5.44 Integration against non-pluripolar measures2111608E9FA1

Proof. For the first statement, add a fixed smooth strictly plurisubharmonic function on each of finitely many coordinate balls, so that the resulting functions are uniformly bounded and plurisubharmonic there. Apply the local lemma on smaller compact balls covering $X$ and sum the integrals; the added functions cancel in the differences.

For the second statement, choose $A>0$ with $A\omega \geq \theta $. If $u_ j\leq C$, the convex chain rule gives $\mathrm{dd}^{\mathrm{c}}\mathrm{e}^{bu_ j}\geq -bA\mathrm{e}^{bC}\omega $, also for unbounded $u_ j$ by bounded truncation; here the exponential is given the value zero at a pole. Thus the functions $\mathrm{e}^{bu_ j}$ are uniformly bounded and quasi-plurisubharmonic, with a common lower curvature bound. They converge in $L^1(\mathrm{d}V)$, because the exponential is Lipschitz with constant $b\mathrm{e}^{bC}$ on $(-\infty ,C]$. The first statement now applies. $\square$

Proof. Since $u_\lambda \leq \varphi +C$ for a constant $C$, the function $u_\lambda $ is unbounded below. Every nonempty strict sublevel set $\{ u_\lambda <t\} $ of the upper semicontinuous function $u_\lambda $ is open, hence of positive smooth volume. Thus $u_\lambda $, and with it $\lambda u_\lambda $, is essentially unbounded below, and both conclusions follow. $\square$

Proof. Choose a Hermitian form $\Omega \geq \theta $, so that $v_ j\in \mathrm{PSH}(X,\Omega )$. By lemma 5.29.2, for every $T>1$,

uniformly in $j$. Since $v_ j\leq 0$ and $1-\mathrm{e}^{-x}\leq x$ for $x\geq 0$, splitting the integral at this sublevel set gives

By lemma 5.29.18, the second term is uniformly small when $T$ is large, and for this fixed $T$ the first term tends to zero as $j\to \infty $. This proves the general assertion; here the exponential is given the value zero at a pole, and pole sets have zero $\mu $-measure.

For the quantitative assertion, lemma 5.26.2, applied in the background $\Omega $, gives a uniform $L^ q$ bound for $-v_ j$, where $q=p/(p-1)$. Hölder's inequality and the bound

give the claimed estimate for every $j$. $\square$

Proof. By lemma 5.25.1, the tail envelopes $A_ j=(\sup _{\ell \geq j}a_\ell )^*$ decrease to $a$. Since $a\leq \max (a_ j,a)\leq A_ j$, monotone convergence in capacity of bounded potentials (lemma 5.29.19) implies

We split $|a_ j-a|=(\alpha _ j-a)+(\alpha _ j-a_ j)$. The integral of the first part against the moving products tends to zero by lemma 5.29.16.

For the second part, put $h_ j=\alpha _ j-a_ j\geq 0$. We compare its integrals against the measures with indices $j$ and $\ell $. The case $k=0$ follows from the assumed $L^1$ convergence. For $k\geq 1$, put $z=w_ j-w_\ell $ and

Factorization and Stokes' theorem on the compact manifold $X$ give

Since $\omega $ is not assumed to be closed, $T$ need not be closed, and the product rule reads

None of the terms involving derivatives of $T$ need vanish.

We estimate the four terms. Since $\mathrm{dd}^{\mathrm{c}}h_ j=\omega _{\alpha _ j}-\omega _{a_ j}$, the absolute contribution of the first is at most

It tends to zero as $j,\ell \to \infty $ by lemma 5.29.16, since $w_ j-w_\ell \to 0$ in capacity. Differentiating each factor shows that $\mathrm{d}T=\mathrm{d}\omega \wedge R$ and $\mathrm d^{\mathrm c}T=\mathrm d^{\mathrm c}\omega \wedge R$, where $R$ is a positive linear combination of mixed products of bidegree $(n-2,n-2)$. Moreover, $\mathrm{dd}^{\mathrm{c}}T$ is a sum of terms consisting of $\mathrm{dd}^{\mathrm{c}}\omega $ times mixed products of degree $n-2$, or of $\mathrm{d}\omega \wedge \mathrm d^{\mathrm c}\omega $ times mixed products of degree $n-3$; terms of negative degree are absent. These smooth torsion factors are bounded in absolute value by fixed multiples of $\omega ^2$ and $\omega ^3$, respectively. As $h_ j$ is uniformly bounded, the contribution of $h_ j\mathrm{dd}^{\mathrm{c}}T$ also tends to zero, by the same capacity estimate.

For the two gradient terms, the pointwise Cauchy–Schwarz inequality for a positive mixed product gives, for example,

The second factor tends to zero, and the first is uniformly bounded. Indeed, if $p$ is any of the uniformly bounded potentials and $S$ any positive mixed product of degree $n-1$ occurring here, then

Expand $\mathrm{dd}^{\mathrm{c}}S$ with all its torsion terms as above, and write $\mathrm{dd}^{\mathrm{c}}p=\omega _ p-\omega $. Boundedness of $p$ and lemma 5.29.16 bound both integrals uniformly. Finally, $\mathrm{d}h_ j\wedge \mathrm d^{\mathrm c}h_ j\leq 2\mathrm{d}\alpha _ j\wedge \mathrm d^{\mathrm c}\alpha _ j+2\mathrm{d}a_ j\wedge \mathrm d^{\mathrm c}a_ j$ as positive quadratic forms, which proves the asserted bound. The other gradient term is estimated in the same way. In dimension one, $T=1$ and all torsion terms in (5.44.8) vanish.

These integration-by-parts and energy identities hold for smooth potentials. They extend to bounded potentials through smooth decreasing approximations in the Hermitian background, using the local Bedford–Taylor chain rule, its gradient products and quasi-continuity; equivalently, they follow from the bounded current calculus of section 5.28. Only the common bound on the potentials enters the estimates.

We have shown that the difference between the integrals of $h_ j$ against the measures with indices $j$ and $\ell $ tends to zero as $j,\ell \to \infty $. For a fixed index $\ell $, the measure $\omega _{w_\ell }^ k\wedge \omega ^{n-k}$ is a fixed finite non-pluripolar measure, and lemma 5.44.2, applied separately to $\alpha _ j\to a$ and $a_ j\to a$ in $L^1(X)$, gives

Choosing first $\ell $ large and then letting $j\to \infty $, we conclude that the integrals of $h_ j$ against $\omega _{w_ j}^ k\wedge \omega ^{n-k}$ tend to zero. Together with the estimate for $\alpha _ j-a$, this proves the result. If the $a_ j$ only have some other fixed lower curvature bound, choose a Hermitian form $\Omega \geq \omega $ such that they are $\Omega $-plurisubharmonic. By section 5.29, convergence in capacity is unchanged by this replacement. Apply the result just proved with $\Omega $ in place of $\omega $; by positivity, the original mixed measures are bounded by their counterparts for $\Omega $. $\square$

Proof. Choose finitely many pairs of concentric coordinate balls $B'_ a\Subset B_ a$ such that the $B'_ a$ cover $X$, and fixed smooth functions $q_ a,r_ a$ on $B_ a$ with $\mathrm{dd}^{\mathrm{c}}q_ a\geq \omega $ and $\mathrm{dd}^{\mathrm{c}}r_ a\geq \theta $, with $r_ a\leq 0$. Then $\psi +q_ a$ is bounded and plurisubharmonic, $w+r_ a$ is negative and plurisubharmonic, and positivity of bounded products gives

The last step is the weighted Chern–Levine–Nirenberg estimate of section 5.27. Summing over the cover and using the normalized $L^1$ bound of theorem 5.25.2 proves the lemma. $\square$

Proof. On each coordinate ball $B_ a$ choose a fixed smooth function $q_ a$ with $\mathrm{dd}^{\mathrm{c}}q_ a\geq \theta $. If $w\in \mathrm{PSH}(X,\theta )$ and $\sup _ Xw=0$, then $w+q_ a$ is plurisubharmonic, and the submean inequality gives

with a constant independent of $w$ and $j$. Integrating and using (5.44.11) proves $A_\theta (\nu _ j)\leq A_\theta (\nu )+C$.

For the second assertion, the tail envelopes

decrease to $z$ by lemma 5.25.1. Enlarge the functions $q_ a$ so that $H_ j+q_ a$, $z_ j+q_ a$ and $z+q_ a$ are all plurisubharmonic. Since $H_ j\geq z_ j$ and $H_ j\geq z$, we have $|z_ j-z|\leq 2H_ j-z_ j-z$, and the identity (5.44.11) and the submean inequality imply

All these integrals are well defined: each function on the larger side of a submean inequality is bounded above, and its negative part is controlled by that of the integrable function on the smaller side plus a constant.

Radial averages of a plurisubharmonic function decrease when the radius decreases. Therefore $S_{a,j}(H_ j+q_ a)$ decreases to $z+q_ a$ on the support of $\eta _ a$: to identify the limit, fix an index $k$, use $H_ j\leq H_ k$ for $j\geq k$, let $j\to \infty $ in the radius, and then let $k\to \infty $; the reverse inequality is the submean inequality. The nonnegative integrand in the second line of the last display is dominated on this compact support by a constant minus $z$, which is $\nu $-integrable, so its integral tends to zero by dominated convergence. The final integral tends to zero by convergence in $L^1(\nu )$. This proves (5.44.13).

Since the exponential is Lipschitz on the common range of the functions, (5.44.13) also gives $\int _ X|\mathrm{e}^{\lambda z_ j}-\mathrm{e}^{\lambda z}|\, \mathrm{d}\nu _ j\to 0$. If $\zeta >0$ is smooth, then $\zeta \mathrm{e}^{\lambda z}=\mathrm{e}^{\lambda z+\log \zeta }$ is bounded and quasi-plurisubharmonic. For any fixed bounded quasi-plurisubharmonic function $F$, the same local submean argument and dominated convergence in (5.44.11) give

Applying this to $F=\zeta \mathrm{e}^{\lambda z}$ gives $\int _ X\zeta \mathrm{e}^{\lambda z_ j}\, \mathrm{d}\nu _ j\to \int _ X\zeta \mathrm{e}^{\lambda z}\, \mathrm{d}\nu $ for every smooth $\zeta >0$. Since every smooth function is a difference of two positive smooth functions, the convergence holds for all smooth test functions, and uniformly bounded masses together with uniform approximation of continuous functions by smooth ones prove weak convergence. $\square$