Proposition 8.11.6390B1CE08171

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 $c_ s^*$. $\square$