On the density of $\mathcal{C}^\infty(\Omega) \cap W^{1,\infty}(\Omega)$ in $W^{1,\infty}(\Omega)$ Let $\Omega$ be an open set of $\mathbb{R}^n$ ($n\ge 1$). We know that the Meyers-Serrin theorem isn't true in $W^{1,\infty}(\Omega)$. But is it true that $\mathcal{C}^\infty(\Omega) \cap W^{1,\infty}(\Omega)$ is dense in $W^{1,\infty}(\Omega)$ for the weak-star topology of $W^{1,\infty}(\Omega)$ ? If not, what is the closure of $\mathcal{C}^\infty(\Omega) \cap W^{1,\infty}(\Omega)$ for the weak-star topology of $W^{1,\infty}(\Omega)$?
 A: I think you can use the Stone-Weierstrass theorem to see that this is true locally for compactly supported ($C^\infty$) functions since you can separate points (Hahn-Banach) and then bootstrap up by partitioning your open set into a countable union of compacts where you are within $\epsilon_n=\epsilon 2^{-n}$ on each compact piece. Since uniform convergence implies convergence in $L^p$ of a compact set, I think you can get in from there.
The general strategy I'm describing is:
1) Break your space up into a countable union of compact sets (a ring decomposition of $\mathbb{R}^n$ is an example for this when you have $\mathbb{R}^n$)
2) Approximate your function (in your norm) by $C^\infty$ functions to within $\epsilon 2^{-n-1}$ or so on the compact set minus say a set of measure $\epsilon 2^{-n-1}$ around the boundary (density using Stone-Weierstrass allows this)
3) Stitch them together using bump functions using the boundary space set as the transition between the $C^\infty$ functions.
4) Add up the errors, receive $\epsilon$-closeness.
