is $H^{1}\left(\Omega\right)\cap L^{\infty}\left(\Omega\right)$ closed in $H^{1}\left(\Omega\right)$?

Question

Let $\Omega$ bounded, open, connected and Lip. domain. Is $H^{1}\left(\Omega\right)\cap L^{\infty}\left(\Omega\right)$ closed in $H^{1}\left(\Omega\right)$? i.e., with the norm of $H^{1}\left(\Omega\right)$. Is important for me because if $H^{1}\left(\Omega\right)\cap L^{\infty}\left(\Omega\right)$ is closed, it is a reflexive Banach space and every bounded sequence contains a subsequence which is weakly convergent.

Answer 1

No. Due to your given geometric conditions we have $C^\infty(\bar{\Omega})$ is dense in $H^1(\Omega)$, and $C^\infty(\bar{\Omega})$ is a subset of $H^1(\Omega)\cap L^\infty(\Omega)$. However, if $d\geq 2$, there exists some $f\in H^1\setminus L^\infty$, so $H^1(\Omega)\cap L^\infty(\Omega)$ is a proper subset of $H^1(\Omega)$, whose closure is $H^1(\Omega)$.