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.

Logarithmic tangent bound and a quadratic entropy envelope

Statement

For $a > 0$ , $\log a \leq a - 1$ , with equality only at $a = 1$ . Put $I (a) = a \log a - a + 1$ . For $0 < a < 1$ , $0 < I (a) < (1 - a)^{2}$ and $I (a) < 1 - a < - \log a$ . For $1 < a < e$ , $I (a) < \log a$ , where $e$ denotes the unique positive number with $\log e = 1$ .

Facts & Assumptions

Given: The variables and hypotheses of the Statement.

[F1]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm gives the order, range and continuity of $\log$ , including $\log 1 = 0$ .

[F2]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t gives $(\log a)' = 1 / a$ .

[F3]

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 converts a derivative sign on an interval into monotonicity.

[F4]

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 differentiability and the sum, scalar-multiple and product rules for differentiable functions on a common domain at a limit point of that domain.

[F5]

For a natural n ≥ 1 the function x ↦ xⁿ is differentiable everywhere with derivative ι(n) x^ n-1; for n = 0 it is the constant 1, with derivative 0; for a natural n ≥ 1 the function x ↦ x⁻ⁿ is differentiable at every x ≠ 0 with derivative -ι(n) x⁻ⁿ⁻¹; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, with exponents $0$ and $1$ , gives the derivatives $0$ and $1$ of the constant function $1$ and the identity function, respectively.

[F6]

A function differentiable at c is continuous at c gives continuity at every point where a function is differentiable and that point is a limit point of its domain.

Proof

Goal: Establish the full Statement.

1.1F1F2F3F4F5F6algebra

On $(0, \infty)$ every point is a limit point of the domain. By F2, F4 and F5, the function $q (a) = a - 1 - \log a$ is differentiable with $q' (a) = 1 - 0 - 1 / a = 1 - 1 / a$ , and F6 makes it continuous there. Its derivative is negative for $0 < a < 1$ and positive for $a > 1$ , and $q (1) = 0$ . Applying F3 on $(0, 1]$ and $[1, \infty)$ therefore gives $q (a) > q (1) = 0$ for $a \neq 1$ . Thus $\log a \leq a - 1$ , with equality exactly at $a = 1$ . The number $e$ exists uniquely by the range and strict increase of $\log$ , and $e > 1$ since $\log 1 = 0 < 1$ .

2.1F1F2F3F4F5F6step 1.1algebra∎

By F2, F4 and F5, the function $I (a) = a \log a - a + 1$ is differentiable on $(0, \infty)$ , with $I' (a) = \log a + a (1 / a) - 1 + 0 = \log a$ , and F6 makes it continuous there. Since $\log a < 0$ on $(0, 1)$ , F3 on $(0, 1]$ gives strict decrease and hence $I (a) > I (1) = 0$ for $0 < a < 1$ . In this range, $(1 - a)^{2} - I (a) = a (a - 1 - \log a) > 0$ by step 1.1, while $I (a) = a \log a + 1 - a < 1 - a$ and step 1.1 gives $1 - a < - \log a$ . For $1 < a < e$ , strict increase of $\log$ gives $0 < \log a < 1$ , so $\log a - I (a) = (a - 1) (1 - \log a) > 0$ . These are all asserted bounds.

Depends on

Used by · 6

Dependency tree · 2 levels

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