Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21
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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.

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Differentiable convex functions are characterized by the gradient inequality

Statement

Let U⊆Rn be open and convex, and let f:U→R be differentiable. Then f is convex if and only if

f(y)≥f(x)+⟨∇f(x),y−x⟩(x,y∈U).

Equivalently, ∇f(x) is a subgradient at every x (Subgradients and the subdifferential of a convex function, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

Facts & Assumptions

Given: The domain and differentiable function in the Statement, with convexity from Convex and strictly convex functions on Euclidean convex sets.

[L1]

The total derivative of a composite is the composite of the total derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Proof

technique · direct
1.1L1givenalgebra

For the forward implication, fix x,y∈U and put d=y−x. For 0<t≤1, convexity gives f(x+td)≤(1−t)f(x)+tf(y), hence f(x+td)−f(x)t≤f(y)−f(x). By [L1] the left side tends to ⟨∇f(x),d⟩ as t↓0, giving the displayed gradient inequality.

2.1assume-hypalgebra∎

For the reverse implication, assume the gradient inequality and take z=(1−t)x+ty. Apply it at z toward x and toward y, multiply the results by 1−t and t, and add. The gradient terms cancel because (1−t)(x−z)+t(y−z)=0, leaving f(z)≤(1−t)f(x)+tf(y). Thus f is convex.

Depends on

Used by

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Dependency tree · two levels

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Sources