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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.

Depends on

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Sources

Semyon Dyatlov, Lecture Notes for 18.155 (current revision) (standard reference, not scraped)
Richard B. Melrose, Differential Analysis, Chapter 3 (standard reference, not scraped)