Lemma 15.20.55D3F7173CA27

Proof. Reduction to one variable. Block-unitary orbits in $U$ are exactly the level sets of $r$: normalize the sum of the squared norms of the two coordinate blocks, and map the two vectors of the same respective lengths by their unitary groups. Thus invariance gives a function $G$ of one real variable.

On a compact $r$-interval, the radial direction annihilates $P,Q$, while $R$ is strictly positive in that direction. Testing positivity of $\omega _0+\mathrm{dd}^{\mathrm{c}}G(r)$ in this direction gives $G''+z\geq 0$ as a one-dimensional distribution. One may obtain this first by smooth approximation of the scalar distribution and then by testing with nonnegative functions. Adding a twice integrated primitive of $z$ makes $G$ convex. A finite bounded convex function on an interval is locally Lipschitz, and its derivative is locally of bounded variation. The quasi-plurisubharmonic representative is the corresponding continuous representative. The horizontal directions give $x+h\geq 0$, $y-h\geq 0$ almost everywhere. Continuity of $x,y$ and the one-sided limits of a function of bounded variation give these bounds for both one-sided representatives. The radial inequality is precisely $\mathrm{d}h+z\, \mathrm{d}r\geq 0$; in particular jumps of $h$ can only be upward.

The radial form of the measure. We next justify the formula for the measure, including possible jumps. Identify $U$ smoothly with $\mathbb R\times Y$, where $Y=\{ r=0\} $, by real scaling of one block. The forms $P,Q$ are independent of $r$ in this identification. So is $\mathrm d^{\mathrm c}r$: in local coordinates it is the angular differential of the relative complex scale plus fixed base terms. Therefore

where $\nu $ is a fixed smooth positive volume form on the compact orbit $Y$, with the orientation chosen by this equality.

For smooth $G$, the rank-one wedge formula gives the radial coefficient of its Monge–Ampère measure, after dividing by $6\nu $, as

This identity passes to a locally Lipschitz $G$ with the preceding lower Hessian bound. Indeed, convolve $G$ on a slightly larger compact interval. The derivatives $h_\epsilon $ are uniformly bounded and converge to $h$ almost everywhere and in every finite $L^ q$ space. Their second derivatives have a uniform lower bound. Consequently the functions $G_\epsilon (r)$ have a common quasi-plurisubharmonic lower Hessian bound on compact subsets of $U$, and converge uniformly to $G(r)$. A fixed smooth local correction and bounded Bedford–Taylor continuity give convergence of their signed products. On the right-hand side of the formula, the two ordinary terms converge in $L^1$, and $\Phi (r,h_\epsilon )\to \Phi (r,h)$ in $L^1$, giving distributional convergence of its derivative. This proves the same formula for the bounded product of $G$.

Continuity of $h$. After the same division by $6\nu $, the right-hand side $f\omega _ X^3$ has radial coefficient $k(r)\, \mathrm{d}r$. This gives (15.20.6), initially for the almost-everywhere representative of $h$. If $h$ jumped at $r_0$, the atom of its left side would be

The right side has no atom. An upward jump of positive size lies in $[-x(r_0),y(r_0)]$, where the integrand is strictly positive in the interior. Since $x+y>0$ everywhere, such an integral is strictly positive. Thus there are no jumps. A locally bounded-variation function whose jumps all vanish has a continuous representative here: its distributional derivative consists of an absolutely continuous measure and a nonnegative singular measure, so its one-sided representatives coincide at every point. We henceforth use this representative. In particular $G$ is $C^1$ and $G'=h$ everywhere.

All coefficients on the right of (15.20.6) are now continuous, so $\Phi (r,h(r))$ is $C^1$. Where $\partial _ s\Phi =A>0$, the implicit function theorem gives $h\in C^1$ and the stated ordinary differential equation. Smoothness then follows by the smooth coefficients of that equation. The function $k$ is smooth, since the original density and its flat joins are smooth. $\square$