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Conjugate duality of H^s and H^{-s}

Statement

Assume Countable Choice and use the first-variable-linear complex inner product conventions. Let $n \geq 1$ , $s \in R$ , and let $H^{s} = H^{s} (R^{n})$ be the real-order Bessel-potential completion with canonical embedding $E_{s} : H^{s} \to S^{'} (R^{n})$ (Real-order Bessel-potential completion H^s, The Bessel completion embeds canonically in tempered distributions). By Real-order H^s as weighted Fourier distributions, for $u \in H^{s}$ the distributional product $⟨ ξ ⟩^{s} F (E_{s} u)$ is the regular distribution of a unique class $g \in L^{2} (R^{n})$ , and $∥ u ∥_{H^{s}} = ∥ g ∥_{2}$ ; likewise $⟨ ξ ⟩^{- s} F (E_{- s} v)$ has a unique $L^{2}$ class $h$ for $v \in H^{- s}$ .

The conjugate dual. A functional $B : H^{s} \to C$ is conjugate-linear when $B (a u + b w) = \overline{a} B (u) + \overline{b} B (w)$ for all $u, w \in H^{s}$ and $a, b \in C$ . It is bounded when $∥ B ∥ : = \sup {∣ B (u) ∣ : ∥ u ∥_{H^{s}} \leq 1} < \infty$ . The set $(H^{s})^{†}$ of bounded conjugate-linear functionals, with this norm, is the conjugate dual of $H^{s}$ . The ordinary dual $(H^{s})^{*}$ is the set of bounded complex-linear functionals with the same norm.

The pairing. For $v \in H^{- s}$ and $u \in H^{s}$ let $g, h \in L^{2}$ be as above and define $A_{v} (u) = \int_{R^{n}} \overline{g (ξ)} h (ξ) d ξ .$ This is the $L^{2}$ pairing of the weighted Fourier classes $⟨ ξ ⟩^{s} F (E_{s} u)$ and $⟨ ξ ⟩^{- s} F (E_{- s} v)$ ; since $F (E_{s} u) = u_{⟨ ξ ⟩^{- s} g}$ and $F (E_{- s} v) = u_{⟨ ξ ⟩^{s} h}$ are regular distributions, the integrand is the pointwise product $\overline{F u} F v$ of their densities, so the displayed integral is what $\int_{R^{n}} \overline{F u (ξ)} F v (ξ) d ξ$ means.

Then:

$A_{v}$ is a bounded conjugate-linear functional on $H^{s}$ , the map $v \mapsto A_{v}$ is complex-linear, and it is isometric: $∥ A_{v} ∥ = ∥ v ∥_{H^{- s}}$ .
$v \mapsto A_{v}$ is a bijection $H^{- s} \to (H^{s})^{†}$ .
The ordinary linear dual is obtained by conjugating this pairing: with $C_{v} (u) : = \overline{A_{v} (u)} = \int_{R^{n}} g (ξ) \overline{h (ξ)} d ξ,$ the map $v \mapsto C_{v}$ is a conjugate-linear isometric bijection $H^{- s} \to (H^{s})^{*}$ .

This pairing is the weighted $L^{2}$ pairing of the two Fourier classes; it extends the $L^{2}$ conjugate pairing on Schwartz tests and is not asserted as a bilinear distribution action on arbitrary pairs of elements of $H^{s}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , the completion $H^{s}$ and its conjugate dual $(H^{s})^{†}$ .

[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 $σ \in R$ the canonical embedding restricts to a bijection $E_{σ} : H^{σ} \to W_{σ}$ onto the set of tempered distributions $u$ for which $⟨ ξ ⟩^{σ} F u = u_{G}$ for a unique $G \in L^{2}$ , with $∥ U ∥_{H^{σ}} = ∥ G ∥_{2}$ (Real-order H^s as weighted Fourier distributions).

[F2]

$H^{s}$ is a complex Hilbert space for the first-variable-linear inner product $(U, V)_{H^{s}} = \int_{R^{n}} (J_{s} U) (ξ) \overline{(J_{s} V) (ξ)} d ξ$ , whose induced norm is the defining completion norm (Every real-order Bessel-potential completion is Hilbert).

[F3]

The weighted Fourier map $J_{σ} : H^{σ} \to L^{2}$ is a surjective linear isometry with $E_{σ} ([u_{j}]) = F^{- 1} (u_{⟨ ξ ⟩^{- σ} J_{σ} [u_{j}]})$ (The Bessel completion embeds canonically in tempered distributions).

[F4]

For a bounded linear functional $B^{'}$ on a complex Hilbert space $H$ there is a unique $w \in H$ with $B^{'} (u) = (u, w)_{H}$ for all $u$ , and $∥ B^{'} ∥ = ∥ w ∥$ (Riesz representation for Hilbert spaces).

[F5]

The complex $L^{2}$ pairing $(f, j) \mapsto \int f \overline{j}$ is first-variable-linear and conjugate-symmetric on classes and satisfies Cauchy–Schwarz $∣ \int f \overline{j} ∣ \leq ∥ f ∥_{2} ∥ j ∥_{2}$ (Complex completeness, density, and inner product: the consumer interface).

Proof

technique · transport the weighted $L^2$ pairing along the surjective Fourier isometries and conjugate the Riesz representation

1.1F1F3F5given

Well-definedness. Let $u \in H^{s}$ , $v \in H^{- s}$ , and let $g, h \in L^{2}$ be the unique classes of [F1], so $g = ⟨ ξ ⟩^{s} F (E_{s} u)$ and $h = ⟨ ξ ⟩^{- s} F (E_{- s} v)$ in the sense of that statement. By [F5] the product $\overline{g} h$ is integrable, $∣ A_{v} (u) ∣ \leq ∥ g ∥_{2} ∥ h ∥_{2}$ , and $A_{v} (u)$ depends only on the classes $g, h$ . Moreover $F (E_{s} u)$ and $F (E_{- s} v)$ are the regular distributions of $⟨ ξ ⟩^{- s} g$ and $⟨ ξ ⟩^{s} h$ by [F3] and [F1], and $\overline{⟨ ξ ⟩^{- s} g} ⟨ ξ ⟩^{s} h = \overline{g} h$ pointwise, so the displayed integral is the density product $\overline{F u} F v$ .

1.2F1F3F5given

Sesquilinearity. If $u \mapsto u^{'} = λ u$ then the class of [F1] is $λ g$ , and $\int \overline{λ g} h = \overline{λ} A_{v} (u)$ ; hence $A_{v}$ is conjugate-linear in $u$ . If $v \mapsto a v + b v^{'}$ then the class is $a h + b h^{'}$ by linearity of $J_{- s}$ [F3], and $\int \overline{g} (a h + b h^{'}) = a A_{v} (u) + b A_{v^{'}} (u)$ ; hence $v \mapsto A_{v} (u)$ is complex-linear for each fixed $u$ .

2.1F1F5step 1.1step 1.2

The bound. For $u \in H^{s}$ and $v \in H^{- s}$ , Cauchy–Schwarz [F5] and the norm identities [F1] give $∣ A_{v} (u) ∣ \leq ∥ g ∥_{2} ∥ h ∥_{2} = ∥ u ∥_{H^{s}} ∥ v ∥_{H^{- s}} .$ Hence $A_{v}$ is a bounded conjugate-linear functional and $∥ A_{v} ∥ \leq ∥ v ∥_{H^{- s}}$ ; combined with step 1.2, $v \mapsto A_{v}$ maps $H^{- s}$ linearly into $(H^{s})^{†}$ .

3.1F1F3F5step 2.1

Attainment and isometry. Let $v \neq 0$ and put $h$ as above, so $h \neq 0$ ; set $g : = h / ∥ h ∥_{2} \in L^{2}$ and $u : = J_{s}^{- 1} g \in H^{s}$ , which exists and has $∥ u ∥_{H^{s}} = ∥ g ∥_{2} = 1$ by the surjective isometry property [F3], with $g$ the class of [F1]. Then $A_{v} (u) = \int \overline{(h / ∥ h ∥_{2})} h = \frac{∥ h ∥_{2}^{2}}{∥ h ∥_{2}} = ∥ h ∥_{2} = ∥ v ∥_{H^{- s}} .$ Thus $∥ A_{v} ∥ \geq ∥ v ∥_{H^{- s}}$ , and with step 2.1, $∥ A_{v} ∥ = ∥ v ∥_{H^{- s}}$ . If $v = 0$ then $h = 0$ , $A_{v} = 0$ and $∥ A_{v} ∥ = 0 = ∥ v ∥_{H^{- s}}$ ; the identity holds in all cases.

4.1step 1.2step 3.1

Injectivity. If $A_{v} = A_{v^{'}}$ then step 1.2 gives $A_{v - v^{'}} = A_{v} - A_{v^{'}} = 0$ , so step 3.1 yields $∥ v - v^{'} ∥_{H^{- s}} = ∥ A_{v - v^{'}} ∥ = 0$ , hence $v = v^{'}$ . Thus $v \mapsto A_{v}$ is injective.

4.2F2F3F4step 2.1step 3.1

Surjectivity and the conjugate dual. Let $B \in (H^{s})^{†}$ and define $B^{'} (u) : = \overline{B (u)}$ . Then $B^{'}$ is complex-linear and $∣ B^{'} (u) ∣ = ∣ B (u) ∣ \leq ∥ B ∥ ∥ u ∥_{H^{s}}$ , so [F4] provides a unique $w \in H^{s}$ with $B^{'} (u) = (u, w)_{H^{s}}$ for all $u$ and $∥ w ∥_{H^{s}} = ∥ B ∥$ . Put $h : = J_{s} w \in L^{2}$ and $v : = J_{- s}^{- 1} h \in H^{- s}$ , which exists by [F3] and has $∥ v ∥_{H^{- s}} = ∥ h ∥_{2} = ∥ w ∥_{H^{s}}$ . For $u \in H^{s}$ with class $g = J_{s} u$ , [F2] gives $(u, w)_{H^{s}} = \int g \overline{h}$ , so $A_{v} (u) = \int \overline{g} h = \overline{\int g \overline{h}} = \overline{(u, w)_{H^{s}}} = \overline{B^{'} (u)} = B (u) .$ Hence $B = A_{v}$ , the map is onto $(H^{s})^{†}$ , and step 3.1 gives $∥ B ∥ = ∥ v ∥_{H^{- s}}$ .

5.1step 1.1step 1.2step 4.1step 4.2

The ordinary dual. Define $C_{v} (u) : = \overline{A_{v} (u)}$ . Conjugation $B \mapsto \overline{B} (\cdot)$ is an isometric bijection from $(H^{s})^{†}$ onto $(H^{s})^{*}$ , because it exchanges conjugate-linear and complex-linear functionals and preserves $∣ \cdot ∣$ pointwise; composing with the bijection $v \mapsto A_{v}$ of steps 4.1 and 4.2, the map $v \mapsto C_{v}$ is an isometric bijection $H^{- s} \to (H^{s})^{*}$ . It is conjugate-linear in $v$ : by step 1.2, $C_{a v + b v^{'}} (u) = \overline{a A_{v} (u) + b A_{v^{'}} (u)} = \overline{a} C_{v} (u) + \overline{b} C_{v^{'}} (u)$ . By the definition of $A_{v}$ in the Statement, $C_{v} (u) = \int g \overline{h}$ , so $C_{v}$ is the weighted Fourier pairing of $u$ with $v$ .

6.1A1step 1.1step 1.2step 3.1step 4.1step 4.2step 5.1∎

Conclusion. Step 1.1 and step 1.2 establish well-definedness and sesquilinearity; step 3.1 gives $∥ A_{v} ∥ = ∥ v ∥_{H^{- s}}$ ; steps 4.1 and 4.2 make $v \mapsto A_{v}$ a bijection onto the conjugate dual; and step 5.1 transfers this to the ordinary dual with the conjugate-linear isometric dependence. This proves statements 1-3 for arbitrary $n \geq 1$ and $s \in R$ . Countable Choice is the stated hypothesis [A1], used through the cited completion, characterization, and Riesz interfaces in those steps.

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Sources

Semyon Dyatlov, Lecture Notes for 18.155, current revision (standard reference, not scraped)
Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (standard reference, not scraped)