Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck pass
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 finite Rényi generators are convex with stated boundary values

Statement

For α>1, the generator hα(t)=tα−1 is convex on (0,∞), satisfies hα(1)=0, has hα(0+)=−1, and has recession limit lim⁡t→∞hα(t)/t=+∞. For 0<α<1, the generator gα(t)=1−tα is convex on (0,∞), satisfies gα(1)=0, has gα(0+)=1, and has recession limit lim⁡t→∞gα(t)/t=0.

Facts & Assumptions

Given: The positive-base power functions in the Statement and the corresponding exponent range.

[F1]
[F3]

A twice-differentiable function on an open interval is convex if and only if its second derivative is nonnegative and Convex and strictly convex functions on Euclidean convex sets: a twice differentiable function on an open interval is convex when its second derivative is nonnegative there.

[F5]

The exponential tends to +∞ at +∞ and to 0 at −∞: exp⁡u→+∞ as u→+∞, and exp⁡u>0.

[F6]

The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x): exp⁡(−u)=1/exp⁡(u) for every real u.

Proof

1.1F1F4F5givenalgebra

Let γ>0. As t→+∞, log⁡t→+∞ because [F4] is increasing and onto. Hence γlog⁡t→+∞, so tγ=exp⁡(γlog⁡t)→+∞ by [F1, F5].

1.2F1F4F5F6givenalgebra

Let γ>0 and put s=1/t. As t↓0, s→+∞. By [F1], [F4], and [F6], tγ=exp⁡(−γlog⁡s)=1/exp⁡(γlog⁡s)→0 by [F5].

2.1F1F2F3step 1.1step 1.2algebra

For α>1, [F2] gives hα′′(t)=α(α−1)tα−2>0, so [F3] proves convexity; also hα(1)=0. By step 1.2, hα(0+)=0α−1=−1. By step 1.1 with γ=α−1, tα−1→+∞, and 1/t≤1 for t≥1, so hα(t)/t=tα−1−1/t→+∞.

3.1F1F2F3step 1.2constructalgebra∎

For 0<α<1, [F2] gives gα′′(t)=α(1−α)tα−2>0, so [F3] proves convexity; also gα(1)=0. By step 1.2, gα(0+)=1. Put u=1/t. As t→+∞, u↓0, and step 1.2 with γ=1−α gives u1−α→0, that is, t−(1−α)→0. For t≥1, 0≤∣gα(t)/t∣=t−(1−α)−1/t≤t−(1−α). Therefore gα(t)/t→0.

Depends on

Used by

Dependency tree · two levels

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Sources