CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Product of two distributions is not canonically defined

Statement refuted

There is no associative commutative differential $C$ -algebra $A$ with all three of the following properties:

there is an injective complex-linear map $J : D^{'} (R) \to A$ ;
a derivation $\partial : A \to A$ satisfies $\partial J (u) = J (u^{'})$ for every distribution $u$ ; and
if $f, g$ are locally integrable piecewise smooth functions and $f g$ is locally integrable, then $J (u_{f}) J (u_{g}) = J (u_{f g})$ for their regular distributions.

Thus an associative commutative product cannot simultaneously extend all such pointwise products, preserve the distributional derivative, and keep the embedding of distributions injective.

Facts & Assumptions

Given: Countable Choice and a hypothetical triple $(A, J, \partial)$ satisfying the three displayed requirements.

[F1]

A locally integrable function $f$ defines the regular distribution $u_{f}$ , and $f \mapsto u_{f}$ is injective (A locally integrable function on $R^{n}$ , Regular distribution from a locally integrable function, Locally integrable functions embed in distributions).

[F2]

Distributional differentiation is defined by $⟨ u^{'}, φ ⟩ = - ⟨ u, φ^{'} ⟩$ (Distributional derivative).

[F3]

The Dirac distribution satisfies $⟨ δ_{0}, φ ⟩ = φ (0)$ (Dirac delta and its derivatives).

[F4]

Products of a distribution with a smooth function already have a canonical meaning, but the Heaviside function used below is not smooth (Multiplication of a distribution by a smooth function).

[F5]

Integration by parts, and hence the endpoint evaluation of an integral of $φ^{'}$ , is valid for compactly supported smooth test functions (Complex integration by parts on intervals and decaying lines).

Counterexample

technique · contradiction from the Heaviside idempotent

1.1

Assume for contradiction that $(A, J, \partial)$ satisfies the three stated requirements. Let $H = 1_{(0, \infty)}$ . It is locally integrable and piecewise smooth; evaluate its derivative on an arbitrary $φ \in D (R)$ .

assume-contraF1F2F5

$⟨ H^{'}, φ ⟩ = - \int_{0}^{\infty} φ^{'} (x) d x = φ (0) = ⟨ δ_{0}, φ ⟩ .$

Thus $H^{'} = δ_{0}$ . [F1, F2, F3, F5]

2.1

Put $h = J (u_{H})$ and $d = J (δ_{0})$ . Since $H^{2} = H$ and $H^{3} = H$ pointwise, property 3 gives $h^{2} = h$ and $h^{3} = h$ . Property 2 and step 1.1 give $\partial h = d$ .

givenstep 1.1

3.1

Apply the derivation to $h^{2} = h$ and $h^{3} = h$ .

givenstep 2.1algebra

$2 h d = d .$

Apply it to $h^{3} = h$ . Associativity, commutativity, and the Leibniz rule give

$3 h^{2} d = d .$

Because $h^{2} = h$ , subtraction of these identities gives $h d = 0$ , and the first identity then gives $d = 0$ . [given, step 2.1, algebra]

4.1

Yet $δ_{0} \neq 0$ : choose a test function with $φ (0) = 1$ and use [F3]. Injectivity of $J$ therefore implies $d = J (δ_{0}) \neq 0$ , contradicting step 3.1.

givenF3step 3.1

5.1

The contradiction concerns only the simultaneous requirements above. Special products, including [F4], and separately chosen nonlinear regularizations are not ruled out. Countable Choice is used only through the published regular-distribution and integration interfaces.

F1F4F5step 4.1discharge-contradiction: step 4.1∎

Depends on

Used by

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Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)