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

Statement

Let $(X, A, μ)$ be a sigma-finite measure space, let $1 \leq p < \infty$ , and let $q$ be conjugate to $p$ . For every bounded linear functional $Λ : L^{p} (μ) \to R$ there exists a unique $g \in L^{q} (μ)$ such that $Λ ([f]) = \int f g d μ ([f] \in L^{p} (μ)) .$ Moreover, $∥ Λ ∥ = ∥ g ∥_{q} .$

Facts & Assumptions

Given: A sigma-finite measure space $(X, A, μ)$ , an exponent $1 \leq p < \infty$ , its conjugate exponent $q$ , and a bounded linear functional $Λ$ on $L^{p} (μ)$ .

[L1]

On a finite measure space, a bounded $L^{p}$ functional has an $L^{q}$ Radon-Nikodym density representing it on all of $L^{p}$ (The Radon-Nikodym density of a bounded $L^{p}$ functional belongs to $L^{q}$ ).

[L2]

On a semifinite measure space, representing $L^{q}$ classes are unique (On a semifinite measure space, a representing $L^{q}$ function is unique).

[L3]

Sigma-finiteness means there is an increasing exhaustion by measurable finite-measure sets, and every sigma-finite measure is semifinite (Finite, sigma-finite, and semifinite measures).

[L4]

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

[L5]

Dominated convergence applies to integrable majorants (Dominated convergence).

[L6]

Monotone convergence applies to increasing nonnegative measurable sequences (Monotone convergence for the integral).

Proof

technique · Restrict the functional to an increasing finite-measure exhaustion, obtain local $L^q$ densities, glue them by uniqueness on overlaps, and pass to the limit using continuity on $L^p$

1.1L1L3givenchooseconstruct

By [L3], choose an increasing sequence $(X_{n})$ of measurable sets with [L1, L3, given, choose, construct] $X = ⋃_{n} X_{n}$ and $μ (X_{n}) < \infty$ for every $n$ . For each $n$ , let $Λ_{n}$ be the bounded functional on $L^{p} (μ ∣_{X_{n}})$ defined by $Λ_{n} ([u]) : = Λ ([\tilde{u}]),$ where $\tilde{u}$ is the extension of $u$ by $0$ outside $X_{n}$ . Then $∥ Λ_{n} ∥ \leq ∥ Λ ∥$ . Applying [L1] on $X_{n}$ , choose $g_{n} \in L^{q} (μ ∣_{X_{n}})$ such that $Λ_{n} ([u]) = \int_{X_{n}} u g_{n} d μ ([u] \in L^{p} (μ ∣_{X_{n}}))$ and $∥ g_{n} ∥_{q} \leq ∥ Λ ∥$ .

2.1L2L3step 1.1

If $m > n$ , then both $g_{n}$ and $g_{m} ∣_{X_{n}}$ represent the same functional [L2, L3, step 1.1] $Λ_{n}$ on the finite measure space $X_{n}$ . Since finite measure implies semifinite, [L2] gives $g_{m} = g_{n} μ -almost everywhere on X_{n} .$ Therefore the local densities are compatible on overlaps.

3.1L6step 1.1step 2.1algebra

Define a measurable function $g$ by setting $g = g_{n}$ almost everywhere on [L6, step 1.1, step 2.1, algebra] each $X_{n}$ . This is consistent by step 2.1. If $q < \infty$ , then $\int_{X_{n}} ∣ g ∣^{q} d μ = \int_{X_{n}} ∣ g_{n} ∣^{q} d μ \leq ∥ Λ ∥^{q},$ so monotone convergence gives $\int_{X} ∣ g ∣^{q} d μ = \lim_{n \to \infty} \int_{X_{n}} ∣ g ∣^{q} d μ \leq ∥ Λ ∥^{q} .$ If $q = \infty$ , then for each $n$ there is a null set $N_{n} \subseteq X_{n}$ with $∣ g_{n} ∣ \leq ∥ Λ ∥$ on $X_{n} ∖ N_{n}$ ; the union $N : = ⋃_{n} N_{n}$ is null, and $∣ g ∣ \leq ∥ Λ ∥$ on $X ∖ N$ . Hence in every case $g \in L^{q} (μ)$ and $∥ g ∥_{q} \leq ∥ Λ ∥$ .

4.1L4L5step 1.1step 3.1givenconstruct

Let $[f] \in L^{p} (μ)$ and choose a representative $w$ . [L4, L5, step 1.1, step 3.1, given, construct] Set $w_{n} : = w 1_{X_{n}}$ . Then $∣ w_{n} - w ∣^{p} = ∣ w ∣^{p} 1_{X ∖ X_{n}} \leq ∣ w ∣^{p},$ and the right-hand side is integrable. Since $w_{n} \to w$ pointwise, [L5] gives $∥ [w_{n}] - [f] ∥_{p}^{p} = \int ∣ w_{n} - w ∣^{p} d μ ⟶ 0.$ For each $n$ , step 1.1 gives $Λ ([w_{n}]) = Λ_{n} ([w ∣_{X_{n}}]) = \int_{X_{n}} w g_{n} d μ = \int w_{n} g d μ .$ By step 3.1 and [L4], the map $I_{g} ([h]) = \int h g d μ$ is a bounded linear functional on $L^{p} (μ)$ . Taking limits in the last display yields $Λ ([f]) = I_{g} ([f]) = \int w g d μ .$ So $g$ represents $Λ$ on all of $L^{p} (μ)$ .

5.1L2L3L4step 4.1∎

Since $Λ = Λ_{g}$ , [L2, L3, L4, step 4.1] [L4] gives $∥ Λ ∥ = ∥ Λ_{g} ∥ = ∥ g ∥_{q} .$ If $h \in L^{q} (μ)$ also represents $Λ$ , then sigma-finite implies semifinite by [L3], so [L2] gives $h = g$ in $L^{q} (μ)$ . Thus the representing class is unique.

Depends on

Used by

Dependency tree · two levels

40 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., Theorem 6.15 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Theorem 15.11 (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 7.14 (standard reference, not scraped)

On a sigma-finite measure space, every bounded linear functional on LpLp is integration against a unique LqLq function