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Every real-order Bessel-potential completion is Hilbert

Statement

Assume Countable Choice. For every $n \geq 1$ and $s \in R$ , the space $H^{s} (R^{n})$ is a complex Hilbert space for the first-variable-linear inner product $(U, V)_{H^{s}} = \int_{R^{n}} (J_{s} U) (ξ) \overline{(J_{s} V) (ξ)} d ξ,$ where $J_{s}$ is the surjective weighted Fourier isometry from The Bessel completion embeds canonically in tempered distributions. Its induced norm is exactly the defining completion norm, and $H^{s}$ is complete.

Facts & Assumptions

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

[A1]

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

[F1]

The weighted Fourier map $J_{s} : H^{s} \to L^{2}$ is a surjective linear isometry (The Bessel completion embeds canonically in tempered distributions).

[F2]

Under Countable Choice, complex $L^{2}$ has the first-variable-linear inner product $\int f \overline{g}$ , its norm is the $L^{2}$ norm, and it is complete (Complex completeness, density, and inner product: the consumer interface).

[F3]

$H^{s}$ is the normed-space completion of Schwartz space with its defining completion norm (Real-order Bessel-potential completion H^s).

[F4]

A complex Hilbert space is a complex inner-product space complete for its induced norm (Hilbert space).

Proof

technique · Pull back the complex $L^2$ inner product along $J_s$

1.1F1F2given

Define $(U, V)_{H^{s}} : = (J_{s} U, J_{s} V)_{L^{2}}$ . The map $J_{s}$ is well-defined and linear by [F1], so this pairing is well-defined; the inner-product properties of the complex $L^{2}$ pairing [F2] give first-variable linearity and conjugate symmetry.

2.1F1F2F3step 1.1

For every $U \in H^{s}$ , $(U, U)_{H^{s}} = ∥ J_{s} U ∥_{2}^{2} \geq 0$ and $(U, U)_{H^{s}}^{1 / 2} = ∥ J_{s} U ∥_{2} = ∥ U ∥_{H^{s}}$ by [F1, F2, F3]. If $(U, U)_{H^{s}} = 0$ , the isometry makes $∥ U ∥_{H^{s}} = 0$ , hence $U = 0$ ; thus the pairing is positive definite and induces exactly the completion norm.

3.1A1F1F2step 2.1

Let $(U_{j})$ be Cauchy in this induced norm. By step 2.1 and [F1], $(J_{s} U_{j})$ is Cauchy in complex $L^{2}$ , so Countable Choice [A1] and [F2] give a limit $g \in L^{2}$ . Surjectivity [F1] gives the unique $U \in H^{s}$ with $J_{s} U = g$ , and the isometry yields $∥ U_{j} - U ∥_{H^{s}} = ∥ J_{s} U_{j} - g ∥_{2} \to 0$ . Thus the induced norm is complete.

4.1F4step 1.1step 2.1step 3.1∎

By [F4], steps 1.1 and 2.1 give a complex inner product whose induced norm is complete by step 3.1. Therefore $H^{s} (R^{n})$ is a complex Hilbert space with the stated inner product and norm.

Depends on

Used by

Conjugate duality of Hˢ and H⁻ˢ Corollary

Dependency tree · two levels

30 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)