Example 11.45.19.3C22B09DCF87 Take $a=1$, $d=1/4$, $q=1$, and let $t$ be the sphere height
\[ t=\frac{|Z_0|^2-|Z_1|^2}{|Z_0|^2+|Z_1|^2},\qquad b=2+t,\qquad H=A=\tfrac 12+|t|. \]
Here $c_0=1$ and $F=\log (2bA)$ is Lipschitz but not $C^1$. The residual has continuous positive coefficients with the same loss of differentiability. Its supporting probability differs from normalized Monge–Ampère volume:
\[ \int _ X\Omega ^2=4,\qquad \int _ Xt\, \mathrm{d}\nu =0,\qquad \int _ Xt\, \frac{\Omega ^2}{4}=\frac5{24}. \]
Even for smooth constant $H$, approximation of the pole by globally smooth potentials need not recover the weak adjoint. For $H=a$, so that $w_0=0$ and $\nu =\mathrm{d}W$, put
\[ \psi _\delta =d\log (\ell +\delta (1-\ell )),\qquad P_\delta =\omega _1+\mathrm{dd}^{\mathrm{c}}(\psi _\delta /d),\qquad 0<\delta \leq 1. \]
The smooth potentials $\psi _\delta $ decrease pointwise to $\psi $, converge in $L^1$ and smoothly off $D$, but the classical adjoint probabilities of $\theta +\mathrm{dd}^{\mathrm{c}}\psi _\delta $ converge to
\[ \frac{a\omega _1+d\delta _ p}{a+d}\wedge \omega _2, \qquad p=\{ Z_0=0\} . \]
This limit assigns mass $d/(a+d)>0$ to $D$, and the weighted currents also retain a nonzero divisor component.
Proof.
The height $t$ is uniform on $[-1,1]$, so $\int A\omega _1=1$ and $\int tA\omega _1=0$. Since $\Omega ^2=2bA\, \mathrm{d}W$, its mass is four and its normalized height moment is
\[ \frac12\int _{-1}^1 t^2(\tfrac 12+|t|)\, \frac{\mathrm{d}t}{2} =\frac5{24}. \]
The weight is $1/(2b)$ and $\nu =A\, \mathrm{d}W$. Moreover, $\mathrm{dd}^{\mathrm{c}}\theta =-2t\, \mathrm{d}W$, so the background has no smooth positive Gauduchon potential perturbation.
For the smoothing family, direct differentiation gives
\[ \frac{P_\delta }{\omega _1} =\frac{4\delta }{((1+\delta )+(1-\delta )t)^2}. \]
It is positive of mass one and converges weakly to $\delta _ p$. Thus the positive smooth forms are $\widehat\Omega _\delta =a\omega _1+dP_\delta +b\omega _2$. Their classical weights and probabilities are
\[ \widehat\rho _\delta =\frac1{2(a+d)b},\qquad \widehat\nu _\delta =\frac{a\omega _1+dP_\delta }{a+d}\wedge \omega _2. \]
The weighted forms are closed, so these are the normalized classical adjoints. Their limits follow from the curve measure convergence. In particular, the weighted-current limit contains $d[D]/(2(a+d)b(p))$. These smoothings fill the pole and do not satisfy the fixed-divisor and uniform-residual hypotheses of proposition 11.45.18.
$\square$
Pome
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