PomeB83D4B8CE619Theorem 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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