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

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The $L^{p}$ norm is the supremum of pairings against unit $L^{q}$ functions

Statement

Let $(X, A, μ)$ be a measure space, let $1 \leq p < \infty$ , and let $q$ be conjugate to $p$ . Assume either $1 < p < \infty$ , or $p = 1$ and $μ$ is sigma-finite. Then for every $f \in L^{p} (μ)$ , $∥ f ∥_{p} = \sup {∣ \int f g d μ ∣ : g \in L^{q} (μ), ∥ g ∥_{q} \leq 1} .$

Facts & Assumptions

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

[L1]

Apply Hölder's inequality to the real nonnegative functions $∣ f ∣$ and $∣ g ∣$ , then use $∣ \int f g d μ ∣ \leq \int ∣ f g ∣ d μ$ . This gives $∣ \int f g d μ ∣ \leq ∥ f ∥_{p} ∥ g ∥_{q}$ for complex or real functions and conjugate exponents, including $p = 1$ , $q = \infty$ (Holder's inequality for integrals, including the endpoint cases).

Proof

technique · Hölder gives the upper bound; the conjugate phase supplies a norming function

1.1L1given

For every $g \in L^{q} (μ)$ with $∥ g ∥_{q} \leq 1$ , [L1] gives $∣ \int f g d μ ∣ \leq ∥ f ∥_{p} ∥ g ∥_{q} \leq ∥ f ∥_{p} .$ Therefore the supremum is at most $∥ f ∥_{p}$ .

1.2givenconstruct

If $f = 0$ , the displayed formula is immediate. Hence assume from now on $f \neq 0$ . For a measurable representative $u$ of $f$ , define its conjugate phase by $θ_{u} (x) : = {\begin{cases} \overline{u (x)} / ∣ u (x) ∣, & u (x) \neq 0, \\ 0, & u (x) = 0. \end{cases}$ Then $∣ θ_{u} ∣ \leq 1$ and $u θ_{u} = ∣ u ∣$ pointwise.

2.1step 1.2givenchooseconstructalgebra

Assume $1 < p < \infty$ and choose a representative $u$ of $f$ . Define $g (x) : = \frac{∣ u (x) ∣^{p - 1} θ_{u} (x)}{∥ f ∥_{p}^{p - 1}} .$ Since $(p - 1) q = p$ , $∣ g ∣^{q} = \frac{∣ u ∣^{p}}{∥ f ∥_{p}^{p}},$ so $∥ g ∥_{q} = 1$ . Also $u θ_{u} = ∣ u ∣$ gives $\int u g d μ = \frac{1}{∥ f ∥_{p}^{p - 1}} \int ∣ u ∣^{p} d μ = ∥ f ∥_{p} .$ Hence the supremum is at least $∥ f ∥_{p}$ .

2.2step 1.2givenchooseconstructalgebra

Assume $p = 1$ and choose a representative $u$ of $f$ . Put $g : = θ_{u}$ . Then $g \in L^{\infty} (μ)$ with $∥ g ∥_{\infty} \leq 1$ , and $\int u g d μ = \int ∣ u ∣ d μ = ∥ f ∥_{1} .$ So the supremum is at least $∥ f ∥_{1}$ .

3.1step 1.1step 2.1step 2.2∎

Step 1.1 gives the upper bound, while step 2.1 or step 2.2 gives an $L^{q}$ function of norm at most one attaining it. Therefore the displayed supremum equals $∥ f ∥_{p}$ .

Depends on

Holder's inequality for integrals, including the endpoint cases

Used by

Dependency tree · two levels

11 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

Richard F. Bass, Real Analysis for Graduate Students, Corollary 15.9 (standard reference, not scraped)
Gerald B. Folland, Real Analysis, 2nd ed., Section 6.2 (standard reference, not scraped)

The LpLp norm is the supremum of pairings against unit LqLq functions