Wikipedia page on Helmholtz decomposition,
https://en.wikipedia.org/wiki/Helmholtz_decomposition
presents an elegant proof which makes use of the fundamental solution/Green function of the Laplace equation. However, in present form it suffers from a serious formal limitation that the vector field $\mathbf{F}$ must have a compact support (in order to be acted upon by the delta function). Since the statement of the theorem usually assumes milder fall-off conditions on the vector field (and in this is form it is most usually used in physics), I was wondering how to fix this type of proof.
Attempt: Suppose that $\Omega = \mathbb{R}^3$. Given a (smooth enough) vector field $\mathbf{F}$ which does not have a compact support, and a point $\mathbf{r} \in \Omega$, one idea is to introduce an auxiliary test function $\varphi(\mathbf{r}')$ which is equal to $1$ on some disc centered at $\mathbf{r}$, and start the proof with $$\mathbf{F}(\mathbf{r}) = \int_\Omega \varphi(\mathbf{r}') \mathbf{F}(\mathbf{r}') \delta(\mathbf{r} - \mathbf{r}') \, \mathrm{d} V' = \dots$$ In the final step of the proof one then has to prove that for any $\epsilon > 0$ there is a test function $\varphi(\mathbf{r}')$ such that $$\left| \int_\Omega \frac{\nabla' \cdot \big( \varphi(\mathbf{r}') \mathbf{F}(\mathbf{r}') - \mathbf{F}(\mathbf{r}') \big)}{|\mathbf{r}'' - \mathbf{r}'|} \, \mathrm{d} V' \right| < \epsilon$$ and $$\left| \int_\Omega \frac{\nabla' \times \big( \varphi(\mathbf{r}') \mathbf{F}(\mathbf{r}') - \mathbf{F}(\mathbf{r}') \big)}{|\mathbf{r}'' - \mathbf{r}'|} \, \mathrm{d} V' \right| < \epsilon$$ hold for all $\mathbf{r}''$ on some open neighbourhood of $\mathbf{r}$. This should be true if $\mathbf{F}$ has fast enough fall-off at infinity.
Does this make any sense? Is there any better/elegant way to polish this proof?