Corollary 7.7.52E8B947536C1

Proof. Let $u_ j=u*\rho _{\varepsilon _ j}$, with $\varepsilon _ j\downarrow 0$ smaller than the distance from $\overline B$ to $\mathbb C^ n\setminus \Omega $, be radial regularizations of $u$; they are smooth plurisubharmonic functions on a neighborhood of $\overline B$ decreasing to $u$ (proposition 5.9.2). Let $H_ j$ be the solution of theorem 7.7.4 with boundary values $u_ j|_{\partial B}$, that is, the upper envelope of the subsolutions for $u_ j|_{\partial B}$. Since $u_ j$ is such a subsolution, $u\le u_ j\le H_ j$ on $B$. Since $u_{j+1}\le u_ j$ on $\partial B$, every subsolution for $u_{j+1}|_{\partial B}$ is one for $u_ j|_{\partial B}$, so the envelopes decrease: $H_{j+1}\le H_ j$. Let $H=\lim _ jH_ j$. The function $H_1$ is subharmonic on $B$ and continuous on $\overline B$ with boundary values $u_1$, so $H_1\le \max _{\partial B}u_1$ by the comparison principle (theorem 3.4.5). Hence $\inf _{\overline B}u\le H\le \max _{\partial B}u_1$, so $H$ is bounded; it is plurisubharmonic by theorem 5.10.1, and theorem 7.3.7 gives $(\mathrm{dd}^{\mathrm{c}}H)^ n=\lim _ j(\mathrm{dd}^{\mathrm{c}}H_ j)^ n=0$. For $x\in \partial B$, since $H\le H_ j$ and $H_ j$ is continuous on $\overline B$ with $H_ j=u_ j$ on $\partial B$,

By proposition 5.9.5, the function $\widehat u$ equal to $\max \{ u,H\} =H$ on $B$ and to $u$ outside $B$ is plurisubharmonic on $\Omega $. It is locally bounded, since $H$ is bounded and $u$ is locally bounded; it equals $u$ outside $B$, satisfies $\widehat u\ge u$ on $B$, and $(\mathrm{dd}^{\mathrm{c}}\widehat u)^ n=(\mathrm{dd}^{\mathrm{c}}H)^ n=0$ on $B$ by locality.

Let $v$ be plurisubharmonic on $B$ with $\varlimsup _{B\ni z\to x}v(z)\le u(x)$ for $x\in \partial B$. Since $u\le u_ j$, the function $v$ is a subsolution for the boundary values $u_ j|_{\partial B}$, so $v\le H_ j$ by theorem 7.7.4. Letting $j\to \infty $ gives $v\le H$. Conversely, $H$ belongs to this class by the displayed estimate, so $H$ is its largest member. If $u\le u'$, the class for $u$ is contained in the class for $u'$, which proves that $\widehat u\le \widehat{u'}$. $\square$