Reason: First publication: proof of the asymptotic lower bound, including the two-step energy bootstrap and the parameter ladder.
Proof
Throughout, "the ledger lemma" is the ledger decomposition lemma, "the energy lemma" is the tracked energy bound lemma, "the cascade lemma" is the block cascade lemma and "the squares theorem" is the completion-of-squares theorem. For an admissible π we write K, tk, Lk, Λ⋆, Υesc, Υlev, Z, Str, P, N, BN, Tknr(s) and CΨ, CN for the corresponding objects of the ledger lemma formed from π for the N-th solution. Since admissibility is exactly the conjunction of the parameter-dependent requirements of the ledger lemma, and the parameter-free standing hypotheses are assumed here, the ledger lemma and everything it adopts — in particular the energy lemma and the cascade lemma — apply to the N-th solution for every admissible π and every N.
Conclusion (a). Fix an admissible π satisfying the absorption condition. Write, as in claim 4 of the energy lemma,
By (I′), κ0≤κ♯ for every N, so the last two summands are at most (4R2Tc⋆+2CDT+2CDG)cQκ♯q0−4N−1, a fixed constant times N−1; choose a natural number N1′ beyond which their sum is at most 1. Then, for N≥N1′, and using ϵ≤1,
RN≤R♯:=2Ctg∨cQ1/2(κ♯)1/2K+2cQ1/2(κ♯)1/2+1,
a real independent of N.
Step 1: a crude N-free bound. The absorption condition is the hypothesis of claim 5 of the energy lemma, and JN≤J♯ by (CB). That claim therefore gives, for every N≥N1′,
Z≤C†♯:=c⋆2(J♯+R♯)+c⋆4128(CnsΛ⋆3/4Υesc)4,
a finite real not depending on N (the right-hand side of claim 5 is nondecreasing in RN and in J♯).
Step 2: absorption at large N. By Step 1 and the monotonicity of x↦x3/4 on [0,∞),
and the right-hand side tends to 0; choose N1≥N1′ beyond which it is at most 1. For N≥N1, claim 4 of the energy lemma and (CB) give
c⋆Z≤J♯+(2Ctg∨Λ⋆+2ϵCS2)Z+1+R♯≤J♯+4c⋆Z+1+R♯,
by the absorption condition. Since Z is finite (claim 1 of the energy lemma), the term 4c⋆Z may be subtracted, giving 43c⋆Z≤J♯+1+R♯, that is Z≤Z♯(π) after substituting the value of R♯. Finally claim 1(b) of the ledger lemma and (I′) give
Str+Z≤(1+2CS2T)Z+2TcQ1/2κ01/2≤W♯(π).
Conclusion (b). Let ε′>0. We choose the entries of π in the following order; at each step the quantities already fixed are treated as constants.
Step 1 (T0, λc, λo). Put λc=λo=T0, so that Λ⋆=max(4Ca2K22T0,T0)=c1T0 with c1=max(4Ca2K22,1), a constant of the common data. By claim 1 of the cascade lemma K is the least natural number with KT0≥T, so K≤TT0−1+1 and K≤(TT0−1+1)1/2, whence
using T0(TT0−1+1)1/2=T01/2(T+T0)1/2. The right-hand side tends to 0 as T0 decreases to 0. Choose T0∈(0,T] so small that
2(Ctg∨+lCZ)Λ⋆Z♯(π)≤6ε′and2Ctg∨Λ⋆≤8c⋆.
This fixes K, the grid tk, and the numbers Z♯(π) and W♯(π), none of which depends on the remaining entries of π.
Step 2 (ϵ). Choose ϵ∈(0,1] with 2ϵ(CS2Z♯(π)+cQ1/2(κ♯)1/2)≤ε′/6 and 2ϵCS2≤c⋆/8. Together with Step 1 this gives the absorption condition 2Ctg∨Λ⋆+2ϵCS2≤c⋆/4. Fix ρG∗>0 for this ϵ as prescribed in the ledger lemma.
Step 3 (η, ϱ). Choose η>0 with ηW♯(π)≤ε′/6, fix a radius ρη>0 for it as in the ledger lemma, and put ϱ=ρη/2.
Step 4 (δcl). Choose δcl∈(0,ρ∗/2].
Step 5 (ε1, q0). All of the following hold as soon as the positive real ε1+q0 is small enough, each condition being satisfied on an interval (0,c] for some c>0 determined by the quantities already fixed: (ε1+q0)2+δcl2≤(ρ∗)2 (possible since δcl≤ρ∗/2), C3(ε1+q0)2≤4r0δcl2, ε1+q0≤εtg — these three are (SM) — together with ε1+q0≤ρG∗, with (ε1+q0)2+ϱ2≤ρη2, which is (SN) and holds as soon as (ε1+q0)2≤43ρη2, and with
Choose ε1>0 and q0>0 accordingly. The tuple π is now admissible and satisfies the absorption condition, and it determines Υesc and Υlev, which are finite positive reals independent of N.
Step 6 (N0). Let N1 be as in conclusion (a) for this π. By claim 1 of the energy lemma, P(N)≤cQκ0q0−4N−2≤cQκ♯q0−4N−2, so NP(N)≤cQκ♯q0−4N−1; and by claim 1(d) of the ledger lemma together with conclusion (a), P≤Λ⋆Z♯(π)ΥlevN−1 for N≥N1. Each of the four quantities
is therefore bounded by a constant independent of N times a strictly negative power of N — for the first and third this uses the substitution of the bound P≤Λ⋆Z♯(π)ΥlevN−1 just displayed, which turns P1/2 into a constant times N−1/2 and P into a constant times N−1 — so all four tend to 0. Moreover P(Ω0)=1 and ∣s0∣≤2N, so E[1Ω0s0⋅Z0s0]=∑γ,δZ0γδE[s0γs0δ] by linearity of the expectation, and by (I) each of the l2 summands converges to Z0γδΠ0γδ. Choose N0≥N1 so that for every N≥N0 the sum of the four displayed quantities is at most ε′/6 and
E[1Ω0s0⋅Z0s0]−γ,δ∑Z0γδΠ0γδ≤6ε′.
Conclusion. Let N≥N0. By conclusion (a), Z≤Z♯(π) and Str+Z≤W♯(π). Every summand of BN is nondecreasing in Z, in Str+Z, in P and in κ0, so, grouping the summands of BN as in Steps 1, 2, 3, 5 and 6 and using the five displayed bounds,
and replacing E[1Ω0s0⋅Z0s0] by ∑γ,δZ0γδΠ0γδ at the cost of a further ε′/6 gives conclusion (b). The displayed sum is nonnegative by claim 2 of the ledger lemma.
Conclusion (c). Let ε′>0 and let π and N0 be as furnished by (b). Discarding the nonnegative filtering sum from the inequality of (b) gives JN≥V0−ε′ for every N≥N0, which is the assertion. □