Let $\operatorname {Cap}=\operatorname {Cap}_\omega $ denote the capacity of section 5.29. Modified comparison bounds the capacity of a thin sublevel set by the Monge–Ampère mass of a slightly larger one. A bound by a power $1+\alpha $ of capacity, $\alpha >0$, makes the resulting iteration converge. For an $L^ p$ density, corollary 5.29.5 supplies the case $\alpha =1$. Iterating the two inequalities gives a lower bound for the capacity of every sublevel set of fixed height above the minimum. Since the capacity of deep sublevel sets of normalized potentials is small by the tail estimate (5.29.4), this bounds the depth of a normalized solution.
Lemma 14.19.14.370AD0860DAD Let $u,v\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$, with $-H\leq v\leq 0$ and $H\geq 1$. Fix $0<\varepsilon <1$, and put
\[ m=\inf _ X(u-(1-\varepsilon )v),\qquad U_ s=\{ u<(1-\varepsilon )v+m+s\} . \]
There are $\delta _\varepsilon >0$, depending only on $(X,\omega ,\varepsilon )$, and $C_ H>0$, depending only on $(n,H)$, such that
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14.19.15
\begin{equation} t^ n\operatorname {Cap}(U_ s)\leq C_ H\int _{U_{s+t}}\omega _ u^ n \qquad (0<s,t,\ s+t<\delta _\varepsilon ). \end{equation}
Every $U_ s$, $s>0$, has positive capacity.
Proof.
Let $B_0$ be the constant of theorem 14.19.6, and set
\[ \delta _\varepsilon =\min \{ 1-\varepsilon ,\varepsilon ^3/(128B_0n^2)\} . \]
For a capacity test $0\leq \rho \leq 1$, put
\[ \tau =\frac{t}{2(1-\varepsilon )(H+1)},\quad w=(1-\tau )v+\tau \rho ,\quad m'=\inf _ X(u-(1-\varepsilon )w). \]
Then $0<\tau <1$ and $0\leq (1-\varepsilon )(w-v)\leq t/2$, so $m-t/2\leq m'\leq m$. Consequently
\[ U_ s\subset V:=\{ u<(1-\varepsilon )w+m'+s+t/2\} \subset U_{s+t}. \]
We apply modified comparison to $u,w$ at height $s+t/2<\delta _\varepsilon $; by the choice of $\delta _\varepsilon $, its inverse factor is at most $(8/7)^ n$. Moreover,
\[ \omega _{(1-\varepsilon )w} \geq (1-\varepsilon )\tau \omega _\rho =\frac{t}{2(H+1)}\omega _\rho . \]
Integrating over $U_ s\subset V$, applying modified comparison on $V$, and taking the supremum over $\rho $, we obtain (14.19.15) with $C_ H=[2(H+1)]^ n(8/7)^ n$. Finally, each $U_ s$ is nonempty by the definition of the infimum, so it has positive smooth volume by lemma 5.24.4, and hence positive capacity, as the constant test function shows.
$\square$
Lemma 14.19.16.9C1A40570E78 In the setting of lemma 14.19.14, suppose that $A,\alpha >0$ and
\[ \omega _ u^ n(E)\leq A\operatorname {Cap}_\omega (E)^{1+\alpha } \qquad (E\subset X\text{ Borel}). \]
For every $0<D<\delta _\varepsilon $,
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14.19.17
\begin{equation} D\leq C_\alpha (C_ HA)^{1/n} \operatorname {Cap}_\omega (U_ D)^{\alpha /n}, \qquad C_\alpha =\frac{2}{1-2^{-\alpha }}. \end{equation}
In particular, when $\alpha =1$,
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14.19.18
\begin{equation} \operatorname {Cap}(U_ D)\geq \frac{D^ n}{4^ nC_ HA}. \end{equation}
Proof.
Put $a(s)=\operatorname {Cap}_\omega (U_ s)^{1/n}>0$ and $K=(C_ HA)^{1/n}$. The thin-sublevel estimate gives
\[ t\, a(s)\leq K a(s+t)^{1+\alpha } \qquad (0<s,t,\ s+t<\delta _\varepsilon ). \]
Suppose $C_\alpha K a(D)^\alpha <D$. Starting with $s_0=D$, set
\[ t_ j=2K a(s_ j)^\alpha ,\qquad s_{j+1}=s_ j-t_ j. \]
If $a(s_ i)\leq 2^{-i}a(D)$ for $0\leq i\leq j$, then
\[ \sum _{i=0}^ jt_ i \leq 2K a(D)^\alpha \sum _{i=0}^ j2^{-i\alpha } <C_\alpha K a(D)^\alpha <D. \]
Thus $s_{j+1}>0$, and the sublevel estimate, applied with $s=s_{j+1}$ and $t=t_ j$, gives $a(s_{j+1})\leq a(s_ j)/2$. Induction justifies every step. The decreasing sequence $s_ j$ has a limit
\[ s_\infty \geq D-C_\alpha K a(D)^\alpha >0. \]
Monotonicity now gives $0<a(s_\infty )\leq 2^{-j}a(D)$ for every $j$, a contradiction. This proves (14.19.17), including the case in which $U_ D=X$.
$\square$
Theorem 14.19.19.EA32B8891E2B There are constants $C_ p,c_ p>0$, depending only on $(X,\omega ,\mathrm{d}V,p)$, such that, whenever $u\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$, $\sup _ Xu=0$, and
\[ \omega _ u^ n=F\, \mathrm{d}V,\qquad F\in L^ p(X,\mathrm{d}V),\quad F\geq 0, \]
we have
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\begin{equation} \| u\| _\infty \leq C_ p(1+\| F\| _ p), \qquad \| F\| _ p\geq c_ p. \end{equation}
Proof.
By corollary 14.19.11, $F$ is not almost everywhere zero. Apply lemma 14.19.16 with $v=0$, $H=1$, $\varepsilon =1/2$, and $A=C\| F\| _ p>0$, where $C$ is the constant of corollary 5.29.5. Fix $D=\delta _{1/2}/2$, and put $M=-\inf _ Xu$. Since $U_ D=\{ u<-M+D\} $, the resulting inequality reads
\[ \operatorname {Cap}(\{ u<-M+D\} )\geq c/\| F\| _ p. \]
If $M>D+1$, (5.29.4) bounds this capacity above by $C/(M-D)$, and the first assertion follows; the case $M\leq D+1$ is immediate. Bounding the same capacity above by $\operatorname {Cap}(X)$ gives the second assertion.
$\square$
Suppose that two uniformly bounded solutions are far apart at some point. The preceding estimates give a fixed positive lower bound for the capacity of a sublevel set of their difference. If their $L^1$ distance is small, this set has small volume, and the uniform $L^ p$ bound for the densities then makes its Monge–Ampère mass small. This contradicts the sublevel estimate (14.19.15), and so upgrades $L^1$ compactness to uniform compactness. Put $p'=p/(p-1)$.
Theorem 14.19.21.BF849D15DF41 Suppose $u_ j\in \mathrm{PSH}(X,\omega )\cap C(X)$, $\sup _ Xu_ j=0$, and
\[ \omega _{u_ j}^ n=F_ j\, \mathrm{d}V,\qquad F_ j\geq 0,\qquad \sup _ j\| F_ j\| _ p<\infty \]
for the fixed $p>1$. If $(u_ j)$ is Cauchy in $L^1(X,\mathrm{d}V)$, then it is Cauchy uniformly. In particular, every such family has a subsequence converging uniformly to a continuous $\omega $-plurisubharmonic function.
Proof.
By theorem 14.19.19, there is a common $H\geq 1$ with $-H\leq u_ j\leq 0$. Suppose that the sequence is not uniformly Cauchy. After exchanging indices if necessary, we find $a>0$ and arbitrarily large $j,k$ with
\[ S:=\inf _ X(u_ k-u_ j)\leq -a. \]
Choose $0<\varepsilon <1$ such that $\varepsilon H<a/4$, and choose a fixed $D>0$ with $2D<\delta _\varepsilon $ and $2D<a/4$. Apply the preceding sublevel lemmas with $u=u_ k$ and $v=u_ j$. Since $v\leq 0$, $m=\inf (u-(1-\varepsilon )v)\leq S$. Thus
\[ U_{2D}\subset \{ u-v<S+2D+\varepsilon H\} \subset \{ |u-v|>a/2\} . \]
The capacity lower bound (14.19.18), with a constant $A$ independent of $j,k$ given by corollary 5.29.5, yields $\operatorname {Cap}(U_ D)\geq c>0$. On the other hand, (14.19.15) and the inequalities of Hölder and Chebyshev give
\[ \begin{split} D^ n\operatorname {Cap}(U_ D) & \leq C_ H\int _{U_{2D}}F_ k\, \mathrm{d}V\\ & \leq C_ H\| F_ k\| _ p \left(\frac{2\| u_ k-u_ j\| _1}{a}\right)^{1/p'} \longrightarrow 0. \end{split} \]
This contradiction proves that the sequence is uniformly Cauchy. For the last assertion, the normalized compactness theorem, theorem 5.25.2, gives an $L^1$-Cauchy subsequence of any such family. Its uniform limit is continuous, and distributional convergence preserves $\omega $-plurisubharmonicity.
$\square$
Pome
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