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Real-order H^s as weighted Fourier distributions

Statement

Assume Countable Choice, let $n \geq 1$ and $s \in R$ , and let $H^{s} (R^{n})$ be the real-order Bessel-potential completion of Real-order Bessel-potential completion H^s with its canonical embedding $E_{s} : H^{s} (R^{n}) \to S^{'} (R^{n})$ (The Bessel completion embeds canonically in tempered distributions). Write $⟨ ξ ⟩ = (1 + ∣ ξ ∣^{2})^{1 / 2}$ for the Japanese bracket, $F$ for the negative-sign $2 π$ -normalized Fourier transform, and $u_{g} (ϕ) = \int_{R^{n}} g (ξ) ϕ (ξ) d ξ$ for the regular tempered distribution of $g \in L_{l o c}^{1} (R^{n})$ . Define $W_{s} = {u \in S^{'} (R^{n}) : ⟨ ξ ⟩^{s} F u = u_{g} in S^{'} (R^{n}) for some g \in L^{2} (R^{n})} .$ Then:

The canonical embedding restricts to a bijection $E_{s} : H^{s} (R^{n}) \to W_{s}$ . Thus, after identifying a completion class with its image under $E_{s}$ , $H^{s} (R^{n})$ is exactly the set of tempered distributions whose bracket-weighted Fourier transform is the regular distribution of an $L^{2}$ class, and that class $g$ is unique.
The norm identity is exact: $∥ U ∥_{H^{s}} = ∥ g ∥_{2} = ∥ ⟨ ξ ⟩^{s} F (E_{s} U) ∥_{2} (U \in H^{s}),$ where the last expression means the $L^{2}$ norm of the unique density $g$ of the distributional product $⟨ ξ ⟩^{s} F (E_{s} U)$ , and the defining completion norm $q_{s} (ϕ) = ∥ ⟨ ξ ⟩^{s} \hat{ϕ} ∥_{2}$ of Real-order Bessel-potential completion H^s is not renormalized.
The product $⟨ ξ ⟩^{s} F u$ is distributional multiplication of $F u$ by the smooth polynomially bounded bracket weight, not an a priori pointwise product; and $H^{s}$ elements are completion classes, not initially assumed to be functions.

Nothing here replaces the bracket weight by the Laplacian weight $(1 + 4 π^{2} ∣ ξ ∣^{2})^{s / 2}$ , and no pointwise value of $g$ at an individual frequency is asserted before $g$ is obtained from the defining condition.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , the Japanese bracket $⟨ ξ ⟩ \geq 1$ , 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]

For every $n \geq 1$ and $s \in R$ , with $M_{s} = {u \in S^{'} (R^{n}) : ⟨ ξ ⟩^{s} F u = u_{g}$ in $S^{'} (R^{n})$ for some $g \in L^{2} (R^{n})}$ , the canonical embedding $E_{s}$ restricts to a bijection $E_{s} : H^{s} (R^{n}) \to M_{s}$ ; the class $g$ is unique; and if $u = E_{s} U$ corresponds to $g$ , then $∥ U ∥_{H^{s}} = ∥ g ∥_{2}$ . The product is multiplication of a tempered distribution by the smooth bracket multiplier (Weighted tempered-distribution characterization of H^s).

[F2]

$H^{s} (R^{n})$ is the normed-space completion of $S$ in the positive-definite norm $q_{s} (ϕ) = ∥ ⟨ ξ ⟩^{s} \hat{ϕ} ∥_{2}$ : its elements are Cauchy-sequence classes $[u_{j}]$ modulo zero limiting distance, $∥ [u_{j}] ∥_{H^{s}} = \lim_{j} q_{s} (u_{j})$ , and the constant-sequence map is the canonical dense linear isometry (Real-order Bessel-potential completion H^s).

[F3]

Weighted Fourier transformation extends to a surjective linear isometry $J_{s} : H^{s} (R^{n}) \to L^{2} (R^{n})$ , $J_{s} ([u_{j}]) = \lim_{j} ⟨ ξ ⟩^{s} F (u_{j})$ , and $E_{s} ([u_{j}]) = F^{- 1} (u_{⟨ ξ ⟩^{- s} J_{s} [u_{j}]})$ defines a well-defined continuous linear injection that is independent of the representing Cauchy sequence (The Bessel completion embeds canonically in tempered distributions).

[F4]

For real $s$ the multipliers $⟨ ξ ⟩^{s}$ and $⟨ ξ ⟩^{- s}$ act continuously and invertibly on $S^{'} (R^{n})$ by transposition, so $⟨ ξ ⟩^{s} F u$ is defined for every tempered distribution $u$ (Real powers of the Japanese bracket act on Schwartz space).

[F5]

Multiplication of a tempered distribution by a smooth polynomially bounded symbol $a$ is $⟨ a u, φ ⟩ = ⟨ u, a φ ⟩$ ; with the regular distribution $⟨ u_{h}, φ ⟩ = \int h φ$ of a locally integrable $h$ (Regular distribution from a locally integrable function) the same display with $u = u_{h}$ gives $a u_{h} = u_{a h}$ ; the bracket weights $⟨ ξ ⟩^{\pm s}$ are such symbols (Smooth polynomially bounded multipliers on schwartz space).

Proof

technique · unfold the two definitions of the same weighted set and transfer the batch-12 bijection, uniqueness and isometry

1.1F1F4given

The set displayed in the statement is exactly the set $M_{s}$ of [F1]: both consist of the tempered $u$ for which $⟨ ξ ⟩^{s} F u = u_{g}$ for some $g \in L^{2}$ , with the same convention that the product is the distributional multiplication of [F4]; hence $W_{s} = M_{s}$ as subsets of $S^{'} (R^{n})$ .

2.1F1F3F5step 1.1

For $U \in H^{s}$ put $g : = J_{s} U \in L^{2}$ and $u : = E_{s} U$ . By [F3] and [F1] one has $F u = u_{⟨ ξ ⟩^{- s} g}$ , and $h : = ⟨ ξ ⟩^{- s} g$ is locally integrable, so [F5] applied with the smooth symbol $a = ⟨ ξ ⟩^{s}$ gives $⟨ ξ ⟩^{s} F u = ⟨ ξ ⟩^{s} u_{⟨ ξ ⟩^{- s} g} = u_{⟨ ξ ⟩^{s} ⟨ ξ ⟩^{- s} g} = u_{g} .$ Hence the unique $L^{2}$ class attached to $u$ by [F1] is $g$ itself, and the batch-12 norm identity gives $∥ U ∥_{H^{s}} = ∥ g ∥_{2} = ∥ ⟨ ξ ⟩^{s} F (E_{s} U) ∥_{2}$ , the last norm being that of the density $g$ of the product.

3.1F2step 2.1

The case $s = 0$ is the instance $⟨ ξ ⟩^{0} = 1$ : the defining condition becomes $F u = u_{g}$ for some $g \in L^{2}$ , the norm identity reads $∥ U ∥_{H^{0}} = ∥ F (E_{0} U) ∥_{2}$ , and the completion norm is $q_{0} (ϕ) = ∥ \hat{ϕ} ∥_{2}$ by [F2]. Nothing in steps 1.1 and 2.1 divides by a vanishing weight or degenerates; the instance is included in the general claims.

4.1F1F4step 1.1step 2.1step 3.1

The bijection. By [F1] the map $E_{s} : H^{s} \to M_{s}$ is a bijection, and by step 1.1 $M_{s} = W_{s}$ ; hence the canonical embedding restricts to the bijection $E_{s} : H^{s} \to W_{s}$ , and uniqueness of the class $g$ is part of [F1]. Reading $E_{s}$ as the canonical identification, $H^{s}$ consists exactly of the tempered distributions whose bracket-weighted Fourier transform is a regular $L^{2}$ distribution, with the exact norm of statement 2. In particular no element of $H^{s}$ is assumed to be a function, and the product is the distributional multiplication of [F4] rather than a pointwise product.

5.1A1F1F2F3step 1.1step 2.1step 3.1step 4.1∎

Conclusion. Step 4.1 gives the set identity, the canonical bijection and uniqueness; step 2.1 gives the exact norm; step 3.1 covers $s = 0$ ; and step 1.1 records the distributional-product convention. This proves statements 1-3 for arbitrary $n \geq 1$ and $s \in R$ . Countable Choice is used exactly through the completion, embedding and characterization interfaces [F1]-[F3], which carry it as their hypothesis.

Depends on

Used by

Conjugate duality of Hˢ and H⁻ˢ Corollary
A Dirac mass has precisely sufficiently negative Sobolev order Example
Bessel potentials shift Sobolev order Lemma

Dependency tree · two levels

29 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, current revision (standard reference, not scraped)
Mark Williams, Notes on Harmonic Analysis (standard reference, not scraped)