Lemma 15.30.1.B30B4EF45772 The function $u_\alpha $ is $\kappa $-plurisubharmonic with supremum zero, and
\[ \mathrm{MA}^{\mathrm{ext}}_\kappa (u_\alpha )=q_\alpha \kappa ^ k, \qquad \int _ Yq_\alpha \kappa ^ k=1. \]
The density $q_\alpha $ is smooth and strictly positive on $Y\setminus \{ o\} $, belongs to $L^1(\kappa ^ k)$, and belongs to no $L^ p(\kappa ^ k)$ with $p>1$. If we set $q_\alpha (o)=+\infty $, then every truncation $\min (q_\alpha ,s)$, $s>0$, is strictly positive and continuous on $Y$.
Proof.
In homogeneous coordinates, $\kappa +\mathrm{dd}^{\mathrm{c}}\psi =\mathrm{dd}^{\mathrm{c}}\log \sum _{i=1}^ k|z_ i|^2\geq 0$ on $Y\setminus \{ o\} $. The function $\chi (t)=1-(1-t)^\alpha $, $t\leq 0$, is increasing and convex, with $0<\chi '\leq \alpha $. Hence, off $o$,
\[ \kappa +\mathrm{dd}^{\mathrm{c}}u_\alpha \geq (1-\chi '(\psi ))\kappa +\chi '(\psi )(\kappa +\mathrm{dd}^{\mathrm{c}}\psi ) \geq (1-\alpha )\kappa . \]
Since $u_\alpha $ is bounded above, it extends across $o$ as a $\kappa $-plurisubharmonic function, with value $-\infty $ at $o$. On the hyperplane $\{ z_0=0\} $ we have $\psi =0$, hence $u_\alpha =0$, so $\sup _ Yu_\alpha =0$.
To compute the measure, work in the affine chart $z_0=1$ and put $r=\sum _{i=1}^ k|z_ i|^2$. On this chart $\kappa =\mathrm{dd}^{\mathrm{c}}\log (1+r)$ and $\kappa +\mathrm{dd}^{\mathrm{c}}u_\alpha =\mathrm{dd}^{\mathrm{c}}F$, where $s(r)=1-\log (r/(1+r))$ and
\[ F(r)=\log (1+r)+1-s(r)^\alpha ,\qquad Q(r)=rF'(r)=\frac{r+\alpha s(r)^{\alpha -1}}{1+r}. \]
For a radial function, the Levi form has the tangential eigenvalue $F'(r)=Q(r)/r$, with multiplicity $k-1$, and the radial eigenvalue
\[ Q'(r)=\frac{1-\alpha s(r)^{\alpha -1}+\alpha (1-\alpha )r^{-1}s(r)^{\alpha -2}}{(1+r)^2}>0. \]
Integrating in the radial variable gives $\int _{\{ a<r<b\} }(\mathrm{dd}^{\mathrm{c}}F)^ k=Q(b)^ k-Q(a)^ k$; the constant of proportionality is fixed by the case $F=\log (1+r)$, for which $(\mathrm{dd}^{\mathrm{c}}F)^ k=\kappa ^ k$ has mass one on the chart and $Q(r)=r/(1+r)$. Since again $Q(0)=0$ and $Q(\infty )=1$, the punctured chart has mass one. The complement of the chart is a hyperplane, which has zero volume, and the non-pluripolar measure does not charge $o$. Hence $\int _ Yq_\alpha \kappa ^ k=1$.
Relative to $\kappa ^ k$, whose tangential and radial eigenvalues are $1/(1+r)$ and $1/(1+r)^2$, the density on this chart is
\[ q_\alpha (r)=(1+r)^{k+1}(Q(r)/r)^{k-1}Q'(r) \asymp r^{-k}(1-\log r)^{-1-k(1-\alpha )}\quad (r\downarrow 0). \]
Near $r=0$ the smooth volume is comparable to $r^{k-1}\mathrm{d}r$ in the radial variable, so $q_\alpha $ is integrable. For $p>1$, the substitution $L=1-\log r$ turns the integral of $q_\alpha ^ p$ into the integral of $\mathrm{e}^{k(p-1)L}$ times a power of $L$, which diverges. The same asymptotics show that $q_\alpha (r)\to +\infty $ uniformly as $r\downarrow 0$, so every truncation $\min (q_\alpha ,s)$ is continuous at $o$. Finally, $q_\alpha \geq (1-\alpha )^ k>0$ by the lower bound $\kappa +\mathrm{dd}^{\mathrm{c}}u_\alpha \geq (1-\alpha )\kappa $.
$\square$
Pome
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