8.11 Local capacities and decreasing productsBA5052EF5C4F

Proof. For one potential and a nonnegative compactly supported cutoff $\chi $, closedness of $\mathrm{dd}^{\mathrm{c}}f$ gives

For $q\geq 1$, the absolute-form estimate 8.2.7 applies at degree $q-1<m$. Nested cutoffs and induction, starting with volume, therefore bound the left side by $C M^ q$; replacing $M$ by $1+M$ includes $M=0$. For mixed factors, positivity of every term in the expansion of $(\sum _ i\mathrm{dd}^{\mathrm{c}}f_ i)^ q\omega ^{n-q}$ bounds the mixed trace by a fixed multiple of the single trace for $\sum _ i f_ i$. This proves the first assertion, including its degree-zero convention.

Replace $f$ by $f+M$, so that $0\leq f\leq 2M$ without changing its gradient or Hessian. The identity

also shows that $f^2$ belongs to the local cone. Wedge with the remaining $q$ Hessians and $\omega ^{n-q-1}$. The first term on the right is nonnegative, and the mixed mass estimate applied to $f^2$ and the other potentials bounds the second term. The degree used is $q+1\leq m$. A cutoff equal to one on $K$ gives the stated local estimate. $\square$

Proof. For $q\leq m-2$, diagonalize $S$ in an $\omega $-unitary frame. Every diagonal coefficient of $S^ q\omega ^{n-q-2}$ is a positive multiple of $S_{q;ij}(\lambda )$ for two omitted coordinates. This is nonnegative by deleted-coordinate positivity, because $q+2\leq m$. The form is a nonnegative sum of simple positive coordinate forms.

For one such simple form, only the remaining complex two-plane contributes to a complementary wedge. The coefficients of $\mathrm d^{\mathrm c}\omega $ are uniformly bounded in unitary frames on the fixed compact set. Its wedge with $\mathrm df$ is consequently bounded by a constant times the gradient norm on that plane. The wedges with $\omega ^2$ and with $\mathrm df\wedge \mathrm d^{\mathrm c}f\wedge \omega $ give, up to fixed positive constants, one and the square of that norm. Pointwise Cauchy–Schwarz over the simple summands, followed by integral Cauchy–Schwarz with weight $a$, proves the estimate. Conjugation gives the other estimate.

For $q=m-1$, at one point choose an $\omega $-unitary coframe $e_ i$ that diagonalizes $S$. Put $\epsilon _ i=\mathrm i e_ i\wedge \bar e_ i$ and $\Omega =\epsilon _1\cdots \epsilon _ n$, so that $\omega =\sum _ i\epsilon _ i$ and $S=\sum _ i\lambda _ i\epsilon _ i$. Writing $B_{ij}=S_{q;ij}(\lambda )$ and $D_ i=S_{q;i}(\lambda )$, ordinary wedge expansion gives

Only the remaining two-plane contributes to its wedge with $\mathrm df\wedge \mathrm d^{\mathrm c}\omega $. If $\partial f=\sum _ i f_ i e_ i$, its absolute density relative to $\Omega $ is bounded by

The constant bounds the tensor coefficients of $\mathrm d^{\mathrm c}\omega $ in all unitary frames on the fixed compact set. Diagonalization is only pointwise; no derivative of this moving frame is taken.

8.2.8 and finite Cauchy–Schwarz bound this sum by

Here $\sum _ iD_ i=(n-q)S_ q(\lambda )$. Moreover

where $\kappa =1/(2\pi )$ by the book's $\mathrm d^{\mathrm c}$ normalization. These two densities are nonnegative. Thus the pointwise torsion density is bounded by a constant times their geometric mean. Multiplication by $a$ and integral Cauchy–Schwarz give the assertion, including the cases where either density vanishes. The other wedge is conjugate up to a fixed sign, which does not affect its absolute value. For $q\geq 1$, use the exact polarization identity

Every nonempty subset sum remains in the cone. Applying the single power estimate to each term and taking absolute values proves the mixed assertion. At $q=0$ use the single term $S^0=1$. No Cauchy–Schwarz assertion for an arbitrary weakly positive $(n-2,n-2)$-form is needed. $\square$

Proof. Write $U=\{ |z|<R\} $ and $|f_ j|\leq M$. Take $\psi =A(|z|^2-R^2)+b$, with $b>M+2$ and then $A$ so large that $\psi <-M-2$ on a neighborhood of $\overline V$. It exceeds $M+2$ on a fixed sufficiently thin boundary collar.

Use the regularized maximum $M_\epsilon $ constructed in 8.2.9. Choose $0<\epsilon _ j\downarrow 0$ with $\epsilon _ j<1$, and put $v_ j=M_{\epsilon _ j}(f_ j,\psi )$. The lemma preserves the local unshifted cone condition. Its two monotonicities give $v_ j\downarrow \max (f,\psi )=v$, and the fixed gaps give the two asserted regions of exact equality. The bounds

give a common bound. In particular all differences $v_ k-v_ j$ have support in one fixed compact subset of $U$. $\square$

Proof. The assertion for $s=0$ follows by bounded convergence for volume. Induct on $s$, assuming it for $s-1$, and put $q=s-1$ and $\theta =\omega ^{n-s}$. For $j\geq k$ let $h=v_ k-v_ j\geq 0$. The support of $h$ lies in one fixed compact set $K_0\Subset U$. Smooth capacity domination and the induction hypothesis give, uniformly over all bounded local smooth mixed potentials,

To see this, split at $h=\delta $. On $\{ h\leq \delta \} $ the mass bound gives $C\delta $. On the remaining set use $h\leq v_ k-v$, the capacity bound and the induction hypothesis. Let $k$ tend to infinity before $\delta \downarrow 0$. For $q=0$ this is the volume estimate. Constants are uniform in $j\geq k$ and all potentials with the fixed common bound.

Consider the real smooth energy

The energies $E_ s(v_ j)$ are bounded: on the fixed collar the integrand is fixed, and the remaining compact set has a uniform smooth mass bound. Since $h$ vanishes near the boundary, integration by parts and closedness of $\mathrm{dd}^{\mathrm{c}}w_ t$ give the exact identity

Here

and the analogous $\mathrm d^{\mathrm c}$ formula holds. A zero scalar coefficient means that its term is absent. The two first-order terms in (8.11.8) tend uniformly to zero by 8.11.4, the uniform local gradient bound, and (8.11.7). The last term has the same property by the absolute-form estimate at degree $q<m$. More explicitly, the first torsion integral is bounded by

The first factor is bounded since $h$ has a common bound and 8.11.1 applies; the second tends to zero by (8.11.7). Conjugation handles the other first-order term. Thus, with $e_ k\downarrow 0$ after taking a tail supremum,

It follows that $E_ s(v_ j)\leq E_ s(v_ k)+e_ k$ for $j\geq k$. Taking a subsequence $k$ approaching the liminf shows that the bounded energy sequence converges. Hence $\Delta _{kj}\to 0$ uniformly for $j\geq k$. Positivity of the binomial expansion $\Lambda _ s(w_ t)\geq (1-t)^ s \Lambda _ s(v_ j)$ then gives

It remains to replace each Hessian factor by a capacity competitor. For any smooth $-1\leq a\leq 0$ in the local cone, put

First-order Stokes, with every metric derivative retained, gives

The second term is $o(1)$ uniformly in $a$ and $j\geq k$. Use finite polarization in 8.11.4; each subset sum has a uniform local gradient bound and its lower trace is controlled by (8.11.7).

The form $\eta _\ell \theta $ is positive of bidegree $(n-1,n-1)$: a complementary nonnegative rank-one form completes a polarized product of $s\leq m$ cone forms. Its associated Hermitian gradient pairing is nonnegative, so Cauchy–Schwarz and the uniform gradient bound for $a$ give

Another exact first-order Stokes identity gives

The dropped term is nonnegative. For the last term the same polarized torsion estimate applies: its first gradient factor is uniformly bounded because $h=v_ k-v_ j$, and its second factor tends to zero by (8.11.7); here $h^2\leq C h$ and the gradient energy of the difference is at most twice the sum of those of $v_ k$ and $v_ j$ for every positive single-power gradient pairing. Starting with (8.11.9) at $I_ s$, finite backward induction gives $I_0\to 0$ uniformly in $a$ and $j\geq k$.

Chebyshev's inequality now yields $c_ s\{ v_ k-v_ j>\delta \} \leq \delta ^{-1}\sup _ a I_0$. These sets increase to $\{ v_ k-v>\delta \} $ as $j\to \infty $. Continuity of $c_ s$ under increasing unions proves the induction.

For any one $0\leq s\leq m$, choose $k_ N$ so that the open sets $G_ N=\{ v_{k_ N}-v>2^{-N}\} $ satisfy $c_ s(G_ N)<\epsilon 2^{-N}$, $N\geq 1$. They are open because $v$ is upper semicontinuous and $v_{k_ N}$ is smooth. On $U\setminus \bigcup _ NG_ N$ these smooth functions converge uniformly to the pointwise limit $v$. Hence $v$ is continuous there. The union has capacity less than $\epsilon $. Boundary localization transfers this conclusion to $f$ on each smaller compact set. Open sets have the same outer capacity, which also proves the assertion for $c_ s^*$. $\square$

Proof. The global decreasing approximation and 8.9.8 provide bounded smooth local approximations to $u+\rho $. The preceding proposition makes its canonical pointwise limit quasicontinuous. Subtracting the smooth $\rho $ preserves quasicontinuity.

For smooth approximations write $\mathrm{dd}^{\mathrm{c}}u_ i=\mathrm{dd}^{\mathrm{c}}(u_ i+\rho )-\mathrm{dd}^{\mathrm{c}}\rho $ and expand the product. Each term has $q\leq p<m$ local cone factors and fixed smooth complementary factors. The absolute-form estimate and (8.11.3), with $s=m-1$, bound each coefficient variation by $C c_{m-1}$ on Borel sets. The constants depend only on the common bound and local data.

Take the iterated limits of 8.9.3 on a slightly larger compact neighborhood of $K$. Lower semicontinuity of signed measure variation on open sets preserves these estimates at each limit. For any open $G\supset E$, intersect with a fixed open neighborhood of $K$ having compact closure in $U$; this gives $|\nu |(E)\leq Cc_{m-1}(G)$. Taking the infimum proves the outer estimate. No simultaneous rough product or rough diagonal polynomial expansion is used. $\square$

Proof. Fix $\phi \in C_ c(U)$ and a compact $K\Subset U$ containing its support. Allocate summable exceptional capacities to all the functions. Their open union $G$ has $c_ s(G)<\epsilon $, and every function is continuous on the compact set $A=K\setminus G$. Dini's theorem gives $b_ i^ j\to b_ i$ uniformly on $A$ for each $i$. Uniform continuity of $F$ on the compact box therefore gives $f_ j\to f$ uniformly on $A$, where $f_ j=F(b_1^ j,\ldots ,b_\ell ^ j)$ and $f=F(b_1,\ldots ,b_\ell )$. Put $M=\| F\| _\infty $. Extend $f|_ A$ to a continuous $g$ on $U$ with $|g|\leq M$; if $A$ is empty, take $g=0$. Then

For fixed $G$, the first two terms tend to zero. The last term is at most $4MC\| \phi \| _\infty \epsilon $. Let $\epsilon $ tend to zero. Weak convergence is applied to $\phi g$ on the whole measure. $\square$

Proof. Work on a compact coordinate test set. Degree zero is constant. Suppose the result holds for $D_{p-1}$, which has measure coefficients since $p-1<m$. The preceding two lemmas, applied coefficientwise with weight $u_ p^ j$, give convergence of $u_ p^ jD_{p-1}(u_1^ j,\ldots ,u_{p-1}^ j)$. Applying $\mathrm{dd}^{\mathrm{c}}$ proves distributional convergence at degree $p$. For $p<m$, the uniform coefficient bounds in 8.9.3 upgrade this to weak convergence of measures, by smooth approximation of compactly supported continuous tests. The same weighted lemma gives the additional weighted conclusion.

For each fixed list choose simultaneous smooth decreasing global approximations. Their smooth raw Hessian wedges are symmetric. The convergence just proved identifies their limit with $D_ p$ in any ordering, proving symmetry without an exchange of previously unproved iterated limits.

The subset expansion for $R_ p$ uses fixed smooth powers of $\omega $, so its distributional convergence and symmetry follow from those of $D_ I$. Its wedge $Q_ p$ is positive with uniform local masses by 8.9.6. Distributional convergence and these mass bounds imply weak convergence as an order-zero positive current, including the Radon measure conclusion at $p=m$. A partition of unity on a finite coordinate cover proves the global assertions. $\square$

Proof. First take smooth potentials, put $\sigma _ i=\omega +\mathrm{dd}^{\mathrm{c}}u_ i$, and choose the fixed local correction $\tau =\mathrm{dd}^{\mathrm{c}}\rho \geq 2\omega $. Then $\beta _ i=\mathrm{dd}^{\mathrm{c}}(u_ i+\rho )=\sigma _ i+(\tau -\omega )$ lies in the same local cone. Since $\tau -\omega \geq 0$, mixed positivity gives

Here $Q_ p\wedge \omega ^{m-p}=\sigma _1\cdots \sigma _ p\omega ^{n-p}$. The last bound follows from smooth mixed-trace domination at $s=m$ for the bounded corrected potentials and a normalized bounded sum with a fixed strictly plurisubharmonic function. Positivity of $Q_ p$ bounds each of its coefficient variations on the compact coordinate set by a fixed multiple of its trace measure.

For bounded potentials choose the smooth decreasing global approximations already established in the chapter. The joint convergence in 8.11.12 identifies their positive-current limit with the previously constructed $Q_ p$. On a compact neighborhood of $K$ inside $U$, lower semicontinuity on open sets preserves trace and coefficient variation estimates. For an arbitrary open $G\supset E$ intersect with a fixed relatively compact neighborhood of $K$ before taking this limit; its capacity is at most $c_ m(G)$. Therefore every such trace or variation measure $\mu $ satisfies $\mu (E)\leq Cc_ m(G)$. Infimize over $G$ to obtain the asserted outer-open bound. The constants of the smooth estimates are independent of the approximation index. The argument does not multiply a rough current of positive order or use an unproved rough diagonal polynomial expansion. $\square$

Proof. Work on a compact coordinate test set. By 8.11.12, every coefficient measure of the positive shifted currents converges weakly and has a uniform local variation bound. 8.11.13 strengthens this to domination by $C c_ m^*$ on Borel sets. 8.11.10 makes the canonical weights $a_ j$ and $a$ quasicontinuous for this capacity. Apply 8.11.11 with $s=m$ to each coefficient, and then use a finite partition of unity. At $p=m$, $Q_ m$ is a positive Radon measure, so this also gives the stated weighted measure convergence. $\square$