Let $\Omega\subset\mathbb{R}^n$ be open and connected. Given any function $V\in L^2(\Omega,\mathbb{R}^n)=:H$, is there an $u\in W^{1,2}(\Omega)$ such that $\nabla u = V$, i.e. $\|\nabla u-V\|_H = 0$? Here $W^{1,2}(\Omega)$ denotes a Sobolev space and the coordinates of the gradient $\nabla u$ are the weak partial derivatives $\partial_i u$ for $1\le i \le n$.
I feel that these question might be quite difficult. I'd be glad for any hints or pointers to literature on this or anything that might help me answer this question.
Thank you!
Symmetry of second partials holds for distributions: for any distribution $f$ we have $\partial_i \partial_j f = \partial_j \partial_i f$. (This is immediate from the corresponding fact for smooth functions, and the definition of the derivative of a distribution.) As such, there is no distribution $f$ on the unit disk with $\nabla f = (-y, x)$, because it would have $\partial_x \partial_y f = 1$ and $\partial_y \partial_x f = -1$.