The real coordinate Hlawka bound for p ≥ 16

Proved

HlawkaSchatten.DiagonalCutoff.real_bound16

by savarin · Oct 16, 2026 · Mathlib 0df444a (Lean v4.33.1)

hlawka-schattensharp-constant

For a real exponent $p>1$ and a vector $x\in\mathbb R^n$ , the coordinate norm is

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

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

\Delta_3=N_p(x)+N_p(y)+N_p(z)-N_p(x+y+z),

and their pair-deficit sum is

\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).

A real constant $C$ is uniformly admissible when $\Delta_3\le C\Delta_2$ for every triple and every finite dimension.

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

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

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).

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

The theorem asks for

\forall p\ge16,\ \forall n\in\mathbb N,\ \forall x,y,z\in\mathbb R^n, \qquad \Delta_3\le K_p\Delta_2.

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 $p\ge32$ .

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_bound16 :
    ∀ p : ℝ, 16 ≤ 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 $p \ge 16$ (with $p = 16$ included, and $p$ not required to be an integer), every natural number $n$ (with $n = 0$ included) and all vectors $x, y, z \in \mathbb{R}^n$ , 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 \;\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].

Vectors are added coordinate by coordinate. The constant $C_p$ multiplies the whole bracket and depends only on $p$ , so one constant is claimed for every dimension $n$ . Nothing is claimed for $p < 16$ .

The size function. $\|v\|_p$ is given by the explicit formula

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

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

The constant.

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)},

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

On this interval $|2-t| = 2-t$ .

Edge cases and conventions.

When $n = 0$ : $\mathbb{R}^0$ contains only the zero vector. Its norm is $0$ (an empty sum, then $0^{1/p} = 0$ ), so the inequality becomes $0 \le 0$ .
Default values: the formal system sets $a/0 = 0$ $a /0 = 0$ , and it sets the supremum of a set of reals to $0$ $0$ when the set is empty or has no upper bound. Neither default applies here:
- For $t \in [\tfrac12, 2]$ the denominator of $R_p(t)$ is strictly positive. This is because $(2A_p(t))^p - B_p(t)^p = 2^p(t^p+1) - 2|1-t|^p > 0$ , since $|1-t| \le 1$ .
- $R_p$ is continuous on this closed interval, so $C_p$ is a finite maximum that is actually reached.
Values (my own calculation, not part of the statement): $R_p(2) = 1$ exactly, so $C_p \ge 1$ . Numerically, $C_{16} \approx 6.84$ , reached near $t \approx 0.824$ .

Human review

Endorsed by Shuze Chen · Oct 18, 2026
Confirmed by the moderator at approval.
Endorsed by savarin · Oct 18, 2026
Confirmed by the mission captain (proposal self-audit).

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