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 $L^{p} (μ)$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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