PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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

Statement

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

Facts & Assumptions

Given: 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 $R^{n}$ .

Proof

technique · direct

1.1

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

L1L3givenalgebra

1.2

For commutativity, fix $x$ where convolution is defined and change [L1, L2, L3, algebra] variables $u : = x - y$ : $(f * g) (x) = \int f (x - y) g (y) d y = \int g (x - u) f (u) d u = (g * f) (x) .$ Associativity is similar: [L2] applies to $∣ f (x - y - z) g (z) h (y) ∣$ , so one may reorder the three integrations and obtain $((f * g) * h) (x) = (f * (g * h)) (x)$ almost everywhere.

L1L2L3algebra

2.1

Therefore convolution is bilinear, commutative, and associative on [step 1.1, step 1.2] $L^{1} (R^{n})$ .

step 1.1step 1.2∎

Depends on

Used by

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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