13.15 Weighted Monge–Ampère measuresA6A92DB9F7DA

Proof. Choose finitely many coordinate balls whose concentric smaller balls, with compact closure in the larger ones, cover $X$. On each larger ball choose a fixed quadratic function $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq \omega $, shifted so that $p=v+q$ is plurisubharmonic with $0\leq p\leq A+C$. Let $\kappa $ be the Euclidean form. In the identity

the integral of the first term against $\kappa ^{n-1}$ is bounded by a fixed multiple of the global trace mass, which is bounded by corollary 11.5.3, and the second term is smooth and fixed. Thus $\int \mathrm{dd}^{\mathrm{c}}p\wedge \kappa ^{n-1}$ is bounded uniformly in $v$ and $A$; in dimension one, Stokes' theorem replaces the trace bound. By (5.27.3), on each smaller ball,

Expand $(\omega +\mathrm{dd}^{\mathrm{c}}v)^ j=(\mathrm{dd}^{\mathrm{c}}p+\omega -\mathrm{dd}^{\mathrm{c}}q)^ j$ and sum over the finite cover. Since the smooth factors have bounded coefficients, the positive mixed masses satisfy

For smooth $v$, the exact form $(\mathrm{dd}^{\mathrm{c}}v)^ n$ integrates to zero, as in the proof of theorem 11.1.5; the regularization of theorem 5.32.1 and the continuity of bounded products give the same identity for bounded $v$. Thus

Since the mixed masses are nonnegative, it follows that

The largest power is $n-2$ when $n\geq 2$; for $n=1$, only the constant term remains. $\square$

Proof. Put $v_ t=\max (u,-t-1)+t+1$. It is $\omega $-plurisubharmonic and takes values in $[0,t+1]$, and by plurifine locality its Monge–Ampère measure equals $\mu _ u$ on $\{ u\geq -t\} \subset \{ u>-t-1\} $. The first estimate therefore follows from proposition 13.15.2, after enlarging the constant. For the second, $v_ t/(1+t)$ is a test function for the capacity, and positivity gives

Restricting this inequality to $E\cap \{ u\geq -t\} $ proves the capacity bound. $\square$

Proof. First suppose that $u$ is bounded. The chain rule (5.27.9) gives

All three summands are positive, and positivity of their bounded mixed products yields the claimed inequality.

For general $u$, use the truncations $u_ t=\max (u,-t)$. The bounded functions $F_ a(u_ t)$ decrease and take values between zero and $a^{-1}$, so their limit $F_ a(u)$ is a bounded $\omega $-plurisubharmonic function. On the plurifinely open set $D_ t=\{ u>-t\} $, it agrees with $F_ a(u_ t)$. Locality and the bounded case therefore give

Let $t\to +\infty $. The pole set is not charged by $\mu _ u$, and it has zero mass for the bounded product on the right. $\square$

Proof. Take $a=p/n$ in proposition 13.15.6. Since $F_{p/n}(u)$ takes values in $[0,n/p]$, proposition 13.15.2 with $A=n/p$ gives the total mass bound, after enlarging $C_\omega $. For the capacity bound put $B=\max (1,n/p)$ and $v=F_{p/n}(u)$. Then $v/B$ is a test function for the capacity, and

Consequently $\nu _{p,u}\leq \mathrm{MA}_\omega (v)\leq B^ n\mathrm{MA}_\omega (v/B)$, which gives the second bound. The measures $\nu _{p,u}$ do not charge pluripolar sets, because $\mu _ u$ does not. $\square$

Proof. Write $s=-u$. Since $w(s)=\int _ s^\infty (-w'(t))\, \mathrm{d}t$ for finite $s$, and since $\mu _ u$ does not charge the pole set $\{ s=+\infty \} $, Tonelli's theorem and lemma 13.15.3 give

Restricting the integral to $\{ s>T\} $ gives the same bound with $B_ w(T)$ in place of $B_ w(0)$, since the inner set is then $\{ T<s\leq t\} $, which is empty unless $t>T$.

For the capacity bound, the second estimate in (13.15.4) gives

Adding the tail estimate for $\{ s>T\} $ proves the asserted bound. Since $B_ w(T)\to 0$, choosing first $T$ large and then the capacity small gives uniform absolute continuity.

To prove continuity, fix a continuous nonincreasing $\eta :[0,+\infty )\to [0,1]$, equal to one on $[0,1]$ and zero on $[2,+\infty )$, and put $\eta _ T(s)=\eta (s/T)$. For fixed $T>0$, set $v_{j,T}=\max (u_ j,-2T)$ and $g_ T(t)=\eta _ T(-t)w(-t)$ on $[-2T,0]$. Since $g_ T(-2T)=0$, locality gives

Truncation preserves convergence in capacity. The functions $g_ T(v_{j,T})$ are uniformly bounded and quasi-continuous, and they converge in capacity to the corresponding function for $u$. The weighted continuity of bounded products in section 5.27, section 5.29 therefore gives weak convergence of the measures in (13.15.10) for each fixed $T$. The measure $(1-\eta _ T(-u_ j))\nu _{w,u_ j}$ is carried by $\{ u_ j<-T\} $, so by the tail bound its integral against a continuous test function $f$ is at most $C_\omega B_ w(T)\| f\| _\infty $, uniformly in $j$; the same holds for $u$. Letting $T\to \infty $ completes the proof. $\square$

Proof. For $w(t)=(1+t)^{-q}$, direct integration gives

Thus proposition 13.15.9 gives the mass and tail estimates. For the capacity bound put $\delta =\operatorname {Cap}_\omega (E)$. If $0<\delta \leq 1$, choose $T=\delta ^{-1/(q+n-r_ n)}\geq 1$ in that theorem. Both $T^ n\delta $ and $T^{r_ n-q}$ equal $\delta ^{(q-r_ n)/(q+n-r_ n)}$, so its two terms have the required bound. If $\delta =0$, let $T\to \infty $; if $\delta >1$, use the total mass bound. $\square$

Proof. Choose $T_0$ such that every tail $\mu _{u_ j}(\{ u_ j<-T\} )$ with $T\geq T_0$ is at most one. By lemma 13.15.3, $\mu _{u_ j}(X)\leq C(1+T_0)^{r_ n}+1$, so the tail condition already gives a uniform bound for the total masses.

With the cutoffs $\eta _ T$ of (13.15.10), the fixed-level argument in the proof of theorem 13.15.9, applied with $w\equiv 1$, gives

The masses of the measures on the left are bounded by the uniform bound for the total masses. Since $\eta _ T(-u)$ increases to one off the pole set as $T\to \infty $, monotone convergence proves $\mu _ u(X)<+\infty $.

For a continuous test function $f:X\to \mathbb R$, the error made by inserting $\eta _ T(-u_ j)$ is at most $\| f\| _\infty \mu _{u_ j}(\{ u_ j<-T\} )$, which is small for large $T$, uniformly in $j$. For the limit potential, the corresponding error tends to zero because $\mu _ u$ is finite and does not charge the pole set. Let first $j\to \infty $ with $T$ fixed, and then $T\to \infty $. $\square$

Proof. Since $\chi \geq 1$, the total masses are at most $C$, and

Thus theorem 13.15.13 applies. For the bound on $J_{\chi ,\omega }(u)$, use (13.15.10) with the compactly supported continuous function $s\mapsto \chi (s)\eta _ T(s)$ in place of $w\eta _ T$. The corresponding weighted integrals converge for each fixed $T$, and they are at most $C$. Let $T\to \infty $ and use monotone convergence. $\square$

Proof. One implication was proved in corollary 13.15.14. Conversely, write the family as $(u_ i)_{i\in I}$. As in the proof of theorem 13.15.13, the tail condition and lemma 13.15.3 give a uniform bound for the total masses. Choose numbers $T_ k\to +\infty $, with $T_{k+1}>T_ k+1$, such that $\mu _{u_ i}(\{ -u_ i>T_ k\} )\leq 2^{-k}$ for every $i\in I$. Let $\rho _ k$ be a continuous nondecreasing function, equal to zero on $[0,T_ k]$ and to one on $[T_ k+1,+\infty )$, and define

The sum is locally finite, so $\chi $ is continuous, nondecreasing and unbounded. Tonelli's theorem gives

so this weight gives a uniform integrability bound for the whole family. $\square$

Proof. The construction in corollary 13.14.23 gives $w_ j=s_ j u_ j$, where $u_ j$ is smooth and $\omega $-plurisubharmonic, $\sup _ Xu_ j=0$ and $0<s_ j<\varepsilon _ j$. The sublevel estimate (5.29.4) gives, for every fixed $\delta >0$ and all sufficiently large $j$,

Put $\mu _ j=(\omega +\mathrm{dd}^{\mathrm{c}}w_ j)^ n$. Fix $T>0$ and write $E_{j,T}=\{ w_ j\geq -T\} $. The lower bound in corollary 13.14.23 gives

Since $w_ j\to 0$ in $L^1$, the smooth volume of $X\setminus E_{j,T}$ tends to zero, and hence $\lambda _{j,T}\to \omega ^ n$ in total variation. For every $p>0$, theorem 13.15.9 gives $\int _ X\mathrm{e}^{pw_ j}\, \mathrm{d}\mu _ j\to V_\omega $, and therefore

Letting $p\downarrow 0$ and combining with the lower bound, we obtain $\mu _ j(E_{j,T})\to V_\omega $. Positivity now yields

Subtracting this shallow mass from $\mu _ j(X)=b$ proves the assertion about the mass on $\{ w_ j<-T\} $.

The positive measures $\mu _ j$ have fixed mass on the compact manifold $X$, so a subsequence converges weakly to a measure $\nu $ of mass $b$. The inequality $\mu _ j\geq (1-\varepsilon _ j)^ n\omega ^ n$ gives $\nu \geq \omega ^ n$. Set $\sigma =\nu -\omega ^ n$. For each $T>0$, subtracting the total-variation limit of the shallow restrictions from $\mu _ j\rightharpoonup \nu $ gives the last assertion along the same subsequence, for every $T$. $\square$

Proof. For $s_ j<1/2$, positivity in the expansion

gives a uniform bound for each $\int _ Yh^{n-k}\wedge H_ j^ k$. In local unitary coordinates, the coefficient matrix of the smooth positive form $H_ j^ k$ on exterior powers is positive semidefinite, so its entries are bounded in absolute value by its trace, up to a dimensional constant. The trace bounds therefore bound every coefficient in total variation, and weak compactness of measures gives a common subsequence and positive limits $K_ k$.

For $n\geq 2$, choose a Gauduchon form $\gamma $. By proposition 11.5.1,

Since $\gamma $ and $h$ are equivalent smooth metrics, it follows that $H_ j\to 0$ in coefficient total variation. For $n=1$ the same conclusion follows from $\int _ YH_ j=s_ j\int _ Yh$. Thus $K_1=0$ in all dimensions. Since $\mathrm{d}H_ j=s_ j\mathrm{d}h$,

as currents, because the preceding trace bounds control multiplication by the fixed smooth form $\mathrm{d}h$. Hence every $K_ k$ is closed.

The smooth form $(\mathrm{dd}^{\mathrm{c}}v_ j)^ k$ is both $\mathrm{d}$-exact and $\mathrm{dd}^{\mathrm{c}}$-exact:

Its integral against either kind of test form $\eta $ in the statement is therefore zero by Stokes' formula. Expanding $\mathrm{dd}^{\mathrm{c}}v_ j=H_ j-s_ jh$ gives

Each integral on the right is bounded by the trace bound for $H_ j^ r$. Taking $k=n$ and $\eta =1$ shows that the positive measures $H_ j^ n$ tend to zero in total variation, which proves $K_ n=0$; the other vanishing integrals follow by weak convergence. Finally, letting $j\to \infty $ in the positive expansion above, the terms with $k=0$ tend to $h^ n$ and those with $k=1$ and $k=n$ vanish in the limit, which proves (13.15.21).

For closed semipositive $\vartheta $, apply the vanishing integrals to $\eta =\vartheta ^{n-k}$. The measure $K_ k\wedge \vartheta ^{n-k}$ is positive and has zero mass. On every relatively compact open set where $\vartheta >0$, it dominates a positive multiple of the trace measure of $K_ k$, so $K_ k$ vanishes there. The decomposition gives the asserted support. In dimension three only $K_2$ occurs, and the same argument applies to the $\mathrm{dd}^{\mathrm{c}}$-closed test form $\vartheta $ itself. $\square$

Proof. Choose $C_0\geq 1$ such that $\theta \leq \Omega :=C_0\omega $. Then $u\in \mathrm{PSH}(X,\Omega )$, and theorem 13.14.3 gives

All the mass, tail and capacity bounds therefore follow from the estimates already proved for $\Omega $. The metric comparison in section 5.29 gives, for every Borel set $E\subset X$,

For continuity, let nonpositive $u_ j,u\in \mathrm{PSH}(X,\theta )$ converge in capacity, and use (13.15.10) with the $\theta $-product. The truncations are bounded and $\Omega $-plurisubharmonic. The weighted capacity continuity of section 5.27, section 5.29 applies term by term to the local expansion of their signed $\theta $-products, with the same bounded quasi-continuous coefficients. Since the cutoff vanishes at the truncation level, the result is exactly the restricted positive measure. The uniform tails obtained from the domination by $\mathrm{MA}^{\mathrm{ext}}_\Omega $ remove the cutoff. Domination also gives the sublevel estimate, so the preceding proofs of the undamped tail and integrability criteria apply without change. $\square$