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Arising as the traces of $H(div; \Omega)$, I am wondering if the space $H^{-1/2}(\partial \Omega)$ has any regularity properties? (Containment in BV would be wonderful, although I doubt it holds.) It seems difficult to find anything in the literature.

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Of course not. For instance, it contains $L^2(\partial\Omega)$. By definition, $H^{-1/2}(\partial\Omega)$ is the dual of $H^{1/2}(\partial\Omega)$. If $\Omega$ is $n$-dimensional $n\ge3$, then $\partial\Omega$ is $(n-1)$-dimensional and $H^{1/2}(\partial\Omega)\subset L^p(\partial\Omega)$ by Sobolev injection, for every $p\le\frac{2(n-1)}{n-3}$ ($<+\infty$ if $n=3$) and therefore $H^{-1/2}(\partial\Omega)$ contains $L^{p'}$. But you cannot say much more. If $s>\frac n2$, then $H^s(\Omega)\subset C^0(\overline\Omega)$ and therefore $H^{\frac12-s}(\partial\Omega)$ is contained in the set of bounded measures. But this does not apply to $H^{-\frac12}$, even if $n=2$ (thanks to A. Rekalo). Actually, a use of the Uniform Boundedness Principle tells us that there exists an element of $H^{-\frac12}(\partial\Omega)$ that is not a bounded measure.

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@Andrey. I acknowledge my mistake about continuity (I'll edit). But I maintain that $H^{-1/2}$ contains $L^2$, just because $L^2\subset H^{1/2}$. –  Denis Serre Dec 3 '12 at 11:41
    
@Denis Serre: Sorry, my reading skills are horrible today. I misread your statement (as if you wrote that $H^{-1/2}$ is contained in $L^2$). Stupid me. –  Andrey Rekalo Dec 3 '12 at 12:26
    
no problem. This happens to me too. And I know that it is easier to see two mistakes when there is at least one for sure. –  Denis Serre Dec 3 '12 at 14:32

Ok, a further question: What if we have $div \phi \equiv 1$ on $\Omega$? (Or up to a set of measure $\epsilon$.) Can we then get extra regularity for the normal trace?

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