A version of the Helmholtz theorem says that, under opportune assumptions on the vector field $\boldsymbol{F}:\mathbb{R}^3\to\mathbb{R}^3$ and on $V\subset\mathbb{R}^3$ the following identity holds: $$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi} \nabla\times\int_V \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi} \nabla\times\int_{\partial V} \frac{\hat{\boldsymbol{n}}(\boldsymbol{x'})\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\sigma'$$$$-\frac{1}{4\pi}\nabla \int_V \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' +\frac{1}{4\pi}\nabla\int_{\partial V} \frac{\boldsymbol{F}(\boldsymbol{x}')\cdot\hat{\boldsymbol{n}}(\boldsymbol{x'})}{\|\boldsymbol{x}-\boldsymbol{x}'\|} d\sigma' .$$
I think (hope) I have proved (here) to myself the validity of such a decomposition for $\boldsymbol{F}\in C^2(\mathbb{R}^3)$ compactly supported.
Are such assumptions consistent with the physical applications of the theorem?
I suspect that $\boldsymbol{F}$ usually is reasonably supposed to be null outside a bounded domain in application of the decomposition to physics, but I would be very grateful to anybody giving a knowledgeable answer on the issue.