False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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FALSE: Holder equality forces the functions themselves to be proportional

Statement

Whenever equality holds in Holder's inequality, the two functions themselves are proportional almost everywhere.

Facts & Assumptions

Given: The endpoint pair $(p, q) = (1, \infty)$ on $[0, 1]$ with Lebesgue measure.

[L1]

Holder's inequality includes the endpoint cases (Holder's inequality for integrals, including the endpoint cases).

[L2]

The strict proportionality criterion on $∣ f ∣^{p}$ and $∣ g ∣^{q}$ was proved only for $1 < p < \infty$ (Equality in Holder's inequality for $1 < p < \infty$ ).

Refutation

Proof technique: Refute at the endpoint $p = 1$ , $q = \infty$ with $f = χ_{A}$ and $g = 1$ on a proper positive-measure subset $A$ . Equality holds, but the functions are not proportional on the whole space.

1.1

Let $A = [0, 1 / 2]$ , let $f : = χ_{A}$ , and let $g : = χ_{[0, 1]}$ . Then [L1, given, algebra] $\int_{0}^{1} ∣ f g ∣ d λ = λ (A) = \frac{1}{2}, ∥ f ∥_{1} = \frac{1}{2}, ∥ g ∥_{\infty} = 1.$ So equality holds in Holder: $\int_{0}^{1} ∣ f g ∣ d λ = ∥ f ∥_{1} ∥ g ∥_{\infty} .$

2.1

There is no constant $c$ with $f = c g$ almost everywhere, because on [L2, step 1.1] $A$ one would need $c = 1$ while on $(1 / 2, 1]$ one would need $c = 0$ . This does not contradict [L2], because [L2] does not cover the endpoint $p = 1$ .

3.1

Thus equality in Holder does not force the functions themselves to be [step 1.1, step 2.1] proportional almost everywhere. ∎

Depends on

Used by

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Sources

Sheldon Axler, Measure, Integration & Real Analysis, Holder's Inequality (standard reference, not scraped)
John K. Hunter, Measure Theory, Section 7.2 (standard reference, not scraped)