CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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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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A convex function on $[0, 1]$ that is discontinuous at the boundary

Statement refuted

Every convex real-valued function on a convex subset of Euclidean space is continuous on its whole domain.

The counterexample below establishes: The function is convex on $[0, 1]$ but is not continuous at $0$ .

Facts & Assumptions

Given: Define $f : [0, 1] \to R$ by $f (0) = 1$ and $f (x) = 0$ for $0 < x \leq 1$ .

[F1]

The function $f : C \to R$ is convex when $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y)$ for all $x, y \in C$ and $t \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

[L1]

Every convex function on an open convex set is continuous on that set (A convex function on an open convex set is continuous).

Counterexample

technique · direct

1.1

A convex combination of two domain points equals zero only when every endpoint having positive weight is zero. In that case [F1] is an equality. Otherwise the left side is zero and the right side is nonnegative, so [F1] again holds.

F1algebra

2.1

For every positive $x$ , $f (x) = 0$ , whereas $f (0) = 1$ , so the right-hand limit at zero is not the function value. The function is convex on $[0, 1]$ but is not continuous at $0$ . This does not contradict [L1], because the domain is not open at zero.

step 1.1L1∎

Depends on

Used by

FALSE: a convex function on a convex set is continuous False statement

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

A convex function on [0,1][0,1] that is discontinuous at the boundary