Fractional Bessel-potential traces on simplex facets

First consider the hyperplane ${\mathbb{R}}^{d-1}\times \{0\}$ and $U\in \mathcal{S}({\mathbb{R}}^d)$. Write $G(x^{\mathrm{\prime }})=U(x^{\mathrm{\prime }},0)$. Fourier inversion in the last coordinate and Fubini on the absolutely integrable Schwartz integrands give $\widehat{G}({\xi }^{\mathrm{\prime }})={\int }_{\mathbb{R}}\widehat{U}({\xi }^{\mathrm{\prime }},\eta )d\eta$.

Apply Cauchy–Schwarz in $\eta$ with weights $⟨({\xi }^{\mathrm{\prime }},\eta ){⟩}^s$ and its reciprocal; this follows by expanding $0\le \mathrm{\parallel }f-\lambda g{\mathrm{\parallel }}_2^2$ and minimizing over the complex scalar $\lambda$. Both weighted factors lie in $L^2(\mathbb{R})$ for fixed ${\xi }^{\mathrm{\prime }}$, the reciprocal factor by [A1] and the weighted Fourier transform because $U$ is Schwartz. The reciprocal-square integral is ${\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^{-s}d\eta =C_s(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2-s}\le C_s,$ by the substitution $\eta =(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}t$ and [A1, F5]. Therefore $\mathrm{\mid }\widehat{G}({\xi }^{\mathrm{\prime }}){\mathrm{\mid }}^2\le C_s{\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^s\mathrm{\mid }\widehat{U}({\xi }^{\mathrm{\prime }},\eta ){\mathrm{\mid }}^{2}d\eta$. Tonelli [F4] and Plancherel [F1] give $\mathrm{\parallel }G{\mathrm{\parallel }}_{L^2({\mathbb{R}}^{d-1})}\le C_s\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s({\mathbb{R}}^d)}$.

Since Schwartz space is dense in the completion [F2], the estimate of step 2.1 makes the restrictions of every Cauchy representative a Cauchy sequence in $L^2$; completeness [F3] defines a bounded trace on the hyperplane. The same estimate applied to the difference of two representatives proves independence of the representative and linearity.

This trace is local. For Schwartz $W$, let $W_t(x^{\mathrm{\prime }}):=W(x^{\mathrm{\prime }},t)$. Partial Plancherel in the tangential variables, Fubini [F4], and the weighted Cauchy--Schwarz estimate of step 2.1 give $\mathrm{\parallel }{W}_t-W_0{\mathrm{\parallel }}_{L^2({\mathbb{R}}^{d-1})}^2\le {\int }_{{\mathbb{R}}^{d-1}}{J}_t({\xi }^{\mathrm{\prime }})G({\xi }^{\mathrm{\prime }})d{\xi }^{\mathrm{\prime }},$ where $J_t({\xi }^{\mathrm{\prime }}):={\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^{-s}\mathrm{\mid }{e}^{2\pi i\eta t}-1{\mathrm{\mid }}^{2}d\eta$ and $G({\xi }^{\mathrm{\prime }}):={\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^s\mathrm{\mid }\widehat{W}({\xi }^{\mathrm{\prime }},\eta ){\mathrm{\mid }}^{2}d\eta$. Choose $0<\delta <\min \{1,2s-1\}$. Since $\mathrm{\mid }{e}^{i\alpha }-1{\mathrm{\mid }}^2\le C_{\delta }\mathrm{\mid }\alpha {\mathrm{\mid }}^{\delta }$, the substitution $\eta =(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}r$ using [F5] gives $J_t({\xi }^{\mathrm{\prime }})\le C_{s,\delta }\mathrm{\mid }t{\mathrm{\mid }}^{\delta }(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{(1-2s+\delta )\mathrm{/}2}\le C_{s,\delta }\mathrm{\mid }t{\mathrm{\mid }}^{\delta }$. Also $\int G({\xi }^{\mathrm{\prime }})d{\xi }^{\mathrm{\prime }}=\mathrm{\parallel }W{\mathrm{\parallel }}_{H^s({\mathbb{R}}^d)}^2$ by [F1, F4]. Hence $\mathrm{\parallel }{W}_t-W_0{\mathrm{\parallel }}_2^2\le C_{s,\delta }\mathrm{\mid }t{\mathrm{\mid }}^{\delta }\mathrm{\parallel }W{\mathrm{\parallel }}_{H^s}^2$. Applying this estimate to differences of two slice parameters, density of Schwartz space and completeness [F3] give every $W\in H^s({\mathbb{R}}^d)$ a continuous $L^2({\mathbb{R}}^{d-1})$ slice representative whose value at $t=0$ is the trace from step 3.1. It agrees with the ordinary sections for almost every $t$: choose Schwartz approximants with summable squared $L^2({\mathbb{R}}^d)$ errors and apply Fubini [F4] to their squared slice errors. If two extensions agree on $K$, their difference vanishes on an interior prism beside every relatively compact patch $E$ of $F$. Fubini [F4] makes almost every interior slice zero on $E$; continuity then makes the trace zero there. A countable exhaustion covers $F$ up to its lower-dimensional boundary, which has surface measure zero, so the trace is independent of the extension. For a translation $b$ and orthogonal map $R$, $\widehat{W(R^{-1}(\cdot -b))}(\xi )=e^{-2\pi ib\cdot \xi }\widehat{W}(R^{-1}\xi )$ by change of variables [F5]; since $\mathrm{\mid }\det R\mathrm{\mid }=1$ and $⟨R\xi ⟩=⟨\xi ⟩$, the whole-space norm is invariant. Thus the hyperplane estimate from step 2.1 applies to the supporting plane of $F$ and then restricts to $F$.