Lemma 8.5.2.1881F6670EF4 For $1\leq a\leq k\leq n$, suppose $u_ j^ a$ and $u^ a$ satisfy the preceding lemma. Let $\theta _ a$ be fixed smooth real $(1,1)$-forms. Then
\[ \bigwedge _{a=1}^ k(\theta _ a+\mathrm{dd}^{\mathrm{c}}u_ j^ a) \longrightarrow \bigwedge _{a=1}^ k(\theta _ a+\mathrm{dd}^{\mathrm{c}}u^ a) \quad \hbox{as currents}. \]
The current on the right is the product of the almost-everywhere coefficients. No closedness or positivity of the $\theta _ a$ is required. For each compactly supported smooth complementary test form $\eta $, the difference of its pairings is bounded by $C_\eta \max _ a\| u_ j^ a-u^ a\| _\infty $.
Proof.
Set $D_ j^ a=\mathrm{dd}^{\mathrm{c}}u_ j^ a$ and $D^ a=\mathrm{dd}^{\mathrm{c}}u^ a$. On a smaller coordinate ball, convolution gives smooth $u_\epsilon ^ a$ whose complex Hessians are uniformly bounded and converge to $H(u^ a)$ in every finite $L^ p$ space. Their coefficient products consequently converge in finite $L^ p$ spaces. In particular the products of the $D^ a$ are closed currents, since their smooth approximants are closed.
Expand the smooth background factors and telescope each remaining product. Every difference has the form
\[ \mathrm{dd}^{\mathrm{c}}(u_ j^ a-u^ a)\wedge T_ j, \]
where $T_ j$ is a product of $D_ j^ b$ and $D^ b$ with $b\ne a$. It has uniformly bounded coefficients and is closed. Local convolution, followed by strong convergence of its finite coefficient products, justifies integration by parts:
\[ \left\langle \mathrm{dd}^{\mathrm{c}}(u_ j^ a-u^ a)\wedge T_ j,\xi \right\rangle =\int (u_ j^ a-u^ a)\, T_ j\wedge \mathrm{dd}^{\mathrm{c}}\xi . \]
Here $\xi $ includes the chosen smooth background factors and the test form, so their derivatives are retained. The right-hand side has the asserted uniform bound. A finite partition of unity proves the claim on $X$. This argument also proves that the coefficient product is the distributional limit; it does not identify a separately defined bounded Hessian operator for all nonsmooth admissible functions.
$\square$
Pome
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