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

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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.

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Convolution on $L^{1} (R^{n})$ is bilinear, commutative, and associative

Statement

Assume the Axiom of Countable Choice. Convolution on $L^{1} (R^{n})$ , for $n \geq 1$ , is bilinear, commutative, and associative.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , and functions in $L^{1} (R^{n})$ for which the displayed algebra laws are to be checked.

[L1]

$L^{1}$ convolution exists almost everywhere and obeys the $L^{1}$ bound (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ ).

[L2]

Tonelli and Fubini justify rearranging absolutely integrable iterated integrals (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product).

[L3]

Convolution is the integral from Convolution of two functions on $R^{n}$ .

[L4]

Translation and reflection preserve Lebesgue measure, and integrals are invariant under measure-preserving maps (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it, Integral invariance under measure-preserving maps).

Proof

technique · direct

1.1L1L3givenalgebra

Bilinearity follows from linearity of the integral in [L3] once [L1] guarantees absolute convergence for almost every $x$ .

1.2L1L3L4algebra

For commutativity, fix $x$ where convolution is defined and use the measure-preserving substitution $u = x - y$ from [L4]: $(f * g) (x) = \int f (x - y) g (y) d y = \int g (x - u) f (u) d u = (g * f) (x) .$ The change of variables also preserves absolute integrability, so the two sides agree almost everywhere as $L^{1}$ classes.

1.3L1L2L4algebra

For associativity, Tonelli and translation invariance give $∭ ∣ f (x - y - z) g (z) h (y) ∣ d x d y d z = ∥ f ∥_{1} ∥ g ∥_{1} ∥ h ∥_{1} < \infty .$ Thus for almost every $x$ the double integral is absolutely convergent; [L2] permits interchange of $y, z$ , and the two orders yield $((f * g) * h) (x)$ and $(f * (g * h)) (x)$ respectively. Representative independence follows from [L1].

2.1step 1.1step 1.2step 1.3∎

Therefore convolution is bilinear, commutative, and associative on $L^{1} (R^{n})$ under the stated Countable Choice.

Depends on

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Sources

Walter Rudin, Real and Complex Analysis, 3rd ed. (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)

Convolution on L1(Rn)L1(Rn) is bilinear, commutative, and associative