Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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.

A differentiable function on an open interval is convex if and only if its derivative is nondecreasing

Statement

For a differentiable f:I→R on an open interval I, f is convex if and only if f′ is nondecreasing on I.

Facts & Assumptions

Given: A differentiable f:I→R on an open interval I.

[L1]

For a convex function, the ordered one-sided-derivative chain holds; at a differentiability point its two one-sided derivatives equal the ordinary derivative (A convex function on an open interval has finite left and right derivatives everywhere, with f−′(u)≤f+′(u)≤(f(v)−f(u))/(v−u)≤f−′(v)≤f+′(v) for u<v).

[L2]

If a differentiable real function has nonnegative derivative on an interval, then it is nondecreasing on that interval (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).

Proof

technique · cases
1.1assume-case forwardL1

Assume f is convex. For u<v, [L1] becomes f′(u)≤f′(v), so f′ is nondecreasing.

1.2assume-case reverseL2algebra

Assume instead that f′ is nondecreasing. For a fixed c, the derivative of g(x)=f(x)−f′(c)x is nonpositive on the left of c and nonnegative on the right; apply [L2] to −g on the left and to g on the right to obtain f(x)≥f(c)+f′(c)(x−c) for every x∈I.

2.1step 1.1step 1.2cases-exhaustive∎

If z=λx+(1−λ)y, multiply the supporting inequalities of step 1.2 at z for x and y by λ and 1−λ and add; this gives f(z)≤λf(x)+(1−λ)f(y). Thus the two cases prove the equivalence.

Depends on

Dependency tree · 2 levels

Sources