TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Equality in Holder's inequality for $1 < p < \infty$

Statement

Let $1 < p < \infty$ , let $q$ be its conjugate exponent, and let $f \in L^{p} (μ)$ and $g \in L^{q} (μ)$ . Then equality holds in Holder's inequality

$\int ∣ f g ∣ d μ = ∥ f ∥_{p} ∥ g ∥_{q}$

if and only if at least one of $f, g$ is zero almost everywhere, or there is a constant $c > 0$ such that

$∣ f ∣^{p} = c ∣ g ∣^{q} μ -almost everywhere .$

Facts & Assumptions

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

[L1]

Holder's inequality for integrals has already been proved (Holder's inequality for integrals, including the endpoint cases).

[L2]

Young's inequality is the scalar step used in that proof (Young's inequality for conjugate real exponents).

[L3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L4]

Membership in $L^{p} (μ)$ and $L^{q} (μ)$ means finiteness of the corresponding power integrals (The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

[L5]

For a weight strictly between zero and one, equality in the two-point exponential convexity inequality holds exactly when its two inputs agree (The two-point convexity inequality for the exponential function).

Proof

technique · Trace where equality can occur in the normalized Young-inequality proof. Equality in Young forces the normalized powers $|f|^p$ and $|g|^q$ to be proportional almost everywhere, and conversely that proportionality makes the inequality an equality

1.1L1L3L4

If $∥ f ∥_{p} = 0$ or $∥ g ∥_{q} = 0$ , then the corresponding function is zero almost everywhere, and Holder's inequality becomes equality with both sides $0$ .

1.2L1L2L3

Assume now that $A : = ∥ f ∥_{p} > 0$ and $B : = ∥ g ∥_{q} > 0$ . The proof of [L1] integrated the nonnegative function [L1, L2, L3] $H : = \frac{∣ f ∣^{p}}{p A^{p}} + \frac{∣ g ∣^{q}}{q B^{q}} - \frac{∣ f g ∣}{A B} .$ If equality holds in Holder, then $\int H d μ = 0$ , so $H = 0$ almost everywhere. Thus equality holds in Young's inequality pointwise almost everywhere for $u = ∣ f ∣ / A$ and $v = ∣ g ∣ / B$ .

2.1L2L5step 1.2algebra

To check the equality condition omitted from [L2]'s Statement, first suppose $u, v > 0$ . Apply [L5] to $\log (u^{p})$ and $\log (v^{q})$ with weights $1 / p$ and $1 / q$ : its geometric side is $u v$ , its arithmetic side is $u^{p} / p + v^{q} / q$ , and equality holds exactly when $u^{p} = v^{q}$ . If one of $u, v$ is zero, equality in Young holds only when both vanish, again exactly when $u^{p} = v^{q}$ . Applying this criterion to step 1.2 gives [step 1.2, L2, L5, algebra] $\frac{∣ f ∣^{p}}{A^{p}} = \frac{∣ g ∣^{q}}{B^{q}} μ -almost everywhere,$ so $∣ f ∣^{p} = (A^{p} / B^{q}) ∣ g ∣^{q}$ almost everywhere.

3.1step 1.1step 2.1L1L4algebra

Conversely, if $∣ f ∣^{p} = c ∣ g ∣^{q}$ almost everywhere for some $c > 0$ and neither function vanishes almost everywhere, integration gives $A^{p} = c B^{q}$ . Thus the normalized powers agree almost everywhere, so the scalar criterion from step 2.1 makes Young an equality almost everywhere and the integrated Holder proof becomes an equality. If either function vanishes almost everywhere, step 1.1 already applies.

4.1step 1.1step 2.1step 3.1∎

Step 1.1 handles the zero-function case, step 2.1 proves the strict necessity, and step 3.1 proves sufficiency. These are exactly the alternatives in the Statement.

Depends on

Used by

Cauchy-Schwarz inequality for L² Corollary
Proportional functions realise the equality case of Holder Example
FALSE: Holder equality forces the functions themselves to be proportional False statement
Equality in Minkowski's inequality for 1 < p < ∞ Theorem

Dependency tree · two levels

27 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

Sheldon Axler, Measure, Integration & Real Analysis, Holder's Inequality (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem 8.6 and converse (standard reference, not scraped)

Equality in Holder's inequality for 1<p<∞1<p<∞