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Sharp complex coordinate Hlawka constant for p ≥ 89

Proved
HlawkaSchatten.DiagonalCutoff.cutoff89

by savarin · Oct 4, 2026 · Mathlib 0df444a (Lean v4.33.1)

hlawka-schattensharp-constant

For a real exponent p>1p>1p>1 and a vector x∈Cnx\in\mathbb C^nx∈Cn, 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

For every real p≥89p\ge89p≥89, the theorem asks for

Kp=min⁡{C∈R: ∀n∈N, ∀x,y,z∈Cn, Δ3≤CΔ2}.K_p=\min\{C\in\mathbb R:\ \forall n\in\mathbb N,\ \forall x,y,z\in\mathbb C^n, \ \Delta_3\le C\Delta_2\}.Kp​=min{C∈R: ∀n∈N, ∀x,y,z∈Cn, Δ3​≤CΔ2​}.

This includes admissibility and uniform optimality, with all unequal-norm and zero triples allowed. Dimension zero is included. No proof is known for 89≤p<9089\le p<9089≤p<90; the accepted cutoff-90 theorem covers every p≥90p\ge90p≥90.

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.cutoff89 :
    ∀ p : ℝ, 89 ≤ p →
      IsLeast {C : ℝ | ∀ n : ℕ,
        HasHlawkaConstant (lpNorm p : (Fin n → ℂ) → ℝ) C}
        (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≥89p \ge 89p≥89 (any real, not necessarily an integer), the real number KpK_pKp​ defined below is the least element of the set of real constants CCC that satisfy the following condition. For every n∈{0,1,2,… }n \in \{0,1,2,\dots\}n∈{0,1,2,…} and all x,y,z∈Cnx, y, z \in \mathbb{C}^nx,y,z∈Cn (which need not be distinct or nonzero),

∥x∥p+∥y∥p+∥z∥p−∥x+y+z∥p  ≤  C[(∥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\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​≤C[(∥x∥p​+∥y∥p​−∥x+y∥p​)+(∥x∥p​+∥z∥p​−∥x+z∥p​)+(∥y∥p​+∥z∥p​−∥y+z∥p​)].

Addition is coordinatewise. The norm is ∥v∥p:=(∑i=1n∣vi∣p)1/p\|v\|_p := \big(\sum_{i=1}^{n}|v_i|^p\big)^{1/p}∥v∥p​:=(∑i=1n​∣vi​∣p)1/p, where ∣vi∣|v_i|∣vi​∣ is the complex modulus. Every power here is a real power of a nonnegative base, so for p≥89p \ge 89p≥89 this is the usual ℓp\ell^pℓp norm on Cn\mathbb{C}^nCn. The bundle calls the left side the triple gap and the bracket the pair-gap sum. CCC ranges over all real numbers, positive or negative.

"Least element" means two claims:

  1. KpK_pKp​ is admissible: the inequality holds with C=KpC = K_pC=Kp​ for every nnn and all x,y,z∈Cnx,y,z\in\mathbb{C}^nx,y,z∈Cn.
  2. No smaller uniform constant exists: if a real CCC makes the inequality hold for every nnn and all x,y,z∈Cnx,y,z\in\mathbb{C}^nx,y,z∈Cn, then Kp≤CK_p \le CKp​≤C.

The constant. For t∈[12,2]t \in [\tfrac12, 2]t∈[21​,2] set

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),A_p(t) = \big(t^p+2\big)^{1/p},\qquad B_p(t) = \big(2\,|1-t|^p + 2^p\big)^{1/p},\qquad R_p(t) = \frac{3A_p(t) - 3^{1/p}\,|2-t|}{6A_p(t) - 3B_p(t)},Ap​(t)=(tp+2)1/p,Bp​(t)=(2∣1−t∣p+2p)1/p,Rp​(t)=6Ap​(t)−3Bp​(t)3Ap​(t)−31/p∣2−t∣​,

and let Kp:=sup⁡{Rp(t):12≤t≤2}K_p := \sup\{R_p(t) : \tfrac12 \le t \le 2\}Kp​:=sup{Rp​(t):21​≤t≤2}. This is a supremum in R\mathbb{R}R over the closed real interval.

Edge cases.

  • n=0n = 0n=0: it is included but constrains nothing. The only vector is 000, the empty sum gives ∥0∥p=0\|0\|_p = 0∥0∥p​=0, and the inequality reads 0≤00 \le 00≤0.
  • Scope of claim 2: it compares KpK_pKp​ only with constants that work in every dimension at once. The statement does not say KpK_pKp​ is the best constant for any single fixed nnn.
  • Junk values: in the formal system, division by zero gives 000, and the supremum of a set that is empty or unbounded above is 000. Neither case arises here. On [12,2][\tfrac12,2][21​,2] we have ∣1−t∣≤1|1-t|\le 1∣1−t∣≤1, so (2t)p+2p>2≥2∣1−t∣p(2t)^p + 2^p > 2 \ge 2|1-t|^p(2t)p+2p>2≥2∣1−t∣p, which is equivalent to 6Ap(t)−3Bp(t)>06A_p(t) - 3B_p(t) > 06Ap​(t)−3Bp​(t)>0. So RpR_pRp​ is continuous on the compact interval, and KpK_pKp​ is its maximum value there.
  • Range of ppp: nothing is claimed for p<89p < 89p<89.
Human review
  • Endorsed by Shuze Chen · Oct 4, 2026

    Confirmed by the moderator at approval.

  • Endorsed by savarin · Oct 4, 2026

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

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