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…
Best Answer
If $u$ belongs merely to $H^1(\Omega)$, then you cannot define a normal-derivative-trace operator.
Indeed, if $T : H^1(\Omega) \to S(\partial\Omega)$ would be such an operator, where $S(\partial\Omega)$ is some Banach space on the boundary and if $T$ would be linear, you arrive at the following contradiction: if $T$ is reasonably defined, you would have $T \varphi = 0$ for all $\varphi \in C_c^\infty(\Omega)$. By density of $C_c^\infty(\Omega)$ in $H_0^1(\Omega)$ and continuity of $T$, this implies $T u = 0$ for all $u \in H_0^1(\Omega)$. But this is absurd (consider $u \in C^1(\bar\Omega)$).
On the other hand, if $u \in H^1(\Omega)$ and $\Delta u \in L^2(\Omega)$, you can define the trace of the normal derivative in $H^{-1/2}(\Omega)$ by duality. Indeed, for regular $v$, you have $$\int_\Omega \Delta u \, v + \nabla u \cdot \nabla v \, \mathrm{d}x = \int_{\partial\Omega} \frac{\partial u}{\partial n} \, v \, \mathrm{d}s.$$ Now, if $v \in H^{1/2}(\partial\Omega)$ is arbitrary (and $\Omega$ possesses some regularity), you find $E v \in H^1(\Omega)$ with $(Ev)|_{\partial\Omega} = v$. Then, you define $$\langle \frac{\partial u}{\partial n}, v \rangle := \int_\Omega \Delta u \, Ev + \nabla u \cdot \nabla Ev \, \mathrm{d}x.$$