5.28 Products with smooth backgrounds5E2B332B89B4

Proof. On the intersection of two coordinate balls, the two corrected potentials differ by a smooth function. The rule $\mathrm{dd}^{\mathrm{c}}(p+f)=\mathrm{dd}^{\mathrm{c}}p+\mathrm{dd}^{\mathrm{c}}f$, for bounded $p$ and smooth $f$ in bounded local products, follows by smooth approximation; together with the binomial formula, it shows that the two expansions agree. The same argument handles a smooth change in the decomposition of a current. Moreover, any two decompositions of the same $(1,1)$-current differ by a smooth function: the difference of the potentials has a smooth Hessian, and taking its local Euclidean trace and applying elliptic regularity shows that it is smooth. Symmetry, multilinearity and locality then follow term by term. Bounded Bedford–Taylor products wedged with smooth forms have locally finite variation and do not charge pluripolar sets, so these properties pass to the finite expansion as well.

It remains to prove positivity, which does not follow from the signs of the summands. Write the factors as $T_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}u_ i$, $1\leq i\leq k$, where each $\theta _ i$ is a smooth real $(1,1)$-form and each $u_ i$ is locally bounded and quasi-plurisubharmonic, and assume $T_ i\geq 0$. Fix $\varepsilon >0$, a point of the coordinate ball and a positive constant form $\kappa $. On a sufficiently small neighborhood of the point, each $\theta _ i$ is close to a constant, hence closed, form $\theta _ i'$:

Then $S_ i=\theta _ i'+\varepsilon \kappa +\mathrm{dd}^{\mathrm{c}}u_ i\geq T_ i$ is a positive closed current there, with a bounded local plurisubharmonic potential, and

where the last term is a smooth semipositive form. Expanding the product of the currents $T_ i+2\varepsilon \kappa $ therefore gives a sum of positive currents, so this product is positive near the chosen point. Such neighborhoods, although they depend on $\varepsilon $, cover the coordinate ball, so the product of the $T_ i+2\varepsilon \kappa $ is positive on the whole ball. By multilinearity, it converges weakly to the product of the $T_ i$ as $\varepsilon \to 0$, which proves positivity. $\square$

Proof. The functions $m_\delta =\max (a+\delta ,b)$ satisfy $b\leq m_\delta \leq b+\delta $ and decrease to $b$ as $\delta \downarrow 0$. By lemma 5.24.1 and proposition 5.28.2, the products $\theta _{m_\delta }^ k\wedge W$ are positive, and by bounded decreasing continuity they converge weakly to $\theta _ b^ k\wedge W$. On the plurifine open set $\{ a+\delta >b\} $, which contains $\{ a=b\} $, we have $m_\delta =a+\delta $, so plurifine locality identifies the restriction of $\theta _{m_\delta }^ k\wedge W$ to this set with that of $\theta _ a^ k\wedge W$. Hence, for every compact set $K\subset \{ a=b\} $ in a coordinate chart, the upper semicontinuity of the mass of compact sets under weak convergence gives

Since the measures are locally finite, inner regularity extends this inequality to all Borel subsets of the contact set, and a countable cover by charts proves the assertion. $\square$

Proof. Work in product coordinates in which $\pi $ is a coordinate projection. On a smaller chart of $Y$, choose smooth quadratic functions $q_ i$ such that $p_ i=u_ i+q_ i$ is plurisubharmonic, and approximate each $p_ i$ by a decreasing sequence of smooth plurisubharmonic functions. Their pullbacks have the same properties on the product chart, and for smooth forms pullback commutes with wedge products. Applying bounded decreasing continuity to the local expansions (5.28.1) on both charts gives the asserted identity, because pullback of currents by the projection is weakly continuous: pairing the pullback with a compactly supported test form amounts to pairing the original current with the integral of the test form over the fiber variables. If $r>\dim Y$, every smooth approximating product vanishes for degree reasons, and so does its limit. $\square$

Proof. On a smaller coordinate ball, add a common smooth quadratic function $q$ such that each $u_ i+q$ is plurisubharmonic, and convolve with a smooth radial probability kernel. The Hessians of the convolutions are uniformly locally bounded and converge to the weak Hessians in every $L^ s$ space with $s<\infty $, so their finite products converge in $L^1$ on smaller balls. Since the potentials converge uniformly, the continuity of bounded Bedford–Taylor products identifies this limit with the pluripotential product. Expanding with respect to the smooth correction, as in (5.28.1), proves the assertion for the original factors. $\square$

Proof. Put $T=\theta +\mathrm{dd}^{\mathrm{c}}u$. Its measurable coefficients are bounded on the compact manifold $X$, so there is $K>0$ such that, almost everywhere, the eigenvalues of $T$ relative to $h$ lie in $[0,K]$. By proposition 5.28.5, their product is the density of $T^ n$ with respect to $h^ n$, and hence it is at least $a$. Each eigenvalue is therefore at least $a/K^{n-1}$ almost everywhere. This inequality of matrices almost everywhere gives the asserted inequality of currents with $\delta =a/K^{n-1}$. $\square$

Proof. Fix $W\Subset U$. Since the forms $\vartheta _ i$ are smooth, there is a constant $C$ such that, for every $x\in W$ and every small ball $B_ r(x)$, the constant forms

satisfy $0\leq S_ i\leq 2Cr\beta $ on $B_ r(x)$, where $S_ i=A_ i-\vartheta _ i$. If $q_ i$ is a quadratic potential for $A_ i$, then $u_ i+q_ i$ is plurisubharmonic on $B_ r(x)$ and

Apply theorem 5.27.5 to these two closed Hessians. After expansion, every term other than $T_1^ k\wedge T_2^{n-k}$ contains some $S_ i$ and is therefore bounded by $C'r\mu $, where

is a locally finite measure.

A finite cover and a partition of unity give, on $W$,

Letting $r\downarrow 0$ proves (5.28.8), and the binomial expansion then gives (5.28.9). $\square$

Proof. Approximate the convolution by finite convex combinations of translates, and apply (5.28.9), extended by induction to finitely many terms. Along these approximations, the background forms converge smoothly and the potentials converge uniformly on compact sets. To pass to the limit, add a common quadratic function and write the difference of two $n$-th powers as a telescoping sum. After pairing with a test form, integration by parts bounds each term by the uniform distance between the potentials times the mass of a mixed current, and successive integration against nested cutoff functions bounds these masses uniformly. Thus the products of the bounded potentials converge, and (5.28.11) follows. $\square$

Proof. Take a nonnegative smooth radial mollifier $\chi _\delta $ supported in the ball of radius $\delta $. Then $T*\chi _\delta \to T_{\mathrm{ac}}$ almost everywhere. Indeed, the convolutions of the absolutely continuous parts converge at Lebesgue points of their densities, while the convolution of the singular part $\lambda _ s$ of a coefficient measure is bounded by $C\delta ^{-2n}|\lambda _ s|(B_\delta )$, which tends to zero almost everywhere by the differentiation theorem for measures. Applying lemma 5.28.10 with $g=F^{1/n}$ and letting $\delta \downarrow 0$ gives $\det _\beta T_{\mathrm{ac}}\geq F$.

For the reverse inequality, write the convolved current as the Hessian of a decreasing regularization of a locally bounded plurisubharmonic function plus a convergent smooth correction. The continuity of bounded products gives $(T*\chi _\delta )^ n\rightharpoonup T^ n$. Fatou's lemma, applied against a nonnegative compactly supported test function, then gives $\det _\beta T_{\mathrm{ac}}\, \beta ^ n\leq T^ n=F\beta ^ n$. $\square$

Proof. The function $G(x,y)=x+\chi (y-x)$, for $y\leq x$, is convex and nondecreasing in each variable: its first derivatives are $1-\chi '$ and $\chi '$, and its Hessian is

For smooth potentials, the chain rule gives the identity. For bounded potentials, add a common quadratic correction and convolve. Convolution preserves the order of the potentials, and since $G(x+r,y+r)=r+G(x,y)$, adding the same correction to both arguments adds it to $G$. The bounded chain rule (5.27.9) and its products with the gradient term give the identity in the limit. Freezing the smooth background as in the proof of proposition 5.28.2 preserves the exact lower bound. Every term on the right is positive, and the positivity of bounded mixed products gives the last assertion.

For a convex weight of class $C^{1,1}$, extend it convexly across zero with derivative in $[0,1]$, and smooth it in one variable. The smooth convex weights and their first derivatives converge uniformly on compact intervals, and the derivatives stay in $[0,1]$. On each smaller coordinate ball, the corresponding potentials converge uniformly to $w$. The continuity of bounded products gives convergence of their top powers, while the weights on the right-hand side converge uniformly against the fixed measure $\theta _ h^ n$. Passing to the limit proves the positivity and the product inequality with the original smooth background. $\square$