Alphabeta Math
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.

Entropy endpoint limits and the two monotone branches

Statement

For a>0, let I(a)=alog⁡a−a+1. Then I′(a)=log⁡a, I′′(a)=1/a>0, I(1)=0, lim⁡a↓0I(a)=1, and I(a)→+∞ as a→+∞. The function strictly decreases on (0,1) with range (0,1) and strictly increases on (1,∞) with range (0,∞). Consequently I(a)=t has exactly two positive solutions for 0<t<1, one in each branch.

Facts & Assumptions

Given: The variables and hypotheses of the Statement.

[F1]

Logarithmic tangent bound and a quadratic entropy envelope gives log⁡y≤y−1 for y>0.

[F2]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm gives continuity, range, strict monotonicity, the product and reciprocal laws, and log⁡1=0.

[F5]

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) attains every value between the endpoint values of a continuous function on a closed real interval.

[F6]

Sums, scalar multiples, products and quotients: (f+g)'(c) = f'(c) + g'(c), (α f)'(c) = α f'(c), (fg)'(c) = f'(c)g(c) + f(c)g'(c), and (f/g)'(c) = (f'(c)g(c) - f(c)g'(c))/g(c)² when g(c) ≠ 0 gives the sum, scalar-multiple and product rules for functions differentiable on a common domain at a limit point of that domain.

[F8]

A function differentiable at c is continuous at c gives continuity at every limit point of the domain where the function is differentiable.

[F9]

Square roots exist: a unique √a ≥ 0 with (√a)² = a; the positives are {x² : x ≠ 0} gives, for each a>0 in R, the unique nonnegative s=a with s2=a; necessarily s>0.

[F10]

The left and right limits of f at c, as limits of the restrictions of f to A ∩ (-∞, c) and A ∩ (c, ∞) defines the right limit at 0 by the condition that for every ε>0 there is δ>0 such that 0<a<δ implies ∣I(a)−1∣<ε. The point 0 is a limit point of (0,∞).

[F11]

Limits at +∞ and -∞, and infinite limits at a point defines divergence to +∞ at +∞ by requiring, for every real B, a real M such that a>M implies I(a)>B. The domain (0,∞) is unbounded above.

[F12]

The derivative f'(c) = lim_x → c f(x) - f(c)/x - c of f : A → ℝ at a point c ∈ A that is a limit point of A, and differentiability on a set defines the derivative at a domain limit point by the limit of the difference quotient, and differentiability on a set pointwise. Here I′′ denotes the derivative of I′.

[F13]

Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of ℝ, with the dictionary to monotone sequences defines strict increase on a domain by x<y⇒f(x)<f(y) and strict decrease by x<y⇒f(x)>f(y); either condition makes the function injective.

[F14]

Continuity of f : A → ℝ at a point of A and on A: the ε-δ condition, its agreement with lim_x → c f(x) = f(c) at a limit point, and continuity at an isolated point defines continuity at c by requiring, for every ε>0, a δ>0 such that domain points x with ∣x−c∣<δ satisfy ∣f(x)−f(c)∣<ε. Continuity on a domain is pointwise continuity, and passes to restrictions of that domain.

Proof

Goal: Establish the full Statement.

1.1F2F3F4F6F7F8F12F13F14algebra

Every point of (0,∞) is a limit point of that domain. The logarithm, the identity and the constant function 1 are differentiable there by F3 and F7. Applying F6 gives I′(a)=log⁡a+a(1/a)−1+0=log⁡a, and F3 then gives I′′(a)=1/a>0. By F8, I is continuous on (0,∞). Since log⁡a<0 for 0<a<1 and log⁡a>0 for a>1, F4 applies on (0,1] and [1,∞), giving strict decrease and strict increase, respectively. Substitution gives I(1)=0.

1.2F1F2F9F10algebrachoose

For 0<a<1, let s=a>0 by F9 and put y=1/s. Since s2=a<1, one has s<1 and y>1. The logarithm laws give log⁡a=2log⁡s=−2log⁡y. Hence F1 and F2 give 0≤−alog⁡a=2alog⁡y≤2a(y−1)≤2s. Given ε>0, take δ=min⁡{1,ε/2,ε2/16}>0. If 0<a<δ, then s2<ε2/16, and the factorization s2−(ε/4)2=(s−ε/4)(s+ε/4) gives s<ε/4. Therefore ∣I(a)−1∣=−alog⁡a+a≤2s+a<ε. This proves the right limit 1 at 0 directly by F10.

1.3F2F11algebrachoose

By the range of log⁡, choose A>0 with log⁡A=2. For every real B, put M=max⁡{A,B,1}. If a>M, monotonicity of log⁡ gives log⁡a≥2, so I(a)≥a+1>B. This is divergence to +∞ at +∞ by F11.

2.1F2F5F13F14step 1.1step 1.2step 1.3choose∎

By step 1.1, I(a)>0 on both branches. If 0<a<1, then log⁡a<0 gives I(a)=1+alog⁡a−a<1. For 0<t<1, the right limit in step 1.2 supplies u∈(0,1) with I(u)>t, while continuity at 1 supplies v∈(u,1) with I(v)<t. The restriction of I to [u,v] is continuous, so F5 supplies a solution in (u,v). For t>0, continuity at 1 supplies u>1 with I(u)<t, and step 1.3 supplies v>u with I(v)>t; F5 on [u,v] again supplies a solution. The strict monotonicity in step 1.1 makes the solution unique on each branch. Thus the branch ranges are exactly (0,1) and (0,∞), and for 0<t<1 there are exactly two positive solutions, since I(1)=0.

Depends on

Used by · 5

Dependency tree · 2 levels

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