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.

  • $\begingroup$ Look at Amrouche's work, I would think. $\endgroup$
    – username
    May 14, 2014 at 13:55

2 Answers 2


I don't know the references. It's hard to find the references for such questions, as they are not often used by others. It is better to derive such results based on the well-known results.

For $1\leq p\leq\infty$, $W^{-m,p}(\Omega)$ is usually defined as the dual space of $W^{m,p'}_0(\Omega)$, the completion of $C^\infty_0(\Omega)$ in $W^{m,p'}(\Omega)$. Therefore, $W^{m,p'}_0(\Omega)$ can be viewed as a closed subspace of $W^{m,p'}_c({\mathbb R}^n)$. By the Hahn-Banach extension theorem, any continuous linear functional on $W^{m,p'}_0(\Omega)$ admits an extension as a continuous linear functional on $W^{m,p'}_c({\mathbb R}^n)$, which means that there exists a function $\widetilde u\in W^{-m,p}({\mathbb R}^n)$ such that $(\widetilde u,v)=(u,v)$ for all $v\in W^{m,p'}_0(\Omega)$ and the given $u\in W^{-m,p'}(\Omega)$. Since $\widetilde u=\sum_{|\alpha|\leq m}\partial^\alpha u_\alpha$ for some $u_\alpha\in L^p({\mathbb R}^n)$, it follows that $u$ can also be represented by $u=\sum_{|\alpha|\leq m}\partial^\alpha u_\alpha$.

Negative real-valued Sobolev spaces also admit such representation. If $s\in(0,1)$ and $m$ is a nonnegative integer, then $$ u\in H^{-m-s}(\Omega) \implies u=\sum_{|\alpha|\leq m+1}\partial^\alpha u_\alpha, $$ where $u_\alpha\in H^{1-s}(\Omega)$.

  • $\begingroup$ For the density result, it seems crucial to consider the dual space of compactly supported function space. Is it still true for $W^{m,p'}$? For the representation result, I have not quite understood. Could you elaborate more? Thanks again! $\endgroup$
    – Hui Zhang
    May 14, 2014 at 14:32
  • $\begingroup$ Hi. The dual space of $W^{m,p'}_0(\Omega)$ can be represented by the derivatives of regular functions. But the functions in the dual space of $W^{m,p'}(\Omega)$ cannot be represented in such a way, their representations may include some boundary terms. $\endgroup$
    – Buyang LI
    May 14, 2014 at 14:47
  • $\begingroup$ Hi, I got a problem for the density result applied to the low dimensional manifold, say $W^{-1/2,p}(\partial\Omega).$ Of which space could I view $W^{1/2,p}(\partial\Omega)$ as a subspace? $\endgroup$
    – Hui Zhang
    May 14, 2014 at 14:53
  • 1
    $\begingroup$ Hi. If $\Omega\subset{\mathbb R}^n$ and $\partial\Omega$ is Lipschitz continuous, then $\partial\Omega$ is locally isomorphic to the unit disk on ${\mathbb R}^{n-1}$ via a Lipschitz continuous bijection (via a partition of unity). Therefore, function spaces such as $H^{-1/2}(\partial\Omega)$ has the same property as $H^{-1/2}({\mathbb R}^{n-1})$. Since the $C^\infty$ functions are dense in $H^{-1/2}({\mathbb R}^{n-1})$, it follows that, via the Lipschitz continuous bijection, the "Lipschitz continuous" functions on $\partial\Omega$ are dense in $H^{-1/2}(\partial\Omega)$. $\endgroup$
    – Buyang LI
    May 15, 2014 at 3:59
  • $\begingroup$ This is a simple argument based on the mapping of $\partial\Omega$ into ${\mathbb R}^{n-1}$. I guess that, by a more delicate argument, $C^\infty({\mathbb R}^n)$ functions are dense in $H^{-1/2}(\partial\Omega)$. $\endgroup$
    – Buyang LI
    May 15, 2014 at 4:03

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.


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