A possible trace (inequality) defined under negative Sobolev scale

Question

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?

Answer 1

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 $u$.

Answer 2

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.

Answer 3

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 $L^2$.