The trace of a one-dimensional Sobolev function is the pair of endpoint values

The operator $T$ is well defined, linear and bounded. By [F1] the representative $u^{\ast }$ and hence the pair $(u^{\ast }(a),u^{\ast }(b))$ depend only on the class, so $T$ is well defined on classes; if $u,v$ are classes with representatives $u^{\ast },v^{\ast }$, then $u^{\ast }+v^{\ast }$ and $\lambda u^{\ast }$ are the absolutely continuous representatives of $u+v$ and $\lambda u$ by [F1]'s uniqueness and the linearity of the fundamental-theorem identity, so $T$ is linear. With $\epsilon :=(b-a)\mathrm{/}2$, [F2] gives for $1<p<\mathrm{\infty }$ the bound $\mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}(\frac{2}{b-a}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}^p+(\frac{b-a}{2}{)}^{p-1}\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^p}^p)$ and the same bound at $b$, hence $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{{\mathbb{K}}^2}\le C(b-a,p)\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(I)}$; for $p=1$, $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le \frac{2}{b-a}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1}+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^1}$ and the same at $b$.

Endpoint vanishing implies membership in $W_0^{1,p}(I)$. Assume $u^{\ast }(a)=u^{\ast }(b)=0$. For $0<\delta <(b-a)\mathrm{/}4$ choose ${\chi }_{\delta }\in C_c^{\mathrm{\infty }}(I)$ with $0\le {\chi }_{\delta }\le 1$, equal to one on $[a+\delta ,b-\delta ]$, vanishing on $(a,a+\delta \mathrm{/}2)\cup (b-\delta \mathrm{/}2,b)$, and $\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{/}\delta$. By [F5], $({\chi }_{\delta }u{)}^{\mathrm{\prime }}={\chi }_{\delta }{u}^{\mathrm{\prime }}+{\chi }_{\delta }^{\mathrm{\prime }}u$. The terms containing $1-{\chi }_{\delta }$ tend to zero in $L^p$ by dominated convergence. Since $u^{\ast }(t)={\int }_a^t{u}^{\mathrm{\prime }}$, [F8] gives $\mathrm{\mid }{u}^{\ast }(t){\mathrm{\mid }}^p\le (t-a{)}^{p-1}{\int }_a^t\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$ (also at $p=1$), hence ${\int }_a^{a+\delta }\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}u{\mathrm{\mid }}^p\le C^p{p}^{-1}{\int }_a^{a+\delta }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p\to 0$; the right endpoint is identical. Thus ${\chi }_{\delta }u\to u$ in $W^{1,p}$. For each fixed $\delta$, [F6] extends ${\chi }_{\delta }u$ by zero to $V_{\delta }\in W^{1,p}(\mathbb{R})$. Choose ${\zeta }_{\delta }\in C_c^{\mathrm{\infty }}(I)$ equal to one near its compact support, and use [F7] to choose ${\psi }_{\delta }\in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\mathrm{\parallel }{\psi }_{\delta }-V_{\delta }{\mathrm{\parallel }}_{W^{1,p}(\mathbb{R})}<\delta \mathrm{/}(1+C_{{\zeta }_{\delta }})$. Then [F5] gives $\mathrm{\parallel }{\zeta }_{\delta }{\psi }_{\delta }-{\chi }_{\delta }u{\mathrm{\parallel }}_{W^{1,p}(I)}<\delta$, and ${\zeta }_{\delta }{\psi }_{\delta }\in C_c^{\mathrm{\infty }}(I)$. Combining these approximations proves membership in the closure [F4].

Membership in $W_0^{1,p}(I)$ implies $Tu=0$. Every $\phi \in C_c^{\mathrm{\infty }}(I)$ vanishes on a neighbourhood of $a$ and of $b$, so its absolutely continuous representative vanishes at both endpoints and $T\phi =(0,0)$. By step 1.1 the map $T$ is bounded, hence continuous; if $u\in W_0^{1,p}(I)$ and ${\phi }_m\in C_c^{\mathrm{\infty }}(I)$ with ${\phi }_m\to u$ by [F4], then $Tu={\lim }_{m}T{\phi }_m=0$.

Conclusion and the two worked functions. Steps 2.1 and 1.2 prove the equivalence $u\in W_0^{1,p}(I)\iff Tu=(0,0)$, and step 1.1 gives well-definedness, linearity and boundedness of $T$; with [F1] this is the assertion that the trace of $u$ is the pair of endpoint values of $u^{\ast }$ and depends only on the class. On $I=(0,1)$: for $u(x)=x(1-x)$ the absolutely continuous representative is $u^{\ast }=u$ with $u^{\ast }(0)=u^{\ast }(1)=0$, so $Tu=(0,0)$ and $u\in W_0^{1,p}(0,1)$ by step 1.2; for $u\equiv 1$ the representative is $u^{\ast }\equiv 1$ with $Tu=(1,1)\mathrm{\ne }(0,0)$, so $u\mathrm{\notin }{W}_0^{1,p}(0,1)$ by step 2.1.