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Why is that not possible (if it is not) to define the trace of a function in a very weak regularity space $H^{-s}(\partial \Omega)$? We usually encounter trace theorem as

$$Tr: H^{s}(\Omega) \rightarrow H^{s- 1/2}(\partial \Omega)$$

$$ \| Tr \, (u) \|_{s - 1/2, \partial \Omega} \leq C \| u \|_{s,\Omega} \,,$$

which is good for $s>1/2$ with suitable regularity of $\partial \Omega$, e.g. $s< 3/2$ if Lipschitz.)

However, I always wonder, if I only aim for a very weak functional, and not expect a well-defined "function", can we just extend the definition of trace of $g$ in $H^{-s}(\partial \Omega)$ as follows?

$$\bigl\langle \frac{\partial f }{\partial \nu} , Tr (g) \bigr\rangle_{\partial \Omega} := \langle \nabla f, \nabla g \rangle_{\Omega} + \langle \Delta f, g \rangle_{\Omega}$$

where now the regularity of test functions $f$ shall be relative high, e.g. $H^{s+2}_0(\Omega)$.

Is there some issue that makes this definition pathological?

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    $\begingroup$ The problem is that if you pick $f\in H_0^{s+2}$, then $\partial f/\partial\nu$ will be zero on the boundary. $\endgroup$ Jun 16 '15 at 10:29
  • $\begingroup$ Thank you very much for your comments. You are right indeed that if we choose this particular space, the Neumann derivative shall be zero, but can we choose any other testing space instead which might legitimize the definition? $\endgroup$
    – mediocre
    Jun 17 '15 at 10:20
  • $\begingroup$ You could just use $C^\infty(\mathbb{R}^n)$ oder $C_c^\infty(\mathbb{R}^n)$ as a test function space. $\endgroup$ Jul 18 '15 at 16:45
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In Lions & Magenes, Non-homogeneous boundary value problems and applications [1], they give a version of the trace inequality for negative Sobolev spaces: Chapter II, Theorem 6.5. However, they only show that $$\operatorname{Tr} : D_A^s(\Omega) \to H^{s-1/2}(\partial \Omega)$$ is bounded, for a function space closely related to $H^s(\Omega)$ but not exactly the same (roughly speaking, it's the subset of $H^s(\Omega)$ consisting of distributions which decay at some rate near the boundary, the necessary rate determined by the elliptic operator $A$). I guess this slight loss is what results from needing to choose the test space carefully.

In the special case where $Au=0$, the $D^s_A$ norm of $u$ coincides with the $H^s$ norm, so it follows from their result that $||\operatorname{Tr} u||_{H^{s-1/2}(\partial \Omega)}\leq C ||u||_{H^s(\Omega)}$ for a constant $C$ independent of $u$.

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Here is the problem in your tentative definition: Say that $\partial\Omega$ is as smooth as you wish. It is correct that given a function $h\in H^s(\partial\Omega)$ you can for instance solve the BVP $\Delta f=0$ in $\Omega$, $\partial f/\partial\nu=h$ on the boundary (actually there is a codimension-one restriction that $\int h=0$, but I see it harmless). You have an estimate $\|f\|_{s+3/2}\le C\|h\|_s$. Then you deduce $$|\langle h,g\rangle_{\partial\Omega}|\le \left|\int_\Omega\nabla f\cdot\nabla g\right|.$$ But what it gives you is $$|\langle h,g\rangle_{\partial\Omega}|\le C\|f\|_{s+3/2}\|\nabla g\|_*$$ where the $*$-norm means that $\nabla g$ is taken in the dual space of $H^{s+1/2}(\Omega)$. If $s>-1/2$, this space is much smaller than $H^{-s-1/2}(\Omega)$.

So your suggestion allows you to define the trace of $g$ in $H^s(\partial\Omega)$ provided $g$ belongs to something like the dual of $H^{s-1/2}(\Omega)$. Somehow, it is just a tautology.

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  • $\begingroup$ How are you defining $H^s(\Omega)$ when $s<0$? $\endgroup$
    – Kweku A
    Apr 23 '19 at 14:03
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    $\begingroup$ @KwekuA When $s=-\sigma<0$, $H^s(\Omega)$ is the dual of the space $H^\sigma_0(\Omega)$, the closure of ${\cal D}(\Omega)$ in $H^\sigma(\Omega)$. Thus $H^s(\Omega)$ ``ignores '' the boundary. $\endgroup$ Apr 23 '19 at 15:07
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The problem is that if you relax the constraints put on $g$, which in this context is an $L^2(\Omega)$ equivalence class of functions, eventually all values on the surface you consider are arbitrary and the integral against a test function, however regular that one may be, is not defined simply because the integrand is not defined. Think of a "function" $f\in L^2$ you can change, at will, all its values on the surface (of measure $0$ in $\Omega$) and still have the same element of $L^2$.

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  • $\begingroup$ It isn't necessary to evaluate functions pointwise to make sense of boundary values. This is the whole point of the trace operator $H^1(\Omega)\to H^{1/2}(\partial\Omega)$ that it can be sensibly defined without point evaluations. $\endgroup$ Jul 18 '15 at 16:48
  • $\begingroup$ @Johannes, I hope we agree that the original poster assumes $s$ to be positive and wants to take $g\in H^{-s}(\Omega)$ (not in $H^{-s}(\partial\Omega)$). Are you suggesting that the expression OP proposes correctly defines a boundary trace $Tr(g)\in H^t(\partial\Omega)$ for some negative $t$? In my opinion, OP's question was why he cannot find trace theorems like such in the literature. $\endgroup$
    – BLM
    Jul 22 '15 at 17:25
  • $\begingroup$ I do not speculate why the OP hasn't found any satisfying theorems since functional analysis is not my speciality. Maybe he just had bad luck with his search terms... In any case, it is still a reasonable question. The RHS of OP's equation is well-defined and linear on any $H^t(\Omega)$ and if one can verify the necessary continuity and density arguments, it can be used to define a trace operator. There are other possible avenues: Consider for example some open neighborhoods of $\partial\Omega$ of width $\epsilon$ (positive measure, so well-defined restrictions exist !) ... $\endgroup$ Jul 22 '15 at 20:38
  • $\begingroup$ ... and then take a limit $\epsilon\to 0$ in some suitable space of distributions on $\partial\Omega$. It seems perfectly reasonable to ask if such a limit exists and I do not see how the negativity of the exponent should cause any trouble. So the result will be a distribution, not a proper function. So what? $\endgroup$ Jul 22 '15 at 20:39
  • $\begingroup$ In order that the RHS of OP's suggested definition defines a trace, the two evaluations, appearing in the RHS, must be defined independently of the LHS. In the original setting of the standard trace theorem, we have two inner products both with only functions in the Hilbert space $L^2(\Omega)$. If you want to relax the constraints on $g$ you should define how to evaluate the $(\nabla f, \nabla g)$ term appearing in the definition for an arbitrary $g$ of the new class. $\endgroup$
    – BLM
    Jul 24 '15 at 17:23

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