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Consider the Hilbert space $L^2([-L,L])$.

Is the Sobolev space $H^2([-L,L])\subset L^2([-L,L])$ dense?

(Maybe this can be seen from the reason why $L^2([-L,L])$ is a Hilbert space?)

mathfemi
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1 Answers1

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Yes, for any bounded domain $\Omega \subset \mathbb{R}^n$, the space $H^2(\Omega) \subset L^2(\Omega)$ is dense. See Theorem 2 in Sec. 5.3.2 in Evans' "Partial Differential Equations", where it is shown that all elements in the Sobolev space $W^{k,p}(\Omega)$ (the space of $k$ times weakly differentiable functions in $L^p$, with derivatives in $L^p$) can be approximated by functions in $C^\infty(\Omega)$.

Jas Ter
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  • I do not see why this implies that $H^2(\Omega)$ is dense in $L^2(\Omega)$. What you wrote means that $C^{\infty}(\Omega)$ is dense in $W^{k,p}(\Omega)$. Do we also have that $C^{\infty}(\Omega)$ is dense in $L^p(\Omega)$? – mathfemi May 08 '17 at 15:59
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    Absolutely! $L^p = W^{0,p}$. The point is that the smooth functions are dense in all of these spaces, so since $C^\infty \subset H^2 \subset L^2$, $H^2$ must be dense in $L^2$. – Jas Ter May 09 '17 at 07:18