LemmaStatement: AI-generatedProof: 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.

The sharp coordinate constant already occurs inside the Bernoulli cube

Statement

Let $n \geq 2$ , $0 < p < 1$ , $d = (\binom{n}{2})$ , and let $T (x)$ be the symmetric matrix with zero diagonal and upper-triangular entries $x \in R^{d}$ . Let $J$ be the $n \times n$ matrix with every entry equal to $1$ . Set $f (x) = ∥ T (x) - p J ∥$ . On the cube $[0, 1]^{d}$ with its ordinary Euclidean coordinate metric, $f$ is convex and has least Lipschitz constant $L_{n} = \sqrt{2 (n - 1) / n} .$ In particular $f$ is not 1-Lipschitz when $n \geq 3$ .

Facts & Assumptions

[F1]

The zero-diagonal symmetric coordinate map gives a convex norm function and the upper Lipschitz bound $L_{n} = \sqrt{2 (n - 1) / n}$ (Independent-entry coordinates, convexity and Lipschitz constants).

[F2]

The operator norm on a Euclidean space is the supremum of the lengths of images of unit vectors (The operator norm is zero on the zero domain and otherwise is max_‖ v‖=1‖ Tv‖).

Proof

Given: The hypotheses and objects in the Statement.

Goal: Establish the asserted conclusions.

1.1F1givenconstruct

The upper bound and convexity follow directly from [F1], restricting its affine norm map to the cube. For sharpness choose $0 < s \leq p$ , take all coordinates of $x$ equal to $p$ , and all coordinates of $y$ equal to $p - s$ . Both points belong to the cube. The corresponding centered matrices are $- p I$ and $- p I - s (J - I)$ .

2.1F2step 1.1algebra∎

The second matrix has eigenvalue $- p - (n - 1) s$ on the constant vectors and eigenvalue $- p + s$ on their orthogonal complement. An orthogonal decomposition into these two spaces shows directly from [F2] that its norm is $p + (n - 1) s$ : the absolute value of its other eigenvalue is at most $p$ , since $s \leq p$ . The first matrix has norm $p$ . Therefore $∣ f (y) - f (x) ∣ / ∥ y - x ∥_{2} = (n - 1) s / (s \sqrt{d}) = L_{n}$ . This proves sharpness, and $L_{n} > 1$ exactly when $n > 2$ .

Depends on

Used by · 1

Convex adjacency norm: sharp coordinate constant and median counterexample

Dependency tree · 2 levels

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