Bessel-Potential Completions and Real-Order Sobolev Spaces

Put $q(\xi )=1+\mathrm{\mid }\xi {\mathrm{\mid }}^2$. Induction on $\mathrm{\mid }\alpha \mathrm{\mid }$, differentiating either the polynomial factor or $q^{s\mathrm{/}2-j}$, expresses each derivative as a finite sum ${\mathrm{\partial }}^{\alpha }{w}_s(\xi )={\sum }_j{P}_{\alpha ,j}(\xi )q(\xi {)}^{s\mathrm{/}2-j},$ where every $P_{\alpha ,j}$ is a polynomial of degree at most $\mathrm{\mid }\alpha \mathrm{\mid }$. Since $q=⟨\xi {⟩}^2$, each term is bounded by a constant times $⟨\xi {⟩}^{s+\mathrm{\mid }\alpha \mathrm{\mid }}$, and hence by $C_{\alpha ,s}⟨\xi {⟩}^{\max (0,s+\mathrm{\mid }\alpha \mathrm{\mid })}$. Enlarging the exponent to an integer gives a polynomial-growth bound for this derivative.

The same induction with $-s$ gives a polynomial-growth bound for every derivative of $w_{-s}$.

The bounds in steps 1.1 and 2.1 meet the hypotheses of [F2], so multiplication by either weight is continuous on Schwartz space. Pointwise $w_s{w}_{-s}=1$, so both compositions on $\mathcal{S}$ are the identity.

For $T\in {\mathcal{S}}^{\mathrm{\prime }}$, transposition defines $⟨w_{\pm s}T,\phi ⟩=⟨T,w_{\pm s}\phi ⟩$. By [F2] these maps are continuous for the weak and strong dual topologies; their compositions evaluate $T$ on $w_s{w}_{-s}\phi =\phi$, so they are inverse on ${\mathcal{S}}^{\mathrm{\prime }}$.

Suppose $q_s(u)=0$. By [F1], $\mathrm{\parallel }{w}_s\widehat{u}{\mathrm{\parallel }}_2=0$, so $w_s\widehat{u}=0$ almost everywhere.

Since $w_s(\xi )=⟨\xi {⟩}^s>0$ at every $\xi$, step 1.1 implies $\widehat{u}=0$ almost everywhere; Plancherel [F2] then gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2=\mathrm{\parallel }\widehat{u}{\mathrm{\parallel }}_2=0$.

If $u(x_0)\mathrm{\ne }0$, continuity from [F3] gives a ball on which $\mathrm{\mid }u\mathrm{\mid }>\mathrm{\mid }u(x_0)\mathrm{\mid }\mathrm{/}2$; its positive Lebesgue measure contradicts $\mathrm{\parallel }u{\mathrm{\parallel }}_2=0$. Therefore the actual Schwartz function vanishes everywhere.

By [F1], $Q_s$ is the complex $L^2$ inner product of the weighted Fourier images, hence is linear in the first variable, conjugate symmetric, and nonnegative on the diagonal; step 3.1 makes it positive definite, and [F1] gives $Q_s(u,u{)}^{1\mathrm{/}2}=q_s(u)$.

Conversely, if $u=0$, its Fourier transform vanishes and the defining formula [F1] gives $q_s(u)=0$; thus the kernel is exactly $\{0\}$.

Fix an arbitrary $h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n;\mathbb{C})$. By [F5], its components and every derivative are continuous with compact support, so each ${\xi }^{\alpha }{\mathrm{\partial }}^{\beta }h$ is bounded and [F6] gives $h\in \mathcal{S}$; [F1] then gives $q=⟨\xi {⟩}^{-s}h\in \mathcal{S}$.

By [F2], $u={\mathcal{F}}^{-1}q$ belongs to Schwartz space, and pointwise $⟨\xi {⟩}^s\mathcal{F}(u)=⟨\xi {⟩}^{s}q=h$. Thus every such $h$ is in the weighted Fourier image.

For any $g\in L^2$ and $\epsilon >0$, [F3] with $p=2$ supplies $h\in C_c^{\mathrm{\infty }}$ with $\mathrm{\parallel }g-h{\mathrm{\parallel }}_2<\epsilon$; step 2.1 puts this same $h$ in the weighted Fourier image, proving that image dense in $L^2$.

For $U=[u_j]$, put $g_j=⟨\xi {⟩}^s\mathcal{F}(u_j)$. The completion norm identity gives $\mathrm{\parallel }{g}_j-g_k{\mathrm{\parallel }}_2=q_s(u_j-u_k)$, 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 $J_{s}U=g$.

Given $g\in L^2$, [F2] makes the weighted Schwartz image dense; for each $j$ choose $u_j\in \mathcal{S}$ with $\mathrm{\parallel }⟨\xi {⟩}^s\mathcal{F}(u_j)-g{\mathrm{\parallel }}_2<2^{-j}$. Countable Choice [A1] selects this sequence.

For $g\in L^2$ define $T_{s}g(\varphi )={\int }_{{\mathbb{R}}^n}g(\xi )⟨\xi {⟩}^{-s}\varphi (\xi )d\xi$. By [F1], $⟨\xi {⟩}^{-s}\varphi \in \mathcal{S}$, and [F6] puts it in $L^2$; the integral is the pairing $(g,\overline{⟨\xi {⟩}^{-s}\varphi }{)}_2$, so [F5] gives absolute convergence independent of the representative of $g$. The function $⟨\xi {⟩}^{-s}g$ is locally integrable because its weight is bounded on compact sets.

Termwise addition and scalar multiplication commute with the $L^2$ limit, and $\mathrm{\parallel }{J}_{s}U{\mathrm{\parallel }}_2={\lim }_j\mathrm{\parallel }{g}_j{\mathrm{\parallel }}_2={\lim }_j{q}_s(u_j)=\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}$; thus $J_s$ is a linear isometry.

The norm identity $q_s(u_j-u_k)=\mathrm{\parallel }⟨\xi {⟩}^s\mathcal{F}(u_j)-⟨\xi {⟩}^s\mathcal{F}(u_k){\mathrm{\parallel }}_2$ 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.

Choose an integer $N>\mathrm{\mid }s\mathrm{\mid }+n\mathrm{/}2$. Polynomial expansion gives $⟨\xi {⟩}^N\mathrm{\mid }\varphi (\xi )\mathrm{\mid }\le C{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le N}{p}_{\alpha 0}(\varphi )$, while dyadic shells show $\int ⟨\xi {⟩}^{-2(N-\mathrm{\mid }s\mathrm{\mid })}d\xi <\mathrm{\infty }$; hence $\mathrm{\parallel }⟨\xi {⟩}^{-s}\varphi {\mathrm{\parallel }}_2\le C^{\mathrm{\prime }}{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le N}{p}_{\alpha 0}(\varphi )$. Cauchy–Schwarz [F5] now bounds $\mathrm{\mid }{T}_{s}g(\varphi )\mathrm{\mid }$ by this finite-seminorm expression times $\mathrm{\parallel }g{\mathrm{\parallel }}_2$, proving temperateness by [F7]. For bounded $B\subset \mathcal{S}$, [F8] and [F11] make the same bound uniform over $\varphi \in B$, so $g\mapsto T_{s}g$ is continuous for both dual topologies.

Define $E_{s}U={\mathcal{F}}^{-1}(T_s(J_{s}U))$. 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.

For the canonical class $i(u)$ of $u\in \mathcal{S}$, $J_s(i(u))=⟨\xi {⟩}^s\widehat{u}$, so $T_s(J_s(i(u)))=u_{\widehat{u}}$. By [F10], $\mathcal{F}{u}_u=u_{\widehat{u}}$; invertibility [F9] gives $E_s(i(u))=u_u$, the functional $\varphi \mapsto \int u\varphi$. 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.

If $E_{s}U=0$ and $g=J_{s}U$, Fourier injectivity [F9] gives $T_{s}g=0$. Multiplication by $⟨\xi {⟩}^s$ is allowed on ${\mathcal{S}}^{\mathrm{\prime }}$ by [F1]; for each $\varphi \in \mathcal{S}$, $⟨⟨\xi {⟩}^s{T}_{s}g,\varphi ⟩=T_{s}g(⟨\xi {⟩}^s\varphi )=\int g\varphi =u_g(\varphi )$. Hence $u_g=0$.

By [F6] and Countable Choice [A1], choose ${\varphi }_j\in \mathcal{S}$ with ${\varphi }_j\to \bar{g}$ in $L^2$. Then $0=u_g({\varphi }_j)=\int g{\varphi }_j$; Cauchy–Schwarz [F5] yields $\int g{\varphi }_j\to \int \mathrm{\mid }g{\mathrm{\mid }}^2$, so $g=0$ in $L^2$. The isometry [F3, step 2.1] gives $U=0$, proving that $E_s$ is injective.

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.

For every $U\in H^s$, $(U,U{)}_{H^s}=\mathrm{\parallel }{J}_{s}U{\mathrm{\parallel }}_2^2\ge 0$ and $(U,U{)}_{H^s}^{1\mathrm{/}2}=\mathrm{\parallel }{J}_{s}U{\mathrm{\parallel }}_2=\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}$ by [F1, F2, F3]. If $(U,U{)}_{H^s}=0$, the isometry makes $\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}=0$, hence $U=0$; thus the pairing is positive definite and induces exactly the completion norm.

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 $\mathrm{\parallel }{U}_j-U{\mathrm{\parallel }}_{H^s}=\mathrm{\parallel }{J}_s{U}_j-g{\mathrm{\parallel }}_2\to 0$. Thus the induced norm is complete.

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

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

For every $\varphi \in \mathcal{S}$, the multiplier action [F1], bilinear pairing [F5], and [F2] give $⟨w_s\mathcal{F}(E_{s}U),\varphi ⟩=⟨u_{w_{-s}g},w_s\varphi ⟩=\int g\varphi =⟨u_g,\varphi ⟩$; hence $w_s\mathcal{F}(E_{s}U)=u_g$ in ${\mathcal{S}}^{\mathrm{\prime }}$, so $E_{s}U\in M_s$, and [F2] gives $\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}=\mathrm{\parallel }g{\mathrm{\parallel }}_2$.

Conversely, let $u\in {\mathcal{S}}^{\mathrm{\prime }}$ and suppose $w_s\mathcal{F}u=u_g$ 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 $\varphi \in \mathcal{S}$, the inverse multiplier action [F1] and bilinear pairing [F5] give $⟨\mathcal{F}u,\varphi ⟩=⟨w_{-s}(w_s\mathcal{F}u),\varphi ⟩=⟨w_{-s}{u}_g,\varphi ⟩=⟨u_g,w_{-s}\varphi ⟩=\int g{w}_{-s}\varphi =⟨u_{w_{-s}g},\varphi ⟩$. By [F2] and step 1.1 this is $⟨\mathcal{F}(E_{s}U),\varphi ⟩$; Fourier injectivity [F3] yields $u=E_{s}U$.

If $h\in L^2$ also satisfies $w_s\mathcal{F}u=u_h$, 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 $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^s}=\mathrm{\parallel }{J}_s^{-1}g{\mathrm{\parallel }}_{H^s}=\mathrm{\parallel }g{\mathrm{\parallel }}_2$. This proves the claimed bijection and norm identity.