I was wondering whether the weak normal derivative is a bounded operator? So let $\Omega$ be bounded and open with $C^1$ boundary. Is $\frac{\partial}{\partial \nu}: H^2(\Omega) \to L^2(\partial\Omega)$ continuous? The weak normal derivative is defined as follows:
$u\in H^1(\Omega)$ with $\Delta u=f \in L^2(\Omega)$. We write that $\partial_n u \in L^2(\partial\Omega)$ if there exists $g\in L^2(\partial\Omega)$ with
\begin{equation*} \int\limits_{\Omega}fv \operatorname{dx}+ \int\limits_{\Omega} \nabla u^T \nabla v \operatorname{dx}= \int\limits_{\partial\Omega}gv \operatorname{d\sigma} \qquad (v\in H^1(\Omega)). \end{equation*} In this case we write $\partial_n u =g$.
The trace operator $$ H^1(\Omega) \ni u \mapsto u|_{\partial \Omega} \in L^2(\partial \Omega) $$ is bounded. For $u \in H^2(\Omega)$, we have $\nabla u \in H^1(\Omega)$, so also $$ H^2(\Omega) \ni u \mapsto \nabla u|_{\partial \Omega} \in L^2(\partial \Omega) $$ is bounded. And of course taking the normal component of a vector field (i.e. $V \mapsto V \cdot \nu$) is bounded on $L^2(\partial \Omega)$.
The normal derivative defined in this way coincides with the weak formulation you described. Gauss-Green formula $$ \int_{\partial \Omega} g \frac{\partial u}{\partial \nu} = \int_\Omega \nabla g \nabla u + \int_\Omega g \Delta u $$ is valid for $g, u \in C^\infty(\bar{\Omega})$. Thanks to regularity of $\Omega$, smooth functions are dense in both $H^1(\Omega)$ and $H^2(\Omega)$, hence by continuity of the trace operator, the formula is also valid for arbitrary $u \in H^2(\Omega)$ and $g \in H^1(\Omega)$.