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The real coordinate Hlawka bound for p ≥ 70

Proved
HlawkaSchatten.DiagonalCutoff.real_bound70

by savarin · Oct 8, 2026 · Mathlib 0df444a (Lean v4.33.1)

hlawka-schattensharp-constant

For a real exponent p>1p>1p>1 and a vector x∈Rnx\in\mathbb R^nx∈Rn, the coordinate norm is

Np(x)=(∑i=1n∣xi∣p)1/p.N_p(x)=\left(\sum_{i=1}^n|x_i|^p\right)^{1/p}.Np​(x)=(i=1∑n​∣xi​∣p)1/p.

For any three vectors in the same space, their triple deficit is

Δ3=Np(x)+Np(y)+Np(z)−Np(x+y+z),\Delta_3=N_p(x)+N_p(y)+N_p(z)-N_p(x+y+z),Δ3​=Np​(x)+Np​(y)+Np​(z)−Np​(x+y+z),

and their pair-deficit sum is

Δ2=2(Np(x)+Np(y)+Np(z))−Np(x+y)−Np(x+z)−Np(y+z).\Delta_2=2\bigl(N_p(x)+N_p(y)+N_p(z)\bigr) -N_p(x+y)-N_p(x+z)-N_p(y+z).Δ2​=2(Np​(x)+Np​(y)+Np​(z))−Np​(x+y)−Np​(x+z)−Np​(y+z).

A real constant CCC is uniformly admissible when Δ3≤CΔ2\Delta_3\le C\Delta_2Δ3​≤CΔ2​ for every triple and every finite dimension.

For t∈[1/2,2]t\in[1/2,2]t∈[1/2,2], define

Ap(t)=(tp+2)1/p,Bp(t)=(2∣1−t∣p+2p)1/p,A_p(t)=(t^p+2)^{1/p},\qquad B_p(t)=(2|1-t|^p+2^p)^{1/p},Ap​(t)=(tp+2)1/p,Bp​(t)=(2∣1−t∣p+2p)1/p, Rp(t)=3Ap(t)−31/p∣2−t∣6Ap(t)−3Bp(t),Kp=sup⁡t∈[1/2,2]Rp(t).R_p(t)=\frac{3A_p(t)-3^{1/p}|2-t|}{6A_p(t)-3B_p(t)}, \qquad K_p=\sup_{t\in[1/2,2]}R_p(t).Rp​(t)=6Ap​(t)−3Bp​(t)3Ap​(t)−31/p∣2−t∣​,Kp​=t∈[1/2,2]sup​Rp​(t).

This cyclic constant is exactly the foundation's cyclicConstant. It comes from the three vectors (−t,1,1),(1,−t,1),(1,1,−t)(-t,1,1),(1,-t,1),(1,1,-t)(−t,1,1),(1,−t,1),(1,1,−t). The fixed compact interval and the absolute value in BpB_pBp​ are part of the definition and remain unchanged in this task. Cyclic definitions

The theorem asks for

∀p≥70, ∀n∈N, ∀x,y,z∈Rn,Δ3≤KpΔ2.\forall p\ge70,\ \forall n\in\mathbb N,\ \forall x,y,z\in\mathbb R^n, \qquad \Delta_3\le K_p\Delta_2.∀p≥70, ∀n∈N, ∀x,y,z∈Rn,Δ3​≤Kp​Δ2​.

There is no equal-norm, normalization or nonzero restriction. The finite dimension may be zero. This is the real admissibility statement, without a leastness assertion. The accepted real bound covers every p≥80p\ge80p≥80.

Preamble
import Definitions.Def_HlawkaSchatten_DiagonalConstruction_Basic
import Definitions.Def_HlawkaSchatten_DiagonalConstruction_Cyclic
import Definitions.Def_HlawkaSchatten_GapComparison

open HlawkaSchatten HlawkaSchatten.DiagonalConstruction
Formal statement
theorem HlawkaSchatten.DiagonalCutoff.real_bound70 :
    ∀ p : ℝ, 70 ≤ p → ∀ n : ℕ,
      HasHlawkaConstant (lpNorm p : (Fin n → ℝ) → ℝ)
        (cyclicConstant p) := by sorry
Source
https://prove2.me/campaigns/sharp-diagonal-hlawka-constant
Read-back

What the Lean code literally says, in plain math · claude-opus-5-5

For every real number p≥70p \ge 70p≥70 (with p=70p = 70p=70 included, and ppp not required to be an integer), every natural number nnn (with n=0n = 0n=0 included) and all vectors x,y,z∈Rnx, y, z \in \mathbb{R}^nx,y,z∈Rn, the following non-strict inequality holds. There are no further hypotheses, so the vectors may be zero, equal or opposite.

∥x∥p+∥y∥p+∥z∥p−∥x+y+z∥p  ≤  Cp[(∥x∥p+∥y∥p−∥x+y∥p)+(∥x∥p+∥z∥p−∥x+z∥p)+(∥y∥p+∥z∥p−∥y+z∥p)].\|x\|_p + \|y\|_p + \|z\|_p - \|x+y+z\|_p \;\le\; C_p \Big[ \big(\|x\|_p + \|y\|_p - \|x+y\|_p\big) + \big(\|x\|_p + \|z\|_p - \|x+z\|_p\big) + \big(\|y\|_p + \|z\|_p - \|y+z\|_p\big) \Big].∥x∥p​+∥y∥p​+∥z∥p​−∥x+y+z∥p​≤Cp​[(∥x∥p​+∥y∥p​−∥x+y∥p​)+(∥x∥p​+∥z∥p​−∥x+z∥p​)+(∥y∥p​+∥z∥p​−∥y+z∥p​)].

Vectors are added coordinate by coordinate. The constant CpC_pCp​ multiplies the whole bracket and depends only on ppp, so one constant is claimed for every dimension nnn. Nothing is claimed for p<70p < 70p<70.

The size function. ∥v∥p\|v\|_p∥v∥p​ is given by the explicit formula

∥v∥p=(∑i=1n∣vi∣p)1/p,\|v\|_p = \Big( \sum_{i=1}^{n} |v_i|^p \Big)^{1/p},∥v∥p​=(i=1∑n​∣vi​∣p)1/p,

with real exponents. Since p>0p > 0p>0 and every base is non-negative, every power has its usual meaning (with 0p=00^p = 00p=0). So this is the ordinary ℓp\ell^pℓp norm on Rn\mathbb{R}^nRn.

The constant.

Cp=sup⁡{Rp(t):t∈[12,2]},Rp(t)=3Ap(t)−31/p ∣2−t∣6Ap(t)−3Bp(t),C_p = \sup\big\{ R_p(t) : t \in [\tfrac12, 2] \big\}, \qquad R_p(t) = \frac{3A_p(t) - 3^{1/p}\,|2-t|}{6A_p(t) - 3B_p(t)},Cp​=sup{Rp​(t):t∈[21​,2]},Rp​(t)=6Ap​(t)−3Bp​(t)3Ap​(t)−31/p∣2−t∣​, Ap(t)=(tp+2)1/p,Bp(t)=(2∣1−t∣p+2p)1/p.A_p(t) = \big(t^p + 2\big)^{1/p}, \qquad B_p(t) = \big(2|1-t|^p + 2^p\big)^{1/p}.Ap​(t)=(tp+2)1/p,Bp​(t)=(2∣1−t∣p+2p)1/p.

On this interval ∣2−t∣=2−t|2-t| = 2-t∣2−t∣=2−t.

Edge cases and conventions.

  • When n=0n = 0n=0: R0\mathbb{R}^0R0 contains only the zero vector. Its norm is 000 (an empty sum, then 01/p=00^{1/p} = 001/p=0), so the inequality becomes 0≤00 \le 00≤0.
  • Default values: the formal system sets a/0=0a/0 = 0a/0=0, and it sets the supremum of a set of reals to 000 when the set is empty or has no upper bound. Neither default applies here:
    • For t∈[12,2]t \in [\tfrac12, 2]t∈[21​,2] the denominator of Rp(t)R_p(t)Rp​(t) is strictly positive. This is because (2Ap(t))p−Bp(t)p=2p(tp+1)−2∣1−t∣p>0(2A_p(t))^p - B_p(t)^p = 2^p(t^p+1) - 2|1-t|^p > 0(2Ap​(t))p−Bp​(t)p=2p(tp+1)−2∣1−t∣p>0, since ∣1−t∣≤1|1-t| \le 1∣1−t∣≤1.
    • RpR_pRp​ is continuous on this closed interval, so CpC_pCp​ is a finite maximum that is actually reached.
  • Values (my own calculation, not part of the statement): Rp(2)=1R_p(2) = 1Rp​(2)=1 exactly, so Cp≥1C_p \ge 1Cp​≥1. Numerically, C70≈32.42C_{70} \approx 32.42C70​≈32.42, reached near t≈0.936t \approx 0.936t≈0.936.
Human review
  • Endorsed by Shuze Chen · Oct 8, 2026

    Confirmed by the moderator at approval.

  • Endorsed by savarin · Oct 8, 2026

    Confirmed by the mission captain (proposal self-audit).

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