Theorem 11.46.8.F273E69013C1 Let $q\ge 0$ be radial, smooth on $\mathbb P^ k\setminus \{ o\} $, including the hyperplane at infinity, and satisfy $\int q\kappa ^ k=1$. Suppose that for some $\delta ,A,r_0>0$,
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11.46.9
\begin{equation} \left|\frac{q(r)}{q_\alpha (r)}-1\right|\le Ar^\delta \quad (0<r<r_0). \end{equation}
Then there is a genuine radial $\kappa $-psh potential $U$, normalized to have value zero on the hyperplane, with
\[ \mathrm{MA}_\kappa (U)=q\kappa ^ k, \qquad U\sim u_\alpha . \]
Its residual $R=U-u_\alpha $ off $o$ extends continuously across $o$, with value $R_0$, and for small $r$ satisfies
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\begin{equation} \begin{aligned} |R-R_0|& \le Cr^\delta L^{\alpha -2},\\ |R'|& \le Cr^{\delta -1}L^{\alpha -2},\qquad |R''|\le Cr^{\delta -2}L^{\alpha -2}. \end{aligned} \end{equation}
It is locally $C^{0,\min \{ 1,2\delta \} }$. For $\delta \ge \frac12$ it extends $C^1$ with zero gradient at $o$; for $\delta \ge 1$ it extends $C^2$ with zero Hessian at $o$. No derivative hypothesis on the error in (11.46.9) is needed for these conclusions. The potential and its residual are smooth on $\mathbb P^ k\setminus \{ o\} $, including every density zero there. The tail condition ensures positive cumulative central mass; local density smoothness alone is not the criterion.
Proof.
Let
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\begin{equation} \begin{aligned} M(r)& =\int _{|z|^2<r}q\kappa ^ k,& Q(r)& =M(r)^{1/k},\\ U'(r)& =\frac{Q(r)-P(r)}r,& U(\infty )& =0. \end{aligned} \end{equation}
The tail condition makes $q$ positive near $o$, so $M(r)>0$ for every $r>0$. One has $M'=q(P^ k)'\ge 0$, $Q'\ge 0$ and $M(\infty )=1$. Thus the positive root defining $Q$ is smooth for every finite $r>0$, even where $q$ vanishes. This proves smoothness across all finite density zeros without a vanishing-order assumption. In $t=1/r$, the tail $1-M$ is the integral from $0$ to $t$ of the smooth density profile times $-\mathrm{d}(1+t)^{-k}/\mathrm{d}t$. Thus $M$ and $Q$ are smooth there, with $Q(0)=1$ in that coordinate. The equation $\mathrm{d}U/\mathrm{d}t=-(Q-P)/t$ is smooth there, since $Q-P$ vanishes at $t=0$. This both defines the normalization and proves smoothness along the full hyperplane.
From (11.46.9) and (11.46.6),
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\begin{equation} |M-Q_\alpha ^ k| \le C\int _0^ r s^{\delta -1}(1-\log s)^{-1-b_\alpha }\, \mathrm{d}s \le Cr^\delta L^{-1-b_\alpha }. \end{equation}
The last inequality follows by writing $s=r\mathrm{e}^{-y}$ and bounding $\int _0^\infty \mathrm{e}^{-\delta y}(L+y)^{-1-b_\alpha }\, \mathrm{d}y$. Dividing by $Q_\alpha ^ k\asymp L^{-b_\alpha }$ shows that $Q/Q_\alpha $ tends to one; the mean value theorem for the positive $k$th root then gives
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\begin{equation} |Q-Q_\alpha |\le Cr^\delta L^{\alpha -2}. \end{equation}
Moreover the exact identity
\[ \frac{Q'}{Q_\alpha '}= \frac q{q_\alpha }\left(\frac{Q_\alpha }{Q}\right)^{k-1} \]
gives $|Q'-Q_\alpha '|\le Cr^{\delta -1}L^{\alpha -2}$. Since $R'=(Q-Q_\alpha )/r$, integration to the origin gives a finite $R_0$ and the first two inequalities in (11.46.10). Differentiating this formula gives the last inequality.
The local potential $\Phi =\log (1+r)+U$ has semipositive Hessian away from $o$ because $Q/r>0$ and $Q'\ge 0$. Also $U=u_\alpha +R_0+o(1)$ near $o$, so $\Phi $ tends to $-\infty $. The local profile has nonnegative increasing slope in $\log r$, so its maxima with its value at $r=\varepsilon $ are plurisubharmonic. They decrease to the locally integrable profile; the local decreasing-limit property gives its genuine plurisubharmonic extension with the canonical pole retained. Thus $U$ is globally $\kappa $-psh. Formula (11.46.4) or the local determinant ratio gives its measure $q\kappa ^ k$ away from $o$; the defining non-pluripolar truncations exclude $o$ and are locally identical to this smooth product elsewhere. The measure has exactly total mass one. The bounded global residual proves the ordinary type.
In real coordinates, with $\rho =|z|=\sqrt r$,
\[ |\nabla R|\le C\rho ^{2\delta -1}L^{\alpha -2}, \qquad |\nabla ^2R|\le C\rho ^{2\delta -2}L^{\alpha -2}. \]
For $2\delta <1$, integration along radial radii bounds differences by $C|\rho _1-\rho _2|^{2\delta }$ and hence by $C|z_1-z_2|^{2\delta }$. For $2\delta \ge 1$ the gradient is bounded, giving local Lipschitz regularity. At $\delta =\frac12$ its bound tends to zero and $R-R_0=o(\rho )$, which give the claimed $C^1$ extension. At $\delta =1$, the Hessian bound tends to zero and $\nabla R=o(\rho )$, giving the $C^2$ extension. Larger $\delta $ only improves these bounds. These arguments justify crossing the pole for $R$; they do not assign a finite value to $U(o)$.
$\square$
Pome
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