Whole-space interpolation of one extension–restriction operator

Use the self-adjoint multiplication operator $M_{1+\mathrm{\mid }\xi {\mathrm{\mid }}^2}$ and the Plancherel unitary to define $B={\mathcal{F}}_2^{-1}{M}_{1+\mathrm{\mid }\xi {\mathrm{\mid }}^2}{\mathcal{F}}_2$ with domain ${\mathcal{F}}_2^{-1}D(M_{1+\mathrm{\mid }\xi {\mathrm{\mid }}^2})$. The transport lemma in [F7], with $U={\mathcal{F}}_2^{-1}$, makes $B$ self-adjoint with spectral PVM $P_B(C)={\mathcal{F}}_2^{-1}{P}_M(C){\mathcal{F}}_2$. Its scalar spectral measure at $F$ is the multiplier's scalar spectral measure at ${\mathcal{F}}_{2}F$. The multiplication model and the calculus domain and norm identities therefore give $D(B^{s\mathrm{/}2})=\{F\in L^2:⟨\xi {⟩}^s{\mathcal{F}}_{2}F\in L^2\}$ and $\mathrm{\parallel }{B}^{s\mathrm{/}2}F{\mathrm{\parallel }}_2=\mathrm{\parallel }⟨\xi {⟩}^s{\mathcal{F}}_{2}F{\mathrm{\parallel }}_2$ for $s\ge 0$. The multiplier is at least one, so $B\ge I$; the calculus is transported first for simple bounded multipliers and then by the bounded-truncation limits in [F7]. By [F2] and Plancherel these domains are exactly the named $H^s$ with their Fourier norms. The whole-space $L^2$ Hilbert space is nonzero for $n\ge 1$, so [F4] applies and gives $[L^2({\mathbb{R}}^n),H^2({\mathbb{R}}^n){]}_{\theta }=H^{2\theta }({\mathbb{R}}^n)$ isometrically in these norms.

The operator $T$ is defined first on $L^2$ as $E{r}_D$. Its $L^2$ bound follows from restriction and the assumed $E$ bound. If $F\in H^2({\mathbb{R}}^n)$, [F2] identifies it with a complex weak $W^{2,2}({\mathbb{R}}^n)$ class with equivalent norm. Apply [F6] with $\mathrm{\Omega }={\mathbb{R}}^n$, $U=D$, $k=2$ and $p=2$: $r_{D}F\in W^{2,2}(D)$ and $\mathrm{\parallel }{r}_{D}F{\mathrm{\parallel }}_{W^{2,2}(D)}\le \mathrm{\parallel }F{\mathrm{\parallel }}_{W^{2,2}({\mathbb{R}}^n)}\le C\mathrm{\parallel }F{\mathrm{\parallel }}_{H^2}$. The assumed endpoint bound for the same $E$ and [F2] now give $TF\in W^{2,2}({\mathbb{R}}^n)=H^2({\mathbb{R}}^n)$ with $\mathrm{\parallel }TF{\mathrm{\parallel }}_{H^2}\le C\mathrm{\parallel }F{\mathrm{\parallel }}_{H^2}$. Thus the two endpoint restrictions are the same complex-linear map. Apply common-operator spectral interpolation with endpoints $0,2$ to obtain $\mathrm{\parallel }TF{\mathrm{\parallel }}_{H^s}\le M_0^{1-s\mathrm{/}2}{M}_2^{s\mathrm{/}2}\mathrm{\parallel }F{\mathrm{\parallel }}_{H^s}$ for $0<s<2$, retaining the already established bounds at $s=0,2$. No domain interpolation assertion is used in obtaining this whole-space bound.

For $s\ge 0$ the weighted norm controls $L^2$, so $r_D:H^s({\mathbb{R}}^n)\to L^2(D)$ is continuous. Its kernel is closed, and its quotient is Banach with the displayed infimum norm by the quotient theorem. All restriction classes are $L^2$ classes, so this also defines their restrictions as distributions and makes the norm definite: $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(D)}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{H^s(D)}$.

If $u=r_{D}F$ for $F\in H^s$, then $Eu=E{r}_{D}F=TF$ as an $L^2$ class, and the whole-space bound makes it an $H^s$ class. If $F$ is replaced by another extension, their restrictions agree in $L^2(D)$, so the value of the already defined $E$ is unchanged. Hence $\mathrm{\parallel }Eu{\mathrm{\parallel }}_{H^s}\le C_s\mathrm{\parallel }F{\mathrm{\parallel }}_{H^s}$ for every extension $F$. Infimizing gives $\mathrm{\parallel }Eu{\mathrm{\parallel }}_{H^s}\le C_s\mathrm{\parallel }u{\mathrm{\parallel }}_{H^s(D)}$, while $r_{D}Eu=u$ follows from its $L^2$ identity. This proves the quotient estimate without choosing a fractional extension operator or assuming a fractional-domain identification.