Proposition 5.9.206C722B7D837

Proof. (1) Let $\Omega _0$ be a connected component of $\Omega $. We show that both conditions are equivalent to $\mathrm{dd}^{\mathrm{c}}u\geq 0$ on $\Omega _0$. Since $u$ is of class $C^2$, integration by parts shows that its distributional derivatives $u_{j\bar k}$ are the continuous functions $\partial _{z_ j}\partial _{\bar z_ k}u$, so for $\xi \in \mathbb {C}^ n$ the distribution $\sum _{j,k}u_{j\bar k}\xi _ j\bar\xi _ k$ is the continuous real function $z\mapsto \sum _{j,k}u_{j\bar k}(z)\xi _ j\bar\xi _ k$. A continuous real function is a positive distribution exactly when it is nonnegative: if it were negative at a point, it would be negative on a ball around that point, and its integral against a nonnegative test function supported in this ball and positive at its center would be negative. Hence, by the matrix criterion of section 5.4, $\mathrm{dd}^{\mathrm{c}}u\geq 0$ on $\Omega _0$ if and only if $(u_{j\bar k}(z))$ is positive semidefinite at every $z\in \Omega _0$.

If $u$ is plurisubharmonic on $\Omega _0$, then $\mathrm{dd}^{\mathrm{c}}u\geq 0$ there by theorem 5.9.1. Conversely, if $\mathrm{dd}^{\mathrm{c}}u\geq 0$ on $\Omega _0$, then by theorem 5.9.1 the class of $u$ has a plurisubharmonic representative $v$ on $\Omega _0$ satisfying the ball-average formula in (3.4.13). Since $v=u$ almost everywhere and $u$ is continuous, $v(x)$ is the limit of the averages of $u$ over the balls $B(x,r)$ as $r\to 0$, that is, $v(x)=u(x)$. Hence $u=v$ is plurisubharmonic on $\Omega _0$.

(2) By theorem 5.9.1, applied on each connected component of $\Omega $, the function $u$ is subharmonic and locally integrable on $\Omega $, and $\mathrm{dd}^{\mathrm{c}}u\geq 0$. By proposition 2.5.3(1), $u*\rho _\varepsilon $ is smooth on $\Omega _\varepsilon $, and its derivatives are the mollifications of the distributional derivatives of $u$. Thus, for $\xi \in \mathbb {C}^ n$, the function $\sum _{j,k}(u*\rho _\varepsilon )_{j\bar k}\xi _ j\bar\xi _ k$ is the mollification of the distribution $\sum _{j,k}u_{j\bar k}\xi _ j\bar\xi _ k$, which is positive by the matrix criterion; so it is nonnegative by proposition 2.5.3(3). By (1), $u*\rho _\varepsilon $ is plurisubharmonic on each connected component of $\Omega _\varepsilon $. Finally, since $u$ is subharmonic on $\Omega $, theorem 3.4.12(2) shows that $u*\rho _\varepsilon (x)\downarrow u(x)$ as $\varepsilon \downarrow 0$, for every $x\in \Omega $. $\square$