The least uniform complex coordinate Hlawka constant for p ≥ 80

Proved

HlawkaSchatten.DiagonalCutoff.cutoff80

by savarin · Oct 6, 2026 · Mathlib 0df444a (Lean v4.33.1)

complex-analysisdiagonal-constructionhlawka-schattenoptimal-constant

For every real exponent p ≥ 80, let N_p(x) be the foundation's finite coordinate p-norm and let K_p be its unchanged cyclicConstant p, defined as the supremum of the cyclic ratio over t ∈ [1/2, 2].

Then K_p is the least real constant C such that the triple deficit N_p(x) + N_p(y) + N_p(z) − N_p(x + y + z) is at most C times the pair-deficit sum 2(N_p(x) + N_p(y) + N_p(z)) − N_p(x + y) − N_p(x + z) − N_p(y + z), for every natural-number dimension n and all complex vectors x, y, z in that dimension.

This combines admissibility and optimality uniformly over all finite complex coordinate dimensions. It includes dimension zero and arbitrary triples with unequal or zero norms. The same shared definitions are used throughout; the claim concerns complex diagonal matrices via their coordinate norms.

No proof is known for 80 ≤ p < 85; the accepted cutoff-85 theorem covers every p ≥ 85.

Preamble

import Definitions.Def_HlawkaSchatten_DiagonalConstruction_Basic
import Definitions.Def_HlawkaSchatten_DiagonalConstruction_Cyclic
import Definitions.Def_HlawkaSchatten_GapComparison
import Mathlib.Analysis.Complex.Circle
import Mathlib.Analysis.Complex.ExponentialBounds
import Mathlib.Analysis.Convex.Deriv
import Mathlib.Analysis.Convex.Function
import Mathlib.Analysis.Convex.Integral
import Mathlib.Analysis.Convex.Jensen
import Mathlib.Analysis.Convex.SpecificFunctions.Basic
import Mathlib.Analysis.Convex.SpecificFunctions.Pow
import Mathlib.Analysis.InnerProductSpace.Basic
import Mathlib.Analysis.InnerProductSpace.Dual
import Mathlib.Analysis.InnerProductSpace.NormPow
import Mathlib.Analysis.Normed.Lp.PiLp
import Mathlib.Analysis.Normed.Module.FiniteDimension
import Mathlib.Analysis.SpecialFunctions.Pow.Continuity
import Mathlib.Data.Fin.VecNotation
import Mathlib.Data.Real.Basic
import Mathlib.Data.Sign.Basic
import Mathlib.LinearAlgebra.Dimension.Finite
import Mathlib.MeasureTheory.Group.Integral
import Mathlib.MeasureTheory.Integral.Bochner.ContinuousLinearMap
import Mathlib.MeasureTheory.Measure.Haar.Basic
import Mathlib.Tactic.Abel
import Mathlib.Tactic.FieldSimp
import Mathlib.Tactic.Linarith
import Mathlib.Tactic.LinearCombination
import Mathlib.Tactic.Module
import Mathlib.Tactic.Positivity
import Mathlib.Tactic.Ring
import Mathlib.Topology.Instances.Sign
import Mathlib.Topology.Order.Compact

/-! # The least dimension-independent complex coordinate Hlawka constant

This packages admissibility and the three-coordinate cyclic obstruction into
one statement, with an explicit lower cutoff on the real exponent.
-/

open HlawkaSchatten
open HlawkaSchatten.DiagonalConstruction

Formal statement

theorem HlawkaSchatten.DiagonalCutoff.cutoff80 :
    ∀ p : ℝ, 80 ≤ 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 $p \ge 80$ , the statement asserts that the number $K_p$ defined below is the least real constant $C$ for which a three-vector inequality for the $p$ -norm holds for all complex vectors in every dimension $n$ at once.

Norm. For $n \in \mathbb{N}$ (including $n = 0$ ) and $x = (x_1, \dots, x_n) \in \mathbb{C}^n$ ,

\|x\|_p = \Big(\sum_{i=1}^{n} |x_i|^p\Big)^{1/p},

where $|x_i|$ is the complex modulus. Vectors are added coordinatewise.

Inequality. For $x, y, z \in \mathbb{C}^n$ , let $T$ be the three-vector triangle deficit and $P$ the sum of the three two-vector deficits:

T(x,y,z) = \|x\|_p + \|y\|_p + \|z\|_p - \|x+y+z\|_p,

P(x,y,z) = \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).

Let $S_p$ be the set of real numbers $C$ such that, for every $n \in \mathbb{N}$ and all $x, y, z \in \mathbb{C}^n$ ,

T(x,y,z) \le C \cdot P(x,y,z).

Constant. For real $t$ , let

A_p(t) = (t^p + 2)^{1/p}, \qquad B_p(t) = \big(2\,|1-t|^p + 2^p\big)^{1/p},

R_p(t) = \frac{3A_p(t) - 3^{1/p}\,|2-t|}{6A_p(t) - 3B_p(t)},

K_p = \sup\{R_p(t) : \tfrac12 \le t \le 2\}.

Claim. For every real $p$ with $80 \le p$ :

$K_p \in S_p$ , meaning $T(x,y,z) \le K_p \cdot P(x,y,z)$ for every $n \in \mathbb{N}$ and all $x, y, z \in \mathbb{C}^n$ ;
$K_p \le C$ for every $C \in S_p$ .

Together these say $K_p = \min S_p$ .

Edge cases.

$p$ can be any real number $\ge 80$ , not only an integer. Nothing is asserted for $p < 80$ .
Every power that enters is a real power of a nonnegative base with a positive exponent. So $\|\cdot\|_p$ is the standard $\ell^p$ expression.
Dimension $n = 0$ is included but imposes nothing. $\mathbb{C}^0$ is a single point of norm $0$ , so the inequality reads $0 \le 0$ .
The vectors $x, y, z$ are unrestricted: they may be zero, repeated or parallel.
Minimality compares $K_p$ only with constants that work in all dimensions at once. For each $C < K_p$ , some $n$ and some $x, y, z \in \mathbb{C}^n$ give $T > C \cdot P$ . The statement does not say which ones, and it says nothing about the best constant in any single fixed dimension.
On $[\tfrac12, 2]$ the denominator of $R_p$ is strictly positive. It is positive exactly when $(2t)^p + 2^p > 2|1-t|^p$ , and there the left side is greater than $2$ while the right side is at most $2$ . So no division-by-zero convention comes into play.
$R_p$ is continuous on $[\tfrac12, 2]$ , so $K_p$ is a genuine supremum and is attained as a maximum. No convention for the supremum of an empty or unbounded set is involved.
On this interval $|2-t| = 2-t$ .

Human review

Endorsed by Shuze Chen · Oct 6, 2026
Confirmed by the moderator at approval.
Endorsed by savarin · Oct 6, 2026
Confirmed by the mission captain (proposal self-audit).

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