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

1.1F1F2given

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.

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

[F1, F2, given]

1.2F2

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]

2.1F1step 1.1algebra

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

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

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

3.1F3step 1.1step 2.1∎

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.

Tempered convolution is smooth with polynomial growth

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources