Lemma 16.27.2A73E0CBD9FAB

Proof. Write $\kappa _ q=\kappa +\mathrm{dd}^{\mathrm{c}}q$. We first prove (16.27.3) when $u=v$ near the poles of $v$ and then reduce to this case; the second assertion follows from the first.

A special case. Assume first that there are $l_0>0$ and an open set $U\supset \{ v<-l_0\} $ on which $u=v$. For $l>l_0$ put $u_ l=\max \{ u,-l\} $ and $v_ l=\max \{ v,-l\} $; these are bounded functions in $\mathrm{PSH}(X,\kappa )$ with $0\leq u_ l-v_ l\leq B$. On $U$ we have $u_ l=v_ l$ and $u=v$, so $\mu _{u_ l}=\mu _{v_ l}$ and $\mu _ u=\mu _ v$ there. The complement of $U$ lies in the plurifine open set $\{ v>-l\} $, on which $u_ l=u$ and $v_ l=v$, because $u\geq v$; by plurifine locality, $\mu _{u_ l}=\mu _ u$ and $\mu _{v_ l}=\mu _ v$ on this set. Hence $\mu _{u_ l}-\mu _{v_ l}=\mu _ u-\mu _ v$. Expanding $\kappa _{u_ l}^ n-\kappa _{v_ l}^ n$ and integrating by parts with lemma 13.5.2, we obtain

For the last inequality write $\mathrm{dd}^{\mathrm{c}}p=(\kappa +\mathrm{dd}^{\mathrm{c}}p)-\kappa $: the first part contributes a nonnegative amount, since $u_ l-v_ l\geq 0$, and the second contributes at least $-B\int _ X\kappa \wedge \kappa _{u_ l}^ j\wedge \kappa _{v_ l}^{n-1-j}=-B\Omega $, by proposition 13.5.1.

The general case. Subtracting a common constant, we may assume $v\leq 0$. For $0<a<1$ put $v_ a=av$ and $u_ a=\max \{ u,av\} $; these belong to $\mathrm{PSH}(X,\kappa )$ because $\kappa +\mathrm{dd}^{\mathrm{c}}(av)=(1-a)\kappa +a\kappa _ v$. Since $v\leq av$, we have $v_ a\leq u_ a\leq v_ a+B$. On the open set $U_ a=\{ v<-B/(1-a)\} $ we have $av-v>B\geq u-v$, so $u_ a=v_ a$ there. With $l_0=aB/(1-a)+1$, the set $\{ v_ a<-l_0\} =\{ v<-l_0/a\} $ is contained in $U_ a$, so the special case applies to $u_ a$ and $v_ a$.

As $a\uparrow 1$, $v_ a$ and $u_ a$ decrease to $v$ and $u$. The masses of $u_ a$ and $v_ a$ agree, and by multilinearity $\int _ X\langle \kappa _{v_ a}^ n\rangle $ is a polynomial in $a$ that tends to $\int _ X\langle \kappa _ v^ n\rangle =\int _ X\langle \kappa _ u^ n\rangle $. Decreasing sequences converge in extended capacity (lemma 13.7.3), so theorem 13.7.5 shows that $\mu _{u_ a}\to \mu _ u$ and $\mu _{v_ a}\to \mu _ v$ weakly along every sequence $a\uparrow 1$. Passing to the limit in (16.27.3) for $u_ a$ and $v_ a$, we obtain (16.27.3) for every continuous, in particular every smooth, $p\in \mathrm{PSH}(X,\kappa )$. For a bounded $p$, take the approximations $p_ j$ of lemma 16.27.1; they converge pointwise with a common bound, and dominated convergence with respect to the finite measures $\mu _ u$ and $\mu _ v$ proves (16.27.3).

The second assertion. Let $\phi $ be a model potential for $\kappa $. Since $V_\kappa =0$, proposition 16.21.3 gives $\mu _\phi \leq \kappa ^ n$. If $\phi -1\leq v\leq \phi $, the first assertion, applied to $\phi \leq v+1\leq \phi +1$, gives for every bounded $p$ with $\sup _ Xp=0$

By proposition 10.3.1, applied on the compact Kähler manifold $(X,\kappa )$ with background $\kappa $, the last integral is bounded by a constant depending only on $X$ and $\kappa $. For an unbounded $p$ with $\sup _ Xp=0$, apply this bound to the truncations $\max \{ p,-l\} $, which are bounded, belong to $\mathrm{PSH}(X,\kappa )$ and have supremum zero, and let $l\to \infty $ by monotone convergence. $\square$