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Theorem 7.7.7BCFDDA1FCB7A

Theorem 7.7.7.BCFDDA1FCB7A A locally bounded plurisubharmonic function $u$ on an open set $\Omega \subset \mathbb C^ n$ is maximal if and only if $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$. In particular, maximality is a local property for locally bounded plurisubharmonic functions.

Proof. Suppose that $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$, and let $v$ be plurisubharmonic on $\Omega $ with $v\le u$ outside a compact set $K$. Choose a bounded open set $D$ with $K\subset D\Subset \Omega $. By upper semicontinuity, $v$ is bounded above on $\overline D$; since $u$ is bounded on $\overline D$, we may choose $M$ with $-M<\inf _ Du$. The function $\max \{ v,-M\} $ is bounded and plurisubharmonic on $D$, and it is at most $u$ on $D\setminus K$; since every point of $D$ near $\partial D$ lies in $D\setminus K$, we get $\varliminf _{D\ni z\to x}(u(z)-\max \{ v(z),-M\} )\ge 0$ for $x\in \partial D$. The last assertion of theorem 7.3.11 therefore gives $\max \{ v,-M\} \le u$ on $D$. Hence $v\le u$ everywhere, and $u$ is maximal.

Conversely, suppose that $u$ is maximal, and let $B\Subset \Omega $ be a ball. The function $\widehat u$ of corollary 7.7.5 is plurisubharmonic, satisfies $\widehat u\ge u$, and equals $u$ outside the compact set $\overline B$. Maximality gives $\widehat u\le u$, so $\widehat u=u$ and $(\mathrm{dd}^{\mathrm{c}}u)^ n=(\mathrm{dd}^{\mathrm{c}}\widehat u)^ n=0$ on $B$. Since the balls $B\Subset \Omega $ cover $\Omega $, and a distribution vanishing on each member of a family of open sets vanishes on their union (lemma 2.5.2), $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$ on $\Omega $. For the last assertion, let every point of $\Omega $ have an open neighborhood $U$ such that $u|_ U$ is maximal on $U$. By the first part applied on $U$, and by locality of the products (proposition 7.2.2), $(\mathrm{dd}^{\mathrm{c}}u)^ n$ vanishes on each such $U$, hence on $\Omega $, and $u$ is maximal on $\Omega $. Conversely, if $u$ is maximal on $\Omega $, then $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$ on every open set $U\subset \Omega $, so $u|_ U$ is maximal on $U$. $\square$

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