TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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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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A convex function on an open interval is differentiable except at at most countably many points

Statement

If $f : I \to R$ is convex on an open interval, then the set of points at which $f$ is not differentiable is at most countable.

Facts & Assumptions

Given: A convex $f : I \to R$ on an open interval.

[L1]

A convex function on an open interval has finite left and right derivatives everywhere, and for $u < v$ they satisfy the ordered one-sided-derivative chain (A convex function on an open interval has finite left and right derivatives everywhere, with $f_{-}^{'} (u) \leq f_{+}^{'} (u) \leq (f (v) - f (u)) / (v - u) \leq f_{-}^{'} (v) \leq f_{+}^{'} (v)$ for $u < v$ ).

[L2]

The discontinuity set of a monotone real function on an interval is at most countable (Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $N$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used).

[L3]

A convex real function is continuous throughout an open interval on which it is convex (A convex real function is Lipschitz on every closed bounded subinterval of the interior of its domain, hence continuous throughout the interior).

Proof

technique · direct

1.1

Put $g (c) : = f_{+}^{'} (c)$ . The order chain in [L1] gives $g (u) \leq g (v)$ whenever $u < v$ , so $g$ is nondecreasing.

1.2

For $u < c$ , [L1] gives $g (u) \leq f_{-}^{'} (c)$ . Conversely, if $a < u < c$ , [L1] gives $s (a, u) \leq f_{-}^{'} (u) \leq g (u)$ . Letting $u \to c^{-}$ , continuity from [L3] gives $\lim_{u \to c^{-}} s (a, u) = s (a, c)$ ; then letting $a \to c^{-}$ gives $\lim_{u \to c^{-}} g (u) = f_{-}^{'} (c)$ . Since $g (c) = f_{+}^{'} (c)$ , nondifferentiability of $f$ at $c$ makes $g$ discontinuous there.

L1L3algebra

2.1

Froda's theorem makes the discontinuity set of $g$ at most countable, and step 1.2 places every nondifferentiability point of $f$ in that set.

L2step 1.1step 1.2∎

Depends on

Used by

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Sources

R. Gardner, Convex Functions, Notes 6.6 (standard reference, not scraped)