Song–Zhang v2: summable budgets and starting depths

Part of the second version of Song–Zhang, Chapter Song–Zhang, second version: technical estimates; the reading order is on the full proofs page.

Overview. This develops the finite-chain argument for Proposition 9.4, corresponding to Propositions 9.23–9.25 of Song & Zhang, 2026. The scalar threshold and degree-sum arguments are combined with the independently reconstructed finite-chain block theorem and the near-unit depth/outer-round lemmas from the small-loss dossier to close the full profile induction.

Dependencies. Use Definition 9.1, Proposition 9.1, and the functions and scalar lemmas in Proposition 9.3. In particular $W_i=\bar t\circ\bar\chi^{\circ i}$ and $\mathcal L_r=\ell_r/\varrho$ . All laws below are centered regular log-concave measures with covariance at most $I$ . No BKL result or KLS conclusion is used.

Uniform block construction¶

The disjoint-degree moment estimate and the full finite-chain block realization are proved in Proposition 10.2. Their constants do not depend on the number of retained coefficient majorants.

Paying the growing depth thresholds¶

Let

\alpha_i={2^{-i}\over16},\quad R_i=\lceil C_R\alpha_i^{-12}\rceil, \quad S_i=1+B\sum_{l<i}2^{-l},\quad A_i=A_0\exp\left(C_A\sum_{l<i}\alpha_l\right).

(141.1)

Proof 141.1

Since $\alpha_i^{-1}=16\cdot2^i$ , the fixed-power inequality for $W_i$ first gives

W_i(C16^p(2^i)^p)\leq W_i(2^i)+b_{C16^p,p}4^{-i}.

(141.4)

The Lipschitz bound at the fixed point four gives $W_i(2^i)\leq4+4^{-i-1}(2^i-4)_+$ , including $i=0$ . This proves the first assertion. Since $\alpha_i^{-12}\geq1$ , $R_i\leq(C_R+1)\alpha_i^{-12}$ . Apply the sum estimate to $R_i+\lceil Ct(x)\rceil$ . Its first term is at most $4+b_12^{-i}$ by the first assertion. The second is at most $W_i(x)+b_24^{-i}$ , since $t(x)\leq C'(x+1)\leq2C'x$ and $\lceil Ct(x)\rceil\leq C''x$ . Because $W_i(x)\geq4$ , taking the maximum and paying the one sum error proves the claim. The order of operations matters: applying Lipschitz directly to $R_i$ would multiply $4^{-i}$ by a quantity of order $2^{12i}$ .

Retaining caps and closing the analytic induction¶

Suppose the profile has been established through stage $i$ , and define

r_l(x)=\max\{R_l,\lceil C_rt(x)\rceil+\lceil D_r/\alpha_l\rceil\} \quad(0\leq l\leq i).

(141.6)

The static transfer Proposition 10.1 gives the fixed valid coefficient majorants

H_l(x)^2=\min\{G_*(x)^2, e^{3\alpha_l}A_l[W_l(r_l(x))+S_l]^{1/3}\}.

(141.7)

Here $G_*$ is the universal original coefficient majorant; its validity comes from the preceding height-reduction argument, not from KLS. Indeed take $\Gamma^2=A_l[W_l(r_l(d))+S_l]^{1/3}/\varrho^2$ at the fixed depth $r_l(d)$ . It is at least one after a fixed choice of $A_0$ . The all-law profile gives the transfer hypothesis at that depth; burn-in bounds $\mathcal L_{r_l(d)}(d)^2$ and $(1+r_l(d)^{-2})^2$ by $e^{\alpha_l}$ each, leaving room in the displayed $e^{3\alpha_l}$ . No comparison with $C_P$ or continuity of the supremum defining $\mathcal A$ is used. Once extracted, these majorants are retained unchanged.

For an additional $0<\epsilon\leq\alpha_i$ , use $K_l=\lceil L\alpha_l^{-2}\rceil$ through $l=i$ , $K_{i+1}=\lceil L\epsilon^{-2}\rceil$ , and $\Xi_l=\lceil4C_{\rm deg}\alpha_l^{-1}K_{l+1}\rceil$ . For $l<i$ , geometric margins give $\Xi_l\leq C\alpha_l^{-3}$ and $r_l(\Xi_l)\leq C'\alpha_l^{-12}$ . Therefore

W_l(r_l(\Xi_l))\leq4+b_{\rm old}2^{-l}.

(141.8)

Taking $B\geq b_{\rm old}$ pays this error from the very next increment of $S$ , and $C_A\geq4$ gives, for $l<i$ ,

(1+\alpha_l)H_l(\Xi_l)^2\leq A_i(4+S_i)^{1/3}.

(141.9)

Indeed $\log(1+\alpha_l)+3\alpha_l\leq4\alpha_l$ and $A_i/A_l\geq e^{C_A\alpha_l}$ . This is a maximum bound on each retained floor, so the number of floors does not enter.

Proof 141.2

Apply Lemma 10.12 with $V=W_i$ , $m=i$ , $\delta=\alpha_i$ , $A=A_i$ and $S=S_i$ . The newest cap is the $H_i$ already extracted above. The earlier floors were bounded individually above by $A_i(4+S_i)^{1/3}$ , which is exactly the retained-floor hypothesis of that lemma. The original-seed floor is constant: $K_0=\lceil C_{\rm cut}\alpha_0^{-2}\rceil$ is fixed. Increase $A_0$ once to bound $C_0G_*(16K_0)^2$ and $C_0$ . The finite-chain theorem is applied with $N=i+1\ge1$ ; its constants are independent of $i$ . Finally $S_i\le1+2B$ , so one universal $b_{\rm pre}$ serves every $i$ . The outer-round proof supplies the coefficient return, initialization, and full terminal-depth induction; none of these is inferred from the block construction alone.

Proof 141.3 (Closure of the outer induction)

The stage-zero profile follows from the initialization proved in the small-loss dossier with $W_0=\bar t$ ; its proof does not require the constant branch of $\widehat W$ . Increase $A_0,C_R$ once accordingly. Use the realized outer-round estimate above at each subsequent stage. For a prescribed $r\geq R_{i+1}$ let $j=\kappa(r)$ , $\epsilon=\alpha_i/(j+1)$ , and take the least odd $Q$ at least $\max\{Q_\epsilon,\alpha_i^{-1}\log(r+1)\}$ . The polynomial threshold calculation in the small-loss argument proves $Q\leq r$ and that this $r$ is admissible, with a universal choice of $C_R$ . Consequently $t^{\circ j}(Q)\leq3<D_i$ , $(r+1)^{1/Q}\leq e^{\alpha_i}$ , and $\epsilon j\leq\alpha_i$ .

Use the product inequality and $j+1\leq\chi(r)$ to obtain

\begin{split} W_i(\epsilon^{-1}) &\leq\max\{W_i(\chi(r)),W_i(\alpha_i^{-1})\}+b4^{-i}\\ &\leq W_{i+1}(r)+b_{\rm ret}2^{-i}. \end{split}

(141.11)

The last step uses threshold absorption, $W_{i+1}\geq4$ , and $W_i\circ\bar\chi=W_{i+1}$ . Choose $B\geq b_{\rm ret}+b_{\rm pre}$ in addition to its earlier requirement. Then $D_i\leq W_{i+1}(r)+S_{i+1}$ . The total multiplicative loss is at most $e^{41\alpha_i}$ , so $C_A\geq42$ covers it. This proves the next profile. All choices are universal: the dependence of $b_{\rm pre}$ and the threshold-absorption constants on $B$ is at most $C\log(e+B)$ , because $S_*=1+2B$ changes only the fixed high-height cutoff, and that cutoff enters a fixed-power height bound. Thus choose $B$ to satisfy $B\ge C\log(e+B)$ and all the finitely many earlier lower bounds; then freeze it before the induction.

The realized recurrences satisfy

\sum_{l<i}\alpha_l\leq1/8,\qquad \sum_{l<i}2^{-l}\leq2,

(141.12)

and therefore the stated bounds on $A_i,S_i$ . Together with the preceding finite induction, these sums prove the claimed bounded-amplitude profiles.

Fences respected. Uniform admissibility is paid before choosing each depth; no estimate for a growing starting depth is suppressed. No uniform coefficient bound is inferred from the target or from BKL.

References¶

Song, Z., & Zhang, X. (2026). An O(1) Bound for the KLS Constant. https://arxiv.org/abs/2610.01447v2