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$\Omega$ is a bounded smooth domain. Take $u \in H^1(\Omega) \cap L^\infty(\Omega)$. Is it possible to find a sequence $u_n$ such that $u_n \in H^1_0(\Omega) \cap L^\infty(\Omega)$ and $u_n \to u$ in $L^\infty(\Omega)$?

It seems so, since we can take a smoothed version of $u\chi_{\Omega_n}$, where $\Omega_n$ is a subset of $\Omega$ that increases to $\Omega$ as $n \to \infty$.

Is it possible to get other norm convergence too? I think not $H^1$ norm.

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    @zaq Would it work for a different $L^p$ norm? – 2017-01-10
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    @zaq If so, then for a subsequence $u_n \to 1$ a.e. Then $u_n \geq 0.5$ a.e. eventually, so we have the same problem, no? So maybe the convergence doesn't hold in any $L^p$ space. – 2017-01-10

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  1. One cannot approximate $u \in H^1(\Omega) \cap L^\infty(\Omega)$ in $H^1$ norm by $H^1_0(\Omega)$ functions unless $u$ is already in $H^1_0(\Omega)$. This is because $H^1_0(\Omega)$ is a closed subspace of $H^1(\Omega)$.

  2. The function $u\equiv 1$ is in $ H^1(\Omega) \cap L^\infty(\Omega)$ but cannot be approximated by $H^1_0(\Omega)$ functions in $L^\infty$ norm. Indeed, if $u_n\to u$ in $L^\infty$, then for sufficiently large $n$ we have $u_n\ge 1/2$ a.e. in $\Omega$, and such a function is not in $H^1_0(\Omega)$.

  3. Approximation is possible with respect to $L^p$ norm for $1\le p<\infty$. Indeed, let $K$ be a compact subset of $\Omega$ such that the measure of $\Omega\setminus K$ is small (it can be made arbitrarily small). Then $u\chi_K$ is close to $u$ in $L^p$ norm. Furthermore, mollification of $u\chi_K$ produces a sequence of smooth functions with compact support in $\Omega$ which converge to $u \chi_K$ in $L^p$.