Proposition 11 — integral of a product with a single-crossing factor
ProvedRandomOrderCRS.proposition_11contention-resolutionintegrationp2o-gran-per-chapterp2o-plan-paperp2o-v1real-analysis
Let be integrable, with and non-increasing. Suppose that there is such that for all and for all . Then
This elementary bound is applied in Lemma 4 to a function that changes sign, where a pointwise comparison is impossible.
Differences from the source:
- encoding: → functions on , with all hypotheses on and integrability as
IntervalIntegrable; reason: values outside do not enter
Preamble
import Mathlib
Formal statement
namespace RandomOrderCRS
/-- Proposition 11, MacRury, Ma & Grammel, arXiv:2209.07520v2, p.21: "Suppose that λ, φ : [0, 1] → ℝ
are integrable, λ ≥ 0, and λ is non-increasing. Moreover, assume that there exists 0 ≤ z_c ≤ 1 such
that φ(z) ≥ 0 for all z ∈ [0, z_c], and φ(z) ≤ 0 for all z ∈ [z_c, 1]. Then, ∫₀¹ λ(z) φ(z) dz ≥
λ(z_c) ∫₀¹ φ(z) dz."
**Formalization Note.** Integrability is `IntervalIntegrable`, so the integrals are not the junk
value `0`. The functions are defined on `ℝ`, and only their values on `[0, 1]` matter. -/
theorem proposition_11 (lam φ : ℝ → ℝ) (hlam : IntervalIntegrable lam MeasureTheory.volume 0 1)
(hφ : IntervalIntegrable φ MeasureTheory.volume 0 1)
(hlam0 : ∀ z ∈ Set.Icc (0 : ℝ) 1, 0 ≤ lam z) (hanti : AntitoneOn lam (Set.Icc (0 : ℝ) 1))
(zc : ℝ) (hzc : zc ∈ Set.Icc (0 : ℝ) 1) (hpos : ∀ z ∈ Set.Icc 0 zc, 0 ≤ φ z)
(hneg : ∀ z ∈ Set.Icc zc 1, φ z ≤ 0) :
lam zc * ∫ z in (0 : ℝ)..1, φ z ≤ ∫ z in (0 : ℝ)..1, lam z * φ z := by sorry
end RandomOrderCRS
Source
MacRury, Ma & Grammel, On (Random-order) Online Contention Resolution Schemes for the Matching Polytope of (Bipartite) Graphs, arXiv:2209.07520v2, Proposition 11, p.21
Human review
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.