Sobolev Traces and Zero Boundary Values

The fundamental-theorem identity and its immediate consequence. By [F2] the representative $u^{\ast }$ is absolutely continuous on the compact interval $[a,a+\epsilon ]\subseteq [a,b]$ and $u^{\ast }(t)-u^{\ast }(a)={\int }_a^t{u}^{\mathrm{\prime }}$ for every $t\in [a,a+\epsilon ]$, so $u^{\ast }(a)-u^{\ast }(t)=-{\int }_a^t{u}^{\mathrm{\prime }}$ and therefore $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le \mathrm{\mid }{u}^{\ast }(t)\mathrm{\mid }+{\int }_a^t\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{u}^{\ast }(t)\mathrm{\mid }+{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$ for every $t\in [a,a+\epsilon ]$.

The case $p=1$. Integrating the pointwise inequality of step 1.1 over $t\in (a,a+\epsilon )$ and dividing by $\epsilon >0$ gives $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le {\epsilon }^{-1}{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }\mathrm{\mid }+{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$, which is (ii) at the left endpoint.

The case $1<p<\mathrm{\infty }$. Raising the pointwise inequality of step 1.1 to the $p$-th power and using $\mathrm{\mid }x+y{\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }x{\mathrm{\mid }}^p+\mathrm{\mid }y{\mathrm{\mid }}^p)$ gives $\mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }{u}^{\ast }(t){\mathrm{\mid }}^p+({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }{)}^p)$ for every $t\in (a,a+\epsilon )$. Integrating in $t$ and dividing by $\epsilon$ yields $\mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}({\epsilon }^{-1}{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p+({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }{)}^p)$, and Holder's inequality [F4] converts the last term into $({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }{)}^p\le {\epsilon }^{p-1}{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$. This is (i) at the left endpoint.

The right endpoint. Put $\tilde{u}(t):=u(a+b-t)$ for $t\in I$; then $t\mapsto u^{\ast }(a+b-t)$ is an absolutely continuous representative of $\tilde{u}$ whose classical derivative exists almost everywhere and equals $-u^{\mathrm{\prime }}(a+b-t)\in L^p(I)$ (the weak derivative class of $\tilde{u}$, by the classical chain rule and [F1]), so $\tilde{u}\in W^{1,p}(I;\mathbb{K})$ with weak derivative $-u^{\mathrm{\prime }}(a+b-\cdot )$, and its absolutely continuous representative takes the value $u^{\ast }(b)$ at $t=a$. Applying steps 2.1 and 2.2 to $\tilde{u}$ on $(a,a+\epsilon )$ gives the same inequalities with $u^{\ast }(b)$ on the left and the integrals over $(b-\epsilon ,b)$ on the right, because $\tilde{u}$ and its weak derivative on $(a,a+\epsilon )$ correspond to $u^{\ast }$ and $u^{\mathrm{\prime }}$ on $(b-\epsilon ,b)$ under the reflection.

Representative independence and the multiplied form. If $v$ is another representative of the class that is absolutely continuous on compact subintervals, then $v-u^{\ast }$ is constant by [F2]'s identity for both representatives with the same weak derivative class; since $v=u^{\ast }$ almost everywhere that constant is $0$ (the interval is nonempty), so the endpoint values of the absolutely continuous representative are determined by the class of $u$. Multiplying (i) and (ii) by $\epsilon$ gives the displayed $C_p$-form $\epsilon \mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p+{\epsilon }^p{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p)$, which at $p=1$ reads $\epsilon \mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le {\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }\mathrm{\mid }+\epsilon {\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$.

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).

The boundary estimate on continuous classes. Let $u\in C(\bar{\mathrm{\Omega }})\cap W^{1,p}(\mathrm{\Omega })$ and let $\{({\mathrm{\Phi }}_j,{\chi }_j)\}$ be the finite atlas and partition of [F3] and [F4], so that $u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}={\sum }_j({\chi }_{j}u){\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$ and the inequality for a sum of $N$ terms gives $\mathrm{\parallel }u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\parallel }}_{L^p(\mathrm{\partial }\mathrm{\Omega })}^p\le N^{p-1}{\sum }_j{\int }_{{\mathbb{R}}^{n-1}}\mid ({\chi }_{j}u)\circ {\mathrm{\Psi }}_j(y){\mid }^p{J}_j(y)dy$, where ${\mathrm{\Psi }}_j(y):={\mathrm{\Phi }}_j^{-1}(y,0)$ and $J_j=\sqrt{1+\mathrm{\mid }D{h}_j{\mathrm{\mid }}^2}$. For each $j$ put $u_j:=({\chi }_{j}u)\circ {\mathrm{\Phi }}_j^{-1}$ on the flattened patch ${\mathrm{\Phi }}_j(W_j)\cap \{t<0\}$ and extend it by zero; because $\mathrm{supp}{\chi }_j$ is compactly contained in $W_j$, the extension is compactly supported and continuous on the closed half-space, and $u_j\in W^{1,p}$ of the half-space with $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{W^{1,p}}\le C_j\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}$ by [F2] and [F7]. Reflecting $t\mapsto -t$ and applying the half-space estimate [F1] to the reflected function, whose boundary value is $y\mapsto ({\chi }_{j}u)\circ {\mathrm{\Psi }}_j(y)$, gives $\int \mathrm{\mid }({\chi }_{j}u)\circ {\mathrm{\Psi }}_j{\mathrm{\mid }}^{p}dy\le C_j^{\mathrm{\prime }}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}^p$. Since $J_j$ is bounded on the compact patch, $\int \mathrm{\mid }({\chi }_{j}u)\circ {\mathrm{\Psi }}_j{\mathrm{\mid }}^p{J}_j(y)dy\le \mathrm{\parallel }{J}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\int \mathrm{\mid }({\chi }_{j}u)\circ {\mathrm{\Psi }}_j{\mathrm{\mid }}^{p}dy$, and summing the finitely many bounds yields $\mathrm{\parallel }u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\parallel }}_{L^p(\mathrm{\partial }\mathrm{\Omega })}\le C(\mathrm{\Omega },p)\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}$.

Construction of $T$ by density. Let $D$ be the class of restrictions to $\mathrm{\Omega }$ of functions in $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$, a linear subspace of $C(\bar{\mathrm{\Omega }})\cap W^{1,p}(\mathrm{\Omega })$ that is dense in $W^{1,p}(\mathrm{\Omega })$ by [F6]. The restriction map $S:D\to L^p(\mathrm{\partial }\mathrm{\Omega })$, $S\phi :=\phi {\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$, is linear and satisfies $\mathrm{\parallel }S\phi {\mathrm{\parallel }}_{L^p(\mathrm{\partial }\mathrm{\Omega })}\le C(\mathrm{\Omega },p)\mathrm{\parallel }\phi {\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}$ by step 1.1. Since $D$ is dense in the complete space $W^{1,p}(\mathrm{\Omega })$ [F9] and carries the subspace norm, the pair $(W^{1,p}(\mathrm{\Omega }),\text{inclusion})$ is a completion of $D$; the completion universal property [F8], applied with the Banach space $L^p(\mathrm{\partial }\mathrm{\Omega })$ [F9], produces a unique bounded linear $T:W^{1,p}(\mathrm{\Omega })\to L^p(\mathrm{\partial }\mathrm{\Omega })$ with $T{\mathrm{\mid }}_D=S$ and $\mathrm{\parallel }T\mathrm{\parallel }\le C(\mathrm{\Omega },p)$.

Agreement, uniqueness and class-dependence. If $u\in C(\bar{\mathrm{\Omega }})\cap W^{1,p}(\mathrm{\Omega })$ and ${\phi }_m\in D$ with ${\phi }_m\to u$ in $W^{1,p}(\mathrm{\Omega })$ (step 1.1 and [F6]), then $T{\phi }_m={\phi }_m{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}\to Tu$ by continuity of $T$, while $\mathrm{\parallel }{\phi }_m{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}-u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\parallel }}_{L^p(\mathrm{\partial }\mathrm{\Omega })}\le C(\mathrm{\Omega },p)\mathrm{\parallel }{\phi }_m-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}\to 0$ by step 1.1 applied to ${\phi }_m-u\in C(\bar{\mathrm{\Omega }})\cap W^{1,p}(\mathrm{\Omega })$; hence $Tu=u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$ in $L^p(\mathrm{\partial }\mathrm{\Omega })$. If $T^{\mathrm{\prime }}$ is another bounded linear operator with the same property on $C(\bar{\mathrm{\Omega }})\cap W^{1,p}(\mathrm{\Omega })$, then $T-T^{\mathrm{\prime }}$ vanishes on the dense subspace $D$ and is bounded, hence zero by the same limiting argument; this is the asserted uniqueness. Finally, if $u=v$ are the same $W^{1,p}$ class in the sense of the quotient representation [F10], then $u-v$ is the zero class, $T(u-v)=0$ because $0\in D$ is a limit of the constant sequence and $S0=0$, so $T$ depends only on the class. The construction is chartwise exactly the transported flat trace, and the chartwise definitions agree because both equal $T$ on the dense class.