Corollary 11.34.24.1AB249BC820A Let $w_ j,w\in \mathrm{PSH}(X,\beta )$ satisfy $|w_ j|,|w|\leq M$ for a fixed $M\geq 1$, and suppose that $w_ j\to w$ in capacity. Put
\[ \begin{gathered} T_ j=\beta +\mathrm{dd}^{\mathrm{c}}w_ j,\qquad T=\beta +\mathrm{dd}^{\mathrm{c}}w,\\ Q_ j=T_ j/4,\quad Q=T/4,\qquad \nu _ j=T_ j^2/4,\quad \nu =T^2/4. \end{gathered} \]
Then $Q_ j\rightharpoonup Q$ as currents and $\nu _ j\rightharpoonup \nu $ as probability measures. More precisely, for real $f\in C^\infty (X)$ choose $C_ f\geq 0$ with $-C_ f\beta \leq \mathrm{dd}^{\mathrm{c}}f\leq C_ f\beta $. For every $\eta >0$, writing $E_{j,\eta }=\{ |w_ j-w|>\eta \} $, one has
5D44ECE3CA31
11.34.25
\begin{equation} \left|\int _ X f\, (\mathrm d\nu _ j-\mathrm d\nu )\right| \leq C_ f\bigl (2\eta +2M^2\operatorname {Cap}_\beta (E_{j,\eta })\bigr ). \end{equation}
If $T_ j$ is smooth and strictly positive, then $Q_ j$ and $\nu _ j$ are its adjoint cofactor and adjoint probability. If $\psi +w$ solves (11.33.1) for some $F\in C(X)$, the limit $\nu $ is the unique member of $\mathcal A_\varphi (F)$.
Proof.
The capacity is the one of section 5.29, and $\beta \leq 2\omega $ gives $\operatorname {Cap}_\beta \leq 4\operatorname {Cap}_\omega $. Set $a_ j=(w_ j+M)/(2M)$ and $S_ j=\beta +\mathrm{dd}^{\mathrm{c}}a_ j$. Then $0\leq a_ j\leq 1$ and
\[ \begin{gathered} S_ j=\left(1-\frac1{2M}\right)\beta +\frac1{2M}T_ j,\\ S_ j^2\geq \left(\frac1M-\frac1{2M^2}\right)\beta \wedge T_ j \geq \frac1{2M}\beta \wedge T_ j. \end{gathered} \]
Positivity of bounded mixed products justifies these inequalities. Thus $\int _ E\beta \wedge T_ j\leq 2M\operatorname {Cap}_\beta (E)$ for every Borel set $E$, and the same bound holds for $T$. By closedness and Stokes' theorem, $\int _ X\beta \wedge (T_ j+T)=8$. Now test the identity
\[ T_ j^2-T^2=\mathrm{dd}^{\mathrm{c}}\bigl ((w_ j-w)(T_ j+T)\bigr ) \]
against $f$. On $X\setminus E_{j,\eta }$ the potential difference has absolute value at most $\eta $, and on $E_{j,\eta }$ at most $2M$. Consequently
\[ \begin{aligned} \left|\int _ Xf(T_ j^2-T^2)\right| & \leq C_ f\left(8\eta +2M\int _{E_{j,\eta }}\beta \wedge (T_ j+T)\right)\\ & \leq C_ f\left(8\eta +8M^2\operatorname {Cap}_\beta (E_{j,\eta })\right). \end{aligned} \]
Dividing by four proves (11.34.25). Letting first $j\to \infty $ with $\eta $ fixed and then $\eta \downarrow 0$ shows that $\int _ Xf\, \mathrm d\nu _ j\to \int _ Xf\, \mathrm d\nu $. All $\nu _ j$ and $\nu $ have mass one by proposition 11.34.18, and smooth functions are uniformly dense in $C(X)$, so $\nu _ j\rightharpoonup \nu $.
The constant test potential in the definition of capacity gives $4\mathrm dV(E)\leq \operatorname {Cap}_\beta (E)$. Splitting the potential difference over the same sets yields
\[ \| w_ j-w\| _{L^1(\mathrm dV)} \leq \eta +\tfrac M2\operatorname {Cap}_\beta (E_{j,\eta }). \]
Taking the limits in the same order proves $L^1$ convergence, and hence convergence of the currents $Q_ j$ by continuity of distributional differentiation. The limiting cofactor satisfies $\mathrm{dd}^{\mathrm{c}}Q=0$, $\int _ X\beta \wedge Q=1$, and $T\wedge Q=\nu $ as a bounded product. For smooth real $h$, the signed measure $2\mathrm{dd}^{\mathrm{c}}h\wedge T$ is the formal linearization of the residual operator at $w$ in the direction $h$, and its integral against the constant $1/4$ vanishes by Stokes' theorem. It is the derivative of the residual measure along $w+th$ whenever these functions remain $\beta $-plurisubharmonic.
If $T_ j$ is smooth and strictly positive, the maximum principle on the connected surface makes its Gauduchon multiplier constant, and the mass four normalizes it to $1/4$. Finally, if $\psi +w$ solves the equation at $F$, then $\nu =T^2/4=\mathrm e^ F\mathrm dV/\int _ X\mathrm e^ F\mathrm dV$, which is the unique supporting probability by proposition 11.34.18.
$\square$
Pome
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