The sup-norm growth inequality via the orthant pigeonhole (fully proven) #
For best simultaneous Diophantine approximation in the sup norm on ℝ^d, the denominators at
least double every 2^d steps:
2 q_N ≤ q_{N + 2^d}.
This is the elementary half of Shutov's L^∞ growth inequality (the full q_{N+2^d} ≥ q_{N+1}+q_N
is Lagarias 1980; the doubling form here already suffices for the gap bound, via
ChevallierGapBound.gap_count_doubling). The proof is the orthant pigeonhole of the Chevallier
survey (§2.4.1, attributed to Lagarias):
- Among the
2^d + 1remainder vectorsε_N, …, ε_{N+2^d}, two share an orthant (a sign pattern in{≥0, <0}^d), since there are only2^dorthants —Fintype.exists_ne_map_eq_of_card_lt. - Two vectors in the same orthant, each of sup-norm
≤ r_N = δ_{q_N}, have difference of sup-norm≤ r_N(coordinatewise, same sign ⟹|a−b| ≤ max(|a|,|b|)). - That difference is the remainder of
m = q_{N+k₂} − q_{N+k₁}(SimApprox.delta_diff_le), soδ_m ≤ r_N; the best-approximation/record property then forcesm ≥ q_N, giving the doubling.
No convex geometry, no kissing number — purely the pigeonhole. Axiom-clean.
Same-orthant sup-norm bound. If u, v : Fin d → ℝ lie in the same orthant
(0 ≤ u k ↔ 0 ≤ v k for every k) and each has sup-norm ≤ r, then ‖u − v‖ ≤ r.
The sup-norm growth inequality, doubling form (orthant pigeonhole). Suppose q enumerates
best-approximation denominators of α in the sup norm, with remainders p attaining the defect
(hattain), defects non-increasing along the sequence (hdec), and each q k a record minimum
(hbsad: every smaller positive denominator has strictly larger defect). Then
2 q_N ≤ q_{N + 2^d}.
This feeds ChevallierGapBound.gap_count_doubling with K = 2^d to give g_∞ ≤ 2^d + 1.
g_∞ ≤ 2^d + 1 from the orthant pigeonhole + Chevallier's Lemma. Assembling
supNorm_growth_doubling (the fully-proven sup-norm growth, K = 2^d) with the doubling gap-count
reduction, the number of distinct nearest-neighbour distances of a sup-norm best-approximation
sequence is at most 2^d + 1. The only remaining hypothesis is Chevallier's Lemma (hg: the
g = n − m / n − m + 1 index dictionary) — the growth geometry is now entirely elementary
(orthant pigeonhole), with no kissing number or convex-geometry input.