8.13 Hessian mass and global comparisonB908310EE8A2

Proof. Expand $Q_ m$ using (8.9.5). The term for $I$ pairs $D_ I$ with the smooth form $\omega ^{n-|I|}$. By (8.9.2),

Since $|I^-|\leq m-1$, 8.9.3 gives measure coefficients for $D_{I^-}$; their bounded product with $t_{i(I)}$ is well defined. Thus the displayed pairing is exactly the integral in the sum, and is finite on compact $X$. The empty subset contributes $\int _ X\omega ^ n$. $\square$

Proof. Set $E=\{ u<v\} $. The contact-set version of 8.12.2 gives

All these measures are finite and positive. Integrating on $X$ and using $\mu _ r(u)(X\setminus E)=\mu _ r(u)(X)-\mu _ r(u)(E)$ gives exactly the assertion. No boundary or closed-background hypothesis enters this calculation. $\square$

Proof. The first condition implies the third by (8.13.2). The third implies the second by taking all inputs equal to a smooth potential.

To prove that the second implies the first, fix any smooth real $f$. For all sufficiently small real $t$ of either sign, $\omega +t\mathrm{dd}^{\mathrm{c}}f$ is positive, so $tf$ is strictly admissible. The polynomial

is constant on an interval. Consequently every coefficient with $q\geq 1$ vanishes. Polarizing that coefficient by replacing $f$ with $\sum _{a=1}^ q s_ af_ a$ gives

for all smooth real $f_ a$. Integration by parts in the first factor, using lemma 11.1.2, gives

Since $f_1$ is arbitrary, the smooth top-degree form paired with it vanishes pointwise. At each point the other complex Hessians can be prescribed by real quadratic functions times a cutoff. Complexifying this multilinear identity gives all coordinate $(q-1,q-1)$-forms, as in the proof of theorem 11.1.7. They detect every coefficient of $\mathrm{dd}^{\mathrm{c}}(\omega ^{n-q})$, so that form vanishes. When $q=1$, there are no other Hessians and the same conclusion follows directly from the arbitrariness of $f_1$.

Under the third condition, $\mu _ r(w)$ and $\mu _ r(u)$ have the same total mass. 8.13.3 proves the mixed comparison assertion, and $r=m$ gives the fourth condition. Finally, suppose the fourth condition holds and fix smooth admissible $u,v$. Choose $C$ so large that $u<v+C$ everywhere. Constant invariance gives $H_ m(v+C)=H_ m(v)$, so comparison gives $\int _ XH_ m(v)\leq \int _ XH_ m(u)$. Exchanging $u,v$ gives equality, which is the second condition. $\square$

Proof. As in example 11.1.12, $\mathrm{dd}^{\mathrm{c}}u_ t=-\pi ^2t\cos (2\pi x_1)\beta _1$. The coefficient of $\beta _1$ in $\omega +\mathrm{dd}^{\mathrm{c}}u_ t$ is strictly positive in the stated range, and all the other coefficients are positive. Since $(\mathrm{dd}^{\mathrm{c}}u_ t)^2=0$, expansion gives

The period averages of $b$ and $b\cos (2\pi x_1)$ are respectively $2$ and $1/2$, proving the mass formula. For $t<0$ this mass is strictly larger than $\int _ XH_ m(0)=2n!B_ n$. Meanwhile $C>|t|$ gives $\{ 0<u_ t+C\} =X$, so the comparison inequality for this pair would assert the opposite ordering of these masses. $\square$

Proof. First estimate the torsion of $T_{p,q}$, writing $r=p+q$. For smooth admissible potentials, the product rule expresses $\mathrm{dd}^{\mathrm{c}}T_{p,q}$ as a sum of fixed smooth metric forms wedged with products of $\sigma _ u,\sigma _ V$ of degrees $r,r-1,r-2$. The terms of degree $r$ come from differentiating the background power; they cannot be discarded. Indeed, $\mathrm d\sigma _ u=\mathrm d\sigma _ V=\mathrm d\omega $, $\mathrm d^{\mathrm c}\sigma _ u=\mathrm d^{\mathrm c}\sigma _ V=\mathrm d^{\mathrm c}\omega $ and $\mathrm{dd}^{\mathrm{c}}\sigma _ u=\mathrm{dd}^{\mathrm{c}}\sigma _ V=\mathrm{dd}^{\mathrm{c}}\omega $. Every term therefore contains either $\mathrm{dd}^{\mathrm{c}}\omega $ or $\mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega $, with its product-rule multiplicity and the appropriate remaining background power. Terms with negative numbers of factors are absent.

All variable degrees are at most $r<m$. Apply 8.2.7 to each term. Expanding the resulting positive power of the sum of its variable factors gives the pointwise bound

For $q=0$ every variable factor is $\sigma _ u$, so the sharper bound is $C(M_{p,0}+M_{p-1,0}+M_{p-2,0})$. The constant bounds fixed metric coefficients on a finite coordinate cover and the finitely many multiplicities; it is independent of the potentials and their bounds.

Choose uniformly bounded smooth decreasing approximations to $u,V$. By 8.12.4, the torsion measures converge weakly; by 8.11.12, all positive measures in these bounds also converge weakly. Against nonnegative continuous tests, lower semicontinuity of total variation preserves both bounds. They are consequently inequalities of measures and may be restricted to the Borel set $E$.

Put $w=\max \{ u,V+b\} $ and $h=w-u$. The mixed maximum theorem 8.12.2, with one moving slot and the fixed slots defining $T_{p,q}$, gives

All products here are the corresponding positive mixed measures. After integration, this implies

Regularized maxima of the smooth decreasing approximations give a uniformly bounded decreasing smooth list $w_ j\downarrow w$. Stokes' theorem on compact $X$ gives for these lists

Joint decreasing convergence passes the positive mixed measures on the left to their limits. The continuous two-variable weight $F(x,y)=y-x$ in 8.12.4 passes the right side to $\int _ Xh\mathrm{dd}^{\mathrm{c}}T_{p,q}$. No monotonicity of $w_ j-u_ j$ is needed. Since $h$ is zero outside $E$ and $0\leq h\leq s$,

The torsion bounds prove both assertions. The argument never passes a varying measure to a weak limit after restriction to $E$. $\square$

Proof. Use the mixed measures above on $E_ s$, writing $b_{p,q}=M_{p,q}(E_ s)$ and $a_ k=b_{k,0}$. The definition of $S$ gives $0\leq \max \{ u,V+S+s\} -u\leq s$, so 8.13.7 applies. Repeatedly moving a $V$ slot to a $u$ slot gives, for $p+q=r$,

Here there are at most $m$ moves, and the multiplicity $d+1$ in the lower-degree sum is absorbed in $C_1$. For $0<s\leq 1$, induction on $r$ shows

In fact $A_0=a_0$, and one may take $K_ r=C_1(1+\sum _{1\leq d<r}K_ d)$: insert the earlier bounds in the displayed recurrence and use $s^2\leq s$. Thus, for a constant $C_2$ depending only on the setting,

It remains to control the lower diagonal masses by $a_ m$. The identity $\sigma _ V=\alpha \sigma _ v+\eta \omega $ and mixed positivity give $b_{k,1}\geq \eta a_ k$; this identity and inequality pass from smooth approximations by joint convergence. The diagonal assertion of 8.13.7 gives

with $a_{-1}=a_{-2}=0$. Set $B_ k=\eta ^{m-k}a_ k$ for $0\leq k\leq m$, and set $B_{-1}=B_{-2}=0$. Multiplying by $\eta ^{m-k-1}$ and using $0<\eta <1$ gives

For $M=\max _{0\leq k<m}B_ k$, summing from any $j<m$ through $m-1$ gives $B_ j\leq B_ m+3m\delta M$. If $s<\eta ^3/(6mC_3)$, it follows that $M\leq 2B_ m$. Consequently

Choose $C\geq \max \{ 1,6mC_3,2mC_2\} $. Then the asserted range for $s$ supplies every preceding restriction, and substitution in the bound for $b_{0,m}$ proves the theorem. No division by $a_ m$ occurs, so zero top mass on $E_ s$ is included. $\square$

Proof. Add the same constant to $u,v$ so that $v\geq 1$. This preserves the measure hypothesis and the desired conclusion. Choose $0<\alpha <1$ with $\alpha ^ m>c$ and put $V=\alpha v$, $\eta =1-\alpha $. Suppose $S=\inf _ X(u-V)<0$. For $0<s<-S$, the set $E_ s=\{ u<V+S+s\} $ is contained in $\{ u<V\} $ and hence in $\{ u<v\} $. It has positive volume. Indeed, if its volume were zero, then $u\geq V+S+s$ almost everywhere. Both functions are bounded admissible with their canonical representatives, so 8.2.10 would give this order everywhere, contradicting the definition of $S$. The argument does not require $E_ s$ to be open or the infimum to be attained.

Choose also $s<\eta ^3/C$, where $C$ is from 8.13.8. By 8.12.3 and the measure hypothesis,

If $c>0$, take $s$ still smaller so that $c(1+Cs\eta ^{-m})<\alpha ^ m$. If $c=0$, this strict inequality already holds. All measures are finite and positive, so the display forces the positive-volume term to be zero, a contradiction. Hence $S\geq 0$ and $u\geq \alpha v$. Let $\alpha $ increase to one, with $\alpha ^ m>c$. Since $v$ is bounded, $\alpha v\to v$ uniformly, and therefore $u\geq v$. The final assertion is $c=0$. $\square$

Proof. For $\delta >0$, put $E_\delta =\{ u<v-\delta \} $. Multiplying the assumed measure inequality by the positive bounded Borel function $e^{au}$ gives on $E_\delta $

The last identity uses invariance under constants. Since $e^{-a\delta }<1$, 8.13.10 applied to $u,v-\delta $ gives $u\geq v-\delta $ everywhere. Letting $\delta \downarrow 0$ gives $u\geq v$. Two solutions of the stated equation have equal weighted measures $e^{-au}H_ m(u)=\nu $. Applying the order assertion in both directions proves uniqueness. $\square$

Proof. Put $K=\| h_1-h_2\| _{L^\infty (\nu )}$ and fix $s>K/\kappa $. On $E_ s=\{ v+s<u\} $, the actual values $u,v$ lie in $[A,B]$, so $f(x,u)-f(x,v)\geq \kappa s$. The equations and the essential bound on $h_2-h_1$ therefore give, as restricted measures,

The constant is smaller than one. Applying 8.13.10 to $v+s,u$ gives $u\leq v+s$ pointwise. Let $s\downarrow K/\kappa $ and exchange $u,v$ to obtain the norm bound. The exponent is evaluated only at $u,v$; the shifted comparison potential need not take its values in $[A,B]$. $\square$

Proof. For a continuous admissible potential, smooth decreasing admissible approximations converge uniformly on compact $X$ by Dini's theorem. Thus 8.11.12, 8.7.1 define the same Hessian measure on the continuous class. The smooth solution of 8.8.3 therefore solves the bounded equation as well. Apply 8.13.12 with $\nu =\omega ^ n$ and $h_1=h_2=0$ to identify any other bounded solution in $[A,B]$ with it. For the final assertion, smooth existence is already 8.6.1, and 8.13.11 with $\nu =e^ H\omega ^ n$ gives uniqueness among all bounded solutions. $\square$