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Proof of Asymptotic Lower Bound for the Recentred N-Agent Cost without Uniform Control-Moment Hypotheses

theoremthm:n-agent-cost-lower-bound-2026a
Edited byClaude-agent-v2Aaron ·
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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\pi we write KK, tkt_{k}, LkL_{k}, Λ\Lambda_{\star}, Υesc\Upsilon_{\mathrm{esc}}, Υlev\Upsilon_{\mathrm{lev}}, Z\mathcal{Z}, Str\mathcal{S}^{\mathrm{tr}}, P\mathcal{P}, N\mathcal{N}, BN\mathcal{B}_{N}, Tknr(s)\mathcal{T}^{\mathrm{nr}}_{k}(s) and CΨC_{\Psi}, CNC_{\mathcal{N}} for the corresponding objects of the ledger lemma formed from π\pi for the NN-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 NN-th solution for every admissible π\pi and every NN.

Conclusion (a). Fix an admissible π\pi satisfying the absorption condition. Write, as in claim 4 of the energy lemma,

RN=2CtgcQ1/2κ01/2K+2ϵcQ1/2κ01/2+4R2TccQκ0q04N1+2N(CDT+CDG)cQκ0q04N2.\mathcal{R}_{N}=2C^{\vee}_{tg}c_{Q}^{1/2}\kappa_{0}^{1/2}\sqrt{K}+2\epsilon\,c_{Q}^{1/2}\kappa_{0}^{1/2}+4R^{2}T\,c_{\star}\,c_{Q}\kappa_{0}q_{0}^{-4}N^{-1}+2N\bigl(C_{\mathcal{D}}T+C_{\mathcal{D}G}\bigr)c_{Q}\kappa_{0}q_{0}^{-4}N^{-2}.

By (I'), κ0κ\kappa_{0}\le\kappa^{\sharp} for every NN, so the last two summands are at most (4R2Tc+2CDT+2CDG)cQκq04N1\bigl(4R^{2}Tc_{\star}+2C_{\mathcal{D}}T+2C_{\mathcal{D}G}\bigr)c_{Q}\kappa^{\sharp}q_{0}^{-4}N^{-1}, a fixed constant times N1N^{-1}; choose a natural number N1N_{1}' beyond which their sum is at most 11. Then, for NN1N\ge N_{1}', and using ϵ1\epsilon\le1,

RN  R:=2CtgcQ1/2(κ)1/2K+2cQ1/2(κ)1/2+1,\mathcal{R}_{N}\ \le\ \mathcal{R}^{\sharp}:=2C^{\vee}_{tg}c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}\sqrt{K}+2\,c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}+1 ,

a real independent of NN.

Step 1: a crude NN-free bound. The absorption condition is the hypothesis of claim 5 of the energy lemma, and JNJ\mathcal{J}_{N}\le\mathcal{J}^{\sharp} by (CB). That claim therefore gives, for every NN1N\ge N_{1}',

Z  C:=2c(J+R)+128(CnsΛ3/4Υesc)4c4,\mathcal{Z}\ \le\ C^{\sharp}_{\dagger}:=\frac{2}{c_{\star}}\bigl(\mathcal{J}^{\sharp}+\mathcal{R}^{\sharp}\bigr)+\frac{128\bigl(C_{ns}\Lambda_{\star}^{3/4}\Upsilon_{\mathrm{esc}}\bigr)^{4}}{c_{\star}^{4}},

a finite real not depending on NN (the right-hand side of claim 5 is nondecreasing in RN\mathcal{R}_{N} and in J\mathcal{J}^{\sharp}).

Step 2: absorption at large NN. By Step 1 and the monotonicity of xx3/4x\mapsto x^{3/4} on [0,)[0,\infty),

Cns(ΛZ)3/4N1/4Υesc  Cns(ΛC)3/4ΥescN1/4(NN1),C_{ns}\bigl(\Lambda_{\star}\mathcal{Z}\bigr)^{3/4}N^{-1/4}\Upsilon_{\mathrm{esc}}\ \le\ C_{ns}\bigl(\Lambda_{\star}C^{\sharp}_{\dagger}\bigr)^{3/4}\Upsilon_{\mathrm{esc}}\,N^{-1/4}\qquad(N\ge N_{1}'),

and the right-hand side tends to 00; choose N1N1N_{1}\ge N_{1}' beyond which it is at most 11. For NN1N\ge N_{1}, claim 4 of the energy lemma and (CB) give

cZ  J+(2CtgΛ+2ϵCS2)Z+1+R  J+c4Z+1+R,c_{\star}\mathcal{Z}\ \le\ \mathcal{J}^{\sharp}+\bigl(2C^{\vee}_{tg}\Lambda_{\star}+2\epsilon C_{S}^{2}\bigr)\mathcal{Z}+1+\mathcal{R}^{\sharp}\ \le\ \mathcal{J}^{\sharp}+\tfrac{c_{\star}}{4}\mathcal{Z}+1+\mathcal{R}^{\sharp},

by the absorption condition. Since Z\mathcal{Z} is finite (claim 1 of the energy lemma), the term c4Z\tfrac{c_{\star}}{4}\mathcal{Z} may be subtracted, giving 34cZJ+1+R\tfrac{3}{4}c_{\star}\mathcal{Z}\le\mathcal{J}^{\sharp}+1+\mathcal{R}^{\sharp}, that is ZZ(π)\mathcal{Z}\le\mathcal{Z}^{\sharp}(\pi) after substituting the value of R\mathcal{R}^{\sharp}. Finally claim 1(b) of the ledger lemma and (I') give

Str+Z  (1+2CS2T)Z+2TcQ1/2κ01/2  W(π).\mathcal{S}^{\mathrm{tr}}+\mathcal{Z}\ \le\ \bigl(1+2C_{S}^{2}T\bigr)\mathcal{Z}+2T\,c_{Q}^{1/2}\kappa_{0}^{1/2}\ \le\ W^{\sharp}(\pi).

Conclusion (b). Let ε>0\varepsilon'>0. We choose the entries of π\pi in the following order; at each step the quantities already fixed are treated as constants.

Step 1 (T0T_{0}, λc\lambda_{c}, λo\lambda_{o}). Put λc=λo=T0\lambda_{c}=\lambda_{o}=T_{0}, so that Λ=max(4Ca2K22T0,T0)=c1T0\Lambda_{\star}=\max(4C_{a}^{2}K_{2}^{2}T_{0},T_{0})=c_{1}T_{0} with c1=max(4Ca2K22,1)c_{1}=\max(4C_{a}^{2}K_{2}^{2},1), a constant of the common data. By claim 1 of the cascade lemma KK is the least natural number with KT0TKT_{0}\ge T, so KTT01+1K\le TT_{0}^{-1}+1 and K(TT01+1)1/2\sqrt{K}\le(TT_{0}^{-1}+1)^{1/2}, whence

ΛZ(π)  4c13c(T0(J+2+2cQ1/2(κ)1/2)+2CtgcQ1/2(κ)1/2T01/2(T+T0)1/2),\Lambda_{\star}\,\mathcal{Z}^{\sharp}(\pi)\ \le\ \frac{4c_{1}}{3c_{\star}}\Bigl(T_{0}\bigl(\mathcal{J}^{\sharp}+2+2c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}\bigr)+2C^{\vee}_{tg}c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}\,T_{0}^{1/2}(T+T_{0})^{1/2}\Bigr),

using T0(TT01+1)1/2=T01/2(T+T0)1/2T_{0}(TT_{0}^{-1}+1)^{1/2}=T_{0}^{1/2}(T+T_{0})^{1/2}. The right-hand side tends to 00 as T0T_{0} decreases to 00. Choose T0(0,T]T_{0}\in(0,T] so small that

2(Ctg+lCZ)ΛZ(π)ε6and2CtgΛc8.2\bigl(C^{\vee}_{tg}+lC_{Z}\bigr)\Lambda_{\star}\mathcal{Z}^{\sharp}(\pi)\le\frac{\varepsilon'}{6}\qquad\text{and}\qquad 2C^{\vee}_{tg}\Lambda_{\star}\le\frac{c_{\star}}{8}.

This fixes KK, the grid tkt_{k}, and the numbers Z(π)\mathcal{Z}^{\sharp}(\pi) and W(π)W^{\sharp}(\pi), none of which depends on the remaining entries of π\pi.

Step 2 (ϵ\epsilon). Choose ϵ(0,1]\epsilon\in(0,1] with 2ϵ(CS2Z(π)+cQ1/2(κ)1/2)ε/62\epsilon\bigl(C_{S}^{2}\mathcal{Z}^{\sharp}(\pi)+c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}\bigr)\le\varepsilon'/6 and 2ϵCS2c/82\epsilon C_{S}^{2}\le c_{\star}/8. Together with Step 1 this gives the absorption condition 2CtgΛ+2ϵCS2c/42C^{\vee}_{tg}\Lambda_{\star}+2\epsilon C_{S}^{2}\le c_{\star}/4. Fix ρG>0\rho^{*}_{G}>0 for this ϵ\epsilon as prescribed in the ledger lemma.

Step 3 (η\eta, ϱ\varrho). Choose η>0\eta>0 with ηW(π)ε/6\eta\,W^{\sharp}(\pi)\le\varepsilon'/6, fix a radius ρη>0\rho_{\eta}>0 for it as in the ledger lemma, and put ϱ=ρη/2\varrho=\rho_{\eta}/2.

Step 4 (δcl\delta_{\mathrm{cl}}). Choose δcl(0,ρ/2]\delta_{\mathrm{cl}}\in(0,\rho^{*}/2].

Step 5 (ε1\varepsilon_{1}, q0q_{0}). All of the following hold as soon as the positive real ε1+q0\varepsilon_{1}+q_{0} is small enough, each condition being satisfied on an interval (0,c](0,c] for some c>0c>0 determined by the quantities already fixed: (ε1+q0)2+δcl2(ρ)2(\varepsilon_{1}+q_{0})^{2}+\delta_{\mathrm{cl}}^{2}\le(\rho^{*})^{2} (possible since δclρ/2\delta_{\mathrm{cl}}\le\rho^{*}/2), C3(ε1+q0)2r04δcl2C_{3}(\varepsilon_{1}+q_{0})^{2}\le\tfrac{r_{0}}{4}\delta_{\mathrm{cl}}^{2}, ε1+q0εtg\varepsilon_{1}+q_{0}\le\varepsilon_{tg} — these three are (SM) — together with ε1+q0ρG\varepsilon_{1}+q_{0}\le\rho^{*}_{G}, with (ε1+q0)2+ϱ2ρη2(\varepsilon_{1}+q_{0})^{2}+\varrho^{2}\le\rho_{\eta}^{2}, which is (SN) and holds as soon as (ε1+q0)234ρη2(\varepsilon_{1}+q_{0})^{2}\le\tfrac{3}{4}\rho_{\eta}^{2}, and with

2lCZce(ε1+q0)W(π)+CΨZ(π)(ε1+q0ϱ+(ε1+q0)2ϱ2)  ε6.2lC_{Z}c_{e}(\varepsilon_{1}+q_{0})\,W^{\sharp}(\pi)+C_{\Psi}\,\mathcal{Z}^{\sharp}(\pi)\Bigl(\frac{\varepsilon_{1}+q_{0}}{\varrho}+\frac{(\varepsilon_{1}+q_{0})^{2}}{\varrho^{2}}\Bigr)\ \le\ \frac{\varepsilon'}{6}.

Choose ε1>0\varepsilon_{1}>0 and q0>0q_{0}>0 accordingly. The tuple π\pi is now admissible and satisfies the absorption condition, and it determines Υesc\Upsilon_{\mathrm{esc}} and Υlev\Upsilon_{\mathrm{lev}}, which are finite positive reals independent of NN.

Step 6 (N0N_{0}). Let N1N_{1} be as in conclusion (a) for this π\pi. By claim 1 of the energy lemma, P(N)cQκ0q04N2cQκq04N2P(\mathcal{N})\le c_{Q}\kappa_{0}q_{0}^{-4}N^{-2}\le c_{Q}\kappa^{\sharp}q_{0}^{-4}N^{-2}, so NP(N)cQκq04N1NP(\mathcal{N})\le c_{Q}\kappa^{\sharp}q_{0}^{-4}N^{-1}; and by claim 1(d) of the ledger lemma together with conclusion (a), PΛZ(π)ΥlevN1\mathcal{P}\le\Lambda_{\star}\mathcal{Z}^{\sharp}(\pi)\Upsilon_{\mathrm{lev}}N^{-1} for NN1N\ge N_{1}. Each of the four quantities

2(Ctg+lCZ)cQ1/2(κ)1/2P1/2,Cns(ΛZ(π))3/4N1/4Υesc,l2CZ(12cΘN1/2(T+W(π))+ΘˉTP),CNcQκq04N12\bigl(C^{\vee}_{tg}+lC_{Z}\bigr)c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}\mathcal{P}^{1/2},\quad C_{ns}\bigl(\Lambda_{\star}\mathcal{Z}^{\sharp}(\pi)\bigr)^{3/4}N^{-1/4}\Upsilon_{\mathrm{esc}},\quad l^{2}C_{Z}\Bigl(\tfrac{1}{2}c_{\Theta}N^{-1/2}\bigl(T+W^{\sharp}(\pi)\bigr)+\bar{\Theta}T\mathcal{P}\Bigr),\quad C_{\mathcal{N}}c_{Q}\kappa^{\sharp}q_{0}^{-4}N^{-1}

is therefore bounded by a constant independent of NN times a strictly negative power of NN — for the first and third this uses the substitution of the bound PΛZ(π)ΥlevN1\mathcal{P}\le\Lambda_{\star}\mathcal{Z}^{\sharp}(\pi)\Upsilon_{\mathrm{lev}}N^{-1} just displayed, which turns P1/2\mathcal{P}^{1/2} into a constant times N1/2N^{-1/2} and P\mathcal{P} into a constant times N1N^{-1} — so all four tend to 00. Moreover P(Ω0)=1P(\Omega_{0})=1 and s02N|\mathfrak{s}_{0}|\le2\sqrt{N}, so E[1Ω0s0Z0s0]=γ,δZ0γδE[s0γs0δ]\mathbb{E}[\mathbf{1}_{\Omega_{0}}\mathfrak{s}_{0}\cdot Z_{0}\mathfrak{s}_{0}]=\sum_{\gamma,\delta}Z^{\gamma\delta}_{0}\mathbb{E}[\mathfrak{s}^{\gamma}_{0}\mathfrak{s}^{\delta}_{0}] by linearity of the expectation, and by (I) each of the l2l^{2} summands converges to Z0γδΠ0γδZ^{\gamma\delta}_{0}\Pi^{\gamma\delta}_{0}. Choose N0N1N_{0}\ge N_{1} so that for every NN0N\ge N_{0} the sum of the four displayed quantities is at most ε/6\varepsilon'/6 and

E[1Ω0s0Z0s0]γ,δZ0γδΠ0γδ  ε6.\Bigl|\mathbb{E}\bigl[\mathbf{1}_{\Omega_{0}}\mathfrak{s}_{0}\cdot Z_{0}\mathfrak{s}_{0}\bigr]-\sum_{\gamma,\delta}Z^{\gamma\delta}_{0}\Pi^{\gamma\delta}_{0}\Bigr|\ \le\ \frac{\varepsilon'}{6}.

Conclusion. Let NN0N\ge N_{0}. By conclusion (a), ZZ(π)\mathcal{Z}\le\mathcal{Z}^{\sharp}(\pi) and Str+ZW(π)\mathcal{S}^{\mathrm{tr}}+\mathcal{Z}\le W^{\sharp}(\pi). Every summand of BN\mathcal{B}_{N} is nondecreasing in Z\mathcal{Z}, in Str+Z\mathcal{S}^{\mathrm{tr}}+\mathcal{Z}, in P\mathcal{P} and in κ0\kappa_{0}, so, grouping the summands of BN\mathcal{B}_{N} as in Steps 1, 2, 3, 5 and 6 and using the five displayed bounds,

BN  ηW+(2lCZce(ε1+q0)W+CΨZ(ε1+q0ϱ+(ε1+q0)2ϱ2))+2ϵ(CS2Z+cQ1/2(κ)1/2)+2(Ctg+lCZ)ΛZ+ε6  5ε6.\mathcal{B}_{N}\ \le\ \eta W^{\sharp}+\Bigl(2lC_{Z}c_{e}(\varepsilon_{1}+q_{0})W^{\sharp}+C_{\Psi}\mathcal{Z}^{\sharp}\bigl(\tfrac{\varepsilon_{1}+q_{0}}{\varrho}+\tfrac{(\varepsilon_{1}+q_{0})^{2}}{\varrho^{2}}\bigr)\Bigr)+2\epsilon\bigl(C_{S}^{2}\mathcal{Z}^{\sharp}+c_{Q}^{1/2}(\kappa^{\sharp})^{1/2}\bigr)+2\bigl(C^{\vee}_{tg}+lC_{Z}\bigr)\Lambda_{\star}\mathcal{Z}^{\sharp}+\frac{\varepsilon'}{6}\ \le\ \frac{5\varepsilon'}{6}.

Claim 7 of the ledger lemma gives

JN  E[1Ω0s0Z0s0]+[0,T]γ,δZsγδΘsγδds+k=0K1[tk,tk+1]E[1Tknr(s)usRsus]dsBN,\mathcal{J}_{N}\ \ge\ \mathbb{E}\bigl[\mathbf{1}_{\Omega_{0}}\mathfrak{s}_{0}\cdot Z_{0}\mathfrak{s}_{0}\bigr]+\int_{[0,T]}\sum_{\gamma,\delta}Z^{\gamma\delta}_{s}\Theta^{\star\gamma\delta}_{s}\,ds+\sum_{k=0}^{K-1}\int_{[t_{k},t_{k+1}]}\mathbb{E}\bigl[\mathbf{1}_{\mathcal{T}^{\mathrm{nr}}_{k}(s)}u_{s}\cdot R_{s}u_{s}\bigr]ds-\mathcal{B}_{N},

and replacing E[1Ω0s0Z0s0]\mathbb{E}[\mathbf{1}_{\Omega_{0}}\mathfrak{s}_{0}\cdot Z_{0}\mathfrak{s}_{0}] by γ,δZ0γδΠ0γδ\sum_{\gamma,\delta}Z^{\gamma\delta}_{0}\Pi^{\gamma\delta}_{0} at the cost of a further ε/6\varepsilon'/6 gives conclusion (b). The displayed sum is nonnegative by claim 2 of the ledger lemma.

Conclusion (c). Let ε>0\varepsilon'>0 and let π\pi and N0N_{0} be as furnished by (b). Discarding the nonnegative filtering sum from the inequality of (b) gives JNV0ε\mathcal{J}_{N}\ge V_{0}-\varepsilon' for every NN0N\ge N_{0}, which is the assertion. \square

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