega \leq B_0\omega ^3. \]
These inequalities are understood after wedging with positive forms of complementary degree; forms of degree larger than $2n$ vanish, and the corresponding terms below are omitted.
Theorem 14.19.6.492E047184B3 Let $u,v\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$ and $0<\varepsilon <1$. For $s>0$, put
\[ m_\varepsilon =\inf _ X\bigl (u-(1-\varepsilon )v\bigr ),\qquad U_{\varepsilon ,s}=\{ u<(1-\varepsilon )v+m_\varepsilon +s\} . \]
If $0<s<\varepsilon ^3/(32B_0n^2)$, then, with $a=B_0n^2s\varepsilon ^{-3}$,
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14.19.7
\begin{equation} (1-4a)^ n\int _{U_{\varepsilon ,s}}\omega _{(1-\varepsilon )v}^{\, n} \leq \int _{U_{\varepsilon ,s}}\omega _ u^ n. \end{equation}
In particular, the reverse factor $(1-4a)^{-n}$ tends to one as $s\downarrow 0$ for fixed $\varepsilon $.
Proof.
Write $v_\varepsilon =(1-\varepsilon )v$, $U=U_{\varepsilon ,s}$ and
\[ \phi =\max (u,v_\varepsilon +m_\varepsilon +s),\qquad T_ k=\omega _ u^ k\wedge \omega _\phi ^{n-k},\qquad I_ k=\int _ U T_ k\quad (0\leq k\leq n). \]
Set $I_{-1}=I_{-2}=0$. Then $0\leq \phi -u\leq s$, this difference vanishes outside $U$, and on $U$ plurifine locality gives $\omega _\phi =\omega _{v_\varepsilon }\geq \varepsilon \omega $ in all bounded products.
By lemma 5.28.3, for every fixed positive mixed product $W$ of $n-1$ bounded-potential factors,
\[ \mathbf1_{\{ u=\phi \} }\omega _ u\wedge W \leq \mathbf1_{\{ u=\phi \} }\omega _\phi \wedge W. \]
For $0\leq k\leq n-1$, apply this with $W=\omega _ u^ k\wedge \omega _\phi ^{n-k-1}$. Since $u=\phi $ on $X\setminus U$, the difference $T_ k-T_{k+1}=(\omega _\phi -\omega _ u)\wedge W$ is nonnegative there. Consequently
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14.19.8
\begin{equation} I_ k-I_{k+1} \leq \int _ X\mathrm{dd}^{\mathrm{c}}(\phi -u)\wedge W =\int _ X(\phi -u)\mathrm{dd}^{\mathrm{c}}W. \end{equation}
This integration by parts is valid for bounded potentials. To see this, approximate $u$ and $\phi $ by decreasing sequences of smooth, uniformly bounded $\omega $-plurisubharmonic functions, using theorem 5.32.1. For smooth functions the identity is ordinary integration by parts. The form $\mathrm{dd}^{\mathrm{c}}W$ expands as below into bounded Hessian products and smooth forms, and bounded Bedford–Taylor convergence, also against bounded quasi-continuous coefficients converging in capacity, passes each term to the limit.
Set $r=n-k-1$. Since $\mathrm{d}\omega _ u=\mathrm{d}\omega _\phi =\mathrm{d}\omega $, expansion gives
\[ \begin{split} \mathrm{dd}^{\mathrm{c}}W={}& k\mathrm{dd}^{\mathrm{c}}\omega \wedge \omega _ u^{k-1}\wedge \omega _\phi ^ r +r\mathrm{dd}^{\mathrm{c}}\omega \wedge \omega _ u^ k\wedge \omega _\phi ^{r-1}\\ & +k(k-1)\mathrm{d}\omega \wedge \mathrm d^{\mathrm c}\omega \wedge \omega _ u^{k-2}\wedge \omega _\phi ^ r\\ & +2kr\, \mathrm{d}\omega \wedge \mathrm d^{\mathrm c}\omega \wedge \omega _ u^{k-1}\wedge \omega _\phi ^{r-1}\\ & +r(r-1)\mathrm{d}\omega \wedge \mathrm d^{\mathrm c}\omega \wedge \omega _ u^ k\wedge \omega _\phi ^{r-2}. \end{split} \]
Terms containing a negative power have coefficient zero and are omitted. On $U$ we may replace each $\omega ^ j$ in an upper bound by $\varepsilon ^{-j}\omega _\phi ^ j$, since $\omega _\phi \geq \varepsilon \omega $ there. Using $\phi -u\leq s$ and $\varepsilon ^{-2}\leq \varepsilon ^{-3}$ in (14.19.8), we obtain
\[ I_ k-I_{k+1}\leq B_0s\varepsilon ^{-3} \bigl (k(k-1)I_{k-2}+k(1+2r)I_{k-1}+r^2I_ k\bigr ). \]
Each coefficient is at most $n^2$, and $a=B_0n^2s\varepsilon ^{-3}$. Hence
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14.19.9
\begin{equation} I_ k-I_{k+1}\leq a(I_{k-2}+I_{k-1}+I_ k). \end{equation}
Put $q=1-4a$; the assumption on $s$ gives $a<1/32$, so $q>7/8$. Suppose inductively that $qI_ j\leq I_{j+1}$ for $j<k$. Then $I_{k-1}\leq q^{-1}I_ k$ and $I_{k-2}\leq q^{-2}I_ k$ whenever the indices exist. Since
\[ 1+q^{-1}+q^{-2}\leq 1+8/7+64/49<4, \]
(14.19.9) gives $qI_ k\leq I_{k+1}$. The case $k=0$ follows from the same formula with the two absent terms set to zero. By induction, $q^ n I_0\leq I_ n$. Since $T_0=\omega _{v_\varepsilon }^ n$ on $U$ and $T_ n=\omega _ u^ n$, this is the theorem. In dimension one all torsion terms vanish and the same argument applies.
$\square$
A strict multiplicative inequality between the Monge–Ampère measures on $\{ u<v\} $ absorbs the error in modified comparison. One first chooses the positivity reserve $\varepsilon $, and then a sublevel set thin enough to preserve the strict inequality.
Theorem 14.19.10.C0DEE578BC5D Let $u,v\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$, and let $0\leq c<1$. If
\[ \mathbf1_{\{ u<v\} }\omega _ u^ n \leq c\, \mathbf1_{\{ u<v\} }\omega _ v^ n, \]
then $u\geq v$. In particular, $\omega _ u^ n(\{ u<v\} )=0$ implies $u\geq v$.
Proof.
Suppose $m=\inf _ X(u-v)<0$. Choose $\varepsilon >0$ so small that
\[ (1-\varepsilon )^ n>c, \qquad 2\varepsilon \| v\| _\infty <-m/4, \]
omitting the second restriction if $v=0$. Let $m_\varepsilon $ and $U_{\varepsilon ,s}$ be defined as in theorem 14.19.6 for this pair $u,v$. Since $|m_\varepsilon -m|\leq \varepsilon \| v\| _\infty $, for all sufficiently small $s>0$ the set $U=U_{\varepsilon ,s}$ is contained in $\{ u<v\} $: there
\[ u-v<m_\varepsilon +s-\varepsilon v \leq m+2\varepsilon \| v\| _\infty +s<0. \]
The set $U$ is nonempty by the definition of the infimum and has positive volume by lemma 5.24.4.
Decrease $s$ further so that theorem 14.19.6 applies and $(1-4B_0n^2s\varepsilon ^{-3})^ n(1-\varepsilon )^ n>c$. Write the first factor as $q^ n$. Positivity of every mixed term gives
\[ \omega _{(1-\varepsilon )v}^ n \geq (1-\varepsilon )^ n\omega _ v^ n+\varepsilon ^ n\omega ^ n. \]
Modified comparison and the assumption would therefore imply
\[ q^ n(1-\varepsilon )^ n\int _ U\omega _ v^ n +q^ n\varepsilon ^ n\int _ U\omega ^ n \leq \int _ U\omega _ u^ n\leq c\int _ U\omega _ v^ n. \]
All integrals are finite, so the choice $q^ n(1-\varepsilon )^ n>c$ and the positive volume of $U$ make this impossible. Hence $m\geq 0$.
$\square$
Corollary 14.19.11.7C5FAB3BA5DA Every bounded $\omega $-plurisubharmonic function has strictly positive Monge–Ampère mass.
Proof.
Let $u$ be a bounded $\omega $-plurisubharmonic function. If $\omega _ u^ n=0$, then theorem 14.19.10, applied with $c=0$ and the constant potential $v=\sup _ Xu+1$, would give $u\geq v$, a contradiction.
$\square$
The positive lower bound in corollary 14.19.11 depends on the potential. A bound that is uniform over all bounded potentials is the condition $v_-(\omega )>0$ of definition 13.14.14.
Corollary 14.19.12.AB62A0258D5D If $u,v\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$ and $\lambda >0$, then
\[ \mathrm{e}^{-\lambda v}\omega _ v^ n\geq \mathrm{e}^{-\lambda u}\omega _ u^ n \quad \Longrightarrow \quad v\leq u. \]
Proof.
For every $t>0$, the assumption gives $\omega _ u^ n\leq \mathrm{e}^{-\lambda t}\omega _ v^ n$ on $\{ u<v-t\} $. Since $\omega _{v-t}=\omega _ v$, theorem 14.19.10, applied to $u$ and $v-t$, gives $u\geq v-t$. Letting $t\downarrow 0$ proves $v\leq u$.
$\square$
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