8.7 Continuous products, envelopes and domination5B2332FC78E2

Proof. 8.6.1, 8.6.2 provide the smooth approximations without using any continuous Hessian product or domination input. For a smooth admissible $u$ put $\sigma =\omega +\mathrm{dd}^{\mathrm{c}}u$. Since $\mathrm d\sigma =\mathrm d\omega $, the coefficients of $\mathrm{dd}^{\mathrm{c}}(\chi \sigma ^{r-1}\wedge \omega ^{n-r})$ are sums of smooth test factors paired with powers $\sigma ^ q$, $q\leq r-1$, where $\chi $ is a fixed cutoff. The absolute-form lemma and integration by parts give the Chern–Levine–Nirenberg bound by induction on $r\leq m$:

At the induction step write $\sigma ^ r=\omega \wedge \sigma ^{r-1} +\mathrm{dd}^{\mathrm{c}}u\wedge \sigma ^{r-1}$, integrate the second term against $\chi $, and move $\mathrm{dd}^{\mathrm{c}}$ to $\chi \sigma ^{r-1}\omega ^{n-r}$. The resulting powers have degrees $r-1,r-2,r-3$, with negative powers absent, and the induction applies on nested slightly larger compact sets. Mixed masses reduce to the single potential average, by positivity of the expansion of its power. Positive mixed currents have total variation controlled by these trace masses.

To prove independence, telescope the difference between two smooth mixed products whose potentials have a common uniform bound. Each term paired with a fixed smooth complementary test is

where $T$ is a product of the remaining admissible factors and $\omega ^{n-m}$. The identity $\mathrm d(\omega +\mathrm{dd}^{\mathrm{c}}u)=\mathrm d\omega $ and its $\mathrm d^{\mathrm c}$ analogue show that derivatives of $T$ introduce only fixed metric derivatives; every term has at most $k-1<m$ admissible factors. The absolute mixed-form bound and the mass estimate just proved bound the variation of $\mathrm{dd}^{\mathrm{c}}(T\wedge \eta )$ uniformly. Consequently the pairing difference is at most $C_\eta \max _ a\| u_ a-v_ a\| _\infty $. The same construction with an arbitrary complementary smooth test also defines the signed products of fewer than $m$ factors without the factor $\omega ^{n-m}$: their variation is bounded by the absolute-form lemma. These products justify the metric-derivative terms above. This proves that all approximations have the same limit and that the products are continuous in the uniform topology. Positivity passes from the smooth mixed products. Using a test supported in an open set on which two limiting potentials coincide proves locality by the same estimate on a smaller neighborhood of its support. $\square$

Proof. Write $M(w)$ for the mixed measure with first potential $w$. Positivity and locality give, for $t>0$,

The first indicators increase to $\mathbf1_{\{ v>u\} }$ as $t\downarrow 0$, and the potentials on the left converge uniformly to $v$. Integrate against nonnegative continuous tests and pass to the limit. Subtract the first restricted measure to obtain the claimed contact inequality. Replacing one factor at a time proves the pure-power claim and the analogous contact inequalities for all mixed powers. $\square$

Proof. For smooth $h$, use the penalty solutions $w_ N$ in 8.6.2. Its maximum argument applies to every continuous admissible competitor $v\leq h$, with an error depending on $h$ and $N$, and not on $v$. Taking their supremum gives

where the error tends to zero. Thus $P(h)$ is their uniform limit and is continuous and admissible. On every compact subset of $\{ P(h)<h\} $ the penalty densities $b e^{N(w_ N-h)}$ tend uniformly to zero. Product continuity implies the contact support assertion. Adding constants to competitors gives the stated Lipschitz bound. Approximate a continuous $h$ uniformly by smooth obstacles. Their envelopes converge uniformly to $P(h)$, proving continuity and admissibility. Any compact subset of $\{ P(h)<h\} $ also lies outside the contact sets of all sufficiently close smooth obstacles; product continuity gives the support assertion for $h$. $\square$

Proof. The solutions $w_ N$ of 8.6.5 are continuous admissible competitors, so $w_ N\leq P(h)$. For each bounded admissible $u\leq h$, the same lemma gives

Let $N\to \infty $ for this fixed obstacle and fixed $b$. Thus $P(h)$ dominates every bounded competitor. Since $P(h)$ itself is a bounded competitor, it equals their supremum. $\square$

Proof. Suppose this integral is zero. For $s>0$ put $w=\max (u,\min _ Xu+s)$ and $U=\{ u<\min _ Xu+s\} $. The open set $U$ is nonempty, $0\leq w-u\leq s$, and $w=u$ outside it. Set $\sigma _ u=\omega +\mathrm{dd}^{\mathrm{c}}u$, $\sigma _ w=\omega +\mathrm{dd}^{\mathrm{c}}w$, and $\mu _ q=\sigma _ u^ q\wedge \sigma _ w^{m-q}\wedge \omega ^{n-m}$. The contact lemma gives $\mathbf1_{X\setminus U}\mu _ q\geq \mathbf1_{X\setminus U}\mu _ m$, for $0\leq q\leq m$. For $q<m$ the telescoping identity is

Integration by parts on $X$ therefore gives

This identity follows first for smooth approximants and then by uniform product continuity. The derivatives of $T_ q$ contain only fixed metric derivatives and products of at most $m-1$ admissible factors, so the absolute-form lemma makes $\mathrm{dd}^{\mathrm{c}}T_ q$ a measure of uniformly controlled variation. Uniform convergence also justifies its continuous weight. On $U$, locality gives $\sigma _ w=\omega $ and thus

The constant depends only on the metric and $m,n$, and not on $u$ or $s$. Consequently

For sufficiently small $s$, absorption and $H_ m(u)=0$ force all these nonnegative integrals to vanish. But $\mu _0=\omega ^ n$ on the nonempty open set $U$, a contradiction. $\square$

Proof. Replace $v$ by $\max (u,v)$, so $u\leq v$. For $b>1$ put $p_ b=P(bu-(b-1)v)$ and $z_ b=b^{-1}p_ b+(1-b^{-1})v\leq u$. On the contact set $C_ b=\{ p_ b=bu-(b-1)v\} $ one has $z_ b=u$; the contact inequality and polarized positivity give

The envelope measure is supported on $C_ b$ and has positive mass. It vanishes on $C_ b\cap \{ u<v\} $ by the hypothesis, so some point of $C_ b\cap \{ u=v\} $ exists. Hence $\sup _ Xp_ b\geq \inf _ Xv$, while $p_ b\leq u$. The estimate (8.7.5) is consequently uniform for $p_ b$. On the open set $E_ a=\{ u<v-a\} $, for $a>0$, one has $p_ b\leq v-ba$, and therefore

Letting $b\to \infty $ gives zero volume for $E_ a$. Since it is open it is empty, and $a\downarrow 0$ gives $u\geq v$. $\square$

Proof. Choose $b>1$ so that $(1-b^{-1})^ m>c$, and put $p=P(bu-(b-1)v)$ and $z=b^{-1}p+(1-b^{-1})v\leq u$. On $C=\{ p=bu-(b-1)v\} $ the contact comparison gives

Restricting to $\{ u<v\} $ and using the hypothesis shows that $H_ m(p)$ vanishes on $C\cap \{ u<v\} $. Its support lies in $C$, where $p<v$ is equivalent to $u<v$. Thus $H_ m(p)$ vanishes on $\{ p<v\} $. The preceding domination proposition gives $p\geq v$. Since $p\leq bu-(b-1)v$, it follows that $u\geq v$. $\square$

Proof. Fix $\delta >0$. On $\{ u<v-\delta \} $, the assumed inequality gives

The constant $e^{-a\delta }$ is strictly smaller than one, so 8.7.8 applied to $u$ and $v-\delta $ gives $u\geq v-\delta $. Letting $\delta \downarrow 0$ proves the assertion. $\square$