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CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A twice-differentiable function on an open interval is convex if and only if its second derivative is nonnegative

Statement

If f:I→Rf:I\to\mathbb R is twice differentiable on an open interval, then ff is convex if and only if f′′(x)≥0f''(x)\ge0 for every x∈Ix\in I.

Facts & Assumptions

Given: A twice differentiable f:I→Rf:I\to\mathbb R on an open interval II.

[L1]

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

[L2]

If a differentiable real function has nonnegative derivative on an interval, then it is nondecreasing on that interval (On an interval II, for ff continuous on II and differentiable at every interior point: f′≥0f' \ge 0 throughout gives ff nondecreasing, f′>0f' > 0 gives ff increasing, f′≤0f' \le 0 and f′<0f' < 0 give the two decreasing forms; conversely a nondecreasing ff has f′≥0f' \ge 0 and a nonincreasing ff has f′≤0f' \le 0 wherever it is differentiable, and no strict converse is claimed).

Proof

technique · cases
1.1

Assume ff is convex. By [L1], f′f' is nondecreasing; its difference quotients on either side of a point are nonnegative, and their common limit f′′(x)f''(x) is therefore nonnegative.

assume-case forwardL1L2
1.2

Assume f′′≥0f''\ge0 on II. Applying [L2] to f′f' shows that f′f' is nondecreasing.

assume-case reverseL2algebra
2.1

By [L1], the conclusion of step 1.2 makes ff convex, while step 1.1 proves the reverse implication.

step 1.1step 1.2cases-exhaustive∎

Depends on

Used by

Dependency tree · next 3 levels

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Sources