CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

How statement and proof provenance work

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.

The half-norm fails the triangle inequality on two indicators

Statement refuted

On every measure space, the functional $∥ \cdot ∥_{1 / 2}$ satisfies the triangle inequality.

Facts & Assumptions

Given: The $0 < p < 1$ non-norm proposition The $p$ -functional need not be a norm for $0 < p < 1$ .

[L1]

The proof of The $p$ -functional need not be a norm for $0 < p < 1$ uses two disjoint equal-mass indicators to violate the triangle inequality.

Counterexample

Proof technique: Take two disjoint indicators of equal positive measure. Then the $1 / 2$ -functional of the sum exceeds the sum of the two $1 / 2$ -functionals.

1.1

On $[0, 1]$ with Lebesgue measure, let [L1, given, algebra] $f : = χ_{[0, 1 / 2]}$ and $g : = χ_{(1 / 2, 1]}$ . Then $∥ f ∥_{1 / 2} = {(\int_{0}^{1 / 2} 1 d λ)}^{2} = \frac{1}{4}, ∥ g ∥_{1 / 2} = \frac{1}{4},$ while $∥ f + g ∥_{1 / 2} = {(\int_{0}^{1} 1 d λ)}^{2} = 1.$

2.1

Therefore [step 1.1, L1] $∥ f + g ∥_{1 / 2} = 1 > \frac{1}{4} + \frac{1}{4} = ∥ f ∥_{1 / 2} + ∥ g ∥_{1 / 2},$ so the triangle inequality fails. This is exactly the phenomenon summarized in [L1]. ∎

Depends on

The $p$-functional need not be a norm for $0 < p < 1$

Used by

FALSE: L^1/2 with its p-functional is a normed space False statement

Dependency tree · two levels

8 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

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem 8.16 (standard reference, not scraped)