Weak normal trace by Green duality

For $w\in H^1(D)$ put $L_v(w):={\int }_D(v\cdot \mathrm{\nabla }w+(\mathrm{div}v)w)$. The two integrals are finite by [F3], and Cauchy–Schwarz in ${\mathbb{R}}^2$ gives $\mathrm{\mid }{L}_v(w)\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2+\mathrm{\parallel }\mathrm{div}v{\mathrm{\parallel }}_2\mathrm{\parallel }w{\mathrm{\parallel }}_2\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H(\mathrm{div};D)}\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(D)}\mathrm{.}$ Thus $L_v$ is a bounded real linear functional on $H^1(D)$.

For every $\varphi \in C_c^{\mathrm{\infty }}(D;\mathbb{R})$, [F1] gives $L_v(\varphi )=0$. If $z\in H_0^1(D)$, then for each $\epsilon >0$, [F2] gives such a $\varphi$ with $\mathrm{\parallel }z-\varphi {\mathrm{\parallel }}_{H^1}<\epsilon$; step 1.1 yields $\mathrm{\mid }{L}_v(z)\mathrm{\mid }=\mathrm{\mid }{L}_v(z-\varphi )\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H(\mathrm{div})}\epsilon$. Since this holds for every $\epsilon >0$, $L_v(z)=0$. Therefore $L_v$ is constant on each coset of $H_0^1(D)$ and defines ${\gamma }_{n}v$ on $T_D$.

For any $[w]\in T_D$, step 2.1 makes $L_v(w)$ the same for every representative. Applying step 1.1 to every representative $w+z$, $z\in H_0^1(D)$, and taking the infimum of their $H^1$ norms gives $\mathrm{\mid }⟨{\gamma }_{n}v,[w]⟩\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H(\mathrm{div};D)}\mathrm{\parallel }[w]{\mathrm{\parallel }}_{T_D}$. Hence ${\gamma }_{n}v$ is bounded with the claimed norm.