# Existence of sequence of $C^{\infty}$ functions to approximate a $W^{1, \infty}$ function

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

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

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.

• 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. – Morgan Jones Jul 3 '20 at 19:13
• 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/… – Morgan Jones Jul 3 '20 at 19:15
• Yes, I meant $m=0$. Your reasoning is correct. – zugzug Jul 3 '20 at 20:29