The half-space trace estimate and the half-space trace operator

The pointwise normal-line identity. Fix a compactly supported $u\in C(\bar{H})\cap W^{1,p}(H)$. By [F1] applied to $\mathrm{\Omega }=H$ there is an ACL representative $u^{\ast }$ of the class of $u$ whose classical last derivative ${\mathrm{\partial }}_t{u}^{\ast }$ exists a.e., lies in $L^p(H)$ and represents ${\mathrm{\partial }}_{n}u$. For a.e. $x^{\mathrm{\prime }}\in {\mathbb{R}}^d$ the section $t\mapsto u^{\ast }(x^{\mathrm{\prime }},t)$ is absolutely continuous on compact subintervals of $(0,\mathrm{\infty })$, and $u^{\ast }=u$ a.e. on $H$; since both the continuous extension of the section (which exists because ${\int }_0^T\mathrm{\mid }{\mathrm{\partial }}_t{u}^{\ast }\mathrm{\mid }dt<\mathrm{\infty }$ for every $T$) and the continuous function $u(x^{\mathrm{\prime }},\cdot )$ agree on a dense set of $t$, they agree everywhere on the line, so the section extends continuously to $t=0$ with value $g(x^{\mathrm{\prime }})=u(x^{\mathrm{\prime }},0)$. Because $u$ has compact support, the section vanishes for large $t$, and the absolutely continuous function $t\mapsto \mathrm{\mid }{u}^{\ast }(x^{\mathrm{\prime }},t){\mathrm{\mid }}^p$ satisfies $\mathrm{\mid }g(x^{\mathrm{\prime }}){\mathrm{\mid }}^p=-{\int }_0^{\mathrm{\infty }}{\mathrm{\partial }}_t\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^{p}dt=-p{\int }_0^{\mathrm{\infty }}\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^{p-2}\mathrm{Re}(\overline{u^{\ast }}{\mathrm{\partial }}_t{u}^{\ast })dt\le p{\int }_0^{\mathrm{\infty }}\mathrm{\mid }u{\mathrm{\mid }}^{p-1}\mathrm{\mid }{\mathrm{\partial }}_{n}u\mathrm{\mid }dt$ for $1<p<\mathrm{\infty }$, while for $p=1$ the absolutely continuous function $t\mapsto \mathrm{\mid }{u}^{\ast }(x^{\mathrm{\prime }},t)\mathrm{\mid }$ has $\mathrm{\mid }g(x^{\mathrm{\prime }})\mathrm{\mid }=-{\int }_0^{\mathrm{\infty }}{\mathrm{\partial }}_t\mathrm{\mid }{u}^{\ast }\mathrm{\mid }dt\le {\int }_0^{\mathrm{\infty }}\mathrm{\mid }{\mathrm{\partial }}_{n}u\mathrm{\mid }dt$.

Density of the smooth restriction class. Put $D:=\{\phi {\mathrm{\mid }}_H:\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n;\mathbb{K})\}$, a linear subspace of $W^{1,p}(H)$, and let $u\in W^{1,p}(H)$ and $\delta >0$. By [F5] there is $Eu\in W^{1,p}({\mathbb{R}}^n)$ with $(Eu){\mathrm{\mid }}_H=u$ a.e.; by [F6] choose $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ with $\mathrm{\parallel }\phi -Eu{\mathrm{\parallel }}_{W^{1,p}({\mathbb{R}}^n)}<\delta$. Then $\phi {\mathrm{\mid }}_H\in D$, and because the restriction to $H$ of an $L^p$ class has $\mathrm{\parallel }\psi {\mathrm{\mid }}_H{\mathrm{\parallel }}_{L^p(H)}\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}$ componentwise, $\mathrm{\parallel }\phi {\mathrm{\mid }}_H-u{\mathrm{\parallel }}_{W^{1,p}(H)}\le \mathrm{\parallel }\phi -Eu{\mathrm{\parallel }}_{W^{1,p}({\mathbb{R}}^n)}<\delta$. Hence $D$ is dense in $W^{1,p}(H)$.

The integrated estimate and the strip form. Integrate the pointwise inequality of step 1.1 over $x^{\mathrm{\prime }}\in {\mathbb{R}}^d$ and use Tonelli [F4] to interchange the $x^{\mathrm{\prime }}$- and $t$-integrals: $\mathrm{\parallel }g{\mathrm{\parallel }}_p^p\le p{\int }_H\mathrm{\mid }u{\mathrm{\mid }}^{p-1}\mathrm{\mid }{\mathrm{\partial }}_{n}u\mathrm{\mid }$. For $p=1$ this gives $\mathrm{\parallel }g{\mathrm{\parallel }}_1\le \mathrm{\parallel }{\mathrm{\partial }}_{n}u{\mathrm{\parallel }}_1\le \mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,1}}$; for $1<p<\mathrm{\infty }$, Holder [F3] with exponents $p^{\mathrm{\prime }}$ and $p$ gives ${\int }_H\mathrm{\mid }u{\mathrm{\mid }}^{p-1}\mathrm{\mid }{\mathrm{\partial }}_{n}u\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}^{p-1}\mathrm{\parallel }{\mathrm{\partial }}_{n}u{\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}}^p$, and therefore $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p}^p\le p\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}^{p-1}\mathrm{\parallel }{\mathrm{\partial }}_{n}u{\mathrm{\parallel }}_{L^p}\le p\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}}^p$. For the strip form, fix $h>0$; for a.e. $x^{\mathrm{\prime }}$ the section of $u^{\ast }$ on $(0,2h)$ is absolutely continuous with $L^p$ derivative, so [F1] in dimension one makes that section an element of $W^{1,p}((0,2h))$ with weak derivative ${\mathrm{\partial }}_{n}u(x^{\mathrm{\prime }},\cdot )$, and [F2] applies with $\epsilon =h$ and gives, after multiplying by $h$, $h\mathrm{\mid }g(x^{\mathrm{\prime }}){\mathrm{\mid }}^p\le C_p({\int }_0^h\mathrm{\mid }u(x^{\mathrm{\prime }},t){\mathrm{\mid }}^{p}dt+h^p{\int }_0^h\mathrm{\mid }{\mathrm{\partial }}_{n}u(x^{\mathrm{\prime }},t){\mathrm{\mid }}^{p}dt)$ with the stated $C_p$.

Construction of $T_{+}$ and agreement with classical restriction. Let $S:D\to L^p({\mathbb{R}}^d)$ be the classical restriction $S\phi :=\phi (\cdot ,0)$, which is linear and, by step 2.1 applied to $\phi \in D\subset C(\bar{H})\cap W^{1,p}(H)$ (each is continuous on $\bar{H}$ with compact support), satisfies $\mathrm{\parallel }S\phi {\mathrm{\parallel }}_p\le p^{1\mathrm{/}p}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{W^{1,p}(H)}$. Since $D$ is dense in $W^{1,p}(H)$ by step 1.2 and carries the subspace norm, and since $W^{1,p}(H)$ is complete by [F8], the pair $(W^{1,p}(H),\text{inclusion})$ is a completion of $D$; by the completion universal property [F9] applied with $Y=L^p({\mathbb{R}}^d)$ (complete by [F7]), $S$ extends uniquely to a bounded linear $T_{+}:W^{1,p}(H)\to L^p({\mathbb{R}}^d)$ with $\mathrm{\parallel }{T}_{+}u{\mathrm{\parallel }}_p\le p^{1\mathrm{/}p}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(H)}$ and $T_{+}{\mathrm{\mid }}_D=S$. If now $u\in C(\bar{H})\cap W^{1,p}(H)$ is compactly supported, choose ${\phi }_m\in D$ with ${\phi }_m\to u$ in $W^{1,p}(H)$; then $T_{+}u={\lim }_{m}S{\phi }_m$ in $L^p$ by continuity, while $\mathrm{\parallel }S{\phi }_m-u(\cdot ,0){\mathrm{\parallel }}_p^p\le p\mathrm{\parallel }{\phi }_m-u{\mathrm{\parallel }}_{W^{1,p}(H)}^p\to 0$ by step 2.1 applied to the compactly supported continuous difference, so $T_{+}u=u(\cdot ,0)$ in $L^p({\mathbb{R}}^d)$.

Uniqueness and class-dependence. If $T^{\mathrm{\prime }}$ is another bounded linear operator on $W^{1,p}(H)$ whose restriction to $D$ is $S$, then $A:=T_{+}-T^{\mathrm{\prime }}$ is a bounded linear operator vanishing on $D$; for $u\in W^{1,p}(H)$ choose ${\phi }_m\in D$ with ${\phi }_m\to u$ (step 1.2), so $\mathrm{\parallel }Au\mathrm{\parallel }={\lim }_m\mathrm{\parallel }A{\phi }_m\mathrm{\parallel }=0$ by boundedness of $A$. Hence $T^{\mathrm{\prime }}=T_{+}$: this is the asserted uniqueness of the bounded extension of classical restriction. Moreover, if $u=v$ in $W^{1,p}(H)$ are the same a.e. class, then $u-v$ is the zero class and linearity gives $T_{+}(u-v)=T_{+}0=0$, because the zero class is the limit of the constant sequence $0\in D$ and $S0=0$; hence $T_{+}u=T_{+}v$ and $T_{+}$ depends only on the class. This proves (i) and (ii).