Lemma 12.14.2C0A6E438A11D

Proof. In the affine chart $Z_0=1$, write $s=|z|^2$. Then $x=s/(1+s)$ and $\omega =\mathrm{dd}^{\mathrm{c}}\log (1+s)$. Polar integration gives

for every integrable function $b$ on $[0,1]$, since the normalized radial density of $\omega ^ m$ in the variable $s$ is $m s^{m-1}(1+s)^{-m-1}\, \mathrm{d}s$. In particular $\int _ B a^{-1}\omega ^ m=mt/(m-1)=1$, and monotone convergence in (12.14.1) gives $H_\epsilon \to 1$.

Define $q_\epsilon \geq 0$ and $\varphi _\epsilon $ on $[0,1]$ by

Near zero,

so $q_\epsilon $ is $x$ times a smooth positive function; moreover $q_\epsilon (1)=1$. Hence both apparent singularities of $\varphi _\epsilon '$, at $0$ and at $1$, are removable, and $\varphi _\epsilon $ is smooth on $[0,1]$. Composing with the smooth function $x$ gives a smooth function on $B$, again denoted $\varphi _\epsilon $.

To compute $A_\epsilon $, recall that for a local potential of the form $F(s)$, with $q=sF'(s)$, the complex Hessian has the tangential eigenvalue $q/s$, of multiplicity $m-1$, and the radial eigenvalue $\mathrm{d}q/\mathrm{d}s$. For $F_\epsilon =\log (1+s)+\varphi _\epsilon (x)$ one finds $sF_\epsilon '=q_\epsilon (x)$. Differentiating the integral defining $q_\epsilon $ shows that both eigenvalues are positive, also at $s=0$ by the smooth factorization above. Their product relative to the Fubini–Study determinant is

At a point of the hyperplane $\{ Z_0=0\} $, the smooth form $A_\epsilon $ is semipositive, as a limit of the positive matrices in the affine chart. Since the determinant just computed extends there with a strictly positive value, every eigenvalue is positive. This proves strict positivity on all of $B$ and (12.14.3).

We now pass to the limit. For $x>0$, $q_\epsilon (x)$ tends to $x^ t$. To bound $q_\epsilon $ uniformly, note that $q_\epsilon ^ m$ is the distribution function of the probability density $\frac{m}{H_\epsilon }\frac{y^{m-1}}{y/t+\epsilon }$ on $[0,1]$. Relative to the probability density $my^{m-1}$, this density has a decreasing weight; relative to $(m-1)y^{m-2}$, its weight is proportional to $y/(y+t\epsilon )$ and hence increasing. If one probability density on $[0,1]$ has a decreasing weight relative to another, its average weight on $[0,x]$ is at least one and its distribution function is the larger of the two; an increasing weight reverses the inequality. Consequently

The function

is bounded and continuous on $[0,1]$, since $t>0$, and smooth on $(0,1]$; its derivative extends smoothly at one. The integrable function $(y^ t-y)/[y(1-y)]$ dominates the derivatives of the functions $\varphi _\epsilon $, so dominated convergence gives uniform convergence $\varphi _\epsilon \to \varphi $ on $B$. Distributional convergence shows that $A$ is a closed positive current. On each coordinate chart, a smooth local potential of $\omega $ plus $\varphi $ is a uniform limit of the corresponding plurisubharmonic functions built with $\varphi _\epsilon $, including at $p$. Hence $\varphi $ is $\omega $-plurisubharmonic. Off $p$, the limit $q=x^ t$ gives

Smoothness of $\varphi $ off $p$ and the same determinant argument at the hyperplane $\{ Z_0=0\} $ prove (12.14.4). $\square$