Proposition 5.40.10B986B6AE211B

Proof. The function $-\log (1-|w|^2)$ is smooth and plurisubharmonic on $|w|<1$: its complex Hessian is

Thus the right-hand side of (5.40.11) is plurisubharmonic; it is negative on $B$ and satisfies the pole condition, so it is a competitor. For the reverse inequality, fix a competitor $u$ and a point $w$ with $|w|<1$. The restriction $u(\cdot ,w)$ is either identically $-\infty $ or subharmonic on the disk of radius $R=(1-|w|^2)^{1/2}$. In the latter case, $u(\zeta ,w)-m\log |\zeta |^2$ is subharmonic on the punctured disk and, by the pole condition, bounded above near zero, so it extends as a subharmonic function across zero.

Since $u\leq 0$, this extension is at most $-m\log r^2$ on every circle $|\zeta |=r<R$. The maximum principle and then $r\uparrow R$ give

On $D$ both sides are $-\infty $. Taking the supremum over $u$ proves the formula. Near a point of $D$, the function $H_ m$ is bounded and the logarithmic term tends to $-\infty $, which proves continuity there as an extended-real function. $\square$