Weak graph limits preserve zero tangential H(curl)

The graph of distributional curl is closed in $L^2(\mathrm{\Omega };{\mathbb{R}}^3)\times L^2(\mathrm{\Omega };{\mathbb{R}}^3)$: if $v_j\to v$ and $\mathrm{curl}{v}_j\to w$ in $L^2$, [F3] gives ${\int }_{\mathrm{\Omega }}w\cdot \phi dx={\int }_{\mathrm{\Omega }}v\cdot \mathrm{curl}\phi dx$ for each test $\phi$, so $\mathrm{curl}v=w$. By [F1] the product is Hilbert, hence its closed graph is Hilbert with the graph inner product. Thus $H(\mathrm{curl};\mathrm{\Omega })$ is a Hilbert space.

The curls are bounded in the separable Hilbert space of [F1]. By [F2] choose a subsequence, still indexed by $n$, with $\mathrm{curl}{u}_n\rightharpoonup w$ weakly in $L^2$. For every test $\phi$, strong convergence of $u_n$ and [F3] give ${\int }_{\mathrm{\Omega }}w\cdot \phi dx={\lim }_n{\int }_{\mathrm{\Omega }}\mathrm{curl}{u}_n\cdot \phi dx={\lim }_n{\int }_{\mathrm{\Omega }}{u}_n\cdot \mathrm{curl}\phi dx={\int }_{\mathrm{\Omega }}u\cdot \mathrm{curl}\phi dx$. Therefore $\mathrm{curl}u=w$ distributionally and $u\in H(\mathrm{curl})$.

The strong convergence of $u_n$ and weak convergence of their curls show that $u_n\rightharpoonup u$ in the graph Hilbert space: for every $v\in H(\mathrm{curl})$, $⟨u_n,v{⟩}_{H(\mathrm{curl})}=(u_n,v{)}_{L^2}+(\mathrm{curl}{u}_n,\mathrm{curl}v{)}_{L^2}⟶(u,v{)}_{L^2}+(w,\mathrm{curl}v{)}_{L^2}=⟨u,v{⟩}_{H(\mathrm{curl})}$. For every $y\in H_0(\mathrm{curl}{)}^{\perp }$, each $⟨u_n,y{⟩}_{H(\mathrm{curl})}=0$; weak convergence gives $⟨u,y{⟩}_{H(\mathrm{curl})}=0$. By [F4], $u\in (H_0(\mathrm{curl}{)}^{\perp }{)}^{\perp }=H_0(\mathrm{curl})$.