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PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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On a semifinite measure space, a representing Lq function is unique

Statement

Let (X,A,μ) be a semifinite measure space, let 1p<, and let q be conjugate to p. If g,hLq(μ) satisfy fgdμ=fhdμ([f]Lp(μ)), then g=h in Lq(μ). Equivalently, on a semifinite measure space a bounded functional on Lp(μ) has at most one representing Lq(μ) class.

Facts & Assumptions

Given: A semifinite measure space (X,A,μ), an exponent 1p<, its conjugate exponent q, and g,hLq(μ) such that fgdμ=fhdμ for every [f]Lp(μ).

[L1]

On a semifinite measure space, Λu=uq(uLq(μ)) (The functional Λg has norm gq; for q= assume μ is semifinite).

[L2]

For finite exponents, the Lq norm is a genuine norm on the quotient space The Lp norm descends to the quotient and makes Lp a normed space for 1p.

[L3]

If u=0, then u0 almost everywhere, so u=0 almost everywhere (The essential supremum is attained as the least essential bound).

Proof

Proof technique: Subtract the two representing functions, observe that the induced pairing is the zero functional, and use the norm formula for Λg to force the difference to be the zero Lq class.

1.1

For every [f]Lp(μ), the assumption gives [given, algebra] Λgh([f])=f(gh)dμ=0. So Λgh is the zero functional on Lp(μ).

givenalgebra
2.1

Applying [L1] to u=gh yields [L1, step 1.1] ghq=Λgh=0.

3.1

If q<, [L2] says that zero Lq norm means the zero class, so [L2, L3, step 2.1] g=h in Lq(μ). If q=, then step 2.1 and [L3] give g=h almost everywhere. In either case the representing class is unique. ∎

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