Keyboard shortcuts

Press ← or → to navigate between chapters

Press ? to show this help

Press Esc to hide this help


slug: nikolov-savov-2023-renewal-minorant-maximality-refutation bibkey: nikolovsavov2024renewal doi: 10.53656/math2024-2-1-pro url: https://arxiv.org/abs/2307.00545v2 triage: theorem motivation_gids:

  • D5/S3/StatisticalMechanics/RandomWalks/RenewalMinorantMaximalityRefutation.result

Refutation of Nikolov–Savov Conjecture 3.7 at k = 4

Problem

N. Nikolov and M. Savov, Properties and conjectures regarding discrete renewal sequences, arXiv:2307.00545v2, §3, Conjecture 3.7, states:

For any k ≥ 3, Q_k is a maximal element in 𝒜_k and it is largest in 𝒜̂_k.

The formal claim settles the first clause:

claim := ∀ k ≥ 3, Q_k ∈ 𝒜_k ∧ ∀ P ∈ 𝒜_k, Q_k ≺ P → P = Q_k.

Here A_k is the simplex of nonnegative first k−1 masses with sum at most one, u is the renewal sequence of Eq. (2.2), Q_k is the product of partial sums in Eq. (2.5), and 𝒜_k is the degree-bounded class of polynomial lower bounds in Eq. (2.7). The hatted class is not used.

Motivation

Issue #12306 preregisters the named conjecture, its quantified claim, and the k = 4 witness. The motivation declaration is the frozen Lean result D5/S3/StatisticalMechanics/RandomWalks/RenewalMinorantMaximalityRefutation.result.

Gap

The source proves the k = 3 case in Proposition 3.8 and leaves the universal maximality assertion open. The bounded literature and repository checks recorded for issue #12306 found no prior proof or refutation of Conjecture 3.7 in the searched scope.

Route

At k = 4 write x = p₁, y = p₂, z = p₃ and w = 1 − x − y − z. Let

P = x²(x + y + z) + xy.

Then P − Q₄ = xyw, so Q₄ ≺ P on A₄, while P has total degree three and differs from Q₄. The renewal masses are u₁ = x, u₂ = x² + y, and u₃ = x³ + 2xy + z. The certificates

u₁ − P = x(z + (1+x)w)
u₂ − P = x²w + y(1−x)
u₃ − P = xy(1−x) + z(1−x²)

are nonnegative on A₄; P ≤ u₁ ≤ u₀ = 1. For n ≥ 4 the renewal recurrence is a convex combination of the preceding four masses, so strong induction gives P ≤ u_n for every n ≥ 1. This proves P ∈ 𝒜₄ and contradicts maximality of Q₄.

Falsifier

A failure of any certificate, of the all-n induction, of the degree bound, of Q₄ ≺ P, or of P ≠ Q₄ would invalidate the refutation. The result does not assert that 𝒜₄ has no maximal element and does not address the largest-in-𝒜̂_k clause.

Evidence

The Lean module is D5/S3/StatisticalMechanics/RandomWalks/RenewalMinorantMaximalityRefutation.lean. Its public settlement is result : ¬ claim; the definitions expose the source simplex, step masses, renewal recurrence, polynomial order, minorant class, product Q, and claim. The all-n minorant argument is the escape witness. Lean checks the theorem with axiom closure [propext, Classical.choice, Quot.sound]. The Scribe result carries an OpenProblemResolutionClaim marked Refuted for this problem.

Triage

theorem; resolution Refuted at k = 4; preregistration #12306. proof_shape: content for result; admission_basis: open-problem-resolution. Information-escape registration is paused under CLAUDE.md section 3.9.

What the settlement shows

  • Proved in this module: Q₄ is not a maximal element of 𝒜₄; the strict improvement is the polynomial xyw, and the all-n renewal minorant bound is supplied by the three certificates and strong induction.
  • Computed: the improvement xyw vanishes on each face x = 0, y = 0, and w = 0, and is positive at the interior point x = y = z = w = 1/4; there P = 7/64 and Q₄ = 6/64.
  • Proved: the k = 3 case in Proposition 3.8 survives this mechanism: at k = 3 the degree bound is two, while the analogous three-coordinate interior factor would require the degree-three term xy(1−x−y). This module does not reprove Proposition 3.8.
  • Open: whether 𝒜₄ has any maximal element remains unsettled. Refuting maximality of Q₄ does not settle existence of another maximal minorant.
  • Open: the separate assertion that Q_k is largest in 𝒜̂_k is not used and remains unsettled.
  • Proved boundary: source results that rely specifically on the universal maximality of Q_k cannot use that clause at k = 4 without an additional hypothesis or replacement minorant. The k = 3 proposition and results independent of Conjecture 3.7 are unaffected by this refutation.

ASSUMED-UNVERIFIED

The literature search is bounded by the sources and queries recorded in issue #12306; it does not establish exhaustive absence of a later settlement. The identification of the Lean definitions with the source notation is documented in the Library note and the Scribe commentary; the hatted class is intentionally outside this result.