The real coordinate Hlawka bound for p ≥ 85

Proved

HlawkaSchatten.DiagonalCutoff.real_bound85

by savarin · Oct 5, 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\ge85,\ \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 $p\ge87$ .

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_bound85 :
    ∀ p : ℝ, 85 ≤ 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 $p \ge 85$ (not necessarily an integer, with no upper limit) and every natural number $n$ (including $n = 0$ ), the following holds for all vectors $x, y, z \in \mathbb{R}^n$ :

\|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].

The inequality is non-strict. $C(p)$ multiplies the whole bracket, which equals $2\|x\|_p + 2\|y\|_p + 2\|z\|_p - \|x+y\|_p - \|x+z\|_p - \|y+z\|_p$ . No condition is placed on $x, y, z$ : they may be zero, equal or collinear. The constant $C(p)$ depends only on $p$ , so the same constant is asserted for every dimension $n$ . The only hypothesis is $p \ge 85$ , and it can be satisfied.

The size functional. Vectors are real $n$ -tuples $x = (x_1, \dots, x_n)$ , added coordinatewise. The size of a vector is given by this explicit formula, which uses powers with real exponents:

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

Because $p \ge 85 > 0$ , every power takes its usual value, including $0^p = 0$ and $0^{1/p} = 0$ . So this is the ordinary $\ell^p$ norm on $\mathbb{R}^n$ . When $n = 0$ the sum is empty and the only vector is the empty tuple, whose size is $0$ . Both sides of the inequality are then $0$ , so it reads $0 \le 0$ .

The constant. For real $t$ , put

A_p(t) = (t^p + 2)^{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)},

and

C(p) = \sup\big\{\, R_p(t) \;:\; \tfrac12 \le t \le 2 \,\big\}.

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

Edge cases. In the formal library, division by zero returns $0$ , and the supremum of a set of reals that is empty or unbounded above is defined to be $0$ . Neither convention comes into play here. For $\tfrac12 \le t \le 2$ we have $|1-t| \le 1$ , so

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

So the denominator $3\big(2A_p(t) - B_p(t)\big)$ is strictly positive. $R_p$ is therefore continuous on the closed interval, and $C(p)$ is a finite maximum that is actually reached.

For orientation only: evaluating the definition numerically gives $C(85) \approx 39.58$ , reached near $t \approx 0.945$ . This is not part of the statement.

Human review

Endorsed by marwahaha · Oct 5, 2026
Endorsed by savarin · Oct 5, 2026
Confirmed by the mission captain (proposal self-audit).
Endorsed by Shuze Chen · Oct 6, 2026
Confirmed by the moderator at approval.

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