Proposition 5.4.4AA6533F0128A

Proof. We first work near a point. Fix $w_0\in D$, a ball $B\Subset \Omega $ centered at $F(w_0)$, and a connected open neighborhood $W$ of $w_0$ with $F(W)\subset B$. The mollifications $u_\varepsilon =u*\rho _\varepsilon $, with the radial mollifier $\rho _\varepsilon $ of section 2.5, are defined on $B$ for small $\varepsilon $, and by (3.4.13) they decrease pointwise to $u$ as $\varepsilon \downarrow 0$. Their complex Hessians are positive semidefinite at every point: for $\xi \in \mathbb C^ n$, the function $\sum _{j,k}(u_\varepsilon )_{j\bar k}\xi _ j\bar\xi _ k$ is the mollification of the positive measure $\sum _{j,k}u_{j\bar k}\xi _ j\bar\xi _ k$ (matrix criterion), hence nonnegative by proposition 2.5.3. Since $F$ is holomorphic, the chain rule gives

so the complex Hessian of $u_\varepsilon \circ F$ is positive semidefinite, and in particular its Laplacian is nonnegative. A smooth function with nonnegative Laplacian is subharmonic, because by proposition 3.1.5 its spherical means increase with the radius and tend to its value at the center. Thus, along a sequence $\varepsilon _ k\downarrow 0$, the smooth subharmonic functions $u_{\varepsilon _ k}\circ F$ decrease to $u\circ F$ on $W$. By theorem 3.4.9, $u\circ F$ is subharmonic on $W$, and it is either identically $-\infty $ there or locally integrable, in which case $u_{\varepsilon _ k}\circ F\to u\circ F$ in $L^1_{\mathrm{loc}}(W)$. In the second case the currents $\mathrm{dd}^{\mathrm{c}}(u_{\varepsilon _ k}\circ F)$, which are positive by the matrix criterion, converge weakly to $\mathrm{dd}^{\mathrm{c}}(u\circ F)$, and positivity passes to the limit; so $\mathrm{dd}^{\mathrm{c}}(u\circ F)\geq 0$ on $W$.

It remains to pass from these neighborhoods to the connected components of $D$. Let $D_0$ be a component and let $A$ be the set of points of $D_0$ having a neighborhood on which $u\circ F\equiv -\infty $. The set $A$ is open. It is also closed in $D_0$: if $w_0\in D_0$ lies in the closure of $A$, the connected neighborhood $W$ of $w_0$ constructed above meets $A$, so $u\circ F=-\infty $ on a nonempty open subset of $W$; since a locally integrable function is finite almost everywhere, the alternative on $W$ forces $u\circ F\equiv -\infty $ on $W$, and $w_0\in A$. Hence either $A=D_0$, or $A=\varnothing $; in the latter case $u\circ F$ is locally integrable and subharmonic with positive complex Hessian near every point of $D_0$, hence on $D_0$. $\square$