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

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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

Proof

technique · cases
1.1

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

assume-case forwardL1
1.2

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.

assume-case reverseL2algebra
2.1

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.

step 1.1step 1.2cases-exhaustive∎

Depends on

Used by

Dependency tree · two levels

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Sources