Theorem 12.10.267449627F15EC

Proof. Put $U=X\setminus Z$ and $A=\omega +\mathrm{dd}^{\mathrm{c}}\Phi _ f$. By theorem 12.10.17, $A$ is smooth on $U$ and $\Phi _ f$ satisfies the equation on $X$; the root $\gamma $ is positive on $U$. Stokes' theorem gives $\int _ XA\wedge T=1$, while attainment gives

The difference $A\wedge \gamma ^{n-1}-(A^ n/\gamma ^ n)^{1/n}\gamma ^ n$ is nonnegative and continuous on $U$ by the arithmetic–geometric mean inequality, and its integral vanishes. The equality case gives $A=r\gamma $ on $U$ with the smooth function $r=\operatorname {tr}_\gamma A/n\geq 0$; where $A^ n=0$, equality forces $A=0$ and $r=0$.

Since $\mathrm{dd}^{\mathrm{c}}\Phi _ f$ is closed,

For a semipositive form $\gamma $ of rank at least two, the real-linear map $W_\gamma :\xi \mapsto \xi \wedge \gamma $ from real one-forms to three-forms is injective at every point. Indeed, write $\gamma =\sum _ k\lambda _ k\, \mathrm dx_ k\wedge \mathrm dy_ k$ at the point in suitable real coordinates $z_ k=x_ k+\mathrm iy_ k$, with $\lambda _ k\geq 0$ and at least two $\lambda _ k$ positive. For each index $j$ choose $k\neq j$ with $\lambda _ k>0$. The coefficient of $\mathrm dx_ j\wedge \mathrm dx_ k\wedge \mathrm dy_ k$ in $\xi \wedge \gamma $ is $\lambda _ k$ times the coefficient of $\mathrm dx_ j$ in $\xi $, and similarly for $\mathrm dy_ j$. Hence $\xi \wedge \gamma =0$ forces $\xi =0$.

Choose an auxiliary Riemannian metric. The left inverse $L_\gamma =(W_\gamma ^*W_\gamma )^{-1}W_\gamma ^*$ is smooth and uniformly bounded on compact $X$. Thus

with smooth one-forms on $X$. In particular $|\mathrm d\log (1+r)|\leq C$ on $U$.

Points of $U$ can be joined by short paths avoiding $Z$. In a coordinate ball, stratify $Z$ into countably many smooth strata of real dimension at most $2n-2$. For a fixed point outside $Z$, the endpoints of segments from it that meet one stratum lie in a cone of dimension at most $2n-1$, which is Lebesgue-null. Given two nearby points outside $Z$, choose a generic intermediate point in a ball of radius comparable to their distance; both connecting segments then avoid $Z$, and their combined length is bounded by a fixed multiple of that distance. Using a finite connected cover by coordinate balls, with intermediate points chosen in the overlaps, we also obtain a uniform bound for the lengths of connecting paths in $U$.

The bound on $\mathrm d\log (1+r)$ now bounds $r$ on $U$, and (12.10.27) bounds its derivative. The short paths therefore give a unique locally Lipschitz extension of $r$ across $Z$. Since $Z$ is Lebesgue-null, the weak derivative of the extension satisfies (12.10.27) on $X$. The right side is continuous, so $r$ is $C^1$, and iterating makes it smooth.

Set $S=r\gamma $. It is smooth and semipositive, and $\mathrm dS=\mathrm d\omega $ on $X$ by density of $U$. The current $R=A-S$ is therefore closed, and it is supported on $Z$. It is also positive: since $A=S$ off $Z$ and $S$ puts no mass on $Z$, the current $R$ is the restriction of the positive current $A$ to $Z$. The support theorem recalled in section 5.38 expresses $R$ as a nonnegative combination of currents of integration along the components of $Z$ of codimension one. Locally $S-\omega $ is smooth and closed, so it has a smooth local $\mathrm{dd}^{\mathrm{c}}$-potential, and the coefficients of $R$ are the generic Lelong numbers of $\Phi _ f$ along these components. All of them vanish, so $\mathrm{dd}^{\mathrm{c}}\Phi _ f=S-\omega $ is smooth. Taking a Euclidean trace and applying elliptic regularity gives a smooth function equal to $\Phi _ f$ almost everywhere, and hence everywhere, since $\Phi _ f$ is its canonical representative. $\square$