Weak tangential trace on a Lipschitz polyhedron

If $v_1-v_2\in H_0^1(\mathrm{\Omega };{\mathbb{R}}^3)$, choose ${\phi }_m\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^3)$ converging to $v_1-v_2$ in $H^1$ by [F2]. The difference of the two proposed pairings is ${\int }_{\mathrm{\Omega }}(\mathrm{curl}u\cdot (v_1-v_2)-u\cdot \mathrm{curl}(v_1-v_2))$. The $L^2$ convergence of both fields and curls permits passage to the limit from ${\phi }_m$, and each approximating pairing is zero by [F1]. Hence the formula is independent of the representative.

For $u,v\in C^1(\bar{\mathrm{\Omega }};{\mathbb{R}}^3)$ as specified in the statement, apply [F4] on every tetrahedron to $u\times v$, using $\mathrm{div}(u\times v)=v\cdot \mathrm{curl}u-u\cdot \mathrm{curl}v$. By [F5], interior face fluxes cancel, while the remaining boundary flux is $n\cdot (u\times v)=(n\times u)\cdot v$; hence the formula is the stated boundary integral.

For real vector fields $a,b\in L^2(\mathrm{\Omega };{\mathbb{R}}^3)$, componentwise use of [F9] gives $\mathrm{\mid }{\int }_{\mathrm{\Omega }}a\cdot b\mathrm{\mid }\le {\sum }_{j=1}^3\mathrm{\parallel }{a}_j{\mathrm{\parallel }}_2\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_2$; applying [F9] again on the three-point counting measure to the component norms bounds this sum by $\mathrm{\parallel }a{\mathrm{\parallel }}_2\mathrm{\parallel }b{\mathrm{\parallel }}_2$. Applying this inequality to $(a,b)=(\mathrm{curl}u,v)$ and $(u,\mathrm{curl}v)$ gives $\mathrm{\mid }⟨{\gamma }_{\tau }u,[v]⟩\mathrm{\mid }\le \mathrm{\parallel }\mathrm{curl}u{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2+\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{curl}v{\mathrm{\parallel }}_2$. A further [F9] application on the two-point counting measure bounds the right side by $\mathrm{\parallel }u{\mathrm{\parallel }}_{H(\mathrm{curl})}(\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\mathrm{curl}v{\mathrm{\parallel }}_2^2{)}^{1\mathrm{/}2}$. The coordinate formula for curl and $(a-b{)}^2\le 2(a^2+b^2)$ give $\mathrm{\parallel }\mathrm{curl}v{\mathrm{\parallel }}_2^2\le 2{\sum }_{\mathrm{\ell },m\in \{x,y,z\}}\mathrm{\parallel }{\mathrm{\partial }}_{\mathrm{\ell }}{v}_m{\mathrm{\parallel }}_2^2$, hence $(\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\mathrm{curl}v{\mathrm{\parallel }}_2^2{)}^{1\mathrm{/}2}\le \sqrt{2}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$. Taking the infimum over all lifts of the quotient class proves boundedness in the quotient norm. Linearity follows from the integral formula.

If $u_m\in C_c^{\mathrm{\infty }}$ converges to $u\in H_0(\mathrm{curl})$ in the graph norm, choose a cutoff $\chi$ equal to one near the compact support of $u_m$ by [F6]. For any $v\in H^1$, $\chi v$ has compact support in $\mathrm{\Omega }$ and its zero extension lies in $H^1({\mathbb{R}}^3)$ by [F7]. Choose another cutoff $\theta \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one on that support. Approximate the zero extension by $g_j\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ using [F7]; then $q_j:=\theta g_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $q_j\to \chi v$ in $H^1$ by [F8], proving $\chi v\in H_0^1$. For each $j$, [F1] applied to $u_m$ and the compactly supported smooth test $q_j$ gives ${\int }_{\mathrm{\Omega }}\mathrm{curl}{u}_m\cdot q_j={\int }_{\mathrm{\Omega }}{u}_m\cdot \mathrm{curl}{q}_j$. Since $H^1$ convergence implies $L^2$ convergence of both $q_j$ and $\mathrm{curl}{q}_j$, Cauchy–Schwarz passes this identity to the limit $q_j\to \chi v$. Because $\chi =1$ and $\mathrm{\nabla }\chi =0$ on the support of $u_m$, the limiting identity is ${\int }_{\mathrm{\Omega }}(\mathrm{curl}{u}_m\cdot v-u_m\cdot \mathrm{curl}v)=0$. Step 2.1 gives ${\gamma }_{\tau }{u}_m\to {\gamma }_{\tau }u$ in $T^{\mathrm{\prime }}$, so the limit trace is zero.