If a $\theta $-plurisubharmonic function exceeds a bounded solution only by a small amount in $L^1$, the excess is also uniformly small. To prove this, we restrict the density of the solution to a strict superlevel set of the difference and construct a subsolution for the restricted density. Domination bounds the height of the superlevel set in terms of the norm of the restricted density, and Hölder's inequality converts its volume into a power of the $L^1$ error.
Throughout this section $\theta $ is semipositive and big, and $p>1$ is fixed.
Theorem 15.4.1.72397EB33655 Fix $L,C_0>0$. Let $u\in \mathrm{PSH}(X,\theta )$ satisfy
\[ -C_0\leq u\leq 0,\qquad \mathrm{MA}_\theta (u)=f\, \mathrm{d}V,\qquad 0\leq f\in L^ p(\mathrm{d}V),\quad \| f\| _ p\leq L. \]
For every
\[ 0<\eta <\frac{p-1}{np+p-1} \]
there is a constant $C$, depending only on the fixed data and $\eta $, such that every $v\in \mathrm{PSH}(X,\theta )$ with $v\leq 0$ satisfies
\[ \sup _ X(v-u)_+ \leq C\left(\int _ X(v-u)_+\, \mathrm{d}V\right)^\eta . \]
The comparison potential $v$ need not be bounded.
Proof.
Replacing $v$ by $\max (v,u)$ leaves both sides unchanged, so we may assume $-C_0\leq u\leq v\leq 0$. Put $I=\int _ X(v-u)\, \mathrm{d}V$. If $I=0$, the two canonical potentials agree everywhere and there is nothing to prove, so we assume $I>0$.
Choose $q\in (1,p)$, to be specified at the end. For $\delta >0$ put
\[ E_\delta =\{ u<v-\delta \} ,\qquad g_\delta =\mathbf1_{E_\delta }f,\qquad m_\delta =\| g_\delta \| _ q. \]
If $m_\delta >0$, apply lemma 15.3.1 to $g_\delta /m_\delta $. It gives $h\in \mathrm{PSH}(X,\theta )$ and $b>0$ such that
\[ -1\leq h\leq 0,\qquad \mathrm{MA}_\theta (h)\geq b\, g_\delta /m_\delta \, \mathrm{d}V. \]
The constant $b$ does not depend on $\delta ,f,u,v$, since the auxiliary density has $L^ q$ norm one.
First suppose $0<m_\delta <b/2$, and take
\[ t=(2m_\delta /b)^{1/n}\in (0,1),\qquad w=(1-t)v+th-\delta -C_0t. \]
Because $h-v\leq C_0$, we have $w\leq v-\delta $, and hence $\{ u<w\} \subset E_\delta $. Positivity of bounded mixed products gives
\[ \mathrm{MA}_\theta (w)\geq t^ n\mathrm{MA}_\theta (h) \geq 2g_\delta \, \mathrm{d}V. \]
On $\{ u<w\} $ the last measure equals $2\mathrm{MA}_\theta (u)$. Bounded strict domination in the semipositive big background (theorem 16.28.56) therefore implies $u\geq w$. Consequently
\[ v-u\leq v-w =\delta +C_0t+t(v-h)\leq \delta +(C_0+1)t. \]
If $m_\delta =0$, the measure of $u$ vanishes on $E_\delta $, and the same domination theorem with coefficient zero, applied to $u$ and $v-\delta $, gives $v-u\leq \delta $. If $m_\delta \geq b/2$, the bound $v-u\leq C_0$ gives the following inequality after an increase of its constant. Thus in every case
E617EAC91766
15.4.2
\begin{equation} \sup _ X(v-u)\leq \delta +C_1m_\delta ^{1/n}. \end{equation}
Hölder's inequality and Chebyshev's inequality give
\[ m_\delta \leq \| f\| _ p\, \mathrm{d}V(E_\delta )^{1/q-1/p} \leq L(I/\delta )^{(p-q)/(pq)}. \]
Set $a=(p-q)/(npq)>0$. For $0<I\leq 1$, choose $\delta =I^{a/(1+a)}$. Substituting in (15.4.2) yields
\[ \sup _ X(v-u)\leq C_2 I^{\eta _ q}, \qquad \eta _ q=\frac{p-q}{npq+p-q}. \]
As $q\downarrow 1$, the exponent $\eta _ q$ tends to $(p-1)/(np+p-1)$. For the prescribed $\eta $ we choose $q>1$ so close to $1$ that $\eta _ q>\eta $; then $I^{\eta _ q}\leq I^\eta $ on $(0,1]$. When $I>1$, the bound $v-u\leq C_0$ gives the same conclusion after increasing $C$. This proves the theorem.
$\square$
Corollary 15.4.3.E429FB866D08 Fix $C_0>0$. Let $u_ j,u\in \mathrm{PSH}(X,\theta )$ satisfy $-C_0\leq u_ j,u\leq 0$ and
\[ \mathrm{MA}_\theta (u_ j)=f_ j\, \mathrm{d}V,\qquad \mathrm{MA}_\theta (u)=f\, \mathrm{d}V, \qquad \sup _ j\| f_ j\| _ p+\| f\| _ p<+\infty . \]
If $u_ j\to u$ in $L^1$, then $u_ j\to u$ uniformly. More precisely, for every exponent $\eta $ as in the theorem,
\[ \| u_ j-u\| _\infty \leq C\| u_ j-u\| _{L^1(\mathrm{d}V)}^\eta . \]
The corollary assumes that the limit $u$ already solves an equation with an $L^ p$ density. In an existence proof, this must be established before the corollary can be applied.
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