False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: the mean value equality holds for vector-valued maps

Statement

False claim. Let $m \in N$ with $m \geq 1$ , let $a < b$ be real, and let $f : [a, b] \to R^{m}$ be continuous on $[a, b]$ and differentiable on $(a, b)$ . Then some $ξ \in (a, b)$ satisfies

$f (b) - f (a) = (b - a) f^{'} (ξ) .$

Facts & Assumptions

Given: The universal equality claim in the Statement.

[L1]

The curve $f (t) = (\cos t, \sin t)$ on $[0, 2 π]$ is continuous on its interval and differentiable in its interior, its endpoint increment is zero, and $∥ f^{'} (t) ∥_{2} = 1$ for every $t \in (0, 2 π)$ ; hence no $ξ \in (0, 2 π)$ satisfies the claimed equality (The circular curve defeats the equality form of the vector-valued mean value theorem).

[L2]

Let $m \in N$ with $m \geq 1$ , let $a < b$ be real, and let $M \geq 0$ be real. If $f : [a, b] \to R^{m}$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $∥ f^{'} (t) ∥_{2} \leq M$ throughout $(a, b)$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ (The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ).

Refutation

technique · direct

1.1

Fact [L1] supplies an instance with $m = 2$ , $a = 0$ , and $b = 2 π$ that satisfies both hypotheses of the false claim but not its conclusion.

2.1

Therefore the universal equality claim is false.

step 1.1

3.1

The failure does not affect the vector-valued mean value inequality: under its derivative-bound hypothesis, the estimate in [L2] remains valid.

L2∎

Remarks

When $m = 1$ , the scalar mean value theorem does give the equality. The circular curve shows that the passage to a vector codomain, not a loss of regularity, is what breaks it.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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

J. Lebl, Basic Analysis II, sections 8.3 and 11.4 (standard reference, not scraped)
University of Toronto MAT237, section 2.1 Differentiation of real-valued functions (standard reference, not scraped)
W. S. Hall and M. L. Newell, The Mean Value Theorem for Vector Valued Functions: A Simple Proof (standard reference, not scraped)