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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 $W^{1, \infty}(\Omega)$?

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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 $$||\nabla{u}-\nabla{u_{n}}|| \rightarrow 0$$

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Let's consider a far simpler problem. Let $m=1$, $n=1,$ $p=\infty$. Let $\Omega$ be any region with a neighborhood around $0$. Prove that the Heaviside step function, $$ H(x)= \begin{cases} 0,&x<0 \\ 1,&x\geq 0 \end{cases} $$ cannot be approximated in the $L^\infty$ norm by smooth functions. The reason why the denseness claim holds for $p=\infty$ is because integrals handle jumps.

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  • $\begingroup$ Thanks for your comment. Did you mean the case $m=0$? I guess by the uniform convergence theorem (en.wikipedia.org/wiki/Uniform_convergence) if we uniformly approximate a function in $L^\infty$ by a sequence of continuous functions then that limit function must also be continuous. The heavyside function is clearly discontinuous at $x=0$ so cannot be approximated in $L^\infty$ by continuous functions. $\endgroup$ – Morgan Jones Jul 3 '20 at 19:13
  • $\begingroup$ I did have some confusion if the $L^\infty$ norm was the same as the uniform norm. It turns out they are for continuous functions over open sets. math.stackexchange.com/questions/618101/… $\endgroup$ – Morgan Jones Jul 3 '20 at 19:15
  • $\begingroup$ Yes, I meant $m=0$. Your reasoning is correct. $\endgroup$ – zugzug Jul 3 '20 at 20:29

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