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
6B6C204B9D9E
15.15.2
\begin{equation} \sup _\Omega (v-u)_+ \leq C\left(\int _\Omega (v-u)_+\, \mathrm{d}V\right)^\gamma . \end{equation}
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).
$\square$
Pome
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