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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3 Answers
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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.
answered Aug 18, 2012 at 23:58
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$\begingroup$ Thanks @timur and Henry. Does this mean that I cannot define $W^{k,p}(\Omega)$ as $H:=$ the completion of $C^{\infty}(\Omega)$ under the norm $$\|f\|=\left( \sum_{|\alpha| \leq k} \int_U \! |D^{\alpha} f|^p \, \mathrm{d}x \right)^{1/p}$$ because then there would be functions in $H$ that have have $\int_{\Omega}=\infty$ and consequently don't belong to $W^{k,p}$. here $\Omega$ is a bounded domain. $\endgroup$ Aug 21, 2012 at 21:52
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$\begingroup$ Then as $C_0^{\infty}(\Omega)$ contains $\widetilde{C}^k(\Omega)$ is true that $W^{k,p}(\Omega)$ is the completion of $C_0^{\infty}(\Omega)$ for $p \in [1,\infty)$. right? $\endgroup$ Aug 21, 2012 at 22:39
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1$\begingroup$ If $C^\infty_0(\Omega)$ denotes smooth functions on $\Omega$ with compact support, then it is not true that $\widetilde{C}^k(\Omega) \subset C^\infty_0(\Omega)$. For example, $\exp(-x^2) \in \widetilde{C}^k(\mathbb{R})$ for any $k$, but $\exp(-x^2)$ is not compactly supported. $\endgroup$ Aug 21, 2012 at 23:28
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$\begingroup$ Thank you Henry, one last question (I hope) , is it then true that $W^{k,p}(\Omega)$ is the completion of $C_0^{\infty}(\Omega)$ ( smooth functions on $\Omega$ with compact support) where $\Omega$ is a bounded domain? I'm particulary interested in the case $p=2$ $\endgroup$ Aug 21, 2012 at 23:41
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$\begingroup$ 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 $W^{k,p}(\mathbb{R}^n)$ =$W_0^{k,p}(\mathbb{R}^n)$. $\endgroup$ Aug 24, 2012 at 1:21
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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/
