TITLE: Negative real order Sobolev spaces: density and representation
QUESTION [8 upvotes]: First, I give my motivation to ask this question. The generalised Neumann trace can be defined as
$$
{}_{H^{-1/2}(\partial\Omega)}\langle\frac{\partial u}{\partial{\mathbf{n}}},v\rangle_{H^{1/2}(\partial\Omega)}
={}_{H^{-1}(\Omega)}\langle\Delta u,v\rangle_{H^1(\Omega)}-\int_{\Omega}\nabla u\cdot\nabla v.
$$
But this involves integral in the volume of $\Omega$ which is not really like a trace to me. In particular, if we substitute this definition to Green's representation formula, we obtain an identity of no use like $0=0.$
Then, I found a theorem in Girault-Raviart's book says $H(\mathrm{div},\Omega)$ always has normal trace in $H^{-1/2}(\partial\Omega)$ by smooth approximation. This is the first time I saw such smooth approximation result of negative order Sobolev spaces. I searched on the web and in books but I found only for $H^{-s}(\mathbb{R}^n)$ that $C_c^\infty(\mathbb{R}^n)$ is dense. I can not find similar results on Lipschitz domains. It would be good to give me a reference.
Second, I also saw a result in McLean's book. That says, for any integer negative order Sobolev space $W^{-m,p}(\Omega)\ni f$ there is a representation
$$
f=\sum_{|\alpha|\leq m}\partial^{\alpha}f_{\alpha} \mbox{ with }f_{\alpha}\in L^p{(\Omega)}.
$$
But he does not say about negative real order Sobolev spaces. I would like a reference on similar results on negative real order Sobolev spaces.

REPLY [2 votes]: There are several equivalent definition of $W^{s,p}({\mathbb R}^n)$ for a general real number $s$. One popular definition is to use the Fourier transform:
$~~~~~$ Let $S({\mathbb R}^n)'$ denote the space of tempered distributions (generalized functions) and let $\langle D\rangle^s$ denote the operator
$$
\langle D\rangle^su=[(1+|\xi|^2)^{s/2}\widehat u(\xi)]^\vee ，
$$
where $\widehat u(\xi)$ is the Fourier transform of $u$ and $^\vee$ denote the inverse Fourier transform. Then one defines
$$W^{s,p}({\mathbb R}^n):=\{u\in S({\mathbb R}^n)':\langle D\rangle^s u\in L^p({\mathbb R}^n)\}$$
The operator $\langle D\rangle^s:W^{s_0,p}({\mathbb R}^n)\rightarrow W^{s_0-s,p}({\mathbb R}^n)$ is bounded for any $s_0,s\in{\mathbb R}$. This is a consequence of this definition itself. 
$~~~~~$ Usually, people use the following argument to prove that $\partial^\alpha$ maps $W^{s,p}({\mathbb R}^n)$ to $W^{s-|\alpha|,p}({\mathbb R}^n)$:
$~~~~~$ For any partial differential operator $\partial^\alpha$, we have
$\partial^\alpha=\langle D\rangle^{|\alpha|}\langle D\rangle^{-|\alpha|}\partial^\alpha$. Since the operator $\langle D\rangle^{|\alpha|}$ maps $W^{s,p}({\mathbb R}^n)$ to $W^{s-|\alpha|,p}({\mathbb R}^n)$, it suffices to prove that the operator $\langle D\rangle^{-|\alpha|}\partial^\alpha$ is a pseudo-differential operator of order $0$, which is bounded on $W^{s,p}({\mathbb R}^n)$ for any $s\in{\mathbb R}$. This is often accomplished by using the theory of singular integrals.