H(div) is Hilbert and H0(div) is closed

Let $(v_n)$ be Cauchy in the graph norm. By [F1] applied componentwise, $v_n\to v$ in $L^2(D;{\mathbb{R}}^d)$ and $\mathrm{div}{v}_n\to g$ in $L^2(D)$ for some $v,g$. For every $\varphi \in C_c^{\mathrm{\infty }}(D;\mathbb{R})$, the distributional identity gives ${\int }_D{v}_n\cdot \mathrm{\nabla }\varphi =-{\int }_D(\mathrm{div}{v}_n)\varphi$. Holder's inequality [F4] passes both pairings to their $L^2$ limits, so ${\int }_{D}v\cdot \mathrm{\nabla }\varphi =-{\int }_{D}g\varphi$. Hence $g=\mathrm{div}v$ distributionally, $v\in H(\mathrm{div};D)$, and [F2] gives graph-norm convergence $v_n\to v$.

The displayed graph inner product is the sum of the real $L^2$ inner products of $v$ and its divergence, so it induces exactly the graph norm. By step 1.1 that norm is complete; therefore H(div) is Hilbert.

By [F3], if $v_n\in H_0(\mathrm{div};D)$ and $v_n\to v$ in H(div), then ${\gamma }_n{v}_n\to {\gamma }_{n}v$ in $T_D^{\mathrm{\prime }}$. Since each ${\gamma }_n{v}_n=0$, the limit trace is zero, so $v\in H_0(\mathrm{div};D)$. This proves closedness; [F5] and step 2.1 show that the restricted inner-product space is complete.