LemmaStatement: AI-adaptedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Cubic estimates used for the order parameter bootstrap

PDF p. 15: Proof of Lemma 2.5, equation (2.11); PDF p. 16: Proof of Lemma 2.5, equation (2.12)

Statement

Assume countable choice. Let $G = (V, E)$ be a finite connected undirected graph, possibly with loops, with $n = ∣ V ∣ > 0$ and symmetric binary adjacency matrix $A$ . Let $d^{*} \geq 0$ and $0 \leq α \leq 1 / 5$ , and suppose $∥ A - (d^{*} / n) J ∥ \leq α d^{*}$ , where $J$ is the all-ones matrix and the norm is the Euclidean operator norm. Let $θ$ be a stable state, meaning a local minimum on the phase torus of $\frac{1}{2} \sum_{x, y} A_{x y} (1 - \cos (θ_{x} - θ_{y}))$ . Put $ρ_{k} = n^{- 1} \sum_{x} e^{i k θ_{x}}$ and rotate all phases together so that $ρ_{1}$ is real and nonnegative. Set $X = ρ_{1}^{2}$ and, for $x > 0$ , $f (x, α) = 2 x + 4 α - 1 - (1 - 2 α^{2} / x)^{2}$ .

Then $X > 0$ , $2 α^{2} / X < 1$ , $f (X, α) \geq 0$ , and $X > 0.431$ . For $x > 0$ , $f (x, α) = 0$ is equivalent to $P_{α} (x) = 0$ , where $P_{α} (x) = x^{3} + (2 α - 1) x^{2} + 2 α^{2} x - 2 α^{4}$ . Its discriminant, defined by the coefficient formula in Root and sign certificates for parameter-dependent polynomials, is $- 4 α^{4} (27 α^{4} + 20 α^{3} + 10 α^{2} - 8 α + 1)$ . It is negative for $0 < α \leq 0.2055$ and zero at $α = 0$ . For $0 < α \leq 1 / 5$ , the cubic has exactly one real root; at $α = 0$ its distinct real roots are $0$ and $1$ , while $f (\cdot, 0)$ has the unique positive root $1$ . Also $f (0.431, α) < 0$ throughout $0 \leq α \leq 1 / 5$ .

Facts & Assumptions

[F1]

Under the stated graph and stability hypotheses, $X \geq (1 + ∣ ρ_{2} ∣^{2}) / 2 - 2 α$ ; the phase rotation preserves the moduli of the order parameters (First and second order parameter inequality — Lemma 2.1, Bounds, saturation, and rotation of order parameters).

[F2]

For $α < 1 / 4$ , the normalized $ρ_{1}$ is positive; when it is positive, $∣ ρ_{2} ∣ \geq 1 - 2 α^{2} / X$ (Positive first order parameter — Corollary 2.2, Lower bound for the second order parameter — Corollary 2.4).

[F3]

Polynomials are continuous, and the intermediate value theorem and derivative monotonicity criterion apply on real intervals under their stated hypotheses (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b] takes every value between f(a) and f(b), On an interval I, for f continuous on I and differentiable at every interior point: f' ≥ 0 throughout gives f nondecreasing, f' > 0 gives f increasing, f' ≤ 0 and f' < 0 give the two decreasing forms; conversely a nondecreasing f has f' ≥ 0 and a nonincreasing f has f' ≤ 0 wherever it is differentiable, and no strict converse is claimed). The coefficient formula for the cubic discriminant is given in Root and sign certificates for parameter-dependent polynomials.

Proof

Given: The graph, stable state, normalization, and parameter range in the Statement, under countable choice.

Goal: Establish the order-parameter estimates and the stated polynomial assertions.

1.1F1F2givenalgebra

By [F2], $X > 0$ . If $X \leq 2 α^{2}$ , then [F1] gives $2 α^{2} \geq X \geq 1 / 2 - 2 α$ . But $2 α^{2} + 2 α \leq 2 / 25 + 2 / 5 = 12 / 25 < 1 / 2$ , a contradiction. Thus $1 - 2 α^{2} / X > 0$ . Squaring the lower bound in [F2] is therefore valid; substituting it in [F1] gives $2 X + 4 α - 1 \geq (1 - 2 α^{2} / X)^{2}$ , or $f (X, α) \geq 0$ .

1.2F3givenalgebra

Multiplication by $x^{2} / 2 > 0$ gives $x^{2} f (x, α) / 2 = P_{α} (x)$ for $x > 0$ . Substitution of the coefficients of $P_{α}$ into the discriminant formula in [F3] gives the displayed expression. Write $q (a) = 27 a^{4} + 20 a^{3} + 10 a^{2} - 8 a + 1$ . For $0 \leq a \leq 411 / 2000 < 21 / 100$ , $q^{'} (a) \leq 108 (21 / 100)^{3} + 60 (21 / 100)^{2} + 20 (21 / 100) - 8 < 0$ . Hence $q (a) \geq q (411 / 2000) = 327454507 / 16000000000000 > 0$ by [F3]. The discriminant therefore has the asserted signs.

2.1F3step 1.2givenalgebra

For $0 < α \leq 1 / 5$ , $P_{α} (0) = - 2 α^{4} < 0$ and $P_{α} (x) \to + \infty$ as $x \to + \infty$ , so [F3] gives a positive real root $r$ . Divide by $x - r$ to write $P_{α} (x) = (x - r) (x^{2} + b x + c)$ with real $b, c$ . Direct expansion of the discriminant formula gives $Δ = (b^{2} - 4 c) (r^{2} + b r + c)^{2}$ . Since step 1.2 gives $Δ < 0$ , $b^{2} - 4 c < 0$ , so the quadratic has no real root by completing the square. Thus $r$ is the unique real root. At $α = 0$ , $P_{0} (x) = x^{2} (x - 1)$ and $f (x, 0) = 2 x - 2$ for $x > 0$ , giving the stated endpoint roots.

3.1F3step 1.1step 1.2step 2.1givenalgebra∎

Put $u = 431 / 1000$ . For $0 \leq a \leq 1 / 5$ , $1 - 2 a^{2} / u \geq 1 - (2 / 25) / u > 0$ , so $\partial f (u, a) / \partial a = 4 + (8 a / u) (1 - 2 a^{2} / u) > 0$ . By [F3], $f (u, a) \leq f (u, 1 / 5) = 2 P_{1 / 5} (u) / u^{2} < 0$ , because $P_{1 / 5} (u) = - 113609 / 1000000000$ . For $a > 0$ , step 2.1 and the intermediate value theorem imply that $P_{a}$ is negative to the left of its unique root: otherwise continuity would produce another root between that point and $u$ or between $u$ and the existing root. Since $f (X, a) \geq 0$ and $f (u, a) < 0$ , $X$ lies to the right of $u$ . For $a = 0$ , step 1.1 gives $f (X, 0) = 2 X - 2 \geq 0$ , so $X \geq 1 > u$ . This proves all claims.

Source notes

The cited article states that the discriminant is strictly negative for $α < 0.2055$ . At the allowed endpoint $α = 0$ the displayed discriminant is zero. The proof above treats that endpoint separately; this correction does not by itself refute the order-parameter conclusion.

Depends on

Used by · 1

Tractable order parameter inequalities — Lemma 2.5

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.

Sources

Expander graphs are globally synchronizing · PDF pp. 15–16, proof of Lemma 2.5, equations (2.11) and (2.12)