Pome10ABC920077EOn a sublevel set just above the infimum of a difference, the maximum changes the lower potential by at most a prescribed small number. Integration by parts then expresses the mass defect through this small difference and a torsion measure. Mixed lower-degree traces control its variation. Rescaling the upper potential adds a positive background term that relates these traces to the top Hessian measure.
For bounded admissible $u,V$, use the notation
These expressions denote the positive mixed measure $Q_ m$ with $p$ slots equal to $u$, $q$ slots equal to $V$ and the remaining slots equal to the constant zero. In particular $M_{0,0}=\omega ^ n$. For $p+q\leq m-1$, put
This denotes $R_{m-1}$ with the same nonconstant slots and zero in its other slots, wedged with $\omega ^{n-m}$.
Lemma 8.13.7.0F8C3F6B0122 Let $u,V$ be bounded admissible, $b$ a real constant and $s>0$. Suppose $0\leq \max \{ u,V+b\} -u\leq s$. Set $E=\{ u<V+b\} $ and $b_{p,q}=M_{p,q}(E)$. There is a constant $C$, depending only on $X,\omega ,m,n$, such that
For $q=0$, the error can instead be bounded by $Cs(b_{p,0}+b_{p-1,0}+b_{p-2,0})$, with negative indices zero.
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 $E$. $\square$
Theorem 8.13.8.B83D4B8CE619 There is $C\geq 1$, depending only on $X,\omega ,m,n$, with the following property. For bounded admissible $u,v$ and $0<\alpha <1$, put $\eta =1-\alpha $, $V=\alpha v$ and $S=\inf _ X(u-V)$. For $0<s<\eta ^3/C$, let $E_ s=\{ u<V+S+s\} $. Then
No differential identity for a power of $\omega $ is assumed.
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. $\square$
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