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Bessel-Potential Completions and Real-Order Sobolev Spaces — Examples

1 · Prerequisites

2 · Summary

These examples check the completion on two basic classes. Every Schwartz function gives its canonical element at every real order, with the stated weighted Fourier norm. At order zero, the canonical distribution embedding identifies the completion with complex $L^{2}$ under the unitary negative-sign $2 π$ transform, with normalization factor one. The first example does not assert the converse characterization of Schwartz space, and the second makes no positive integer derivative-norm comparison.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Every Schwartz function belongs to every real-order H^s

Statement

Assume Countable Choice. For every $n \geq 1$ , real $s$ , and actual Schwartz function $u \in S (R^{n})$ , its canonical completion class is sent by $E_{s}$ to the usual regular distribution of $u$ . In particular $S (R^{n}) \subset H^{s} (R^{n})$ for every real $s$ , with $∥ u ∥_{H^{s}} = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{2} < \infty .$ This inclusion does not assert that the intersection of all real-order $H^{s}$ spaces is exactly Schwartz space.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , and $u \in S (R^{n})$ .

[A1]

Countable Choice is used by the metric completion construction defining $H^{s}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Multiplication by $⟨ ξ ⟩^{s}$ maps Schwartz space continuously to itself (Real powers of the Japanese bracket act on Schwartz space).

[F2]

Fourier transformation is an automorphism of Schwartz space (Fourier transform is a topological automorphism of Schwartz space).

[F3]

Every Schwartz function determines a complex $L^{2}$ class (Schwartz space is dense in L2).

[F4]

The candidate norm is $q_{s} (u) = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{2}$ (Weighted Fourier candidate norm on Schwartz space).

[F5]

The canonical constant-sequence map embeds Schwartz space linearly and isometrically into its completion $H^{s}$ (Real-order Bessel-potential completion H^s).

[F6]

The canonical embedding sends a Schwartz class to its usual regular distribution (The Bessel completion embeds canonically in tempered distributions).

Proof

technique · Apply the weighted Fourier norm directly to a Schwartz function

1.1F1F2F3F4given

By [F2], $\hat{u} \in S$ ; then [F1] gives $f (ξ) = ⟨ ξ ⟩^{s} \hat{u} (ξ) \in S$ . Thus [F3] gives $f \in L^{2}$ and the defining integral calculation is $q_{s} (u)^{2} = \int_{R^{n}} ∣ ⟨ ξ ⟩^{s} \hat{u} (ξ) ∣^{2} d ξ = ∥ f ∥_{2}^{2} < \infty$ by [F4].

2.1A1F5F6step 1.1∎

Under the stated Countable Choice assumption [A1], [F5] places the constant sequence $i (u) = [(u, u, \dots)]$ in $H^{s}$ with $∥ i (u) ∥_{H^{s}} = q_{s} (u)$ ; [F6] gives $E_{s} (i (u)) = u_{u}$ , the regular distribution of $u$ . Step 1.1 supplies the finite norm, proving the asserted inclusion and formula for every real $s$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

The zero-order Bessel completion is exactly L2

Statement

Assume Countable Choice and use the negative-sign $2 π$ Fourier convention. For $n \geq 1$ , let $J_{0} : H^{0} (R^{n}) \to L^{2} (R^{n})$ be the surjective weighted-transform isometry from the completion theorem, and let $E_{0} : H^{0} (R^{n}) \to S^{'} (R^{n})$ be its canonical distribution embedding. The map $I_{0} = F_{2}^{- 1} \circ J_{0} : H^{0} (R^{n}) ⟶ L^{2} (R^{n})$ is a surjective linear isometry, agrees with the identity on canonical Schwartz classes, and satisfies $E_{0} (U) = u_{I_{0} U}$ for every $U \in H^{0}$ . Consequently $E_{0}$ identifies $H^{0}$ with precisely the regular distributions of complex $L^{2}$ classes, and $∥ U ∥_{H^{0}} = ∥ I_{0} U ∥_{2}, q_{0} (u) = ∥ u ∥_{2} (u \in S) .$ The normalization factor in both norm identities is exactly one.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , and the fixed negative-sign $2 π$ Fourier transform.

[A1]

Countable Choice holds for the countable approximations and completion interfaces used by the cited Plancherel and Bessel-completion results (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The candidate norm is $q_{s} (u) = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{2}$ (Weighted Fourier candidate norm on Schwartz space).

[F2]

$H^{s}$ is the norm completion of Schwartz space with its canonical dense constant-sequence map (Real-order Bessel-potential completion H^s).

[F3]

$J_{s}$ is a surjective linear isometry and the embedding formula is $E_{s} (U) = F^{- 1} (u_{⟨ ξ ⟩^{- s} J_{s} U})$ ; $E_{s}$ is injective (The Bessel completion embeds canonically in tempered distributions).

[F4]

The Plancherel extension $F_{2}$ is a surjective complex-linear isometry extending Fourier transformation on Schwartz space (Plancherel theorem).

[F5]

For $f \in L^{2}$ , the distributional transform satisfies $F u_{f} = u_{F_{2} f}$ (Fourier transform agrees with l one and plancherel transforms).

[F6]

Schwartz classes are dense in complex $L^{2}$ (Schwartz space is dense in L2).

[F7]

Fourier transformation is injective on $S^{'}$ because it is a topological automorphism (Fourier transform is a topological automorphism of tempered distributions).

Proof

technique · Compose the two unitary identifications at order zero and check their distributional meaning

1.1F1F4given

For $u \in S$ , $⟨ ξ ⟩^{0} = 1$ , so [F1] gives $q_{0} (u) = ∥ \hat{u} ∥_{2}$ . The extension property and isometry in [F4] give $∥ \hat{u} ∥_{2} = ∥ u ∥_{2}$ , proving the exact factor-one norm identity on Schwartz space.

1.2A1F2F3F4F6given

Define $I_{0} = F_{2}^{- 1} \circ J_{0}$ . Under the inherited Countable Choice assumption [A1], [F3] and [F4] supply two surjective linear isometries, so $I_{0}$ is a surjective linear isometry. If $i (u)$ is the canonical constant-sequence class of $u \in S$ , then $J_{0} i (u) = \hat{u}$ and [F4] gives $I_{0} i (u) = u$ as an $L^{2}$ class. By [F6], this canonical copy of Schwartz space is dense in the target.

2.1A1F3F4F5F7step 1.2step 1.1∎

Given $U \in H^{0}$ , put $f = I_{0} U$ , so $J_{0} U = F_{2} f$ . At $s = 0$ , [F3] gives $F (E_{0} U) = u_{J_{0} U}$ , while [F5] gives $F (u_{f}) = u_{F_{2} f} = u_{J_{0} U}$ . Injectivity [F7] yields $E_{0} U = u_{f}$ . Since $I_{0}$ is onto, every regular distribution $u_{f}$ with $f \in L^{2}$ occurs as an $E_{0}$ image; injectivity of $E_{0}$ in [F3] makes this identification unique. The isometry of $I_{0}$ gives $∥ U ∥_{H^{0}} = ∥ I_{0} U ∥_{2}$ , and step 1.1 gives the Schwartz norm formula.

Bessel-Potential Completions and Real-Order Sobolev Spaces — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Every Schwartz function belongs to every real-order H^s

Statement

Facts & Assumptions

Proof

The zero-order Bessel completion is exactly L2

Statement

Facts & Assumptions

Proof

Sources