Corollary 13.29.7.21F64887A53E Suppose that smooth real $(1,1)$-forms $\theta _ j$ on $Y$ of positive area converge uniformly to a smooth form $\theta _0$ of positive area. Put $\lambda _ j=A_{\theta _ j}/A_{\theta _0}$, and let $F_{j0}:\mathcal M(\theta _0)\to \mathcal M(\theta _ j)$ be the correspondence of theorem 13.29.3. For every $t>0$,
0B648D2C5A55
13.29.8
\begin{equation} \begin{split} \sup _{p\in \mathcal M(\theta _0)} \| F_{j0}(p)-p\| _{L^1(Y)}& \longrightarrow 0,\\ \sup _{p\in \mathcal M(\theta _0)} \operatorname {Cap}_\eta (\{ |F_{j0}(p)-p|>t\} )& \longrightarrow 0. \end{split} \end{equation}
More generally, let $q_ j\in \mathcal M(\theta _ j)$ have polar measures $\nu _ j$, and let $p=p_{\theta _0}(\mu )$. If $\lambda _ j^{-1}\nu _ j\to \mu $ in total variation, then $q_ j\to p$ in $L^1$ and in capacity.
Proof.
The inverse of the scalar Laplacian for $\eta $ on functions of mean zero has a symmetric Green kernel $G$. Set
\[ h_ j(x)=\int _ YG(x,y)\, (\theta _ j-\lambda _ j\theta _0)(y). \]
The integrand form has integral zero, so $\mathrm{dd}^{\mathrm{c}}h_ j=\theta _ j-\lambda _ j\theta _0$, and elliptic regularity makes $h_ j$ smooth. Symmetry and the uniform $L^1$ bound from subsection 20.3.9 give
\[ \| h_ j\| _{L^\infty (Y)} \leq C\left\| \frac{\theta _ j-\lambda _ j\theta _0}{\eta }\right\| _{L^\infty (Y)} \longrightarrow 0. \]
Here $C$ depends only on $\eta $; the symmetric kernel may be made nonpositive by subtracting a constant, which does not change $h_ j$.
By (13.29.6), applied with $h=h_ j$, we have outside the common poles
\[ |F_{j0}(p)-p|\leq \operatorname {osc}_ Yh_ j+|\lambda _ j-1|\, |p|. \]
Normalized compactness bounds $\| p\| _{L^1}$ uniformly over $\mathcal M(\theta _0)$, which proves the first limit. Choose $C_0>0$ with $\theta _0\leq C_0\eta $. Then $p/C_0$ is $\eta $-plurisubharmonic with supremum zero, so the uniform sublevel estimate (5.29.4) bounds the capacity of $\{ |\lambda _ j-1|\, |p|>t/2\} $, uniformly in $p$; since pole sets have zero capacity, this proves the second limit.
For the last assertion put $p_ j=F_{0j}(q_ j)$, so that $q_ j=F_{j0}(p_ j)$. The polar measure of $p_ j$ is $\lambda _ j^{-1}\nu _ j$, so proposition 13.29.1 gives $p_ j\to p$ in $L^1$ and in capacity. The two uniform limits in (13.29.8), applied to $p_ j$, give $q_ j-p_ j\to 0$ in the same senses. The triangle inequality in $L^1$ and the subadditivity of capacity complete the proof.
$\square$
Pome
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