17.2 Integration by parts for bounded differencesF694A6981C1D

Proof. We write the difference of the two integrals as

and show that both terms tend to zero. For the first term fix $\varepsilon >0$ and put $A_ j=\{ |f_ j-f|>\varepsilon \} $. Splitting the integral over $A_ j$ and its complement gives

Since $\operatorname {Cap}_\omega (A_ j)\to 0$, uniform absolute continuity gives $\nu _ j(A_ j)\to 0$. Thus the upper limit of the first term is at most $\varepsilon \sup _ k\nu _ k(X)$, and $\varepsilon $ is arbitrary.

For the second term let $\eta >0$, and choose an open set $G$ with $\operatorname {Cap}_\omega (G)<\eta $ such that the restriction of $f$ to the compact set $X\setminus G$ is continuous. By lemma 5.2.7, there is a metric $d$ on $X$ which induces its topology, and by lemma 7.5.3, applied in the metric space $(X,d)$, the restriction of $f$ to $X\setminus G$ extends to a continuous function $g$ on $X$ with $|g|\leq B$. Since $X$ is compact, $g$ has compact support, so weak convergence of measures (see the discussion following definition 5.3.5) gives $\int _ Xg\, \nu _ j\to \int _ Xg\, \nu $. Since $f=g$ outside $G$, we have $|\int _ Xf\, \lambda -\int _ Xg\, \lambda |\leq 2B\lambda (G)$ for each $\lambda \in \{ \nu ,\nu _1,\nu _2,\ldots \} $. The upper limit of the second term is therefore at most $4B$ times the supremum of $\lambda (G)$ over the family, which tends to zero as $\eta \downarrow 0$ by uniform absolute continuity. $\square$

Proof. If the potentials were bounded, (17.2.4) would follow from Stokes' formula. We reduce to that case by truncation in three steps, assuming first that $\theta $ is a Kähler form, and remove this assumption in a fourth step.

Step 1: agreement on deep sublevel sets. Let $\theta =\kappa $ be a Kähler form, and assume that $\varphi _1\leq \varphi _2$, $\psi _1\leq \psi _2$, and that there is $L>0$ such that $\varphi _1=\varphi _2$ on $U=\{ \varphi _1<-L\} $ and $\psi _1=\psi _2$ on $V=\{ \psi _1<-L\} $. These sets are open, because $\varphi _1$ and $\psi _1$ are upper semicontinuous. For an integer $l>L$ put $\varphi _ j^ l=\max \{ \varphi _ j,-l\} $ and $\psi _ j^ l=\max \{ \psi _ j,-l\} $. These are bounded $\kappa $-psh functions; let $a^ l=\varphi _1^ l-\varphi _2^ l$, $b^ l=\psi _1^ l-\psi _2^ l$, and let $S_ j^ l$ be defined as $S_ j$ with the truncated functions. The algebraic identity $x^ n-y^ n=(x-y)\sum _ kx^ ky^{n-1-k}$ for the commuting Bedford–Taylor products gives

Since $\Sigma ^ l$ is a closed positive current, Stokes' formula for bounded potentials (lemma 13.5.2) yields $\int _ Xa^ l\, \mathrm{dd}^{\mathrm{c}}b^ l\wedge \Sigma ^ l=\int _ Xb^ l\, \mathrm{dd}^{\mathrm{c}}a^ l\wedge \Sigma ^ l$, which is (17.2.4) for the truncated functions.

We show that both sides of this identity coincide with the corresponding sides of (17.2.4). On $U$ we have $a^ l=a=0$. On the open set $V$ we have $\psi _1^ l=\psi _2^ l$ and $\psi _1=\psi _2$, so $\mu _{\psi _1^ l}=\mu _{\psi _2^ l}$ and $\mu _{\psi _1}=\mu _{\psi _2}$ there, by locality of Bedford–Taylor and non-pluripolar products. Hence both left sides are integrals over the closed set $F=X\setminus (U\cup V)$. On $F$ we have $\varphi _2\geq \varphi _1\geq -L$ and $\psi _2\geq \psi _1\geq -L$. Thus $F$ is contained in the plurifine open set where all four potentials exceed $-l$; there every truncation coincides with its potential, so the bounded products of the truncations coincide with the corresponding non-pluripolar products, by plurifine locality (proposition 13.2.6). Consequently the two left sides are equal. The right sides are treated in the same way with the roles of $U$ and $V$ exchanged: $b^ l=b=0$ on $V$, while $S_1^ l=S_2^ l$ and $S_1=S_2$ on $U$. This proves (17.2.4) under the assumptions of this step.

Step 2: a Kähler background with a margin. Let $\theta =\kappa $ be Kähler, and suppose that all four potentials are $\lambda \kappa $-psh for some $0<\lambda <1$. Adding a constant to one of the four potentials changes neither side of (17.2.4). Indeed, the measures are unchanged, and a constant added to $a$ or $b$ contributes nothing, since $\int _ X(\mu _{\psi _1}-\mu _{\psi _2})=0$ and $\int _ X(S_1-S_2)=0$ by corollary 13.8.8. Subtracting suitable constants from $\varphi _1$ and $\psi _1$, and then a common constant from each pair, we may therefore assume that all four potentials are nonpositive and that

for a constant $B\geq 1$. Take a sequence $\varepsilon \downarrow 0$ with $\lambda (1+\varepsilon )<1$ and put

Since $(1+\varepsilon )\varphi _2$ is $(1+\varepsilon )\lambda \kappa $-psh, the function $\varphi _{2,\varepsilon }$ is $\kappa $-psh, and $\varphi _1\leq \varphi _{2,\varepsilon }\leq \varphi _2\leq \varphi _1+B$ because $\varphi _2\leq 0$. Where $\varphi _1<-B-B/\varepsilon $,

so $\varphi _{2,\varepsilon }=\varphi _1$ there. The same holds for the $\psi $'s. Step 1 therefore applies to the pairs $(\varphi _1,\varphi _{2,\varepsilon })$ and $(\psi _1,\psi _{2,\varepsilon })$ and gives (17.2.4) with $a_\varepsilon =\varphi _1-\varphi _{2,\varepsilon }$ and $b_\varepsilon =\psi _1-\psi _{2,\varepsilon }$ in place of $a,b$, and with the corresponding measures.

Step 3: the limit $\varepsilon \downarrow 0$. As $\varepsilon \downarrow 0$, the functions $(1+\varepsilon )\varphi _2$ increase to $\varphi _2$, so $\varphi _{2,\varepsilon }$ increases to $\varphi _{2,0}=\max \{ \varphi _1,\varphi _2\} =\varphi _2$; likewise $\psi _{2,\varepsilon }\uparrow \psi _{2,0}=\psi _2$. By corollary 13.8.9, every measure in the identity of Step 2 converges weakly to the corresponding measure in (17.2.4). To pass to the limit in the integrals we apply lemma 17.2.2 to each of the finitely many terms, each of which is the integral of a bounded weight against a positive measure.

We first check uniform absolute continuity. For $\varepsilon \geq 0$ put

Since $-B\leq \varphi _{2,\varepsilon }-\varphi _2\leq 0$ and $-B\leq \psi _{2,\varepsilon }-\psi _2\leq 0$, we have $\chi \leq \zeta _\varepsilon \leq \chi +B$ for all $\varepsilon \geq 0$. Each of the four potentials is $\lambda \kappa $-psh, so $\mathrm{dd}^{\mathrm{c}}\chi \geq -\frac45\lambda \kappa $, and $\kappa +\mathrm{dd}^{\mathrm{c}}\chi $ is the sum of $(1-4\lambda /5)\kappa $ and the closed positive current $\frac45\lambda \kappa +\mathrm{dd}^{\mathrm{c}}\chi $. In particular $\chi $ is $\kappa $-psh, and expanding $\langle (\kappa +\mathrm{dd}^{\mathrm{c}}\chi )^ n\rangle $ by multilinearity and compatibility with smooth factors (proposition 13.2.6, proposition 13.4.3) gives $\int _ X\langle (\kappa +\mathrm{dd}^{\mathrm{c}}\chi )^ n\rangle \geq (1-4\lambda /5)^ n\int _ X\kappa ^ n>0$.

Now let $\mu $ refer to the background $\kappa $. By the identity

and multilinearity, each measure occurring in the identity of Step 2 (for $\varepsilon >0$) or in (17.2.4) (for $\varepsilon =0$) is a non-pluripolar product of $n$ of the currents on the right, and hence is at most $5^ n\mu _{\zeta _\varepsilon }$. The potential $q=B^{-1}\zeta _\varepsilon +(1-B^{-1})\chi $ satisfies $\chi \leq q\leq \chi +1$, so $q-1$ is a competitor for $\operatorname {Cap}_{\kappa ,\chi }$, and $\mu _ q\geq B^{-n}\mu _{\zeta _\varepsilon }$ by multilinearity, since $B\geq 1$. Hence $\mu _{\zeta _\varepsilon }(A)\leq B^ n\operatorname {Cap}_{\kappa ,\chi }(A)$ for every Borel set $A$, and every measure in question, for every $\varepsilon \geq 0$, is bounded by $5^ nB^ n\operatorname {Cap}_{\kappa ,\chi }$; in particular its total mass is at most $5^ nB^ n\operatorname {Cap}_{\kappa ,\chi }(X)=5^ nB^ n\int _ X\langle (\kappa +\mathrm{dd}^{\mathrm{c}}\chi )^ n\rangle \leq 5^ nB^ n\int _ X\kappa ^ n$, by proposition 16.25.6, theorem 13.5.4. Since $\chi $ has positive mass, theorem 16.27.17 bounds $\operatorname {Cap}_{\kappa ,\chi }(A)$ by a continuous nondecreasing function of $\operatorname {Cap}_\omega (A)$ vanishing at zero. Uniform absolute continuity follows.

Next we check the weights. Assign the value zero to $a_\varepsilon ,b_\varepsilon ,a,b$ on the pole sets; these are Borel functions with values in $[-B,0]$. Given $\eta >0$, proposition 16.27.18(3), applied to each of the four potentials with the bound $\eta /4$, and the countable subadditivity of $\operatorname {Cap}_\omega $ (proposition 16.25.6) give an open set $G$ with $\operatorname {Cap}_\omega (G)<\eta $ which contains the pole sets of the four potentials and such that the four potentials are finite and continuous on the compact set $X\setminus G$. Then $a$ and $b$ are continuous on $X\setminus G$, so they satisfy the continuity hypothesis of lemma 17.2.2. Moreover the four potentials are bounded in absolute value on $X\setminus G$ by a constant $C_\eta >0$. Since $(1+\varepsilon )\varphi _2\leq \varphi _{2,\varepsilon }\leq \varphi _2$, we have on $X\setminus G$

and likewise $|b_\varepsilon -b|\leq \varepsilon C_\eta $. Hence, for every $\delta >0$ and $\varepsilon <\delta /C_\eta $, the sets $\{ |a_\varepsilon -a|>\delta \} $ and $\{ |b_\varepsilon -b|>\delta \} $ lie in $G$ and have capacity less than $\eta $. Since $\eta $ is arbitrary, $a_\varepsilon \to a$ and $b_\varepsilon \to b$ in capacity. Lemma 17.2.2 now passes the identity of Step 2 to the limit, which is (17.2.4).

Step 4: an arbitrary background. For all sufficiently large $s>0$ we have $s\omega \geq \theta $ and the form $\kappa _ s=\theta +s\omega $ is Kähler. Each $q\in \mathrm{PSH}(X,\theta )$ is then $\tfrac 12\kappa _ s$-psh, because $\tfrac 12\kappa _ s+\mathrm{dd}^{\mathrm{c}}q=\theta _ q+\tfrac 12(s\omega -\theta )\geq 0$. By Steps 2 and 3 with $\lambda =\frac12$, (17.2.4) holds in the background $\kappa _ s$ for all large $s$. Since $\kappa _ s+\mathrm{dd}^{\mathrm{c}}q=\theta _ q+s\omega $, multilinearity and compatibility with smooth factors (proposition 13.2.6, proposition 13.4.3) show that every measure in the identity is a polynomial in $s$ whose coefficients are finite measures of the form $\langle \theta _{q_1}\wedge \cdots \wedge \theta _{q_ k}\rangle \wedge \omega ^{n-k}$; the weights $a,b$ do not depend on $s$. Both sides of the identity are therefore polynomials in $s$, and since they agree for all large $s$, their coefficients agree. The constant coefficients are the two sides of (17.2.4) in the background $\theta $. $\square$

Proof. Step 1: a repeated factor and a common background. Let $(\varphi _1,\varphi _2)$ and $(\psi _1,\psi _2)$ be presentations of $a$ and $b$, and suppose that these four functions and a further potential $\eta $ all belong to $\mathrm{PSH}(X,\theta )$ for one smooth closed real $(1,1)$-form $\theta $. For $0\leq t\leq 1$ the functions $\psi _ j^ t=(1-t)\eta +t\psi _ j$ are $\theta $-psh, and $\psi _1^ t-\psi _2^ t=tb$ is bounded. Lemma 17.2.3, applied to $\varphi _1,\varphi _2,\psi _1^ t,\psi _2^ t$, gives

Since $\theta _{\psi _ j^ t}=(1-t)\theta _\eta +t\theta _{\psi _ j}$, multilinearity expands every measure here as a polynomial in $t$ with finite measures as coefficients, so both sides are polynomials in $t$. We compare their coefficients of $t$. On the left this coefficient is

where we used that (17.2.1) may be computed in any smooth background with respect to which the presentation is psh. On the right it is $\int _ Xb(S_1^0-S_2^0)$, and $S_ j^0=n\langle \theta _{\varphi _ j}\wedge \theta _\eta ^{n-1}\rangle $, so it equals $n\int _ Xb\, \mathrm{dd}^{\mathrm{c}}a\wedge \langle \theta _\eta ^{n-1}\rangle $. This proves (17.2.6) when $T_1=\cdots =T_{n-1}=\theta _\eta $ and all potentials share the background $\theta $. For $n=1$ there are no currents $T_ j$, and (17.2.6) is (17.2.4) for a Kähler background with respect to which the four functions are psh.

Step 2: arbitrary currents. Let $n>1$. By theorem 10.2.15, write $T_ j=\theta _ j+\mathrm{dd}^{\mathrm{c}}\eta _ j$ with $\theta _ j$ smooth closed and $\eta _ j\in \mathrm{PSH}(X,\theta _ j)$. For positive numbers $r_1,\ldots ,r_{n-1}$ with $\sum _ jr_ j=1$ put $\theta _ r=\sum _ jr_ j\theta _ j$ and $\eta _ r=\sum _ jr_ j\eta _ j$, so that $\sum _ jr_ jT_ j=\theta _ r+\mathrm{dd}^{\mathrm{c}}\eta _ r$. The potential $\eta _ r$ is $(\theta _ r+s\omega )$-psh for every $s\geq 0$, since $\theta _ r+s\omega +\mathrm{dd}^{\mathrm{c}}\eta _ r=\sum _ jr_ jT_ j+s\omega \geq 0$. Let $\kappa $ be a multiple of $\omega $ such that the presentations of $a$ and $b$ are $\kappa $-psh. Since the forms $\theta _ r$ are convex combinations of the finitely many smooth forms $\theta _ j$, there is $s_0$ with $\theta _ r+s\omega \geq \kappa $ for all $r$ and all $s\geq s_0$, and then the presentations are $(\theta _ r+s\omega )$-psh as well. Step 1 applied to the current $\sum _ jr_ jT_ j+s\omega =(\theta _ r+s\omega )+\mathrm{dd}^{\mathrm{c}}\eta _ r$ gives

where, by the independence of (17.2.1) from the background, the Hessian factors may be computed in one fixed Kähler background. Expanding the repeated current by multilinearity, both sides become polynomials in $s$, and comparing constant coefficients gives

Both sides are homogeneous polynomials of degree $n-1$ in $(r_1,\ldots ,r_{n-1})$. Since they agree on the open simplex, homogeneity makes them agree on the open positive cone of $\mathbb R^{n-1}$. Hence their coefficients agree: a polynomial that vanishes on a nonempty open set is zero, because all its partial derivatives vanish at a point $r_0$ of that set and it equals its finite Taylor expansion at $r_0$. The coefficient of $r_1\cdots r_{n-1}$ is $(n-1)!$ times the corresponding side of (17.2.6). $\square$

Proof. Put $\lambda _ u=\langle \theta _ u\wedge T_1\wedge \cdots \wedge T_{n-1}\rangle $ and define $\lambda _ v$ in the same way, so that $\mathrm{dd}^{\mathrm{c}}(u-v)\wedge R=\lambda _ u-\lambda _ v$ by (17.2.1). Since $u-v$ is bounded, corollary 13.8.8, applied after writing $T_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}\eta _ i$ with smooth $\theta _ i$ and $\eta _ i\in \mathrm{PSH}(X,\theta _ i)$ (theorem 10.2.15), gives $\lambda _ u(X)=\lambda _ v(X)$. Hence the integral does not change if a constant is added to $u$, and we may assume $u\leq v$. We must show that $\int _ X(v-u)\lambda _ v\leq \int _ X(v-u)\lambda _ u$.

Let $t\geq 0$ and $\varepsilon >0$, and put $q=\max \{ u,v-t-\varepsilon \} $. Then $v-t-\varepsilon \leq q\leq v$, so $\lambda _ q(X)=\lambda _ v(X)=\lambda _ u(X)$ by the same corollary. The sets $O_1=\{ u<v-t-\varepsilon \} $ and $O_2=\{ u>v-t-\varepsilon \} $ are disjoint and plurifine open, by proposition 5.47.25 applied in charts after adding the same smooth local potential of $\theta $ to $u$ and $v$; neither of them meets the common pole set of $u$ and $v$. Moreover $q=v-t-\varepsilon $ on $O_1$ and $q=u$ on $O_2$, so plurifine locality (proposition 13.2.6) gives $\lambda _ q=\lambda _ v$ on $O_1$ and $\lambda _ q=\lambda _ u$ on $O_2$, and hence

Since pole sets carry no mass, it follows that

As $\varepsilon \downarrow 0$ the sets $\{ u<v-t-\varepsilon \} $ increase to $\{ u<v-t\} $, and therefore

On the common finite locus the bounded function $v-u\geq 0$ equals $\int _0^\infty \mathbf1_{\{ u<v-t\} }\, \mathrm dt$. Integrating the last inequality in $t$ and using Tonelli's theorem, we obtain $\int _ X(v-u)\lambda _ v\leq \int _ X(v-u)\lambda _ u$, as required. $\square$