Sobolev spaces over closed domains.

Question

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?

Answer 1

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 $W^{m,p}(\Omega)$.