TheoremStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

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Every $g \in L^{q} (μ)$ defines a bounded linear functional on $L^{p} (μ)$

Statement

Let $(X, A, μ)$ be a measure space, let $1 \leq p < \infty$ , and let $q$ be conjugate to $p$ . For each $g \in L^{q} (μ)$ , the formula $Λ_{g} ([f]) : = \int f g d μ$ defines a bounded linear functional on $L^{p} (μ)$ , and $∣ Λ_{g} ([f]) ∣ \leq ∥ [f] ∥_{p} ∥ g ∥_{q} ([f] \in L^{p} (μ)) .$ In particular, $∥ Λ_{g} ∥ \leq ∥ g ∥_{q} .$

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , an exponent $1 \leq p < \infty$ , its conjugate exponent $q$ , an element $g \in L^{q} (μ)$ , and an element $[f] \in L^{p} (μ)$ .

[L1]

Elements of $L^{p} (μ)$ are almost-everywhere equivalence classes of measurable representatives (The space $L^{p} (μ)$ as the quotient by null functions).

[L2]

Holder's inequality gives $\int ∣ u v ∣ d μ \leq ∥ u ∥_{p} ∥ v ∥_{q}$ for measurable representatives $u \in L^{p} (μ)$ and $v \in L^{q} (μ)$ when both exponents are finite, with the corresponding $L^{1}$ -- $L^{\infty}$ representative form at the endpoint (Holder's inequality for integrals, including the endpoint cases).

[L3]

Two integrable functions have the same integral whenever they are equal almost everywhere (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[L4]

The Lebesgue integral is linear on $L^{1} (μ)$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · Use Holder to make the pairing integrable, then use almost-everywhere invariance of the integral to descend from representatives to $L^p$ classes

1.1L1L2given

Choose measurable representatives $u$ of $[f]$ and $v$ of $g$ . Then $u \in L^{p} (μ)$ ; if $q < \infty$ then $v \in L^{q} (μ)$ , while if $q = \infty$ then $v$ is an essentially bounded representative. Thus [L2] gives $u v \in L^{1} (μ)$ and $∣ \int u v d μ ∣ \leq ∥ u ∥_{p} ∥ v ∥_{q} = ∥ [f] ∥_{p} ∥ g ∥_{q} .$

2.1L1L3step 1.1

If $u^{'}$ is another representative of $[f]$ , then $u = u^{'}$ almost everywhere. Hence $u v = u^{'} v$ almost everywhere. Since both products are integrable by step 1.1, [L3] gives $\int u v d μ = \int u^{'} v d μ,$ so $Λ_{g} ([f])$ is well defined on the class $[f]$ .

2.2L1L3step 1.1

If $v^{'}$ is another representative of $g$ , then $v = v^{'}$ almost everywhere. Hence $u v = u v^{'}$ almost everywhere. Since both products are integrable by step 1.1, [L3] gives $\int u v d μ = \int u v^{'} d μ,$ so $Λ_{g} ([f])$ is independent of the chosen representative of $g$ .

3.1step 1.1step 2.1step 2.2L4given

Let $[f], [h] \in L^{p} (μ)$ and let $a, b \in C$ . Choose representatives $u, w$ . By step 2.1 we may compute with representatives, and complex linearity in [L4] gives $Λ_{g} (a [f] + b [h]) = \int (a u + b w) v d μ = a \int u v d μ + b \int w v d μ = a Λ_{g} ([f]) + b Λ_{g} ([h]) .$ The estimate from step 1.1 also shows $∣ Λ_{g} ([f]) ∣ \leq ∥ [f] ∥_{p} ∥ g ∥_{q},$ so $Λ_{g}$ is complex linear and bounded.

4.1step 3.1given∎

The estimate in step 3.1 holds for every $[f]$ with $∥ [f] ∥_{p} \leq 1$ . Taking the supremum over the unit ball gives $∥ Λ_{g} ∥ \leq ∥ g ∥_{q} .$ Thus $Λ_{g}$ is a bounded linear functional on $L^{p} (μ)$ .

Depends on

Used by

Dependency tree · two levels

21 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, Proposition 15.10 (standard reference, not scraped)
John K. Hunter, Measure Theory, Proposition 7.13 (standard reference, not scraped)

Every g∈Lq(μ)g∈Lq(μ) defines a bounded linear functional on Lp(μ)Lp(μ)