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Counting measure specializes the representation theorem to $ℓ^{p}$ and $ℓ^{q}$

Statement

Let $1 \leq p < \infty$ and let $q$ be conjugate to $p$ . Every bounded linear functional $Λ : ℓ^{p} \to R$ is of the form $Λ (a) = \sum_{n = 0}^{\infty} a_{n} b_{n}$ for a unique sequence $b \in ℓ^{q}$ . Moreover, $∥ Λ ∥ = ∥ b ∥_{ℓ^{q}} .$

Facts & Assumptions

Given: An exponent $1 \leq p < \infty$ , its conjugate exponent $q$ , and a bounded linear functional $Λ$ on $ℓ^{p}$ .

[L1]

On a sigma-finite measure space, every bounded linear functional on $L^{p}$ is integration against a unique $L^{q}$ function (On a sigma-finite measure space, every bounded linear functional on $L^{p}$ is integration against a unique $L^{q}$ function).

[L2]

On counting measure over $N$ , the spaces $L^{p}$ and $ℓ^{p}$ coincide, the $L^{q}$ density becomes a sequence, and the integral becomes the series pairing ( $ℓ^{p}$ is the $L^{p}$ 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

1.1L2given

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

2.1L1L2step 1.1choose∎

Applying [L1], choose $g \in L^{q} (#)$ such that [L1, L2, step 1.1, choose] $Λ (a) = \int a g d # (a \in ℓ^{p}) .$ By [L2], writing $b_{n} : = g (n)$ turns $g$ into a sequence $b = (b_{n}) \in ℓ^{q}$ , and the integral identity becomes $Λ (a) = \sum_{n = 0}^{\infty} a_{n} b_{n} .$ The same translation in [L2] turns uniqueness and norm equality from [L1] into uniqueness of $b$ and $∥ Λ ∥ = ∥ b ∥_{ℓ^{q}}$ .

Depends on

Used by

A coercive functional need not attain without weak lower semicontinuity Counterexample
A minimising sequence need not converge strongly Counterexample
A norm-closed nonconvex set need not be weakly closed Counterexample
c₀ is not reflexive Counterexample
Ell-one and ell-infinity are not reflexive Counterexample
Injective transpose does not imply surjectivity Counterexample
The canonical bidual map of c0 misses the constant sequence Counterexample
The transpose of an injective map need not have norm-dense range Counterexample
Coordinate functionals on sequence spaces Example
Every bounded linear functional on ℓ² is summation against a unique ℓ² sequence Example
Reflexivity of ℓᵖ and Lᵖ Example
Transposes of the right and left shifts Example
Dual and bidual models for James space Lemma
The published Lp duality theorem in abstract notation Remark

Dependency tree · two levels

16 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)
Richard F. Bass, Real Analysis for Graduate Students, Section 15.4 (standard reference, not scraped)

Counting measure specializes the representation theorem to ℓpℓp and ℓqℓq