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CorollaryStatement: AI-adaptedProof: AI-generatedprecheck 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→R is twice differentiable on an open interval, then f is convex if and only if f′′(x)≥0 for every x∈I.

Facts & Assumptions

Given: A twice differentiable f:I→R on an open interval I.

[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 I, for f continuous on I and differentiable at every interior point: f′≥0 throughout gives f nondecreasing, f′>0 gives f increasing, f′≤0 and f′<0 give the two decreasing forms; conversely a nondecreasing f has f′≥0 and a nonincreasing f has f′≤0 wherever it is differentiable, and no strict converse is claimed).

Proof

technique · cases
1.1assume-case forwardL1L2

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

1.2assume-case reverseL2algebra

Assume f′′≥0 on I. Applying [L2] to f′ shows that f′ is nondecreasing.

2.1step 1.1step 1.2cases-exhaustive∎

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

Depends on

Dependency tree · 2 levels

Sources