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The subsolutions of lemma 15.14.1 stay within a bounded distance of a given reference. With them, the proof of theorem 15.4.1 extends to unbounded potentials lying above $\rho $, and the resulting estimate is uniform over all references between $\rho $ and zero. For a fixed reference, it upgrades $L^1$ compactness of solutions to uniform convergence of their differences on $\Omega $.
Theorem 15.15.1.668D07ADFF44 Fix $C_0,L>0$, $p>1$, and
\[ 0<\gamma <\frac{p-1}{np+p-1}. \]
Let $\eta \in \mathrm{PSH}(X,\theta )$ satisfy $\rho \leq \eta \leq 0$, and suppose that $u\in \mathrm{PSH}(X,\theta )$ satisfies
\[ \eta -C_0\leq u\leq \eta ,\qquad \mathrm{MA}_\theta (u)=f\, \mathrm{d}V,\qquad 0\leq f\in L^ p(\mathrm{d}V), \quad \| f\| _ p\leq L. \]
There is a constant $C$, depending only on the fixed geometry and on $\rho ,C_0,L,p,\gamma $, such that every $v\in \mathrm{PSH}(X,\theta )$ with $v\leq \eta $ satisfies
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15.15.2
\begin{equation} \sup _\Omega (v-u)_+ \leq C\left(\int _\Omega (v-u)_+\, \mathrm{d}V\right)^\gamma . \end{equation}
In particular, the constant does not depend on the reference $\eta $, and the comparison potential $v$ need not have the ordinary singularity type of $\eta $. The difference $v-u$ is evaluated on $\Omega $, where $u$ and $\eta $ are locally bounded; since $Z$ has zero volume, the integral is the same as over $X$. Since $Z$ is pluripolar, the resulting inequality $v\leq u+C\bigl (\int _\Omega (v-u)_+\, \mathrm{d}V\bigr )^\gamma $ on $\Omega $ extends to the canonical representatives on all of $X$.
Proof.
Replacing $v$ by $\max (v,u)$, we may assume $\eta -C_0\leq u\leq v\leq \eta $, so that $0\leq v-u\leq C_0$ on $\Omega $. Fix $q\in (1,p)$ and $\delta >0$, and put
\[ E_\delta =\{ u<v-\delta \} \cap \Omega ,\qquad g_\delta =\mathbf1_{E_\delta }f,\qquad m_\delta =\| g_\delta \| _ q. \]
When $m_\delta >0$, lemma 15.14.1, followed by subtracting one from its potential, gives $h\in \mathrm{PSH}(X,\theta )$ and a fixed $b>0$ such that
\[ \eta -1\leq h\leq \eta ,\qquad \mathrm{MA}_\theta (h)\geq b\, g_\delta /m_\delta \, \mathrm{d}V. \]
The constant $b$ is uniform in $\eta $, because the auxiliary density has $L^ q$ norm one and the strict barrier is fixed.
If $0<m_\delta <b/2$, set
\[ t=(2m_\delta /b)^{1/n},\qquad w=(1-t)v+th-\delta -C_0t. \]
Since $h-v\leq C_0$, we have $w\leq v-\delta $, so the set $\{ u<w\} \cap \Omega $ is contained in $E_\delta $. On this set the local expansion into positive mixed products gives $\mathrm{MA}_\theta (w)\geq t^ n\mathrm{MA}_\theta (h)\geq 2\mathrm{MA}_\theta (u)$. Both potentials are bounded below by $\rho $ minus a constant, and $w\leq u+C_0$, so theorem 15.13.2 applies and gives $u\geq w$. It follows that
\[ v-u\leq \delta +C_0t+t(v-h)\leq \delta +(C_0+1)t. \]
If $m_\delta =0$, we apply the same domination theorem with coefficient zero to $u$ and $v-\delta $. If $m_\delta \geq b/2$, we use $v-u\leq C_0$. In all cases,
\[ \sup _\Omega (v-u)\leq \delta +C_1m_\delta ^{1/n}. \]
Write $I=\int _\Omega (v-u)\, \mathrm{d}V$. If $I=0$, the canonical potentials agree. Otherwise Hölder's and Chebyshev's inequalities give
\[ m_\delta \leq L(I/\delta )^{(p-q)/(pq)}. \]
For $0<I\leq 1$, the choice $\delta =I^{(p-q)/(npq+p-q)}$ yields the exponent $\gamma _ q=(p-q)/(npq+p-q)$. We choose $q>1$ so close to one that $\gamma _ q>\gamma $. The case $I>1$ follows from the bound $v-u\leq C_0$. This proves (15.15.2).
$\square$
Corollary 15.15.3.131F17206F6A Fix $\eta \in \mathrm{PSH}(X,\theta )$ with $\rho \leq \eta \leq 0$, constants $C_0,L>0$, and $p>1$. Consider the family of potentials $u\in \mathrm{PSH}(X,\theta )$ satisfying
\[ \eta -C_0\leq u\leq \eta ,\qquad \mathrm{MA}_\theta (u)=f_ u\, \mathrm{d}V,\qquad 0\leq f_ u\in L^ p(\mathrm{d}V), \quad \| f_ u\| _ p\leq L. \]
This family is compact for uniform convergence of differences on $\Omega $. If $u_ j$ belong to the family and $u_ j\to u$ in $L^1$, then $u$ belongs to the family and
\[ \sup _\Omega |u_ j-u|\leq C\| u_ j-u\| _{L^1(\mathrm{d}V)}^\gamma \longrightarrow 0 \]
for every exponent $\gamma $ as in theorem 15.15.1. Moreover, $f_{u_ j}$ converges weakly to $f_ u$ in $L^ p$.
Proof.
Applying the theorem in both orders to $u_ j$ and $u_ k$, with the common reference $\eta $, we see that an $L^1$-Cauchy sequence in the family is uniformly Cauchy in differences on $\Omega $. Its uniform limit on $\Omega $ is locally $\theta $-plurisubharmonic and agrees almost everywhere, hence everywhere for the canonical representatives, with the $L^1$ limit $u$. The bounds $\eta -C_0\leq u\leq \eta $ hold on all of $X$. Letting $k\to \infty $ in the pairwise estimate proves the quantitative bound, before any equation for $u$ is used.
The densities form a bounded set in the reflexive space $L^ p$. Let $f$ be the weak limit of a subsequence $f_{u_{j_ k}}$. Since the potentials are locally bounded and converge locally uniformly on $\Omega $, Bedford–Taylor continuity, which holds also for the smooth nonclosed background, gives $\mathrm{MA}_\theta (u)=f\, \mathrm{d}V$ on $\Omega $. Neither measure charges the analytic set $Z$, so this equality holds on $X$. Positivity and weak lower semicontinuity of the norm give $f\geq 0$ and $\| f\| _ p\leq L$. Thus every weak cluster point equals the density $f_ u$ of $\mathrm{MA}_\theta (u)$, which proves weak convergence of the whole sequence. Finally, the common bounds by $\eta $ and $\eta -C_0$ give $L^1$ compactness of the family, by the compactness of quasi-plurisubharmonic functions, and the preceding argument upgrades each convergent subsequence to uniform convergence of differences on $\Omega $.
$\square$
For solutions with minimal singularities, normalized by $\sup _ Xu=0$, we take $\eta =V_\theta $. Then $\rho \leq V_\theta \leq 0$, and every normalized comparison potential $v\in \mathrm{PSH}(X,\theta )$ satisfies $v\leq V_\theta $, as the theorem requires.
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