Surjectivity of curl 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?
 A: 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.
A: Update : I deleted the previous answer, as it was irrelevant. Here is a second attempt, hopefully better than the first one. Morally speaking, it is often (if not always) possible to choose a Schwartz anticurl.
We will still work in Fourier space, but more symbolically this time. Let 
$$\hat{w}(k) = \frac{k \wedge \hat{v}(k)}{|k|^2} $$
be our first guess. We know that it has a singularity near $0$, but if this singularity may be carried by a gradient, that is, if we have a decomposition
$$\hat{w}(k) = \hat{w}_{smooth}(k) + k \hat{f}(k) $$
with some irrelevant $\hat{f}$ and a Schwartz $\hat{w}_{smooth}$, we are done.
Because $v$ is divergence free, a few computations using that $k \cdot \hat{v}(k) = 0$ lead to : 
$$\hat{w}(k) = \frac{1}{k_2} 
\left( \begin{array}{c}
\hat{v}_3 \\
0 \\
- \hat{v}_1 \\
\end{array} \right) + k \left( \frac{k_3}{k_1} \hat{v}_1 - \frac{k_1}{k_2}\hat{v}_3 \right).$$
And indeed, using again that $k \cdot \hat{v}(k) = 0$, you may check that $k \wedge$ (the left term) is, up to a sign we don't care about, $\hat{v}(k)$.
So, a sufficient condition for $\hat{w}_{smooth}$ to be... well, smooth is for both $\hat{v}_1$ and $\hat{v}_3$ to be divisible by $k_2$, which is almost given by the divergence free condition, but not exactly. In particular, such a condition implies immediately that $v$ has zero mean, but is much, much weaker than flatness.
Notive that we may reason symetrically with the other variables, leading to a divisibility condition with a "OR", not an "AND".
If you try to cook up a counterexample by negating one divisibility condition, then surely you will end up with something like 
$$\hat{v}(k) = k \wedge e_1 $$
(up to a fast decreasing factor to make it Schwartz) where $e_1$ is some fixed vector in $\mathbb{R}^3$, which obviously has a Schwartz anticurl.
If someone is able to prove or disprove that a Schwartz vector field satisfying $k \cdot \hat{v}(k)$ always satisfies the "OR divisibility" condition, this would be a definitive answer.
