Let $\Omega$ be a bounded domain, $u_k\in H^1(\Omega)$, $k=1,2,..$ and the normal derivative $\frac{\partial u_k}{\partial \mathbf{n}}\in L^2(\partial\Omega)$. As $k$ goes to infinity, $u_k$ converges weakly in $H^1(\Omega)$ to $u\in H^1(\Omega)$, and the normal derivative $\frac{\partial u_k}{\partial \mathbf{n}}$ converges weakly in $L^2(\partial\Omega)$ to $g\in L^2(\partial\Omega)$.
Given all the above, can we conclude that $\frac{\partial u}{\partial \mathbf{n}}=g$?
If needed, we can also assume that $u_k$ satisfies $\int_{\Omega} \nabla u_k\cdot \nabla \varphi-u_k\varphi = \int_{\partial\Omega}\frac{\partial u_k}{\partial \mathbf{n}}\varphi$, $\forall \varphi\in H^1(\Omega)$.
A candidate proof using the addition assumption of variational equation was as follows. Passing to the limit of the variational form, we get $$(1)\quad\int_{\Omega} \nabla u\cdot \nabla \varphi-u\varphi = \int_{\partial\Omega} g\varphi,\quad \forall \varphi\in H^1(\Omega).$$ If the following equality holds: $$ (2)\quad \int_{\Omega} \nabla u\cdot \nabla \varphi = \int_{\Omega} (-\Delta u) \varphi + \int_{\partial\Omega} \frac{\partial u}{\partial \mathbf{n}}\varphi,\quad \forall \varphi\in H^1(\Omega), $$ then (1) can be substituted to $$ (3)\quad \int_{\Omega} (-\Delta u - u) \varphi = \int_{\partial\Omega} (g-\frac{\partial u}{\partial \mathbf{n}})\varphi,\quad \forall \varphi\in H^1(\Omega). $$ Taking $\varphi\in C_c^{\infty}(\Omega)$ in (3), we get $-\Delta u - u=0$ and hence $g=\frac{\partial u}{\partial \mathbf{n}}$.
However, I am not sure about (2). In what sense is $-\Delta u$ defined in (2)? If $-\Delta u$ is defined in $\mathcal{D}'(\Omega)=(C_c^{\infty}(\Omega))'$, then $-\Delta u$ may not be applicable to $\varphi\in H^1(\Omega)$ and formal integration by part $\int_{\Omega}\nabla u\cdot\nabla \phi=\int_{\Omega}(-\Delta u)\phi$ can not generate a boundary term for $\phi\in C_c^{\infty}(\Omega)$.
(2) can be justified in this way(?). Taking $\varphi\in C_c^{\infty}(\Omega)$ in (1), we get $-\Delta u - u=0$ in $D'(\Omega)$. Since $u\in H^1(\Omega)\subset L^2(\Omega)$, we get $\Delta u\in L^2(\Omega)$. In this case, (2) is justified.