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.

Inward comparison and small-shift limits of entropy roots

Statement

Under $0 < ε < 1$ and $1 + ε < γ < 1 + ε^{- 2}$ , put $u = 1 / γ$ , $t = (1 + ε) / γ$ , $c_{\pm} (γ) = a_{\pm} (u) - 1$ , $c_{-} (γ, ε) = a_{-} (t) - 1 + ε$ , and $c_{+} (γ, ε) = a_{+} (t) - 1 - ε$ . Then $- 1 < c_{-} (γ) < c_{-} (γ, ε) < 0 < c_{+} (γ, ε) < c_{+} (γ)$ . For each fixed $γ > 1$ , $c_{\pm} (γ, ε) \to c_{\pm} (γ)$ as $ε ↓ 0$ within the admissible range.

Facts & Assumptions

Given: The variables and hypotheses of the Statement.

[F1]

Logarithmic tangent bound and a quadratic entropy envelope gives $- \log x > I (x)$ for $0 < x < 1$ and $\log y > I (y)$ for $1 < y < e$ .

[F2]

Entropy endpoint limits and the two monotone branches gives the strict branch order and $I (e) = 1$ , where $\log e = 1$ .

[F3]

Differentiable entropy inverse branches and shifted contracts gives the continuous inverse functions $a_{\pm}$ and the shifted solution formulas.

[F4]

Parameter bounds giving the signs of shifted entropy roots gives the signs and admissibility of the two shifted solutions.

[F5]

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 gives monotonicity from a derivative bound on a closed interval.

Proof

Goal: Establish the full Statement.

1.1F1F2F5algebra

Put $x = a_{-} (u) \in (0, 1)$ . If $x \leq ε$ , then $a_{-} (t) + ε > ε \geq x$ . If $x > ε$ , on $[x - ε, x]$ the derivative of $I (z) - z \log x$ is $\log z - \log x \leq 0$ . Thus $I (x - ε) \geq I (x) - ε \log x > (1 + ε) I (x) = t$ . Since $I$ strictly decreases on its lower branch, $x - ε < a_{-} (t)$ , again giving $x < a_{-} (t) + ε$ . Therefore $c_{-} (γ) < c_{-} (γ, ε)$ .

2.1F1F2F4F5step 1.1algebra

Put $y = a_{+} (u)$ . Since $u < 1 = I (e)$ , $1 < y < e$ . On $[y, y + ε]$ the derivative of $I (z) - z \log y$ is $\log z - \log y \geq 0$ . Therefore $I (y + ε) \geq I (y) + ε \log y > (1 + ε) I (y) = t$ . Strict increase on the upper branch gives $a_{+} (t) < y + ε$ , hence $c_{+} (γ, ε) < c_{+} (γ)$ . Together with F4 and $x > 0$ this is the entire strict comparison.

3.1F3algebra∎

For fixed $γ > 1$ , all sufficiently small positive $ε$ obey the hypotheses: take $ε < \min {1, γ - 1, (γ - 1)^{- 1 / 2}}$ . Then $t \to u \in (0, 1)$ , and continuity of both inverses gives $a_{\pm} (t) \to a_{\pm} (u)$ . The additive shifts tend to zero, proving both limits.

Depends on

Used by · 1

Entropy rates, inverse branches and shifted root contracts

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.