Lemma 14.15.1.6E3495B1A225 Local uniform limits of admissible maximal functions are admissible and maximal. If $v$ is maximal on a bounded open set $D\subset \mathbb C^ n$, $h$ is smooth and admissible on $D$, both functions are continuous on $\overline D$, and $h\leq v$ on $\partial D$, then $h\leq v$ on $D$.
Lemma 14.15.16E3495B1A225
Proof. Let $v_ j$ be admissible maximal functions converging locally uniformly to $v$. Distributional positivity is preserved under this convergence. For maximality, suppose that a smooth function $h$ touches $v$ from below at $z_0$ and $P_ n(H(h)(z_0))>0$. Choose a small ball on which $v-h\geq 0$. Replacing $h$ by $h-\delta |z-z_0|^2$, with $\delta >0$ sufficiently small, keeps its matrix strictly positive near $z_0$ and makes the minimum strict. Uniform convergence then produces interior minimum points of $v_ j-h+\delta |z-z_0|^2$ tending to $z_0$. Adding the minimum value to the test function gives a function touching $v_ j$ from below with strictly positive matrix, a contradiction.
For comparison, choose $R$ with $\overline D\subset \{ |z|\leq R\} $. If $h>v$ somewhere, then for small $\delta >0$ the smooth function
still exceeds $v$ at some interior point and is at most $v$ on the boundary. Hence the negative minimum of $v-h_\delta $ is attained in the interior. But $P_ n(H(h_\delta ))=P_ n(H(h))+\delta I>0$ there, which contradicts maximality after the minimum value is added to $h_\delta $. No boundary regularity of $D$ is needed for this argument. $\square$
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