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The Bessel completion embeds canonically in tempered distributions

Statement

Assume Countable Choice. For every $n \geq 1$ and $s \in R$ , weighted Fourier transformation extends from Schwartz space to a surjective linear isometry $J_{s} : H^{s} (R^{n}) ⟶ L^{2} (R^{n}), J_{s} ([u_{j}]) = \lim_{j \to \infty} ⟨ ξ ⟩^{s} F (u_{j})$ in $L^{2}$ . If $g = J_{s} ([u_{j}])$ , then $E_{s} ([u_{j}]) = F^{- 1} (u_{⟨ ξ ⟩^{- s} g})$ where $u_{h}$ denotes the functional $ϕ \mapsto \int h (ξ) ϕ (ξ) d ξ$ whenever this integral defines a tempered distribution. This is a well-defined continuous linear injection $H^{s} (R^{n}) \to S^{'} (R^{n})$ for both weak and strong dual topologies. It sends the canonical Schwartz class to its usual regular distribution and is independent of the representing Cauchy sequence.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , and a completion class $U \in H^{s} (R^{n})$ .

[A1]

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

[F1]

Both bracket powers are inverse continuous multipliers on Schwartz space and act invertibly on $S^{'}$ (Real powers of the Japanese bracket act on Schwartz space).

[F2]

The weighted Fourier image of Schwartz space is dense in complex $L^{2}$ (Weighted Fourier transforms of Schwartz functions are dense in L2).

[F3]

$H^{s}$ consists of norm-Cauchy Schwartz sequences modulo zero limiting distance, with the limiting norm and canonical dense constant-sequence map (Real-order Bessel-potential completion H^s).

[F4]

Complex $L^{2}$ is complete under Countable Choice (Complex completeness, density, and inner product: the consumer interface).

[F5]

The first-variable-linear complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz (Complex completeness, density, and inner product: the consumer interface).

[F6]

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

[F7]

A tempered distribution is a continuous complex-linear functional on Schwartz space, with bilinear test pairing (Tempered distribution).

[F8]

The weak topology tests individual Schwartz functions; the strong topology tests bounded subsets of Schwartz space, bounded in every Schwartz seminorm (Weak and strong topologies on tempered distributions).

[F9]

Fourier transformation is a topological automorphism of $S^{'}$ for both weak and strong topologies (Fourier transform is a topological automorphism of tempered distributions).

[F10]

The distributional Fourier transform agrees with the unitary Plancherel transform on regular $L^{2}$ distributions (Fourier transform agrees with l one and plancherel transforms).

[F11]

Schwartz seminorms are $p_{α β} (ϕ) = \sup_{ξ} ∣ ξ^{α} \partial^{β} ϕ (ξ) ∣$ (Schwartz space and its seminorms).

Proof

1.1F3F4given

For $U = [u_{j}]$ , put $g_{j} = ⟨ ξ ⟩^{s} F (u_{j})$ . The completion norm identity gives $∥ g_{j} - g_{k} ∥_{2} = q_{s} (u_{j} - u_{k})$ , so $(g_{j})$ is Cauchy; by [F4] it has an $L^{2}$ limit $g$ . Equivalent Cauchy sequences have difference norm tending to zero, hence the same limit. Define $J_{s} U = g$ .

1.2A1F2given

Given $g \in L^{2}$ , [F2] makes the weighted Schwartz image dense; for each $j$ choose $u_{j} \in S$ with $∥ ⟨ ξ ⟩^{s} F (u_{j}) - g ∥_{2} < 2^{- j}$ . Countable Choice [A1] selects this sequence.

1.3F1F5F6F7

For $g \in L^{2}$ define $T_{s} g (ϕ) = \int_{R^{n}} g (ξ) ⟨ ξ ⟩^{- s} ϕ (ξ) d ξ$ . By [F1], $⟨ ξ ⟩^{- s} ϕ \in S$ , and [F6] puts it in $L^{2}$ ; the integral is the pairing $(g, \overline{⟨ ξ ⟩^{- s} ϕ})_{2}$ , so [F5] gives absolute convergence independent of the representative of $g$ . The function $⟨ ξ ⟩^{- s} g$ is locally integrable because its weight is bounded on compact sets.

2.1F3step 1.1

Termwise addition and scalar multiplication commute with the $L^{2}$ limit, and $∥ J_{s} U ∥_{2} = \lim_{j} ∥ g_{j} ∥_{2} = \lim_{j} q_{s} (u_{j}) = ∥ U ∥_{H^{s}}$ ; thus $J_{s}$ is a linear isometry.

2.2F3step 1.2

The norm identity $q_{s} (u_{j} - u_{k}) = ∥ ⟨ ξ ⟩^{s} F (u_{j}) - ⟨ ξ ⟩^{s} F (u_{k}) ∥_{2}$ makes $(u_{j})$ Cauchy in the Schwartz norm $q_{s}$ . Its completion class $U = [u_{j}]$ satisfies $J_{s} U = g$ , so $J_{s}$ is onto.

2.3F5F7F8F11step 1.3

Choose an integer $N > ∣ s ∣ + n / 2$ . Polynomial expansion gives $⟨ ξ ⟩^{N} ∣ ϕ (ξ) ∣ \leq C \sum_{∣ α ∣ \leq N} p_{α 0} (ϕ)$ , while dyadic shells show $\int ⟨ ξ ⟩^{- 2 (N - ∣ s ∣)} d ξ < \infty$ ; hence $∥ ⟨ ξ ⟩^{- s} ϕ ∥_{2} \leq C^{'} \sum_{∣ α ∣ \leq N} p_{α 0} (ϕ)$ . Cauchy–Schwarz [F5] now bounds $∣ T_{s} g (ϕ) ∣$ by this finite-seminorm expression times $∥ g ∥_{2}$ , proving temperateness by [F7]. For bounded $B \subset S$ , [F8] and [F11] make the same bound uniform over $ϕ \in B$ , so $g \mapsto T_{s} g$ is continuous for both dual topologies.

3.1F3F8F9step 2.1step 2.3

Define $E_{s} U = F^{- 1} (T_{s} (J_{s} U))$ . It is linear and continuous for weak and strong dual topologies by the isometry [F3, step 2.1], the uniform estimate in step 2.3, and the continuous inverse Fourier transform [F9]; it depends only on $U$ because $J_{s}$ is well-defined.

4.1F1F3F8F9F10step 1.3step 3.1

For the canonical class $i (u)$ of $u \in S$ , $J_{s} (i (u)) = ⟨ ξ ⟩^{s} \hat{u}$ , so $T_{s} (J_{s} (i (u))) = u_{\hat{u}}$ . By [F10], $F u_{u} = u_{\hat{u}}$ ; invertibility [F9] gives $E_{s} (i (u)) = u_{u}$ , the functional $ϕ \mapsto \int u ϕ$ . If $U = [u_{j}]$ , then $d (i (u_{j}), U) = \lim_{k} q_{s} (u_{j} - u_{k}) \to 0$ by the Cauchy condition, so step 3.1 gives $u_{u_{j}} \to E_{s} U$ in both topologies and the map is independent of the representing sequence.

4.2F1F9step 1.3step 3.1

If $E_{s} U = 0$ and $g = J_{s} U$ , Fourier injectivity [F9] gives $T_{s} g = 0$ . Multiplication by $⟨ ξ ⟩^{s}$ is allowed on $S^{'}$ by [F1]; for each $ϕ \in S$ , $⟨ ⟨ ξ ⟩^{s} T_{s} g, ϕ ⟩ = T_{s} g (⟨ ξ ⟩^{s} ϕ) = \int g ϕ = u_{g} (ϕ)$ . Hence $u_{g} = 0$ .

5.1A1F3F5F6step 4.2∎

By [F6] and Countable Choice [A1], choose $ϕ_{j} \in S$ with $ϕ_{j} \to \overline{g}$ in $L^{2}$ . Then $0 = u_{g} (ϕ_{j}) = \int g ϕ_{j}$ ; Cauchy–Schwarz [F5] yields $\int g ϕ_{j} \to \int ∣ g ∣^{2}$ , so $g = 0$ in $L^{2}$ . The isometry [F3, step 2.1] gives $U = 0$ , proving that $E_{s}$ is injective.

Depends on

Used by

Conjugate duality of Hˢ and H⁻ˢ Corollary
Every real-order Bessel-potential completion is Hilbert Corollary
Every Schwartz function belongs to every real-order Hˢ Example
The zero-order Bessel completion is exactly L2 Example
Integer-order W^k,2 and Hᵏ agree with equivalent norms Theorem
Real-order Hˢ as weighted Fourier distributions Theorem
Weighted tempered-distribution characterization of Hˢ Theorem

Dependency tree · two levels

39 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)
Richard B. Melrose, Differential Analysis, Chapter 3 (standard reference, not scraped)