Pome

Theorem 15.6.102C5F4AFC7FF

Theorem 15.6.1.02C5F4AFC7FF Let $p>1$ and let $u\in \mathrm{PSH}(X,\theta )\cap L^\infty (X)$ satisfy

\[ \mathrm{MA}_\theta (u)=f\, \mathrm{d}V,\qquad 0\leq f\in L^ p(\mathrm{d}V). \]

Then $u$ is continuous on every open set $U\subseteq X$ on which $\rho $ is locally bounded.

Proof. After adding constants we may assume $-C_0\leq u\leq 0$ and $\rho \leq -C_0-1$. Choose

\[ 0<\eta <\frac{p-1}{np+p-1},\qquad \alpha =2\eta <1, \]

and set, for small $\delta >0$,

\[ c=\delta ^\alpha +\epsilon (\delta ),\qquad e=Ac+\epsilon (\delta ),\qquad s=e/\kappa ,\qquad v_\delta =\frac{\Phi _{c,\delta }+s\rho }{1+s}. \]

Since $s\kappa =e$, the curvature bounds for $\Phi _{c,\delta }$ and $\rho $ give $v_\delta \in \mathrm{PSH}(X,\theta )$. Moreover

\[ \rho \leq -C_0-1\leq \Phi _{c,\delta } \leq R_\delta u\leq 0 \]

for small $\delta $: the lower bound follows from $\Phi _{c,\delta }\geq u-K(\delta ^2+\delta )$, exactly as in the proof of theorem 15.5.1. Hence $v_\delta \leq \Phi _{c,\delta }\leq R_\delta u\leq 0$.

Theorem 15.4.1 allows an unbounded comparison potential, so it applies to $v_\delta $. Together with the $L^1$ smoothing estimate, it gives

\[ \sup _ X(v_\delta -u)_+ \leq C\left(\int _ X(v_\delta -u)_+\, \mathrm{d}V\right)^\eta \leq C\delta ^\alpha . \]

Fix $M$ and consider the points where $\rho \geq -M$. Since $s\leq Cc$ and $c\to 0$, the identity

\[ \Phi _{c,\delta }-u=(1+s)(v_\delta -u)+s(u-\rho ) \]

gives $\Phi _{c,\delta }-u\leq C_ Mc$ there.

At a radius $t$ where the infimum defining $\Phi _{c,\delta }$ is attained, the same lower estimate yields

\[ -c\log (t/\delta ) \leq \Phi _{c,\delta }-u+K(\delta ^2+\delta ) \leq C_ Mc. \]

Thus $t\geq a_ M\delta $ with $a_ M>0$ independent of the point and of $\delta $. Since $R_ tu+Kt^2$ is increasing in $t$, we obtain

\[ R_{a_ M\delta }u+K(a_ M\delta )^2 \leq R_ tu+Kt^2 \leq \Phi _{c,\delta }+K\delta ^2+K\delta . \]

This upper bound and the lower bound $R_ su\geq u-Ks^2$ imply

\[ \sup _{\{ \rho \geq -M\} } |R_{a_ M\delta }u-u| \leq C_ M\bigl (\delta ^\alpha +\epsilon (\delta )\bigr ) \longrightarrow 0. \]

On every compact subset of $U$, the function $\rho $ is bounded below. The smooth functions $R_{a_ M\delta }u$ therefore converge uniformly to $u$ on each such compact set, which proves that $u$ is continuous on $U$. $\square$

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