Proposition 5.46.74A6AA259F9D8

Proof. Locally, adding a smooth quadratic function to $u$ and subtracting a constant gives a negative plurisubharmonic function $a$ with pole set $S=\{ u=-\infty \} $. Put $\Theta =\mathrm{dd}^{\mathrm{c}}a$. Since smooth forms give $S$ zero mass, the polar current equals $\mathbf1_ S\Theta $ and is positive. In dimension one it is closed for degree reasons.

Now suppose that $n\geq 2$. Choose a smooth convex nondecreasing function $\chi $ on $\mathbb R$ with $\chi (t)=0$ for $t\leq 1/2$ and $\chi (1)=1$, and a smooth function $0\leq \vartheta \leq 1$ that vanishes near zero and equals one near one. Set

These functions are bounded and plurisubharmonic, with $0\leq w_ k\leq 1$. They vanish on $S$ and tend to one at every point outside $S$. Hence $\vartheta (w_ k)\Theta $ converges locally, coefficientwise in total variation, to $\mathbf1_{X\setminus S}\Theta $. Moreover, $\vartheta '(w_ k)$ vanishes on $S$ and tends to zero everywhere.

Fix a constant strongly positive $(n-2,n-2)$-form $\beta $ and put $\Xi =\Theta \wedge \beta $. The bounded Bedford–Taylor chain rule gives

Indeed, approximate $w_ k$ on a smaller ball by a decreasing sequence of smooth plurisubharmonic functions. Dominated convergence gives convergence of the corresponding $\vartheta $-weighted currents, while the convergence of bounded Bedford–Taylor energies and the Cauchy–Schwarz inequality give convergence of the differentiated expressions; this proves the chain rule for $w_ k$. For a smooth compactly supported test one-form $\alpha $, choose a nonnegative smooth cutoff $\tau $ equal to one near its support. Since $w_ k\geq 0$ and $\Xi $ is closed and positive, bounded integration by parts gives

Here $C$ depends only on the local mass of $\Xi $ and the fixed cutoff. The Cauchy–Schwarz inequality for a positive current now yields

where $\kappa $ is the Euclidean form; the limit follows by dominated convergence with respect to the fixed locally finite trace measure. Hence $\mathrm d(\vartheta (w_ k)\Xi )\to 0$, and $\mathbf1_{X\setminus S}\Theta \wedge \beta $ is closed. Since constant strongly positive forms span all constant forms of bidegree $(n-2,n-2)$, the current $\mathbf1_{X\setminus S}\Theta $ is closed. Subtracting it from $\Theta $ shows that $\mathbf1_ S\Theta $ is closed. The local currents agree on overlaps and give the asserted global current. $\square$