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I am not mathematician (engineer) and would like to know whether the closure of the well-known sobolev space $H^1(\Omega)$ is equal to $L^2(\Omega)$.

My attempt: $H^1(\Omega)$ is dense in $L^2(\Omega)$ (because $C_0^\infty$ is contained in $H^1(\Omega)$) and $H^1(\Omega)\subset L^2(\Omega)$ therefore $L^2(\Omega)$ is the closure. I am not sure if this is the right way to prove the statement

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    Yes, your proof works. – 2017-01-16
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    You should make clear that it is the closure with respect to the $L^2$ norm. – 2017-01-16
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    And therefore $H^1(\Omega)$ is not closed subset of $L^2(\Omega)$ (w.r.t $L^2$-norm) since a closed set = its closure. – 2017-01-16

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