Helmholtz (-Hodge) decomposition commonly used in physics includes decomposition of a (sufficiently smooth) vector field $F = -\mathrm{grad}(U) + \mathrm{curl}(W)$ on bounded simply connected domain $\Omega \subseteq \mathbb{R}^3$ (with smooth boundary), with scalar $U$ and vector field $W$ which are explicitly given by integrals on Wiki page <https://en.wikipedia.org/wiki/Helmholtz_decomposition>


The proof presented on the same Wiki page (an early instance of which appeared in Aris: "Vectors, Tensors, and the Basic Equations of Fluid Mechanics"; see also Griffiths: "Introduction to Electrodynamics", Appendix C) silently assumes, among other things, that vector field $F$ is test (smooth and compactly supported), whereas one would like to have decomposition for vector fields which are merely $C^2(\Omega) \cap C^0(\overline{\Omega})$.


Is there some standard reference where the decomposition, with explicit expressions for $U$ and $W$, is proven in this modest setting?



Proof of Helmholtz-Hodge decomposition, poor man's version
For avoiding $\delta$, you should be able to just do as follows: 


Let $G(x)$ denote the Newton potential $\frac{1}{4\pi |x|}$, and $G\_y(x) = G(x-y)$


Let $r'(x)$, for $x \in \Omega$, denote $\frac12 d(x, \Omega^c)$. 


Given $y\in \Omega$, consider the integral for $\lambda\in (0,1)$


$$ \tilde{F}(\lambda,y) = - \int\_{\partial B(y,\lambda r'(y))} \partial\_\nu G\_y(z) F(z) ~\mathrm{d}\sigma(z) $$
Here $\partial\_\nu$ is the derivative in the direction of the outward unit normal, and $\sigma$ is the induced surface measure. 


As $F$ is $C^2$, by Taylor's theorem we know that $F(z) = F(y) + O(|z - y|)$. Therefore we have
$$ \lim\_{\lambda \to 0} \tilde{F}(\lambda,y) = F(y)$$
uniformly. 


Applying divergence theorem to the integral defining $\tilde{F}$ you get


$$ \tilde{F}(\lambda,y) = \int\_{\Omega \setminus B(y,\lambda r'(y))} (\nabla G\_y(z) \cdot \nabla) F(z) ~\mathrm{d}z - \int\_{\partial\Omega} \partial\_n G\_y(z) F(z) ~\mathrm{d}\sigma(z). $$ 


The volume integral you can use the vector triple product formula to write


$$ (\nabla G\_y \cdot \nabla)F = (\nabla\cdot F) \nabla G\_y + \nabla \times (F \times \nabla G\_y ) $$


Since $\nabla G\_y$ blows up no faster than $1 / |z-y|^2$, which is integrable in $\mathbb{R}^3$, we can take the limit $\lambda \to 0$ of the integral. This gives 


$$ F(y) = \int\_{\Omega} (\nabla \cdot F) \nabla G\_y + \nabla \times (F \times \nabla G\_y) ~\mathrm{d}z - \int\_{\partial\Omega} \partial\_n G\_y F ~\mathrm{d}\sigma $$


after which you can continue following exactly the derivation on Wikipedia. 


