13.14 Non-pluripolar Monge–Ampère measuresB724B74AE68C

Proof. Work in a coordinate ball with a fixed positive constant form $\kappa $, and fix $\epsilon >0$. Every point has a smaller neighborhood carrying a constant real form $\theta '$ such that

Take a smooth local potential $r$ for $\theta '+\epsilon \kappa $. Then $u+r$ is plurisubharmonic. On a relatively compact smaller neighborhood choose $k>t+\sup |r|$ and set

This function is locally bounded and equals $u_ t=u$ on $D_ t$. Moreover,

is the sum of a positive closed current with bounded potential and a smooth positive form, so its bounded products are positive. By plurifine locality, the restriction of $(\theta +\mathrm{dd}^{\mathrm{c}}u_ t+2\epsilon \kappa )^ n$ to $D_ t$ is positive on this neighborhood.

These neighborhoods cover the coordinate ball. Now let $\epsilon \downarrow 0$. The product with the fixed bounded potential $u_ t$ is a polynomial in $\epsilon $ whose coefficients have locally finite variation, so it converges in local total variation, also after restriction to $D_ t$. This proves the first inequality. The same construction, applied to each factor, gives the positivity of mixed products.

For the second inequality, expand the difference of the two powers by the telescoping identity, using the smooth positive form $\Omega -\theta $. Every summand is a mixed product of factors that are positive on $D_ t$, as just proved. The difference is therefore nonnegative there. $\square$

Proof. For $s>t$, plurifine locality on $D_ t\subset D_ s$ gives

Together with lemma 13.14.2, this proves that the measures increase. Their increasing limit is countably additive by monotone convergence, and it vanishes on pluripolar sets because each bounded product does. The comparison with $\Omega $ passes to the limit. By the compatibility just proved,

and the mass of this measure is bounded by that of the bounded product $(\Omega +\mathrm{dd}^{\mathrm{c}}u_ t)^ n$. The sets $D_ t$ exhaust the complement of $\{ u=-\infty \} $, which has zero mass, so the limit is $\sigma $-finite.

For a smooth real function $f$, consider $\theta '=\theta +\mathrm{dd}^{\mathrm{c}}f$ and $u'=u-f$. On a common superlevel set of $u$ and $u'$, sufficiently deep truncations of $u$ and $u'$ define the same current with bounded potentials, so their restricted products agree. These sets exhaust the complement of the common pole set, which proves invariance under this change of decomposition. By the elliptic-regularity argument in the proof of proposition 5.28.2, any two smooth decompositions of the same current differ in this way, and adding a constant is a special case. $\square$

Proof. Fix a coordinate ball with a constant positive form $\kappa $, a level $t>0$, and $\epsilon >0$. On smaller neighborhoods, freeze $\theta $ to a constant form $\theta '$ with $-\epsilon \kappa \leq \theta -\theta '\leq \epsilon \kappa $, and choose a quadratic function $r$ with $\mathrm{dd}^{\mathrm{c}}r=\theta '+\epsilon \kappa $. Then $u+r$ and $v+r$ are plurisubharmonic. For a large integer $k$ put

These functions are bounded on relatively compact neighborhoods, satisfy $a_ k\leq b_ k$, and

with the same inequality for $b_ k$. We may therefore apply the top-degree case of lemma 5.28.3 with background $\theta +2\epsilon \kappa $.

On $D_ t=\{ u>-t\} $, both $u$ and $v$ are larger than $-t$. Taking $k>t+\sup |r|$ makes $a_ k=u$ and $b_ k=v$ on $D_ t$. Restricting the bounded inequality to $\{ u=v\} \cap D_ t$ and using locality, we obtain the contact inequality for the background $\theta +2\epsilon \kappa $ on that set. The smaller neighborhoods cover the original ball, so the restricted inequality holds throughout it.

Choose a Hermitian form $\Omega \geq \theta $. The truncations $u_ t=\max (u,-t)$ and $v_ t=\max (v,-t)$ are bounded and $\Omega $-plurisubharmonic, so their signed products with background $\theta $ are defined. On $D_ t$, the products used above are the restrictions of $(\theta +\mathrm{dd}^{\mathrm{c}}u_ t+2\epsilon \kappa )^ n$ and $(\theta +\mathrm{dd}^{\mathrm{c}}v_ t+2\epsilon \kappa )^ n$. These are polynomials in $\epsilon $ whose coefficients have locally finite variation, so they converge in local total variation as $\epsilon \downarrow 0$, also after restriction to $\{ u=v\} \cap D_ t$. This proves the inequality for the background $\theta $ on each finite level. Finally let $t\to +\infty $: the common pole set has zero extended mass, and the sets $\{ u=v\} \cap D_ t$ exhaust its complement in the contact set. $\square$

Proof. Subtracting a common constant, we may assume that all the potentials are nonpositive. Choose a Hermitian form $\Omega \geq \theta $. For $T>0$, let $g_ T:[-\infty ,0]\to [0,1]$ be a continuous nondecreasing cutoff, equal to zero on $[-\infty ,-2T]$ and to one on $[-T,0]$, chosen so that $g_ T$ increases pointwise to one on $(-\infty ,0]$ as $T\to \infty $. Put $v_ j=\max (u_ j,-2T)$ and $v=\max (u,-2T)$. These are uniformly bounded $\Omega $-plurisubharmonic functions, and they converge in capacity.

Fix $T$. Expanding $(\theta +\mathrm{dd}^{\mathrm{c}}v_ j)^ n$ by means of $\theta +\mathrm{dd}^{\mathrm{c}}v_ j=(\Omega +\mathrm{dd}^{\mathrm{c}}v_ j)+(\theta -\Omega )$, and bounding each smooth factor by a constant multiple of $\Omega $, we see that the total variations of the signed terms are controlled by positive bounded mixed measures. By lemma 5.29.16, these have a common capacity bound. Bounded Bedford–Taylor continuity gives weak convergence term by term. It also allows the weights $b_ jg_ T(v_ j)$: their differences from the limiting weight tend to zero in capacity, so the common capacity bound makes the corresponding integrals tend to zero. For the fixed limiting weight, remove an open set of arbitrarily small capacity, extend the weight continuously from its complement, and control the discarded part by the same bound. Since the cutoff vanishes at the truncation level $-2T$, locality gives

The first integral is nonnegative and at most the corresponding integral without the cutoff, although the bounded product $(\theta +\mathrm{dd}^{\mathrm{c}}v_ j)^ n$ may be signed. Hence the limit is bounded by the lower limit in the assertion. Letting $T\to \infty $ and using monotone convergence completes the proof. $\square$

Proof. The cases $t=0$ and $t=1$ are the hypotheses. Let $0<t<1$, translate $u,v$ to be nonpositive, and put $u_ k=\max (u,-k)$ and $v_ k=\max (v,-k)$. Since $\theta $ is semipositive, these are bounded $\theta $-plurisubharmonic functions, and by locality

Apply lemma 5.28.7 to each mixed term in the positive expansion of $\theta _{tu_ k+(1-t)v_ k}^{n}$. On $E_ k=\{ u>-k\} \cap \{ v>-k\} $ this gives the density on the right of (13.14.11). On the same set, plurifine locality identifies the left-hand side with the extended measure of $tu+(1-t)v$. As $k\to \infty $, the sets $E_ k$ exhaust the complement of the two pole sets, which have zero smooth volume. Monotone convergence on Borel sets proves the inequality. $\square$

Proof. For bounded $u$, lemma 5.28.7, applied in local coordinates with $T_1=\omega +\mathrm{dd}^{\mathrm{c}}u$, $T_2=\gamma $ and $k=1$, gives the asserted inequality.

In general, subtract a constant so that $u\leq 0$, and put $u_ k=\max (u,-k)$. Positivity and locality give

so the bounded case applies to $u_ k$ with the density $\mathbf1_{\{ u>-k\} }g$. Since $u_ k\to u$ in $L^1$, the positive measures $(\omega +\mathrm{dd}^{\mathrm{c}}u_ k)\wedge \gamma ^{n-1}$ converge weakly to $(\omega +\mathrm{dd}^{\mathrm{c}}u)\wedge \gamma ^{n-1}$. The densities on the right converge in $L^1$, because $g^{1/n}$ is integrable and the pole set has zero smooth volume. Passing to the limit proves the assertion. $\square$

Proof. For another Hermitian form $\eta $, choose $C>0$ such that $\eta \leq C\omega $. If $v$ is a bounded $\eta $-plurisubharmonic function, then $v/C$ is $\omega $-plurisubharmonic and

By positivity of mixed bounded products, $\int _ X(\eta +\mathrm{dd}^{\mathrm{c}}v)^ n\leq C^ n v_+(\omega )$. Interchanging the two forms proves metric independence.

For the last assertion, approximate a bounded potential by theorem 5.32.1. The smooth approximants decrease and remain uniformly bounded, so the continuity of bounded products along decreasing sequences gives convergence of their total masses. The reverse inequality between the two suprema is immediate. $\square$

Proof. The comparison in theorem 13.14.3, applied with the Hermitian form $C\omega $, gives

Since $u_ t/C$ is bounded and $\omega $-plurisubharmonic, the right-hand side has mass at most $C^ n v_+(\omega )$. Let $t\to \infty $. $\square$

Proof. Set $a_ j=2^{-j}$ for $j\geq 1$. By proposition 13.14.15, we can choose smooth $\omega $-plurisubharmonic functions $u_ j$ with $\sup _ Xu_ j=0$ and

The partial sums $U_ N=\sum _{j\leq N}a_ ju_ j$ decrease and satisfy

Their $L^1$ norms are bounded by (5.26.1), so their limit $U$ is $\omega $-plurisubharmonic and nonpositive.

We first prove that the mass is infinite. Fix $j$ and write $v_ j=U-a_ ju_ j$. Passing to the limit in the remaining partial sums gives $(1-a_ j)\omega +\mathrm{dd}^{\mathrm{c}}v_ j\geq 0$. For $t>0$, put

The function $v_{j,t}$ is a bounded potential for the positive background $(1-a_ j)\omega $, so $W_{j,t}$ is bounded and $\omega $-plurisubharmonic.

On $D_ t=\{ U>-t\} $, the inequality $u_ j\leq 0$ gives $v_ j=U-a_ ju_ j\geq U>-t$, and hence $W_{j,t}=U=\max (U,-t)$ there. Plurifine locality and positivity of bounded mixed products yield

The sets $D_ t$ exhaust $X\setminus \{ U=-\infty \} $. This pole set is pluripolar, hence has zero smooth volume, so the smooth measure $\omega _{u_ j}^ n$ gives it no mass. Letting $t\to +\infty $, we obtain

Since $j$ is arbitrary, the mass is infinite.

For the integrability, fix $p>0$, let $\alpha $ be the exponent of lemma 5.26.2, and take $N$ so large that $p\delta \leq \alpha $, where $\delta =\sum _{j>N}a_ j=2^{-N}$. Write $U=A_ N+V_ N$ with $A_ N=\sum _{j\leq N}a_ ju_ j$ smooth. For a finite tail ending at $m>N$, Jensen's inequality for the convex exponential function, with the unused weight assigned to the value zero, gives

Because $u_ j\leq 0$ and $p\delta \leq \alpha $, the integrals of the exponentials on the right are uniformly bounded. Integrating and letting $m\to +\infty $ by monotone convergence, we obtain $\int _ X\mathrm{e}^{-pV_ N}\mathrm{d}V<+\infty $. Multiplication by the bounded smooth function $\mathrm{e}^{-pA_ N}$ proves $\int _ X\mathrm{e}^{-pU}\mathrm{d}V<+\infty $.

Consequently, the constant germ $1$ is locally integrable against each weight $\mathrm{e}^{-pU}$, and every multiplier ideal $\mathcal I(pU)$ is the whole structure sheaf. $\square$

Proof. The first assertion follows from proposition 13.14.16, theorem 13.14.18. For the second, take $\varepsilon U$, $0<\varepsilon \leq 1$, with $U$ as in theorem 13.14.18. It is $\omega $-plurisubharmonic and has the required exponential integrability. On $D_ t=\{ U>-t\} $, its truncation at level $\varepsilon t$ is $\varepsilon \max (U,-t)$, and

Positivity of mixed products bounds the restricted measure of $\varepsilon U$ below by $\varepsilon ^ n$ times that of $U$. Its limiting mass is therefore infinite. Finally, $\| \varepsilon U\| _{L^1}\to 0$ as $\varepsilon \to 0$. $\square$

Proof. Write $T_ j=\omega +\mathrm{dd}^{\mathrm{c}}u_ j$ for the smooth positive forms in that construction, and put

The smooth positive forms $S_ N$ increase pointwise, and the Gauduchon identity gives

By monotone convergence of their traces, the series of matrices has a finite positive limit $S$ almost everywhere. The coefficients of $S$ are integrable, because in a $\gamma $-unitary frame the absolute value of every coefficient of a positive matrix is bounded by its trace. Positivity and the same trace identity give

In particular, $S_ N\to S$ in the sense of distributions. Since

in the sense of distributions, by $L^1$ convergence of the potentials, we obtain $T=S$. The triangle inequality now gives

To compare $\mathrm{MA}^{\mathrm{ext}}_\omega (U)$ with the determinant, fix $N$ and set $V_ N=U-U_ N$. The remaining partial sums give $\delta _ N\omega +\mathrm{dd}^{\mathrm{c}}V_ N\geq 0$ and $V_ N\leq 0$. For $t>0$, the function $W_{N,t}=U_ N+\max (V_ N,-t)$ is bounded, and its current is the sum of two positive terms:

On $D_ t=\{ U>-t\} $, the inequality $U_ N\leq 0$ gives $V_ N\geq U>-t$, and consequently $W_{N,t}=U=\max (U,-t)$ there. Locality and positivity of bounded products imply

The smooth measure $S_ N^ n$ gives no mass to the pluripolar pole set of $U$, so letting $t\to \infty $ proves $\mathrm{MA}^{\mathrm{ext}}_\omega (U)\geq S_ N^ n$. The positive matrices $S_ N$ increase to $T$ almost everywhere, so their determinants increase to $\det _\gamma T$. Monotone convergence on every Borel set gives the asserted inequality between $\mathrm{MA}^{\mathrm{ext}}_\omega (U)$ and the determinant measure.

Finally, for every $N\geq j$, positivity gives $S_ N\geq a_ jT_ j$. Hence both the measurable limit and the smooth approximating forms satisfy

The first bound holds for every $j$, and the second for every $N\geq j$; together they prove the two assertions about infinite mass. $\square$

Proof. Take $w_\varepsilon =\varepsilon U$, with $U$ and $T=\omega +\mathrm{dd}^{\mathrm{c}}U$ as in proposition 13.14.21. Its current is

The lower bound and the two $L^1$ convergence assertions follow immediately. The full trace measures have integrable densities, and their distance from $\omega \wedge \gamma ^{n-1}$ in total variation is at most

Every multiplier ideal is trivial by the exponential integrability of $U$ proved in theorem 13.14.18. Since $T_\varepsilon \geq \varepsilon T$ almost everywhere, the determinant integral of $T_\varepsilon $ is at least $\varepsilon ^ n$ times that of $T$, hence infinite. The extended mass of $w_\varepsilon $ is infinite by the proof of corollary 13.14.19. $\square$

Proof. By proposition 13.14.15, we can choose a smooth $u_ j\in \mathrm{PSH}(X,\omega )$ with $\sup _ Xu_ j=0$ and

Put $T_ j=\omega +\mathrm{dd}^{\mathrm{c}}u_ j$. The polynomial

satisfies $F_ j(0)<b_ j$, and also $F_ j(\varepsilon _ j)\geq \varepsilon _ j^ n\int _ XT_ j^ n>b_ j$ by positivity of the mixed terms. The intermediate value theorem gives $s_ j\in (0,\varepsilon _ j)$ with $F_ j(s_ j)=b_ j$. Set $w_ j=s_ ju_ j$. Then $\sup _ Xw_ j=0$, $\omega +\mathrm{dd}^{\mathrm{c}}w_ j=(1-s_ j)\omega +s_ jT_ j\geq (1-\varepsilon _ j)\omega $, and $w_ j$ has the prescribed mass $b_ j$.

The Gauduchon identity gives $\| T_ j\| _{L^1(\gamma )}=\| \omega \| _{L^1(\gamma )}=M$. Hence

Since the absolute value of the trace of a Hermitian matrix is bounded by the trace of its absolute value, the same bound holds for the total variation of $\mathrm{dd}^{\mathrm{c}}w_ j\wedge \gamma ^{n-1}$. With $K_\omega $ as in (5.26.1), we also have $\| w_ j\| _{L^1(\mathrm{d}V)}\leq K_\omega \varepsilon _ j$.

Finally, let $\alpha ,C$ be the constants of lemma 5.26.2 for $(X,\omega ,\mathrm{d}V)$, and fix $p>0$. For all sufficiently large $j$, one has $ps_ j\leq \alpha $. Convexity of the exponential function, applied with weight $ps_ j/\alpha $, gives

Integrating, we bound the $L^1(\mathrm{d}V)$ norm of $\mathrm{e}^{-pw_ j}-1$ by $pC\varepsilon _ j/\alpha $, which proves the last convergence. $\square$