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

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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.
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The functional $f \mapsto \int_{0}^{1 / 2} f$ on $L^{p} [0, 1]$ has norm $2^{- 1 / q}$

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $1 \leq p < \infty$ and let $q$ be conjugate to $p$ . On $[0, 1]$ with Lebesgue measure, define $Λ ([f]) : = \int_{0}^{1 / 2} f (x) d x .$ Then $Λ$ is a bounded linear functional on $L^{p} [0, 1]$ and $∥ Λ ∥ = 2^{- 1 / q},$ with the endpoint convention $2^{- 1 / \infty} = 1$ when $p = 1$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and an exponent $1 \leq p < \infty$ with conjugate exponent $q$ .

[L1]

The interval $[0, 1 / 2]$ has Lebesgue measure $1 / 2$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[L2]

If $g \in L^{q} (μ)$ , then the pairing functional $Λ_{g}$ has norm $∥ g ∥_{q}$ in the ranges treated on the A page (The functional $Λ_{g}$ has norm $∥ g ∥_{q}$ ; for $q = \infty$ assume $μ$ is semifinite).

Verification

technique · Represent the functional by the indicator of $[0,1/2]$, compute that indicator's $L^q$ norm from its measure, and invoke the norm formula for $\Lambda_g$

1.1givenconstruct

Let $g : = 1_{[0, 1 / 2]}$ . [given, construct] Then $Λ ([f]) = \int_{0}^{1} f (x) g (x) d x = Λ_{g} ([f]),$ so $Λ$ is exactly the pairing functional associated to $g$ .

1.2L1given

By [L1], if $q < \infty$ then $∥ g ∥_{q}^{q} = \int_{0}^{1} ∣ g ∣^{q} d x = \int_{0}^{1 / 2} 1 d x = \frac{1}{2},$ so $∥ g ∥_{q} = 2^{- 1 / q} .$

1.3L1given

If $q = \infty$ , then $∣ g ∣ \leq 1$ everywhere and $g = 1$ on a set of positive measure, so $∥ g ∥_{\infty} = 1 = 2^{- 1 / \infty}$ .

2.1L2step 1.1step 1.2step 1.3∎

Applying [L2] to step 1.1 and steps 1.2-1.3 gives $∥ Λ ∥ = ∥ Λ_{g} ∥ = ∥ g ∥_{q} = 2^{- 1 / q} .$

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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

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

The functional f↦∫01/2ff↦∫01/2​f on Lp[0,1]Lp[0,1] has norm 2−1/q2−1/q