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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...

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    Can you help me understand the intuition that $u\in H^2(\Omega)$ should be required? My intuition is that $\|\partial_\nu u\|^2_{L^2(\partial \Omega} \leq \int_{\partial\Omega} |\nabla u||n| ds \leq \|\nabla u\|^2_{L^2} |\partial \Omega|$ so it seems like $\partial_\nu u \in L^2(\partial \Omega).$ – 2017-02-01
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    Hmm. However on the other hand, the trace of $u$ on the boundary should live in $H^{1/2}$ and so a derivative would live in $H^{-1/2}$... So at least one (or both) of my intuitions is wrong! – 2017-02-01
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    @Matt I edited my question in order to answer to your comment – 2017-02-01
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    Yes, I agree with your intuition. The trace theorem can do a bit better than you have written, but yes, we will lose some regularity of $u$ once we restrict to the boundary (this is the obvious thing I missed in my first comment!) Probably we can compute by hand a counter example where $\partial_\nu u \not \in L^2(\partial \Omega)$ by taking any $u \in H^1\setminus H^2$. – 2017-02-01
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    Sounds reasonable! – 2017-02-01

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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 $u$, 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 $$\langle \frac{\partial u}{\partial n}, v \rangle := \int_\Omega \Delta u \, Ev + \nabla u \cdot \nabla Ev \, \mathrm{d}x.$$