If you restrict a $H^s$-function $f$ to a submanifold of codimension $k$ you get a $H^{s-k/2}$ function. Now we work in $H^m(B_1)\subset \mathcal D'(\mathbb R^n)$ (any $m$). So (up to the choice of a subsequence) $u_n\to u$ in $\sigma(H^1(B_1),H^{-1}(B_1))$ with respect to the $L^2$-duality. Integration by parts and restriction gives
$\nabla u_n \to \nabla u$ in $\sigma(H^{0},H^0)$, thus But only if $\nabla u_n|_{\partial B_1} \to \nabla u|_{\partial B_1}$ s$is large enough (vector valued now) in$\sigma(H^{-1/2}(\partial B_1)^n,H^{1/2}(\partial B_1)^n)$Sobolev lemma threshold). Since Thus the unit normal$n$is smoothrest of my answer does not work, and I deleted itis in$H^{1/2}(\partial B_1)^n$, thus $$\int_{\partial B_1} \nabla u_n\cdot n vol \to \int_{\partial B_1} \nabla u\cdot n vol.$$. 1 If you restrict a$H^s$-function$f$to a submanifold of codimension$k$you get a$H^{s-k/2}$function. Now we work in$H^m(B_1)\subset \mathcal D'(\mathbb R^n)$(any$m$). So (up to the choice of a subsequence)$u_n\to u$in$\sigma(H^1(B_1),H^{-1}(B_1))$with respect to the$L^2$-duality. Integration by parts and restriction gives:$\nabla u_n \to \nabla u$in$\sigma(H^{0},H^0)$, thus$\nabla u_n|_{\partial B_1} \to \nabla u|_{\partial B_1}$(vector valued now) in$\sigma(H^{-1/2}(\partial B_1)^n,H^{1/2}(\partial B_1)^n)$. Since the unit normal$n$is smooth, it is in$H^{1/2}(\partial B_1)^n\$, thus $$\int_{\partial B_1} \nabla u_n\cdot n vol \to \int_{\partial B_1} \nabla u\cdot n vol.$$