The Sobolev space $W^{1,\infty}(\mathbb R^n,\mathbb R^n)$ is not dense in $L^\infty(\mathbb R^n,\mathbb R^n)$. In fact the functions in $W^{1,\infty}(\mathbb R^n,\mathbb R^n)$ are Lipshitz, and not even continuous functions are dense in $L^\infty$.


But, when $n>1$, could $X := \{ B \in L^\infty(\mathbb R^n,\mathbb R^n): \nabla \cdot B \in L^\infty(\mathbb R^n,\mathbb R^n) \}$ 
be dense in $L^\infty(\mathbb R^n,\mathbb R^n)$? 


$\nabla \cdot B$ denotes the divergence operator.



Is $X = \{ B \in L^\infty(\mathbb R^n,\mathbb R^n): \nabla \cdot B \in L^\infty(\mathbb R^n,\mathbb R^n) \}$ a dense subspace?
Here is a sketch that I think could be turned into a counterexample.


Take $n=2$, let $\hat{r}$ be the outward-pointing radial unit vector field, and set $F = \cos(\pi r^2) \hat{r}$. Let $G$ be a vector field with $|\nabla \cdot G |\le M$. Now let $A\_n$ be the annulus defined by $\sqrt{2n-1} \le r \le \sqrt{2 n}$. You may verify that the outward flux of $F$ across $\partial A\_n$ is of order $\sqrt{n}$. On the other hand, noting that $A\_n$ has area $\pi$ for every $n$ and using Green's theorem, the outward flux of $G$ across $\partial A\_n$ is at most $M \pi$. This seems incompatible with $F$ and $G$ being uniformly close.


It's not quite a counterexample because $G$ need not satisfy the hypotheses of Green's theorem. In particular, you could mess with its value on the boundary of all the $A\_n$ (which is a null set) to get the flux to come out right. But by fuzzing the annuli a little bit, you ought to get an uncountable family of annuli where the flux of $F$ is getting large on all of them, but the flux of $G$ has to remain bounded on "almost every" of them.


