# Sobolev spaces over closed domains.

I am currently working through books on Sobolev spaces and I notice that these spaces are almost always defined over open domains, i.e. we look at $W^{m,p}(\Omega)$, where $\Omega$ is open. Because these spaces are equivalence classes and ignore sets of measure $0$, my intuition tells me that we should have $W^{m,p}(\Omega) = W^{m,p}(\bar{\Omega})$ and that all the results given for the space $W^{m,p}(\Omega)$ could equally be given for $W^{m,p}(\bar{\Omega})$. Is my intuition on this correct?

No, this intuition is not correct, because of the issue of domains with cracks. For an example, consider the unit disk $D\subset\mathbb{R}^2$, and the domain $\Omega$ which is obtained from $D$ by removing several straight lines. The domain $\Omega$ is not connected, even though $\bar{\Omega} = D$. This means that there can be piecewise constant functions in $W^{m,p}(\Omega)$ even though this wouldn't be allowed for functions in $W^{m,p}(D)$.
The issue is that the space of test functions on $\Omega$ and $D$ are very different, so they give rise to different notions of distributional derivative and thus different Sobolev spaces. In particular, there are no test functions (smooth functions with compact support) on $\Omega$ which are nonzero on the "cracks" of the domain, so it cannot detect the discontinuities that may form for functions in $W^{m,p}(\Omega)$.
• Thanks for the explanation. I am essentially restricting my work to intervals in $\mathbb{R}^{+}$. Are there similar examples that may cause issues if we consider the Sobolev spaces over $I=(0,a)$ and $\bar{I}=[0,a]$? – User817374 Aug 9 '15 at 16:10
• I don't think that will cause any issues. I believe the spaces $W^{m,p}(I)$ and $W^{m,p}(\bar{I})$ are isomorphic, and here's roughly why. Though the space of distributions over $I$ and over $\bar{I}$ are different, the only difference is the behavior at the boundary. Distributions over $I$ can blow up arbitrarily quickly near the boundary, while this can't happen for distributions over $\bar{I}$. I think the integrability requirements built into $W^{m,p}(I)$ rule out any differences though. – felipeh Aug 9 '15 at 16:18
• I have been thinking about this statement today (hence why I haven't closed the question). It seems to me that since $\bar{I}$ is closed, $C_{0}^{\infty}(\bar{I})$ would contain the constant function $f(x) = 1$, where as $C_{0}^{\infty}({I})$ would not. Thus there would be functions considered weakly differentiable in the space over ${I}$ but not $\bar{I}$? How would one go about proving that all such functions are excluded by the integrability criteria? – User817374 Aug 10 '15 at 20:44
• I realise that a major difference may be in the functions that are locally integrable over open and closed intervals. Adams gives the definition of local integrability as " $u \in L^{1}_{loc}(\Omega)$ if $u \in L^{1}(A)$ for every open A compactly contained in \Omega". If $\Omega$ is closed, then can we not choose $A=\Omega \setminus \Gamma \Omega$ and have that $L^{1}_{loc}(\Omega) = L^{1}(\Omega)$? – User817374 Aug 10 '15 at 21:26