8.12 Locality and maxima of bounded potentials927D22F5110F

Proof. Choose smooth uniformly bounded decreasing approximations of all the potentials by 8.6.6. Construct $w_ j$ by 8.2.9. Work on one compact coordinate test set. Fix $\delta >0$ and a smooth $\chi _\delta :\mathbb R\to [0,1]$ that vanishes for $s\leq \delta /2$ and equals one for $s\geq \delta $. For $j$ large enough that $\epsilon _ j<\delta /2$, ordinary smooth locality gives

Indeed, wherever the weight is nonzero, $w_ j=u_ j$ on an open neighborhood. The equality elsewhere follows from its zero weight.

Both sequences of positive currents converge by 8.11.12; every coefficient variation is uniformly dominated by $C c_ m^*$ by 8.11.13. The potentials and their limits are quasicontinuous by 8.11.10. Apply 8.11.11 with $s=m$ and $F(x,y)=\chi _\delta (x-y)$ to the real and imaginary parts of each coefficient. Passing to the limit gives the same identity with $u,v,w,t_2,\ldots ,t_ p$. Choose the cutoffs so that $\chi _\delta (s)\to \mathbf1_{\{ s>0\} }$ pointwise as $\delta \downarrow 0$. Dominated convergence against the fixed finite coefficient variations on the test set gives the asserted restriction. A partition of unity proves it globally. $\square$

Proof. Symmetry and 8.12.1, applied to the $r$ slots, give $\mu _ r(w)=\mu _ r(v)$ on $\{ u<v\} $ and $\mu _ r(w)=\mu _ r(u)$ on $\{ u>v\} $. To include the contact set, put $w_\epsilon =\max \{ u+\epsilon ,v\} $. Strict locality and invariance under addition of constants give

The invariance follows from the smooth products and their joint decreasing limits. As $\epsilon \downarrow 0$, $w_\epsilon \downarrow w$, so the left side converges weakly to $\mu _ r(w)$. Against any nonnegative continuous compactly supported test function, the right side converges by dominated convergence for the fixed measure $\mu _ r(u)$ to $\mathbf1_{\{ u\geq v\} }\mu _ r(u)$. This proves the measure inequality on the contact set as well as on $\{ u>v\} $. On its disjoint complement $\{ u<v\} $, the already proved equality supplies $\mu _ r(v)$. Restricting the positive measures to these two Borel sets and adding gives the assertion. Exchanging $u,v$ gives its other form. $\square$

Proof. Take smooth decreasing uniformly bounded admissible $v_ j\downarrow v$. Cone convexity makes $\alpha v_ j+b$ admissible because

For $\alpha >0$, an upper test $q$ for $\alpha v+b$ gives the upper test $(q-b)/\alpha $ for $v$. The same convex combination of forms therefore proves admissibility of the limit. For $\alpha =0$ the potential is constant. The pointwise decreasing limit has the displayed canonical affine values. Expand the smooth top product. Every mixed term is positive by polarized Gårding positivity. Keeping the two extreme terms gives

Joint decreasing convergence from 8.11.12 passes this inequality against each nonnegative continuous test to the asserted inequality of positive Radon measures. $\square$

Proof. The subset definition and the torsion formulas in 8.9.7 give, in distributions,

The sum with two hats is absent when $k=1$, and the last term is absent when $\ell =1$. Every factor $R_ r$ has degree $r\leq k<m$, so its coefficients are signed measures. Expanding each $R_ r$ over subsets expresses the displayed measure as fixed smooth forms wedged with raw currents $D_ p$ of degrees $p\leq m-1$. 8.11.10 therefore bounds its variation locally by $C c_{m-1}^*$, uniformly under common bounds on the potentials. Joint convergence follows from 8.11.12 applied to this expansion. All the canonical weights are quasicontinuous for $c_{m-1}^*$. 8.11.11, with $s=m-1$, thus proves the weighted assertion coefficientwise on a finite coordinate cover. A partition of unity gives it on $X$. The term involving $R_ k\mathrm{dd}^{\mathrm{c}}(\omega ^\ell )$ retains degree $m-1$; the argument requires no coefficient estimate in degree $m$. $\square$

Proof. Suppose instead that $H_ m(u)=0$. Fix $s>0$, set $c=\inf _ Xu+s$, and put $w=\max \{ u,c\} $. The upper semicontinuity of $u$ makes $U=\{ u<c\} $ open, and the definition of the infimum makes it nonempty. Moreover $0\leq w-u\leq s$ and this difference is zero outside $U$. The maximum is bounded admissible with its canonical pointwise values by 8.2.9.

For $0\leq q\leq m$, define the positive Radon measures

Thus $\mu _ m=H_ m(u)=0$. For $q<m$, let $T_ q$ be the sum of $m-q$ currents with $q+r$ slots equal to $u$ and $m-q-1-r$ slots equal to $w$, for $0\leq r<m-q$, each of degree $m-1$ and wedged with $\omega ^{n-m}$. In smooth notation this is

Here each displayed product denotes the symmetric shifted current $R_{m-1}$ of (8.9.5), wedged with $\omega ^{n-m}$.

Choose smooth bounded admissible $u_ j\downarrow u$ and the regularized maxima $w_ j=M_{\epsilon _ j}(u_ j,c)\downarrow w$, where $\epsilon _ j\downarrow 0$, as in 8.2.9. Both lists have a common bound. The smooth polynomial identity and Stokes' theorem on the compact manifold give

Joint decreasing convergence passes the left side to its limit. The preceding lemma, with $F(x,y)=y-x$ and the decreasing lists $u_ j,w_ j$, passes the right side to its limit. Hence

The difference of the two decreasing sequences need not itself decrease; the continuous two-variable weight is essential here.

On the open set $U$, the potential $w$ is constant. Locality of the raw recursion therefore sets every factor $\mathrm{dd}^{\mathrm{c}}w$ to zero there, so that its shifted factors become $\omega $. The formula in the preceding lemma expresses $\mathrm{dd}^{\mathrm{c}}T_ q|_ U$ as finitely many smooth metric-derivative forms wedged with powers of $\omega +\mathrm{dd}^{\mathrm{c}}u$ of degrees at most $m-1$. Use smooth admissible approximations to $u$ and the constant $c$ in these local expressions; their limits are the same restricted currents. For those smooth approximations, the absolute mixed-form estimate 8.2.7 bounds each such term by a constant times the corresponding positive trace, and therefore by $C\sum _{r=0}^{m-1}\mu _ r$ on $U$ after taking joint limits. More explicitly, a term with $r$ factors is bounded by $C(\omega +\mathrm{dd}^{\mathrm{c}}u)^ r\omega ^{n-r}$; this is $C\mu _ r$ on $U$. Against compactly supported nonnegative tests in $U$, weak convergence and lower semicontinuity of total variation preserve the estimate. Exhaustion gives the measure inequality

Here $C$ depends only on $\omega ,m,n$, not on $u$ or $s$; a fixed finite coordinate cover bounds all the smooth metric coefficients.

Since $w-u$ vanishes outside $U$, positivity gives

Summing these $m$ inequalities and choosing $s>0$ with $mCs<1$ forces $\sum _{r=0}^{m-1}\int _ U\mu _ r=0$. But locality gives $\mu _0=\omega ^ n$ on the nonempty open set $U$, whose volume is positive. This contradiction proves the result. $\square$