CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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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 $f:I\to\mathbb R$ on an interval is a global minimum: $f(c)\le f(y)$ for every $y\in I$ .

Facts & Assumptions

Given: A convex $f:I\to\mathbb R$ on an interval and a local minimum $c\in I$ .

[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 $f(c)\le f(x)$ whenever $x\in I$ and $|x-c|<r$ (Local (relative) maximum and minimum of $f : A \to \mathbb{R}$ 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 $y\in I$ has $f(y)<f(c)$ .

assume-contraL1L2

2.1

Choose $0<\lambda<\min\{1,r/|y-c|\}$ when $y\ne c$ and put $z=(1-\lambda)c+\lambda y$ ; then $z\in I$ and $|z-c|<r$ .

step 1.1L2algebra

3.1

Convexity gives $f(z)\le(1-\lambda)f(c)+\lambda f(y)<f(c)$ , contradicting the local-minimum inequality.

step 1.1step 2.1discharge-contradiction∎

Depends on

Used by

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Sources

S. Boyd and L. Vandenberghe, Convex Optimization, §3.1 (standard reference, not scraped)