Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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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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Every local minimum of a convex function on an interval is a global minimum

Statement

Every local minimum c of a convex function f:I→R on an interval is a global minimum: f(c)≤f(y) for every y∈I.

Facts & Assumptions

Given: A convex f:I→R on an interval and a local minimum c∈I.

[L1]

A function is convex when the convex-combination inequality holds for every weight in [0,1] (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval).

[L2]

To have a local minimum at c means that some radius r>0 satisfies f(c)≤f(x) whenever x∈I and ∣x−c∣<r (Local (relative) maximum and minimum of f:A→R at a point, the strict forms, and what it means for the point to be interior to A).

Proof

technique · contradiction
1.1

Fix r>0 as in [L2], and suppose for contradiction that some y∈I has f(y)<f(c).

assume-contraL1L2
2.1

Choose 0<λ<min⁡{1,r/∣y−c∣} when y≠c and put z=(1−λ)c+λy; then z∈I and ∣z−c∣<r.

step 1.1L2algebra
3.1

Convexity gives f(z)≤(1−λ)f(c)+λf(y)<f(c), contradicting the local-minimum inequality.

step 1.1step 2.1discharge-contradiction∎

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