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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Proposition 6.4 (pp. 12--13) of the retained PDF, read in the canonical conversion and checked against the text layer; held by its library card, Korsky 2026, resolution. The inputs are Lemma 6.2 and Lemma 6.3.

Standing. Author-recorded reconstruction; not an independent review; changes no status and assigns no tier. The source is an unrefereed preprint; the implied constants are absolute and kept implicit as in the source, except where the reconstruction names one.

Definitions

Notation as on the Lemma 6.2 page: Δt(x,D)\Delta_t(x,D), Zt(D)Z_t(D), identity (6.4), hypothesis (6.1) with A≥1A\ge1. Put zt(D)=Zt(D)/Dz_t(D)=Z_t(D)/D for D>0D>0.

Statement (Proposition 6.4, p. 12)

There are absolute constants C2,C3>0C_2,C_3>0 with the following property. Suppose that (6.1) holds with A≥1A\ge1 and r≥C2Ar\ge C_2A, and put

Λ=log⁡(r/A),S=ArΛ2.\Lambda=\log(r/A),\qquad S=\frac{\sqrt{Ar}}{\Lambda^2}.

Then, for all sufficiently large integers nn,

sup⁡0≤D≤S ∫T∣Nn((x,x+D/n])−D∣ dx ≤ C3A.(6.7)\sup_{0\le D\le S}\ \int_{\mathbb T}\Bigl|N_n\bigl((x,x+D/n]\bigr)-D\Bigr|\,dx \ \le\ C_3A . \tag{6.7}

The time threshold may depend on rr, AA and the sequence, but not on DD.

Proof

Put θ=A/r\theta=\sqrt{A/r} and K=ArK=\sqrt{Ar}, and take C2C_2 large enough that θ\theta and θΛ\theta\Lambda are small (both tend to 00 as r/A→∞r/A\to\infty).

The scale chain. Take K=D0<D1<⋯<Dh=rK=D_0<D_1<\cdots<D_h=r with Di+1=2DiD_{i+1}=2D_i except that the last step is shortened to end at rr; adjacent scales satisfy K≤D≤E≤2DK\le D\le E\le2D, and h≤log⁡2(r/K)+1=O(Λ)h\le\log_2(r/K)+1=O(\Lambda). For an adjacent pair D<ED<E apply Lemma 6.2 with k=⌈D/K⌉k=\lceil D/K\rceil, so D/K≤k≤2D/KD/K\le k\le2D/K and

q=Ekr ≤ 2Kr=2θ<1,kAD ≤ 2AK=2θ.q=\frac E{kr}\ \le\ \frac{2K}r=2\theta<1,\qquad \frac{kA}D\ \le\ \frac{2A}K=2\theta .

Dividing (6.5) by DD and using Zt+(E)=E zt+(E)Z_{t_+}(E)=E\,z_{t_+}(E),

zt(D) ≤ (1+q) z(1+q)t(E)+q+8kAD+4krtD ≤ (1+Cθ) z(1+q)t(E)+Cθ+4krtD(6.8)z_t(D)\ \le\ (1+q)\,z_{(1+q)t}(E)+q+\frac{8kA}D+\frac{4kr}{tD} \ \le\ (1+C\theta)\,z_{(1+q)t}(E)+C\theta+\frac{4kr}{tD} \tag{6.8}

with the absolute constant C=18C=18.

Iteration. Lemma 6.3 gives zt(r)=Zt(r)/r≤A/r=θ2z_t(r)=Z_t(r)/r\le A/r=\theta^2 at every late time. Starting at scale KK and time tt, apply (6.8) along the chain, the time being multiplied by 1+qi≤1+2θ1+q_i\le1+2\theta at the ii-th step. With h=O(Λ)h=O(\Lambda) steps and θΛ\theta\Lambda small, (1+Cθ)h≤exp⁡(Cθh)=O(1)(1+C\theta)^h\le\exp(C\theta h)=O(1), so

zt(K) ≤ O(1)⋅θ2+O(hθ)+O(1)∑i<h4kirtiDi ≤ O(θΛ)+O(hrKt),z_t(K)\ \le\ O(1)\cdot\theta^2+O(h\theta)+O(1)\sum_{i<h}\frac{4k_ir}{t_iD_i} \ \le\ O(\theta\Lambda)+O\Bigl(\frac{hr}{Kt}\Bigr),

using ki≤2Di/Kk_i\le2D_i/K and ti≥tt_i\ge t for the last sum. For fixed rr and AA the last term tends to 00 as t→∞t\to\infty, so for every sufficiently large tt,

zt(K) ≤ C′θΛ(6.9)z_t(K)\ \le\ C'\theta\Lambda \tag{6.9}

with an absolute constant C′C'.

Descent to short intervals. Apply Lemma 6.2 once more with E=KE=K and k=1k=1, so q=K/r=θq=K/r=\theta: for 0<D≤S0<D\le S,

Zt(D) ≤ (1+θ)DK Z(1+θ)t(K)+θD+8A+4rt ≤ (1+θ)C′DθΛ+θD+8A+4rt,Z_t(D)\ \le\ \frac{(1+\theta)D}K\,Z_{(1+\theta)t}(K)+\theta D+8A+\frac{4r}t \ \le\ (1+\theta)C'D\theta\Lambda+\theta D+8A+\frac{4r}t ,

by (6.9) at the time (1+θ)t(1+\theta)t. Since θ≤θΛ\theta\le\theta\Lambda, this is at most 8A+C′′DθΛ+4r/t8A+C''D\theta\Lambda+4r/t with C′′C'' absolute, and the definition of SS gives DθΛ≤SθΛ=Kθ/Λ=A/Λ≤AD\theta\Lambda\le S\theta\Lambda=K\theta/\Lambda=A/\Lambda\le A. So Zt(D)≤(8+C′′)A+4r/tZ_t(D)\le(8+C'')A+4r/t.

Integer times. At an integer time nn large enough that 4r/n≤A4r/n\le A, identity (6.4) gives

∫T∣Δn(x,D)∣ dx=2Zn(D) ≤ 2(9+C′′)A,\int_{\mathbb T}\bigl|\Delta_n(x,D)\bigr|\,dx=2Z_n(D)\ \le\ 2(9+C'')A ,

which is (6.7) with C3=2(9+C′′)C_3=2(9+C''). The case D=0D=0 is trivial. The threshold on nn comes from (6.9) at the time (1+θ)n(1+\theta)n, from 4r/n≤A4r/n\le A, from the finitely many chain comparisons behind (6.9), and from the descent comparison, whose transport error 8A+4r/t8A+4r/t is free of DD and whose only DD-dependent largeness requirement (Lemma 6.2 page, end of proof) is that the arc of length D/nD/n be shorter than 11; identity (6.4) needs the same. Since D≤SD\le S, any n>Sn>S meets both at once, so one late time serves every 0≤D≤S0\le D\le S.

Role in the argument

Localizing the points of a moving short interval, with insertion time as a second coordinate, turns (6.7) into a planar L1L^1 discrepancy bound that contradicts Halász's theorem for large SS; this is Lemma 7.2, and the assembly is on the Theorem 1.1 page.