Weak H(div) normal trace surjects onto the scalar trace dual

The space $H^1(D)$ is Hilbert for $(z,w{)}_{H^1}=(z,w{)}_{L^2}+(\mathrm{\nabla }z,\mathrm{\nabla }w{)}_{L^2}$. Indeed, a Cauchy sequence of functions and their weak derivatives converges componentwise in $L^2$ by [F2]; testing against $C_c^{\mathrm{\infty }}(D)$ and passing both integrals to the limits shows the derivative limits are the weak derivatives of the function limit. The quotient map $q:H^1(D)\to T_D$ has norm at most one, so $\mathrm{\Lambda }(w):=L(qw)$ is bounded on $H^1(D)$ with $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\le \mathrm{\parallel }L\mathrm{\parallel }$. By [F3] choose its Riesz representative $z\in H^1(D)$, so $\mathrm{\Lambda }(w)=(\mathrm{\nabla }z,\mathrm{\nabla }w)+(z,w)$.

For every $\varphi \in C_c^{\mathrm{\infty }}(D)$, $q\varphi =0$ because $\varphi \in H_0^1(D)$; hence $0=\mathrm{\Lambda }(\varphi )=(\mathrm{\nabla }z,\mathrm{\nabla }\varphi )+(z,\varphi )$. By the distributional definition of divergence, this means $\mathrm{div}(\mathrm{\nabla }z)=z$. Since $z\in L^2(D)$ and $\mathrm{\nabla }z\in L^2(D{)}^d$, [F5] gives $\tau :=\mathrm{\nabla }z\in H(\mathrm{div};D)$, with $\mathrm{\parallel }\tau {\mathrm{\parallel }}_{H(\mathrm{div})}^2=\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2+\mathrm{\parallel }z{\mathrm{\parallel }}_2^2=\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}^2=\mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^2\le \mathrm{\parallel }L{\mathrm{\parallel }}^2$.

For any $[w]\in T_D$, [F4] and step 2.1 give $⟨{\gamma }_n\tau ,[w]⟩=(\mathrm{\nabla }z,\mathrm{\nabla }w)+(z,w)=\mathrm{\Lambda }(w)=L([w])$. Thus ${\gamma }_n\tau =L$, and step 2.1 gives the claimed norm bound. Since $L$ was arbitrary, ${\gamma }_n$ is onto.