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.

Vanishing numerical condition for fixed positive Laplacian margins

Statement

Fix $c^{-} > - 1$ with $c^{-} < 0$ , and $c^{+} > 0$ . For $d \to \infty$ and $α = d^{- 1 / 4}$ , eventually $0 < α \leq 1 / 5$ and $\frac{32 α (1 + 4 c^{+} - 3 c^{-})}{(1 + c^{-})^{2}} \log ⁣ (\frac{1 + c^{+} + α}{2 α}) < 1.$ In particular $α \log (1 / α) \to 0$ .

Facts & Assumptions

[F1]

For every $q > 0$ , $\log d / d^{q} \to 0$ (The logarithm grows more slowly than every positive real power). The exponential maps $R$ bijectively onto $(0, \infty)$ , with inverse logarithm (The natural logarithm as the inverse of the exponential function); logarithm product identities follow by applying this inverse to The exponential addition formula exp(x+y)=exp(x)exp(y).

[F2]

For $a > 0$ and $x \in R$ , real powers mean $a^{x} : = \exp (x \log a)$ (Real powers for positive bases, with the zero-base positive-exponent convention).

[F3]

On $(0, \infty)$ the logarithm is differentiable with $\log' (x) = 1 / x$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t). Differentiability at a limit point implies continuity there (A function differentiable at c is continuous at c). A continuous function on an interval with strictly positive derivative at every interior point is strictly increasing (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).

[F4]

Limits at $+ \infty$ have the real-tolerance meaning in Limits at +∞ and -∞, and infinite limits at a point: $f (d) \to 0$ means that for every $η > 0$ there is a threshold beyond which $∣ f (d) ∣ < η$ .

Proof

Given: The fixed constants and $d \to \infty$ .

Goal: Prove the numerical inequality and the stated limit.

1.1F1F2givenalgebra

Eventually $d > 0$ . Positivity of the exponential and its addition formula give $\exp (0) = 1$ and $\exp (- t) = 1 / \exp (t)$ . Applying the inverse logarithm to $\exp (\log u + \log v) = u v$ gives $\log (u v) = \log u + \log v$ for $u, v > 0$ , and $\log 1 = 0$ . By [F2], $α = \exp (- \frac{1}{4} \log d) > 0$ , $1 / α = d^{1 / 4} = \exp (\frac{1}{4} \log d)$ , and $\log (1 / α) = \frac{1}{4} \log d$ . Put $β (d) = \log d / d^{1 / 4}$ , so $β (d) \to 0$ by [F1] with $q = 1 / 4$ , and $α \log (1 / α) = β (d) / 4$ .

2.1F1F3F4step 1.1

Every point of $(0, \infty)$ is a limit point and interior point of this interval. By [F3], the logarithm is continuous there and its derivative $1 / x$ is strictly positive, so the monotonicity theorem applies. For $d \geq \exp (1)$ , it gives $\log d \geq \log (\exp (1)) = 1$ , and hence $0 < α \leq α \log d = β (d)$ . For every $η > 0$ , step 1.1 and [F4] give $β (d) < η$ eventually; thus $0 < α < η$ eventually. This proves $α \to 0$ , and in particular eventually $0 < α \leq 1 / 5$ .

3.1F1step 1.1step 2.1givenalgebra

Set $K = 32 (1 + 4 c^{+} - 3 c^{-}) / (1 + c^{-})^{2}$ and $C = \log ((2 + c^{+}) / 2)$ . These are finite fixed positive constants: $- 1 < c^{-} < 0$ , $c^{+} > 0$ , and monotonicity gives $C > \log 1 = 0$ . For $0 < α \leq 1 / 5$ , the ratio $(1 + c^{+} + α) / (2 α)$ exceeds one, and, since $α \leq 1$ , logarithm monotonicity and the product identity give $0 < \log ((1 + c^{+} + α) / (2 α)) \leq C + \log (1 / α) = C + \frac{1}{4} \log d$ . Thus for sufficiently large $d$ , step 2.1 yields $0 < K α \log ((1 + c^{+} + α) / (2 α)) \leq K (C α + β (d) / 4) \leq K (C + 1 / 4) β (d)$ .

4.1F4step 1.1step 2.1step 3.1∎

For any $η > 0$ , step 1.1 and [F4] give $β (d) < η / (K (C + 1 / 4))$ eventually. The bound in step 3.1 therefore makes the numerical expression less than $η$ eventually; taking $η = 1$ proves the strict inequality. Also, for any $η > 0$ , eventually $0 \leq β (d) < 4 η$ , so the identity in step 1.1 gives $∣ α \log (1 / α) ∣ < η$ . By [F4] this proves the stated final limit.

Depends on

Used by · 1

Erdős–Rényi global synchronization — Theorem 1.2

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.