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? 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.
 A: 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.
