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Is $H^1_0(\Omega)$ dense in $L^2(\Omega)$ for bounded domains? It is true for $H^1$ functions of course but what about this subset?

Sorry for the elementary question but I never see this so I think the answer is it's not.

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    $\begingroup$ $C_0^\infty(\Omega) \subset H_0^1(\Omega) \subset L^2(\Omega)$. $\endgroup$ – Siminore Apr 12 '13 at 10:24
  • $\begingroup$ as $C^\infty_0$ is dense in $L^2$, so is $H^1_0$. How stupid of me. Thanks $\endgroup$ – michael_faber Apr 12 '13 at 10:46
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I am still interested in the question: is this how things work?

1) $\overline{C_0^{\infty}(\Omega)}^{||\cdot||_{2}} = L^2(\Omega)$

2) $\overline{C_0^{\infty}(\Omega)}^{||\cdot||_{H^1}} = H_0^1(\Omega)$

3) $C_0^{\infty} \subset H^1_0(\Omega) \subset L^2(\Omega)$ and taking the closure with respect the $L^2$-norm we get $L^2(\Omega) \subset \overline{H^1_0(\Omega)}^{||\cdot||_2} \subset L^2(\Omega)$.

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