CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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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]

For every $g \in L^{q} (μ)$ , the pairing functional satisfies $∥ Λ_{g} ∥ = ∥ g ∥_{q}$ in the ranges covered by this page (The functional $Λ_{g}$ has norm $∥ g ∥_{q}$ ; for $q = \infty$ assume $μ$ is semifinite).

Proof

technique · Holder gives the universal upper bound. Equality comes from the explicit norming function $|f|^{p-1}\operatorname{sgn} f$ for $p>1$ and the sign function for $p=1$

1.1

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

L1given

1.2

If $f = 0$ , the displayed formula is immediate. Hence assume from now on $f \neq 0$ .

given

1.3

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

givenchooseconstructalgebra

1.4

Assume $p = 1$ and choose a representative $u$ of $f$ . [given, choose, construct, algebra] Put $g : = sgn u .$ 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 $∥ f ∥_{1}$ .

givenchooseconstructalgebra

2.1

Step 1.1 gives the upper bound, while step 1.3 or step 1.4 gives a unit [step 1.1, step 1.3, step 1.4] $L^{q}$ function attaining it. Therefore the displayed supremum equals $∥ f ∥_{p}$ .

step 1.1step 1.3step 1.4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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