Theorem 5.29.23.FD7A4AE0DAC6 Let $\sigma _ j$ be smooth real $(1,1)$-forms on $Y$ that converge uniformly to a smooth form $\sigma _0$, and suppose that
\[ A_ j=\int _ Y\sigma _ j\geq 0,\qquad \int _ Y\sigma _0=0. \]
Let $u_0$ be the normalized $\sigma _0$-plurisubharmonic potential of lemma 5.29.22. Then, for every $t>0$,
\[ \begin{aligned} \sup _{u\in \mathrm{PSH}(Y,\sigma _ j),\ \sup _ Yu=0}\| u-u_0\| _{L^1(Y)}& \longrightarrow 0,\\ \sup _{u\in \mathrm{PSH}(Y,\sigma _ j),\ \sup _ Yu=0} \operatorname {Cap}_\eta (\{ |u-u_0|>t\} )& \longrightarrow 0. \end{aligned} \]
Moreover, the curvature measures $\sigma _ j+\mathrm{dd}^{\mathrm{c}}u$ of these potentials tend to zero in total variation, uniformly in $u$, since each has total variation $A_ j$.
Proof.
Choose $C>0$ with $\sigma _ j\leq C\eta $ for every $j$. Every sequence of normalized $\sigma _ j$-plurisubharmonic functions then lies in the compact family of normalized $C\eta $-plurisubharmonic functions of theorem 5.25.2. Every $L^1$ limit $u$ of a subsequence is normalized and, by passing to the limit in the sense of distributions, satisfies $\sigma _0+\mathrm{dd}^{\mathrm{c}}u\geq 0$. The preceding lemma gives $u=u_0$. If the first uniform limit failed, a sequence violating it would have a subsequence converging to this same limit, a contradiction.
For a potential $u$ in the $j$-th normalized family, put $T=\sigma _ j+\mathrm{dd}^{\mathrm{c}}u$. Positivity and Stokes' theorem give $\| T\| _{\mathrm{TV}}=A_ j$, and $A_ j\to \int _ Y\sigma _0=0$. Since $\sigma _0+\mathrm{dd}^{\mathrm{c}}u_0=0$,
\[ \| \mathrm{dd}^{\mathrm{c}}(u-u_0)\| _{\mathrm{TV}} \leq A_ j+\| \sigma _ j-\sigma _0\| _{\mathrm{TV}}\longrightarrow 0 \]
uniformly over that family. Apply lemma 5.27.11 on a fixed finite cover of $Y$ by concentric coordinate disks. The first limit and the bound above make each local capacity tend to zero uniformly, and the comparison between local and global capacity at the beginning of this section proves the second limit. The identity $\| T\| _{\mathrm{TV}}=A_ j$ proves the final assertion.
$\square$
Pome
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