CorollaryStatement: AI-adaptedProof: AI-generatedprecheck pass

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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.

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

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

1.1L1

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

2.1L2step 1.1∎

Because $2 > 1$ , the strict-exponent equality criterion at $p = q = 2$ says that equality in step 1.1 holds exactly when at least one of $f, g$ is zero almost everywhere, or $∣ f ∣^{2} = c ∣ g ∣^{2}$ almost everywhere for some constant $c > 0$ . These are precisely the alternatives in the Statement.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Cauchy-Schwarz inequality for L2L2

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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