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I have a short question about density of spaces. Consider: $C_c^{\infty}(0,1)=\{f\in C^{\infty}(0,1); supp(f)\subset (0,1)\;\text{compact}\}, $ $H_0^1(0,1)=\overline{C_c^{\infty}(0,1)}^{\|\cdot\|_{1,2}} $ with norm $\|u\|_{1,2}=(\|u\|_{L^2}^2+\sum\limits_{|\alpha|\le 1}\|D^{\alpha}u\|_{L^2}^2)^{\frac{1}{2}}$. I know, that $C_c^{\infty}(0,1)$ is dense regarding the $L^2$-norm in the Hilbert space $L^2(0,1)$ of measurable square-integrable functions and it is $C_c^{\infty}(0,1)\subseteq H_0^1(0,1)\subseteq L^2(0,1)$. Is it true if I conclude: $\overline{C_c^{\infty}(0,1)}^{\|\cdot\|_{L^2}} \subseteq \overline{H_0^1(0,1)}^{\|\cdot\|_{L^2}}\subseteq \overline{L^2(0,1)}^{\|\cdot\|_{L^2}} =L^2(0,1)$ and it follows: $ H_0^1(0,1)$ is dense in $L^2(0,1)$ regarding the $L^2$-norm? And generally, consider the Banach space $W^{k,p}(0,1)=\{f\in L^p(0,1);\; f$ has a weak derivative $D^{\alpha}f\in L^p(0,1)$ for all $|\alpha|\le k\}$, $k\in\mathbb{N}$, $1\le p<\infty$ endowed with the norm $\|u\|_{k,p}=(\|u\|_{L^p}^p+\sum\limits_{|\alpha|\le k}\|D^{\alpha}u\|_{L^p}^p)^{\frac{1}{p}}$ (So called Sobolev spaces). Is $\overline{W^{k,p}(0,1)}^{\|\cdot\|_{L^p}}=L^p(0,1)$ for all $k\in\mathbb{N}$ and $1\le p <\infty$? Or is it completely false? Regards.

A supplement: Related to the last question I think: $\overline{W^{k,p}(0,1)}^{\|\cdot\|_{L^p}}=L^p(0,1)$ for all $k\in\mathbb{N}$ and $1\le p <\infty$ has to be true because of the reason mentioned above: $C_c^{\infty}(0,1)$ is dense in $L^P(0,1)$ reagrding the $L^p$-norm and it is $C_c^{\infty}(0,1)\subseteq W^{k,p}(0,1)$. But I nowhere find the result on internet

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  • $\begingroup$ Yes, the statement and the proposed proof are correct. $\endgroup$ – PhoemueX Feb 18 '15 at 23:33
  • $\begingroup$ Thanks! Do you mean all statements, including $\overline{W^{k,p}(0,1)}^{\|\cdot\|_{L^p}}=L^p(0,1)$ for all $k\in\mathbb{N}$ and $1\le p <\infty$ or do you relate to the statement "$H_0^1(0,1)$ is dense in $L^2(0,1)$ regarding the $L^2$-Norm? $\endgroup$ – NewBunny Feb 18 '15 at 23:43
  • $\begingroup$ $\overline{W^{k,p}(0,1)}^{\|\cdot\|_{L^p}}=L^p(0,1)$ for all $k\in\mathbb{N}$ and $1\le p <\infty$ has to be true because of the reason mentioned above: $C_c^{\infty}(0,1)$ is dense in $L^P(0,1)$ reagrding the $L^p$-norm and it is $C_c^{\infty}(0,1)\subseteq W^{k,p}(0,1)$. But I nowhere find the result on internet $\endgroup$ – NewBunny Feb 19 '15 at 6:51
  • $\begingroup$ I meant all your statements :) It is probably not mentioned anywhere, because (as you noted yourself) it is a direct consequence of the density of $C_c^\infty$, which is a (much) stronger claim. Thus, there is not that much of a point in stating the density of $W_0^{k,p}$. $\endgroup$ – PhoemueX Feb 19 '15 at 7:45
  • $\begingroup$ ah ok! thank you very much! $\endgroup$ – NewBunny Feb 19 '15 at 8:15

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