All sums below are finite; C depends only on K. Put
S=∑j=0m−1bj, L1=∑j=1mlj and
D1=∑j=1mDj. The first telescope is
D1≤L1+z(bm+1−b1)≤L1, since bm+1≤b1. On this block the terminal coherent sum is
at most γS, and all convolution source indices are less than m.
Thus
L1≤2+K/z+γS+(K/z)D1. Choose z≥2K. Absorption gives
D1,L1≤C(1+γS) and
∑j=2mlj≤C/z+CγS.
It also gives the boundary estimate
z(b1−bm+1)≤L1.
For δj=bj−bj−1, the defect inequality implies
δj+1≥δj−lj/z. Since
∑j=1mδj=0, some δj is nonnegative; propagating
from that index gives δm+1≥−L1/z.
On the other hand, b2m≤1 and
δ2m+1≤2b2m/z≤2/z.
Telescoping the defect inequality over the second block and adding the
first block yields
Dtot:=j=1∑2mDj≤j=1∑2mlj+2+L1. The coherent sum through 2m is at most K/z+2γS, because
∑j=02m−1bj≤2S. A second absorption proves
Dtot≤C+CγS,j=2∑2mlj≤C/z+CγS.(G1) Fix 1≤q<m, and extend
gq(j)=min{j,q}(m−max{j,q})/m by zero outside 1≤j<m.
It is nonnegative and globally one-Lipschitz. The discrete second
differences give exactly
j∑gq(j)(bj+1−2bj+bj−1)=1−bq. For E=∑jgq(j)lj and F=∑jgq(j)Dj this gives
F≤E+z(1−bq)≤E+z.
The term j=1 contributes at most two. The early coherent terms
contribute at most K/z, since gq(j)≤j; the later terms at
most γmS. Reindex the convolution and use
gq(j+s)≤gq(j)+s to obtain
E≤2+K/z+γmS+(K/z)F+(K/z)D1. By the preceding estimates, absorption gives E≤C+CγmS.
Since F≥0, the Green identity implies
bq≤1+C/z+CγmS/z. Setting
M=max0≤q≤mbq and using S≤mM yields
M≤1+C/z+Cc0M. Fix z0 large and then c0 small so
M≤2. The shift inequality propagates this upper bound to all indices.
For the lower bound use the weight
n(j)=#{k∈{2,…,m+1}:k≤j≤m+k−1}. It is one-Lipschitz, supported in [2,2m], and bounded by
min{j−1,m}. With En=∑jn(j)lj, Fn=∑jn(j)Dj,
telescoping each window gives
Fn≤En+z(b2m+1−2bm+1+b1)≤En+L1. The second inequality uses b2m+1≤bm+1 and the first
boundary estimate. The coherent contribution is at most
K/z+2γmS. Convolution reindexing, now using (G1), gives
En≤K/z+2γmS+(K/z)Fn+(K/z)Dtot≤C/z+CγmS+(K/z)En. Therefore En≤C/z+CγmS.
Apply the Dirichlet Green identity on [q,q+m] at m.
Both endpoint values are at most bq, since bq+m≤bq;
the lower bound Δ2bj≥−lj/z therefore implies
1=bm≤bq+z−1j=q+1∑q+m−1Gq,m(j)lj, where Gq,m(j)=(min{m,j}−q)(q+m−max{m,j})/m.
For j≤m, this is at most j−q≤j−1=n(j); for j>m
it is at most q+m−j≤2m−j≤n(j).
Thus bq≥1−C/z2−CγmS/z≥1/2 by S≤2m
and the same universal choices. Endpoints already equal one.
Finally replace the m windows in n by the J windows starting
at 2,…,J+1. The new weight is still one-Lipschitz, at most
min{j−1,J}, and supported inside [2,2m]. Its boundary telescope is
z(bm+J+1−bm+1−bJ+1+b1)≤z(b1−bm+1)≤L1,
using bm+J+1≤bJ+1. Its coherent sum is at most
K/z+2γJm, now that bj≤2. The identical reindexing and
absorption give the asserted window estimate. This proves the lemma.