# Normal derivative of a $H^1$- Sobolev function

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...

• 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 $\partial_\nu u \in L^2(\partial \Omega).$ – Matt Feb 1 '17 at 16:36
• 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! – Matt Feb 1 '17 at 16:41
• @Matt I edited my question in order to answer to your comment – avati91 Feb 1 '17 at 16:47
• 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 $u \in H^1\setminus H^2$. – Matt Feb 1 '17 at 16:51
• Sounds reasonable! – avati91 Feb 1 '17 at 16:53

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 $$\langle \frac{\partial u}{\partial n}, v \rangle := \int_\Omega \Delta u \, Ev + \nabla u \cdot \nabla Ev \, \mathrm{d}x.$$
• In fact, we only need $u \in H^1(\Omega) \cap H(div,\Omega)$ to define the trace of normal derivative. This is an important result of Lions and Magenes (Non-Homogeneous Boundary Value Problems and Applications Vol. 1). – macnguyen Nov 9 '20 at 22:59
• What do you mean by $H(div,\Omega)$? Typically, this is the space of vector-valued $u \in L^2(\Omega)^n$ with $div(u) \in L^2(\Omega)$. However, in the question, $u$ is a scalar-valued function. – gerw Nov 10 '20 at 8:09
• My bad, it should be $u \in H^1(\Omega)$ and $\nabla u \in H(div, \Omega)$. What I was trying to say is, we don't actually need anything related to second derivative in order to define the normal trace. This turns out to be more convenient, especially when choosing suitable spaces for weak formulation of certain PDEs. – macnguyen Nov 11 '20 at 16:59
• @macnguyen: But $\nabla u \in H(div,\Omega)$ is actually the very same as $\Delta u \in L^2(\Omega)$. – gerw Nov 11 '20 at 19:12