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CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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Counting measure specializes the representation theorem to p and q

Statement

Let 1p< and let q be conjugate to p. Every bounded linear functional Λ:pR is of the form Λ(a)=n=0anbn for a unique sequence bq. Moreover, Λ=bq.

Facts & Assumptions

Given: An exponent 1p<, its conjugate exponent q, and a bounded linear functional Λ on p.

[L1]

On a sigma-finite measure space, every bounded linear functional on Lp is integration against a unique Lq function (On a sigma-finite measure space, every bounded linear functional on Lp is integration against a unique Lq function).

[L2]

On counting measure over N, the spaces Lp and p coincide, the Lq density becomes a sequence, and the integral becomes the series pairing (p is the Lp space of counting measure).

Proof

technique · View $\ell^p$ as $L^p$ of counting measure on $\mathbb N$, note that counting measure on $\mathbb N$ is sigma-finite by finite initial segments, and translate the representing $L^q$ function back to a sequence
1.1

Counting measure on N is sigma-finite because [L2, given] N=n=0{0,1,,n}, and each initial segment has finite counting measure. By [L2], we may therefore regard Λ as a bounded linear functional on Lp(#).

L2given
2.1

Applying [L1], choose gLq(#) such that [L1, L2, step 1.1, choose] Λ(a)=agd#(ap). By [L2], writing bn:=g(n) turns g into a sequence b=(bn)q, and the integral identity becomes Λ(a)=n=0anbn. The same translation in [L2] turns uniqueness and norm equality from [L1] into uniqueness of b and Λ=bq.

L1L2step 1.1choose

Depends on

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