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I'm reading a theorem which says that $C^\infty$ intersection with $W^{k,p}$ is dense in $W^{k,p}$. I don't understand why they take the intersection. Isn't it $C^\infty$ a subspace of $W^{k,p}$? Thanks.

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You don't say what space the functions are on, but here let's consider functions on $\mathbb{R}$. Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be given by $f \equiv 1$. Then $f \in C^\infty(\mathbb{R})$ but certainly $f \notin L^p(\mathbb{R})$ for any $p \ge 1$ since $\left(\int_\mathbb{R} |f|^p \right)^{1/p} = \left(\int_\mathbb{R} 1\right)^{1/p} = \infty.$ But if $f \in W^{k,p}(\mathbb{R})$, then in particular $f \in L^p(\mathbb{R})$. Therefore we see that it is not necessarily true that every smooth function lies in $W^{k,p}$.


$W^{k,p}(\Omega)$ is not the completion of $C^\infty(\Omega)$ for general $\Omega$ since, as was shown above, $C^\infty(\Omega)$ isn't contained in $W^{k,p}(\Omega)$. Define a norm $\|f\|_{k,p} = \left( \sum_{|\alpha| \leq k} \|D^{\alpha} f\|_p^p \right)^{1/p}.$ and write $\widetilde{C}^k(\Omega) = \{f \in C^k(\Omega) : \|f\|_{k,p} < \infty\}.$ Then $W^{k,p}(\Omega)$ is the completion of $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$ as long as $p \in [1, \infty)$. Look up the Meyers-Serrin theorem for more details.

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    Ok, I think I have the answer. Taking a constant function that will belong to $W^{k,p}(\Omega)$ I see that $C_0^{\infty}(\Omega)$ is not dense. Actually the definition of $W_0^{k,p}(\Omega)$ is precisely the completion of $C_0^{k}(\Omega)$ . Also it is true that $W^{k,p}(\mathbb{R}^n)$ =$W_0^{k,p}(\mathbb{R}^n)$.2012-08-24
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You need to take intersection since smooth functions can grow arbitrarily fast near the boundary. It is another matter to talk about $W^{k,p}_0$.

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In this webpage they give an answer for your problem, it is used to prove the density of $C^{\infty}_c$

http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/