CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Cauchy-Schwarz inequality for $L^{2}$

Statement

If $f, g \in L^{2} (μ)$ , then

$\int ∣ f g ∣ d μ \leq ∥ f ∥_{2} ∥ g ∥_{2} .$

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

$∣ f ∣^{2} = c ∣ g ∣^{2} μ -almost everywhere .$

Facts & Assumptions

Given: Functions $f, g \in L^{2} (μ)$ .

[L1]

Holder's inequality holds for conjugate exponents (Holder's inequality for integrals, including the endpoint cases).

[L2]

The strict-exponent equality criterion for Holder has already been proved (Equality in Holder's inequality for $1 < p < \infty$ ).

Proof

Proof technique: Specialize Holder to $p = q = 2$ , and inherit the equality clause from the strict-exponent equality theorem.

1.1

The exponent $2$ is conjugate to itself, so [L1] with $p = q = 2$ gives [L1] $\int ∣ f g ∣ d μ \leq ∥ f ∥_{2} ∥ g ∥_{2} .$

2.1

Because $2 > 1$ , the equality clause is exactly the specialization of [L2] to [L2, step 1.1] $p = q = 2$ . ∎

Depends on

Used by

Finite counting measure recovers finite Holder and implies the signed Cauchy-Schwarz inequality Remark

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

Sheldon Axler, Measure, Integration & Real Analysis, Section 7A (standard reference, not scraped)
John K. Hunter, Measure Theory, Section 7.2 (standard reference, not scraped)

Cauchy-Schwarz inequality for L2L2