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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

EDIT:

$\nabla u_n \to \nabla u$ 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 $$\int_{\partial B_1} \nabla u_n\cdot n vol \to \int_{\partial B_1} \nabla u\cdot n vol.$$.

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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 $$\int_{\partial B_1} \nabla u_n\cdot n vol \to \int_{\partial B_1} \nabla u\cdot n vol.$$