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

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Convolution of two tempered distributions need not exist

Statement refuted

Every pair of tempered distributions has an ordinary convolution.

Already on $R^{n}$ , for $n \geq 1$ , the two constant tempered distributions $1$ and $1$ do not have an ordinary convolution.

Facts & Assumptions

Given: $n \geq 1$ .

[F1]

A function of polynomial growth defines a tempered distribution (Polynomial growth functions define tempered distributions).

[F2]

Convolution is canonically defined when one distribution has compact support; the constants in this example have no compact support (Convolution of distributions when one has compact support).

Counterexample

technique · divergence along the fibers of the addition map

1.1F1F2

The constant function $1$ has polynomial growth of order zero, so each factor defines a tempered distribution by [F1]. Neither factor is compactly supported, so [F2] does not itself define their convolution.

1.2choose

Choose a nonnegative $ψ \in D (R^{n})$ with $c : = \int_{R^{n}} ψ (z) d z > 0$ . The formal distributional convolution pairing would require the following addition-pullback integral.

$\int_{R^{n}} \int_{R^{n}} ψ (x + y) d y d x .$

For the cube $Q_{R} = [- R, R]^{n}$ , translation in the inner integral gives

$\int_{Q_{R}} \int_{R^{n}} ψ (x + y) d y d x = ∣ Q_{R} ∣ c = (2 R)^{n} c .$

The quantities on the right tend to $+ \infty$ . Equivalently, $ψ (x + y)$ is not compactly supported on $R^{2 n}$ : every nonempty addition fiber has infinite volume. [given, algebra]

2.1F2step 1.2∎

Hence the ordinary integral construction does not produce a finite pairing even on this one nonnegative test function, and $1 * 1$ is undefined as an ordinary distributional convolution. This does not say that no separately chosen regularization can assign an object to the pair; such an assignment is additional structure, not the ordinary convolution supplied by [F2].

Depends on

Used by

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Dependency tree · two levels

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Sources

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