PomeLemma 5.29.22.E8D07C38BC7C For a smooth real $(1,1)$-form $\sigma $ on $Y$, put $A=\int _ Y\sigma $. There is a smooth real function $h$ with $\sigma =A\eta +\mathrm{dd}^{\mathrm{c}}h$. The class $\mathrm{PSH}(Y,\sigma )$ is nonempty if and only if $A\geq 0$. If $A=0$, it consists of the functions $-h+c$, $c\in \mathbb R$, and its unique member with supremum zero is $u_\sigma =-h+\min _ Yh$.
5.29.21 Vanishing area on curvesB092BD05E335
On a compact curve, the integral of the background is the total mass of every curvature current. When this integral vanishes, positivity forces the current to vanish. Normalized compactness and the local estimate of capacity by total variation make this uniqueness stable as the area tends to zero.
Throughout this subsection, $Y$ is a compact connected Riemann surface and $\eta $ is a smooth positive $(1,1)$-form on $Y$ with $\int _ Y\eta =1$, which we use for $L^1$ norms and for the capacity $\operatorname {Cap}_\eta $ on $Y$.
Proof. The smooth form $\sigma -A\eta $ has integral zero, so the Poisson equation on the compact connected surface $Y$ has a smooth solution $h$, unique up to an additive constant. If $u\in \mathrm{PSH}(Y,\sigma )$, positivity and Stokes' theorem give
\[ 0\leq (\sigma +\mathrm{dd}^{\mathrm{c}}u)(Y)=A. \]
Conversely, $\sigma +\mathrm{dd}^{\mathrm{c}}(-h)=A\eta $, so $-h\in \mathrm{PSH}(Y,\sigma )$ when $A\geq 0$. If $A=0$, every positive curvature measure has total mass zero and therefore vanishes. Thus $\mathrm{dd}^{\mathrm{c}}(u+h)=0$ for every $u\in \mathrm{PSH}(Y,\sigma )$, and elliptic regularity and the uniqueness of harmonic functions on $Y$ show that $u+h$ is constant. Normalizing the supremum gives the stated formula. $\square$
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. $\square$
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