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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!

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This is definitely false in the smooth category, e.g. $V(x,y) = (-y,x)$ is not the gradient of any $C^2$ function on the unit disk. I'd expect that it isn't the gradient of any $W^{1,2}$ function either. –  Nate Eldredge Nov 1 '12 at 16:18
    
This is exactly the counterexample I had in mind in the smooth case. I could disprove it in the $W^{1,2}$-case however... –  Sh4pe Nov 1 '12 at 16:21
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In fact, since $V$ is a smooth function with all derivatives bounded, if $\nabla f = V$ weakly, the Sobolev embedding theorem should guarantee that $f$ is smooth, and sophomore calculus rules this out. There is probably a direct way to see it too. –  Nate Eldredge Nov 1 '12 at 16:27
    
if $u_i=\partial_i f$ for any distribution $f$ then certainly $\partial_j u_i=\partial_i u_j$ (equality of distributions). –  user8268 Nov 1 '12 at 20:11
    
@NateEldredge Yes, thank you. Do you want to write a quick answer to this question? Then I'll upvote it and accept it as the correct answer... –  Sh4pe Nov 2 '12 at 10:03

1 Answer 1

up vote 2 down vote accepted

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 $\partial_y \partial_x f = -1$.

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