TITLE: Surjectivity of curl
QUESTION [14 upvotes]: Let: $\mathbb R^3\ni x\mapsto v(x)\in\mathbb R^3$ be a vector field with null divergence belonging to the Schwartz class such that
$$
\int_{\mathbb R^3} v(x) dx=0.
$$
Is it true that there exists a vector field $w$ in the Schwartz class such that $$\text{curl } w=v\quad?$$
In other words, this  is a regularity question for a Poincaré lemma: let $u$
be a closed two-form on $\mathbb R^3$. Then,
there exists a one-form $a$ such that $u=da$. If $u$ is smooth, $a$ can be chosen smooth; the above question can be reformulated: if $u$ belongs to the Schwartz class, is it possible to choose $a$ in the Schwartz class?

REPLY [3 votes]: I think that the answer is Yes.
1st step. Because Fourier transform is an automorphism of the Schwartz class, the problem is equivalent to show that every vector field $v(x)\in{\mathcal S}({\mathbb R}^3)^3$ verifying $x\cdot v(x)\equiv0$ can be ``divided'', that is can be written 
$$v(x)=x\wedge w(x), \qquad w\in{\mathcal S}({\mathbb R}^3)^3.$$
2nd step. The jets at the origin. Let 
$$v(x)\sim \sum_kv^k(x)$$
be the formal power series of $v$ at the origin, with $v_k$ a homogeneous polynomial vector field of degree $k$. Each $v_k$ satifies $x\cdot v_k(x)=0$.
It is an algebraic fact that there exists a polynomial vector field $w_k$, homogeneous of degree $k-1$, such that $v_k=x\wedge w_k$. Choose a  $\phi\in{\mathcal D}({\mathbb R}^3)$ be such that $\phi\equiv1$ in $B(0;1)$, and form 
$$w^0(x)=\sum_k\phi(kx)w_k(x).$$
This is a locally finite series, therefore convergent to a ${\cal C}^\infty$-function away from the origin, compactly supported. It is actually ${\cal C}^\infty$ everywhere, because $w^0$ differs from a ${\cal C}^k$ field by an $O(|x|^{k+1})$. Therefore $w^0$ is in the Schwarz class. Then the jet of $v^0:=x\wedge w^0$ equals that of $v$ at the origin. 
3rd step. Away from the origin. Define $v^1:=v-v^0$, which is of Schwartz class and is flat at the origin. Define $$w^1=\frac{x}{|x|^2}\wedge v^1$$
is of Schwarz class and satisfies $v^1=x\wedge w^1$.
Finally, $w=w^0+w^1$ is the solution of the problem.