TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Tempered convolution is smooth with polynomial growth

Statement

For $u \in S^{'} (R^{n})$ and $φ \in S (R^{n})$ , the function $u * φ$ is smooth and every derivative has polynomial growth. For each multi-index $γ$ ,

$\partial^{γ} (u * φ) = (\partial^{γ} u) * φ = u * (\partial^{γ} φ) .$

Consequently $u * φ$ , interpreted as a regular distribution, belongs to $S^{'} (R^{n})$ .

Facts & Assumptions

Given: $u \in S^{'}$ and $φ \in S$ , with convolution as in Convolution of a tempered distribution with a schwartz function.

[F1]

There are $C, N, M$ giving a finite rectangular seminorm estimate for $u$ (Finite seminorm bound characterizes tempered distributions).

[F2]

Translation, reflection, and differentiation are continuous on Schwartz space (Basic operations are continuous on Schwartz space).

[F3]

Smooth pointwise-polynomial-growth functions define regular tempered distributions (Polynomial growth functions define tempered distributions).

Proof

technique · differentiate translated tests in Schwartz seminorms

1.1

For every $x$ and coordinate $j$ , Taylor's integral remainder and $1 + ∣ y ∣ \leq (1 + ∣ x ∣) (1 + ∣ x - y ∣)$ show that the $x_{j}$ -difference quotients of $y \mapsto φ (x - y)$ converge in every Schwartz seminorm to $y \mapsto \partial_{j} φ (x - y)$ . Continuity of $u$ permits differentiation of the scalar pairing, and iteration gives every multi-index derivative.

F1F2given

$\partial^{γ} (u * φ) (x) = ⟨ u_{y}, \partial^{γ} φ (x - y) ⟩ .$

[F1, F2, given]

1.2

Apply the definition of distributional differentiation to the translated test.

$⟨ \partial^{γ} u_{y}, φ (x - y) ⟩ = (- 1)^{∣ γ ∣} ⟨ u_{y}, \partial_{y}^{γ} φ (x - y) ⟩ = ⟨ u_{y}, \partial^{γ} φ (x - y) ⟩ .$

Together with step 1.1 this proves both derivative identities, including $γ = 0$ . [F2, step 1.1]

2.1

Apply [F1] to the translated test in step 1.1. Write $z = x - y$ and expand $y^{α} = (x - z)^{α}$ .

F1step 1.1algebra

$\sup_{y} ∣ y^{α} \partial_{y}^{β} \partial^{γ} φ (x - y) ∣ \leq C_{φ, N, M, γ} (1 + ∣ x ∣)^{N} .$

Hence $∣ \partial^{γ} (u * φ) (x) ∣ \leq C_{γ} (1 + ∣ x ∣)^{N}$ . [F1, algebra]

3.1

Step 1.1 gives smoothness, and step 2.1 gives pointwise polynomial growth for every derivative. In particular the function is locally integrable and [F3] makes its regular distribution tempered. If $u = 0$ or $φ = 0$ , all formulas reduce to zero. No parameter integral or choice axiom is used.

F3step 1.1step 2.1∎

Depends on

Used by

Fourier transform converts allowed tempered convolutions to products Theorem

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)
Radu Gelca, Functional Analysis (standard reference, not scraped)