The Meyer Serrin Theorem states that the space $C^\infty(\Omega) \cap W^{m,p}(\Omega)$ is dense in $W^{m,p}(\Omega)$ where $\Omega \subset \mathbb{R}^n$ is some open set and $1 \le p < \infty$. I am interested in the case when $p= \infty$, where in general the Meyer Serrin Theorem does not hold. However does the $p = \infty$ case hold under the stronger assumption $\Omega$ is bounded and of finite measure? To be more precise I would like to know if the following statement is true:
Let $\Omega \subset \mathbb{R}^n$ be a bounded open set of finite measure and $u(x)$ Lipshitz continuous (so $u \in W^{1, \infty}(\Omega)$). Then there exists a sequence of functions $u_i \in C^\infty(\Omega)$ such that $\lim_{i \to \infty} ||u_i- u||_{W^{1,\infty}(\Omega)}=0$.
It seems that this result is true as indicated in Exercise 11.31 from the book “A first course in Sobolev spaces" by Giovanni Leoni. This exercise has been considered on stack Exchange before but I am still not convinced that the my above statement is correct. The stack exchange questions can be found at:
Use $C^\infty$ function to approximate $W^{1,\infty}$ function in finite domain
Why $C^{\infty}(\Omega) \cap W^{1, \infty}(\Omega)$ isn't dense in $W^{1, \infty}(\Omega)$?
Best Answer
Note that if we have a sequence of $u_{n} \in C^{1}(\Omega)$ converging to some function $u$ with respect to the $||\;||_{W^{1,\infty}}(\Omega)$ - norm, then by completeness of the space $X = (C^{1}(\Omega),||\;||_{W^{1,\infty}(\Omega)})$, the limit $u$ is also in $X$.
But $W^{1,\infty}(\Omega)$ does not only contain $C^{1}$-functions - it is easy to construct a Lipschitz function that does not have a continuous derivative.
Thus, the statement is wrong - we can´t even hope to approximate by $C^{1}$-functions.
Note that the other questions you linked handle a slightly different (weaker) problem: $||\nabla{u_{n}}|| \rightarrow ||\nabla{u}||$ does not imply that $$||\nabla{u}-\nabla{u_{n}}|| \rightarrow 0$$