Lemma 8.3.189F810E43B3E

Proof. The distributional inequalities pass to local uniform limits. If a lower test $h$ for the limit at $z_0$ has Hessian in the interior, subtract $\delta |z-z_0|^2$, for sufficiently small $\delta >0$, to make the contact strict while retaining interior admissibility near $z_0$. Interior minimum points for the approximants then tend to $z_0$. Adding the minimum value produces lower tests for them with Hessian in the interior, contradicting their maximality.

For comparison choose $R$ with $\overline D\subset \{ |z|\leq R\} $. If $h>v$ somewhere, the function $h_\delta =h+\delta (|z|^2-R^2)$ still exceeds $v$ somewhere for small $\delta >0$ and is at most $v$ on the boundary. The negative minimum of $v-h_\delta $ is attained inside $D$. Since $H(h_\delta )=H(h)+\delta I$ belongs to the interior, adding the minimum value gives a forbidden lower test. This argument requires no regularity of the boundary. $\square$