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Is $C^{k+1}(\Omega)\cap W^{k,p}(\Omega) $ dense in $W^{k,p} (\Omega)$? Assuming $\Omega$ has a $C^1$ boundary (or Lipschitz continuous boundary, if allowed?). I know the standard results on smooth compactly supported functions being dense in Sobolev spaces on $\mathcal{R}^n$. Also, on a bounded domain what is the minimum regularity/conditions needed on continuous functions, intersection with $W^{k,p}$ taken as needed, to generate a density result for $W^{k,p}$. Thanks, Sandy

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  • $\begingroup$ @Kakashi a new question posted here. $\endgroup$
    – sandy kuks
    Commented Dec 27, 2022 at 21:53
  • $\begingroup$ Thanks for previous answer/comments @blamethelag, a new question posted. $\endgroup$
    – sandy kuks
    Commented Dec 27, 2022 at 21:54

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please do not tag me when you post a new question. Classical questions about Sobolev spaces are discussed in the three books of Brézis, Evans, Leoni, which should encompass all the information you need.

Here are some density results for Sobolev functions, of course I skipped $p = \infty$ (can you tell why?)

  • $W_0^{k,p}(\Omega)$ is defined as the closure of $C_c^\infty(\Omega)$ for the $W^{k,p}(\Omega)$ norm, hence the density. There might be a touchy point in this definition when $p = \infty$ (can you tell why?).

  • On a bounded non empty $\Omega$, $C_c^\infty(\Omega)$ is never dense in $W^{1,p}(\Omega)$. This can be shown by the Poincaré inequality.

  • $C_c^\infty(\mathbb R^d)$ is dense in $W^{k,p}(\mathbb R^d)$ (can you show it?)

  • The Meyer Serrin density theorem: $C^\infty(\Omega) \cap W^{k,p}(\Omega)$ dense in $W^{k,p}(\Omega)$.

  • The Friedrich density theorem: If $u \in W^{k,p}(\Omega)$ there is $(u_j)$ a sequence of $C_c^\infty(\mathbb R^d)$ such that $u_j \rightarrow u$ in $W^{k-1,p}(\Omega)$ and forall $\omega \subset \subset \Omega$, forall $|\alpha | \leq k$, $\partial ^\alpha u_j \rightarrow \partial ^\alpha u$ in $L^p(\omega)$.

  • Density up to extension : if $\Omega$ is bounded, $\partial \Omega$ is Lipschitz, then $C_c^\infty(\mathbb R^d)$ is dense in $W^{k,p}(\Omega)$ (can you show it?). Actually you can weaken the regularity of $\Omega$ to be open with boundary of class $C^0$ (see Leoni, theorem 11.35).

Observe that the Meyer Serrin's theorem shows that $C(\Omega) \cap W^{k,p}(\Omega)$ is dense in $W^{k,p}(\Omega)$. Now a good question is to know when $C(\overline \Omega) \cap W^{k,p}(\Omega)$ is dense in $W^{k,p}(\Omega)$. The result in Leoni's book tells us that a sufficent condition is $\partial \Omega$ to be continuous. In fact the exercise 11.48 tells us that it's pretty sharp: there exists $\Omega \subset \mathbb R^2$ open bounded non empty for which $C(\overline \Omega) \cap W^{k,p}(\Omega)$ is never dense in $W^{k,p}(\Omega)$

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  • $\begingroup$ Thanks @blamethelag great points. Do you think the bullet point on the Friedrich density theorem can be further specialized to say something about a sequence $u_j$ $\in$ $C_c^k(\mathbb{R}^n)\C_c^{\infty}(\mathbb{R}^n)$ converging in $W^{k,p} (\Omega)$ to the $u$ $\in$ $W^{k,p}(\Omega)$. Basically, I am looking for just the "right" degree regularity? $\endgroup$
    – sandy kuks
    Commented Dec 28, 2022 at 20:11
  • $\begingroup$ And, if possible in the above comment ideally if we could say something about a sequence $u_j$ $\in$ $C_c^k(\mathbf{R}^n)$ \ $C_c^{k+1}(\mathbf{R}^n)$....? $\endgroup$
    – sandy kuks
    Commented Dec 28, 2022 at 20:29
  • $\begingroup$ Why do you want so specific results? I suggest you start by understanding the mentionned theory. Your questions are both solved. $\endgroup$ Commented Dec 29, 2022 at 0:24
  • $\begingroup$ thanks I am aware of the standard stuff, and am aware of references/material you pointed. Unfortunately, in your answer the very specific question I asked was not directly answered. $\endgroup$
    – sandy kuks
    Commented Dec 29, 2022 at 0:29
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If you want to approximate a Sobolev function $f$ in $W^{k,p} (\Omega)$ by continuous functions on $\Omega$, you can choose a smooth function $g$ that approximates $f$ in the $L^p$ norm.

We can be sure of that because the approximation property of smooth functions, which states that for any $\epsilon>0$, there exists a smooth function $g$ such that $|f-g|_p < \epsilon$.

So choose a smooth function $h$ that approximates $g$ in the $W^{k,p}$ norm. You can do this by using the density of the smooth functions in the Sobolev space $W^{k,p} (\Omega)$, because for any $\epsilon>0$, there exists a smooth function $h$ such that $|g-h|_{W^k_p} < \epsilon$.

Therefore function $h$ is a continuous function on $\Omega$ that approximates $f$ in the $W^{k,p}$ norm.

You could also use the fact that the Sobolev space $W^{k,p} (\Omega)$ is compactly embedded in the space $C^0(\Omega)$ of the continuous functions on $\Omega$, which means that any sequence of functions in $W^{k,p} (\Omega)$ has a convergent subsequence in $C^0(\Omega)$. So you can use a diagonalization argument to construct a sequence of continuous functions that converges to $f$ in the $W^{k,p}$ norm...If you wanted to get fancy.

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