Proposition 14.19.23.3B70BD39881D Fix $A,\alpha >0$ and $H\geq 1$. Let $u,v\in \mathrm{PSH}(X,\omega )\cap C(X)$ satisfy $\sup _ Xu=0$ and $-H\leq v\leq 0$, and suppose
\[ \mu (E)\leq A\operatorname {Cap}_\omega (E)^{1+\alpha }, \qquad \mu =\omega _ u^ n, \qquad E\subset X\text{ Borel}. \]
There is a constant $C$, depending only on $(X,\omega ),A,\alpha ,H$, such that
B0DB641874CE
14.19.24
\begin{equation} \sup _ X(v-u)_+ \leq C\left(\int _ X(v-u)_+\, \mathrm d\mu \right)^{\sigma _\alpha }, \qquad \sigma _\alpha =\frac{\alpha }{3n(1+\alpha )+\alpha }. \end{equation}
For the square-capacity bound, $\alpha =1$, the exponent is $1/(6n+1)$.
Proof.
Write $L=\int _ X(v-u)_+\, \mathrm d\mu $ and $S=\sup _ X(v-u)_+$. Apply lemma 14.19.16 with competitor zero, $\varepsilon =1/2$ and one fixed $0<D<\delta _{1/2}$. It bounds $\operatorname {Cap}_\omega \{ u<\inf _ Xu+D\} $ below by a positive constant. The tail estimate (5.29.4) therefore bounds $-\inf _ Xu$ in terms of $(X,\omega ),A,\alpha $. Consequently $S$ has a fixed upper bound.
For $0<\varepsilon \leq 1/2$, choose a fixed small $c>0$ and put $D=c\varepsilon ^3$. The formula for $\delta _\varepsilon $ in lemma 14.19.14 allows $c$ to be chosen so that $2D<\delta _\varepsilon $ and $4D\leq \varepsilon $. With
\[ m=\inf _ X(u-(1-\varepsilon )v),\qquad U_ s=\{ u<(1-\varepsilon )v+m+s\} , \]
we have $m\leq -S$ when $S>0$, since $v\leq 0$. On $U_{2D}$,
\[ v-u>S-\varepsilon H-2D. \]
If $S>2\varepsilon H+4D$, this is greater than $S/2$. The backwards and thin-sublevel estimates give
\[ c_0D^{n(1+1/\alpha )} \leq D^ n\operatorname {Cap}_\omega (U_ D) \leq C_ H\mu (U_{2D})\leq \frac{2C_ HL}{S}, \]
where $c_0>0$ depends only on the stated data. Hence $S\leq CL\varepsilon ^{-3n(1+1/\alpha )}$ in this case. In the other case, $S\leq (2H+1)\varepsilon $.
For sufficiently small $L>0$, choose $\varepsilon =L^{\sigma _\alpha }$. The two alternatives give (14.19.24). The remaining positive values of $L$ are covered by the fixed upper bound for $S$. If $L=0$, the first alternative is impossible for every small $\varepsilon $, and the second gives $S=0$ on letting $\varepsilon \downarrow 0$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.