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Theorem 13.3.18A125B338F90

Theorem 13.3.1.8A125B338F90 Every locally finite non-pluripolar product is closed.

Proof. Closedness is a local property, so we may shrink $U$ and subtract constants to assume that $u_ j\leq 0$ for every $j$. If $p=n$ the product has top degree and is closed for degree reasons, so let $p<n$. Fix a constant strongly positive $(n-p-1,n-p-1)$-form $\gamma $ and put

\[ \Theta _ k=\gamma \wedge \bigwedge _ j\mathrm{dd}^{\mathrm{c}}u_{j,k},\qquad S_\gamma =\gamma \wedge \Bigl \langle \bigwedge _ j\mathrm{dd}^{\mathrm{c}}u_ j\Bigr \rangle . \]

The $\Theta _ k$ are closed positive currents of bidimension $(1,1)$. By lemma 13.2.2, $\mathbf1_{O_ k}\Theta _ k=\mathbf1_{O_ k}S_\gamma $ increases to $S_\gamma $, and in particular these restrictions have uniformly bounded mass on compact sets. We shall prove that $\mathrm dS_\gamma =0$.

Cutoff functions. Put $u=\sum _ ju_ j\leq 0$ and $w_ k=\max \{ 1+u/k,0\} $. By proposition 5.9.5, $w_ k$ is plurisubharmonic, and $0\leq w_ k\leq 1$. As $k\to \infty $, $w_ k$ increases to $1$ at every point where $u>-\infty $, because $u/k$ increases to $0$ there, while $w_ k=0$ on the pole set of $u$. Moreover $w_ k=0$ outside $O_ k$: if $u_ j\leq -k$ for some $j$, then $u\leq u_ j\leq -k$, since all $u_ i\leq 0$. Choose a smooth function $\zeta $ on $\mathbb R$ with $0\leq \zeta \leq 1$, $\zeta =0$ on $(-\infty ,1/2]$ and $\zeta =1$ near $1$. Then $\zeta (w_ k)$ and $\zeta '(w_ k)$ vanish outside $O_ k$, so that $\zeta (w_ k)\Theta _ k=\zeta (w_ k)S_\gamma $. At every point where $u>-\infty $ we have $\zeta (w_ k)\to 1$ and $\zeta '(w_ k)\to 0$, and $S_\gamma $ gives the pole set of $u$ zero mass. Dominated convergence therefore gives

\[ \zeta (w_ k)\Theta _ k\longrightarrow S_\gamma , \qquad \zeta '(w_ k)\longrightarrow 0 \quad S_\gamma \text{-almost everywhere}. \]

Gradient bound. Let $\eta $ be a compactly supported smooth real one-form, and let $\tau \geq 0$ be a smooth compactly supported function equal to one near $\operatorname{Supp}\eta $. By (7.14.4), the gradient measure of $w_ k$ with respect to $\Theta _ k$ is $E_{\Theta _ k}(w_ k)=\frac12\mathrm{dd}^{\mathrm{c}}(w_ k^2\Theta _ k)-w_ k\mathrm{dd}^{\mathrm{c}}w_ k\wedge \Theta _ k$, and the last term is a positive measure because $w_ k\geq 0$. Hence

\[ \int \tau E_{\Theta _ k}(w_ k) \leq \tfrac 12\int w_ k^2\mathrm{dd}^{\mathrm{c}}\tau \wedge \Theta _ k \leq C\int _{\operatorname{Supp}\tau } \mathbf1_{O_ k}\Theta _ k\wedge \beta \leq C'. \]

The second inequality uses $w_ k=0$ outside $O_ k$, $w_ k\leq 1$, and the bound of the coefficients of $\mathrm{dd}^{\mathrm{c}}\tau \wedge \Theta _ k$ by a constant times the trace measure (theorem 5.3.7). The constant $C'$ is independent of $k$ because $\mathbf1_{O_ k}\Theta _ k\leq S_\gamma $ and $S_\gamma $ is locally finite.

Conclusion. By the chain rule (7.14.5), $\mathrm d(\zeta (w_ k)\Theta _ k)=\zeta '(w_ k)\, \mathrm dw_ k\wedge \Theta _ k$. We apply (7.14.6) with $a=\zeta '(w_ k)$. The constant $C_\eta $ there depends only on $\eta $, so it is the same for every $k$. The first factor on the right is bounded by the gradient bound, and the second tends to zero, since

\[ |\zeta '(w_ k)|^2\Theta _ k\wedge \beta =|\zeta '(w_ k)|^2\mathbf1_{O_ k}\Theta _ k\wedge \beta \leq |\zeta '(w_ k)|^2 S_\gamma \wedge \beta \]

and the right side tends to zero on $\operatorname{Supp}\eta $ by dominated convergence. Hence the pairing of $\mathrm d(\zeta (w_ k)\Theta _ k)$ with $\eta $ tends to zero. Since $\zeta (w_ k)\Theta _ k\to S_\gamma $, the pairing of $\mathrm dS_\gamma $ with $\eta $ vanishes, so $\mathrm dS_\gamma =0$.

Finally, $\gamma \wedge \mathrm d\langle \bigwedge _ j\mathrm{dd}^{\mathrm{c}}u_ j\rangle =\mathrm dS_\gamma =0$ for every constant strongly positive $\gamma $. These forms span the constant real $(n-p-1,n-p-1)$-forms (proposition 5.3.2), and their products with smooth functions and the coordinate one-forms $\mathrm dz_ i,\mathrm d\bar z_ i$ span the test forms of the two bidegrees against which $\mathrm d\langle \bigwedge _ j\mathrm{dd}^{\mathrm{c}}u_ j\rangle $ is evaluated. Therefore the non-pluripolar product is closed. $\square$

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