ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

A power function on $(0, 1]$ realizes the duality norm on the unit interval

Example

Let $1 < p < \infty$ with conjugate exponent $q$ , and choose $a$ with $0 < a < 1 / q$ . On $(0, 1]$ with Lebesgue measure, define $g (x) : = x^{- a}, f (x) : = (1 - a q)^{1 / p} x^{- a (q - 1)} .$ Then $g \in L^{q} (0, 1)$ , the class $[f]$ has $∥ [f] ∥_{p} = 1$ , and $\int_{0}^{1} f (x) g (x) d x = ∥ g ∥_{q} = ∥ Λ_{g} ∥ = (1 - a q)^{- 1 / q} .$ So this explicit power pair realizes the norm of the duality functional.

Facts & Assumptions

Given: An exponent $1 < p < \infty$ , its conjugate exponent $q$ , and a real parameter $a$ with $0 < a < 1 / q$ .

[L1]

The pairing functional $Λ_{g}$ has norm $∥ g ∥_{q}$ (The functional $Λ_{g}$ has norm $∥ g ∥_{q}$ ; for $q = \infty$ assume $μ$ is semifinite).

Verification

technique · Take $g(x)=x^{-a}$ with $0<a<1/q$, normalize the extremizer $|g|^{q-1}$ explicitly, and compute both norms by one-variable power integrals

1.1givenalgebra

Since $a q < 1$ , one has $1 - a q > 0$ and therefore $\int_{0}^{1} x^{- a q} d x = {[\frac{x^{1 - a q}}{1 - a q}]}_{0}^{1} = (1 - a q)^{- 1} .$ Hence $g \in L^{q} (0, 1)$ and $∥ g ∥_{q}^{q} = \int_{0}^{1} x^{- a q} d x = (1 - a q)^{- 1}, ∥ g ∥_{q} = (1 - a q)^{- 1 / q} .$

2.1step 1.1givenalgebra

The chosen function $f$ satisfies $∣ f (x) ∣^{p} = (1 - a q) x^{- a q} .$ Step 1.1 therefore gives $∥ [f] ∥_{p}^{p} = (1 - a q) \int_{0}^{1} x^{- a q} d x = 1,$ so $∥ [f] ∥_{p} = 1$ .

3.1L1step 1.1givenalgebra∎

Also $f (x) g (x) = (1 - a q)^{1 / p} x^{- a q},$ so another use of step 1.1 gives $\int_{0}^{1} f (x) g (x) d x = (1 - a q)^{1 / p} \int_{0}^{1} x^{- a q} d x = (1 - a q)^{1 / p - 1} = (1 - a q)^{- 1 / q} .$ By step 1.1 and [L1], this equals $∥ g ∥_{q} = ∥ Λ_{g} ∥$ .

Depends on

The functional $\Lambda_g$ has norm $\|g\|_q$; for $q=\infty$ assume $\mu$ is semifinite

Used by

Nothing in the library uses this result yet.

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

John K. Hunter, Measure Theory, Proposition 7.13 (standard reference, not scraped)
Gerald B. Folland, Real Analysis, 2nd ed., Section 6.2 (standard reference, not scraped)

A power function on (0,1](0,1] realizes the duality norm on the unit interval