Decomposition of a vectorial field in free-curl and free-divergence fields Is it always possible to do that decomposition? I'm asking it because Helmholtz theorem says a field on $\mathbb{R}^3$ that vanishes at infinity ($r\to \infty$) can be decomposed univocally into a gradient and a curl. 
But I also know, for example, that a constant field $\mathbf{E}$ on $\mathbb{R}^3$ is a gradient (not univocally definied): $\mathbf{E}(x+y+z+\mbox{constant})$. And the electric field is $-\nabla G+ d\mathbf{A}/dt$, where $\mathbf{A}$ can be (Coulomb Gauge) free-divergence.
So, is it always possible to do the decomposition of a (regular, of course) field on $\mathbb{R}^3$ into two fields, free-curl and free-divergence? 
And on a limited domain?
 A: The splitting of a vector field 
$$\tag{1}\vec{V}~=~\vec{V}_{\parallel}+\vec{V}_{\perp}$$ 
into a curl-free part, 
$$\tag{2}\vec{\nabla}\times\vec{V}_{\parallel}~=~\vec{0},$$ 
and a divergence-free part, 
$$\tag{3}\vec{\nabla}\cdot\vec{V}_{\perp}~=~0,$$ 
is given as
$$\begin{align}\tag{4}\vec{V}_{\parallel} &:=\vec{\nabla}(\Delta^{-1}(\vec{\nabla}\cdot\vec{V})),\\ \tag{5}\vec{V}_{\perp} &:=\vec{V}-\vec{V}_{\parallel}.\end{align}$$
Here $\Delta:=\vec{\nabla}\cdot\vec{\nabla}$ is the Laplacian, and $\Delta^{-1}$ is a right inverse $\Delta\circ\Delta^{-1} = {\rm id}$. 
The operators $\Delta$ and $\Delta^{-1}$ take scalars $f:\mathbb{R}^3\to\mathbb{R}$ into scalars. Potential problems are related to zero-modes and whether the inverse $\Delta^{-1}$ is well-defined. For sufficiently well-behaved scalars $f$ (and implicit choice of boundary conditions), the following integral formula applies
$$\tag{6}(\Delta^{-1}f)(\vec{x}) 
~=~ - \iiint_{\vec{y}\neq\vec{x}}\frac{d^3y}{4\pi}
\frac{f(\vec{y})}{|\vec{y}-\vec{x}|},$$
cf.  Poisson equation.
A: As long as the field can be Fourier transformed,
$$\tilde{\mathbf F}(\mathbf k) = \frac1{(2\pi)^{3/2}} \iiint e^{-i\mathbf k\cdot\mathbf r} \mathbf F(\mathbf r) d^3\mathbf r, $$
we can separate $\tilde{\mathbf F}$ into the longitudinal and traverse parts
$$ \tilde{\mathbf F}(\mathbf k) = \tilde{\mathbf F}_\parallel(\mathbf k) + \tilde{\mathbf F}_\perp(\mathbf k)$$
where
$$ \tilde{\mathbf F}_\parallel(\mathbf k) = \hat{\mathbf k} (\hat{\mathbf k}\cdot\tilde{\mathbf F}(\mathbf k)).$$
This makes
$$ \mathbf k\cdot\tilde{\mathbf F}_\perp(\mathbf k) = 0, \quad \mathbf k\times\tilde{\mathbf F}_\parallel(\mathbf k) = 0,$$ which is equivalent to ($\mathbf k \mapsto -i\nabla$)
$$ \nabla\cdot{\mathbf F}_\perp(\mathbf r) = 0, \quad \nabla\times{\mathbf F}_\parallel(\mathbf r) = 0,$$
when performing the inverse Fourier transform (again requiring it works). Thus shows $\mathbf F$ is split into a divergence-free part $\mathbf F_\perp$ and curl-free part $\mathbf F_\parallel$.
