Proposition 5.40.12.1709375DAB71 For $t>0$, define
\[ \Phi _ t(\zeta ,w)=\min \{ -m\log |\zeta |^2,t\} , \qquad V_ t=P_ B\Phi _ t, \]
with $\Phi _ t=t$ on $D$. Then $0\leq V_ t\leq t$, and
F4D9A718C648
5.40.13
\begin{equation} -m\log \bigl (1-(1-\mathrm e^{-t/m})|w|^2\bigr ) \leq V_ t(\zeta ,w)\leq H_ m(w). \end{equation}
In particular, $V_ t\uparrow H_ m$ pointwise on all of $B$, including $D$. The convergence is uniform on every compact subset, with
1BF247F761E6
5.40.14
\begin{equation} 0\leq H_ m-V_ t \leq \frac{m\mathrm e^{-t/m}R^2}{1-R^2} \qquad \text{where }|w|\leq R<1. \end{equation}
Proof.
The obstacle is continuous, nonnegative and at most $t$. Zero is a competitor, so the upper-envelope theorem gives $V_ t\in \mathrm{PSH}(B)$ with $0\leq V_ t\leq \Phi _ t\leq t$. Consequently $m\log |\zeta |^2+V_ t$ is a competitor for $G_ m$, and proposition 5.40.10 gives $V_ t\leq H_ m$ off $D$. At a point $(0,w)$, the submean inequality on any sufficiently small circle in the normal line gives the same bound, since $H_ m$ does not depend on $\zeta $.
For the lower bound, set $a_ t=1-\mathrm e^{-t/m}$ and $F_ t(w)=-m\log (1-a_ t|w|^2)$. The Hessian calculation above, with $|w|^2$ replaced by $a_ t|w|^2$, shows that $F_ t$ is plurisubharmonic. Moreover, $0\leq F_ t\leq t$, and $F_ t\leq H_ m\leq -m\log |\zeta |^2$ on $B\setminus D$ because $|\zeta |^2<1-|w|^2$. Thus $F_ t\leq \Phi _ t$ everywhere, and hence $F_ t\leq V_ t$. Finally,
\[ H_ m-F_ t =m\log \left(1+\frac{\mathrm e^{-t/m}|w|^2}{1-|w|^2}\right) \leq \frac{m\mathrm e^{-t/m}|w|^2}{1-|w|^2}. \]
This proves (5.40.14) and the convergence; the monotonicity of $V_ t$ in $t$ follows from that of $\Phi _ t$.
$\square$
Pome
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