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In the Navier-Stokes Equations by Constantin and Foias, there is a proposition says that

Let $\Omega$ satisfy the segment property. Then $C_0^\infty(\Bbb{R}^n)$ is dense in $W^{m,p}(\Omega)$, for $1\leq p<\infty$.

$\Omega$ is an open subset of $\Bbb{R}^n$. There is no explanation for the notation $C_0^\infty(\Bbb{R}^n)$ in that old book, which I think means the space of smooth functions with compact support.

The domain of functions in $C_0^\infty(\Bbb{R}^n)$ is $\Bbb{R}^n$. But the domain of functions in $W^{m,p}(\Omega)$ is $\Omega$.

Here is my question:

In what sense is $C_0^\infty(\Bbb{R}^n)$ a subspace of $W^{m,p}(\Omega)$?

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    $\begingroup$ You have a restriction $\rho \colon C_0^{\infty}(\mathbb{R}^n) \to C^{\infty}(\Omega)$. The image of that restriction is contained in $W^{m,p}(\Omega)$ and dense in it. $\endgroup$ – Daniel Fischer Nov 7 '15 at 21:59
  • $\begingroup$ Is it the same as "$C_c^\infty(\Omega)$ is dense in $W^{m,p}(\Omega)$?" $\endgroup$ – Jack Nov 7 '15 at 22:07
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    $\begingroup$ No. For nice $\Omega$ and sufficiently large $m$ (can't remember what the bound was, could have been $m > n/2$ or so), functions in $W^{m,p}(\Omega)$ have "well-defined boundary values" in a certain sense. Functions in $C_c^{\infty}(\Omega)$ of course have boundary values $0$, but not all functions in $W^{m,p}(\Omega)$ vanish on the boundary, and the restrictions of $f\in C_0^{\infty}(\mathbb{R}^n)$ to $\Omega$ typically don't. The closure of $C_c^{\infty}(\Omega)$ in $W^{m,p}(\Omega)$ is often denoted by $W^{m,p}_0(\Omega)$. Usually, that's a proper subspace of $W^{m,p}(\Omega)$. $\endgroup$ – Daniel Fischer Nov 7 '15 at 22:19
  • $\begingroup$ Ah. Then the restriction $C_0^\infty(\Bbb{R}^n)\to C^\infty(\Omega)$ is also different from the restriction $C^\infty(\Bbb{R}^n)\to C^\infty(\Omega)$? $\endgroup$ – Jack Nov 7 '15 at 23:20
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    $\begingroup$ The two restrictions have different domains, in that sense they are different, but "in principle" they are the same. On the smaller of the two spaces, they coincide. $\endgroup$ – Daniel Fischer Nov 8 '15 at 9:57
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I would express the statement as: every function in $W^{m,p}(\Omega)$ can be approximated, in the norm, by functions that are $C^\infty$ smooth on all of $\mathbb{R}^n$, with compact support.

The validity of this statement depends on the geometry of $\Omega$; I don't know what is meant by the segment property but it must be something that enables the extension of Sobolev functions to functions on $\mathbb{R}^n$ in the same Sobolev space (i.e., $\Omega$ is a Sobolev extension domain). Once such an extension is achieved, the extended function can be multiplied by a smooth cut-off to make it compactly supported without changing the values on $\Omega$. Then the standard density result applies.

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