Let $\Omega \subset \mathbb R^n$ a domain. We define $W_0^{1,p}(\Omega )$ to be the closure of $\mathcal C^\infty _0(\Omega )$ (i.e. function $\mathcal C^\infty $ compact supported) in $W^{1,p}_0(\Omega )$. I know that $\mathcal C^{\infty }_0(\Omega )$ is dense in $W^{1,p}(\Omega )$. Doesn't it should mean that $W_0^{1,p}(\Omega )=W^{1,p}(\Omega )$ ? The thing is if it would be true, to introduce $W_0^{1,p}(\Omega )$ wouldn't be useful, that's why I have doubt about this result...
Do we have that $\overline{\mathcal C_0^\infty (\Omega )}=W^{1,p}(\Omega )$?
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sobolev-spaces
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0@OpenBall: Could you be more clear please ? I don't really understand. – 2017-01-07
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0Assuming $\Omega\neq\mathbb{R}^n$, then no. – 2017-01-07
1 Answers
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$C_0^\infty(\Omega)$ is NOT dense in $W^{1,p}(\Omega)$.
For example, the function $f(x)=1$, for $x\in(0,1)$, CANNOT be approximated by functions in $C_0^\infty(0,1)$, in the $W^{1,2}-$norm, since the $W^{1,2}-$norm is stronger than the uniform norm on $(0,1)$. Thus the elements of the closure of $C_0^\infty(0,1)$, with $W^{1,2}-$norm, extend continuously in $[0,1]$, and vanish at $x=0$ and $x=1$.