Let $u\in H^1(\Omega)$, where $\Omega$ is a bounded open set of $\mathbb{R}^n$ with Lipschitz boundary.
We denote the outward unit normal as $n$, defined a.e. on $\partial\Omega$, and the normal derivative of $u$ as
$$ \frac{\partial u}{\partial n}:=\nabla u\cdot n. $$
Which space does the normal derivative belong to?
Is it possible to show $\frac{\partial u}{\partial n}\in L^2(\partial\Omega)$?
I think it's not possible if we don't require at least that $u\in H^2(\Omega)$. Indeed it is easy to get
$$ \|\frac{\partial u}{\partial n}\|_{L^2(\partial \Omega)}\le \|\nabla u \|_{L^2(\partial \Omega)}. $$
By the Trace theorem, we know that $\nabla u \in L^2(\partial\Omega)$ if $\nabla u\in H^1(\Omega)$, i.e. $u\in H^2(\Omega)$.
Note that my notation is quite messy when I deal with the norm of the gradient...