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Weighted tempered-distribution characterization of H^s

Statement

Assume Countable Choice. For every $n \geq 1$ and $s \in R$ , define $M_{s} = {u \in S^{'} (R^{n}) : ⟨ ξ ⟩^{s} F u = u_{g} in S^{'} (R^{n}) for some g \in L^{2} (R^{n})},$ where $u_{g} (ϕ) = \int_{R^{n}} g (ξ) ϕ (ξ) d ξ$ is the regular tempered distribution. The canonical embedding $E_{s}$ restricts to a bijection $E_{s} : H^{s} (R^{n}) ⟶ M_{s} .$ Thus, after identifying $H^{s}$ with its image under $E_{s}$ , it is exactly the space described by the weighted tempered-distribution condition. The class $g \in L^{2}$ is unique, and if $u = E_{s} U$ corresponds to $g$ , then $∥ u ∥_{H^{s}} : = ∥ U ∥_{H^{s}} = ∥ g ∥_{2} .$ The product $⟨ ξ ⟩^{s} F u$ is multiplication of a tempered distribution by the smooth Japanese-bracket multiplier, not an a priori pointwise product.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , and the canonical embedding $E_{s} : H^{s} (R^{n}) \to S^{'} (R^{n})$ .

[A1]

Countable Choice permits one selection from each nonempty set in a countable family (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The multipliers $w_{s} (ξ) = ⟨ ξ ⟩^{s}$ and $w_{- s} (ξ) = ⟨ ξ ⟩^{- s}$ act continuously and inversely on $S^{'}$ (Real powers of the Japanese bracket act on Schwartz space).

[F2]

The map $J_{s} : H^{s} \to L^{2}$ is a surjective linear isometry, and $E_{s} U = F^{- 1} (u_{w_{- s} J_{s} U})$ defines an injective canonical embedding (The Bessel completion embeds canonically in tempered distributions).

[F3]

Fourier transformation is an automorphism of $S^{'}$ with inverse $F^{- 1}$ (Fourier transform is a topological automorphism of tempered distributions).

[F4]

Each complex $L^{2}$ class defines the regular tempered distribution $u_{g} (ϕ) = \int g ϕ$ (Polynomial growth functions define tempered distributions).

[F5]

Elements of $S^{'}$ are continuous complex-linear functionals on $S$ , with bilinear test pairing (Tempered distribution).

Proof

technique · Cancel the inverse bracket weights and use the completed Fourier isometry

1.1F2F3given

Let $U \in H^{s}$ and put $g = J_{s} U$ . By [F2], $E_{s} U = F^{- 1} (u_{w_{- s} g})$ ; Fourier inversion [F3] gives $F (E_{s} U) = u_{w_{- s} g}$ .

2.1F1F2F4F5step 1.1

For every $ϕ \in S$ , the multiplier action [F1], bilinear pairing [F5], and [F2] give $⟨ w_{s} F (E_{s} U), ϕ ⟩ = ⟨ u_{w_{- s} g}, w_{s} ϕ ⟩ = \int g ϕ = ⟨ u_{g}, ϕ ⟩$ ; hence $w_{s} F (E_{s} U) = u_{g}$ in $S^{'}$ , so $E_{s} U \in M_{s}$ , and [F2] gives $∥ U ∥_{H^{s}} = ∥ g ∥_{2}$ .

2.2A1F1F2F3F5step 1.1

Conversely, let $u \in S^{'}$ and suppose $w_{s} F u = u_{g}$ for some $g \in L^{2}$ . Countable Choice [A1] is the inherited hypothesis for [F2]; its bijection $J_{s}$ gives the unique $U = J_{s}^{- 1} g$ . For every $ϕ \in S$ , the inverse multiplier action [F1] and bilinear pairing [F5] give $⟨ F u, ϕ ⟩ = ⟨ w_{- s} (w_{s} F u), ϕ ⟩ = ⟨ w_{- s} u_{g}, ϕ ⟩ = ⟨ u_{g}, w_{- s} ϕ ⟩ = \int g w_{- s} ϕ = ⟨ u_{w_{- s} g}, ϕ ⟩$ . By [F2] and step 1.1 this is $⟨ F (E_{s} U), ϕ ⟩$ ; Fourier injectivity [F3] yields $u = E_{s} U$ .

3.1F2step 2.2∎

If $h \in L^{2}$ also satisfies $w_{s} F u = u_{h}$ , applying step 2.2 to both $g$ and $h$ gives $E_{s} (J_{s}^{- 1} g) = u = E_{s} (J_{s}^{- 1} h)$ . Injectivity of $E_{s}$ [F2] yields $J_{s}^{- 1} g = J_{s}^{- 1} h$ , hence $g = h$ ; the isometry [F2] gives $∥ u ∥_{H^{s}} = ∥ J_{s}^{- 1} g ∥_{H^{s}} = ∥ g ∥_{2}$ . This proves the claimed bijection and norm identity.

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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)