I am trying to prove the Helmholtz decomposition theorem which states that given a smooth vector field $\mathbf{F}$, there are a scalar field $\phi$ and a vector field $\mathbf{G}$ such that \begin{equation} \mathbf{F}=\nabla \phi +\nabla \times \mathbf{G} \end{equation} However, during the proof I questioned myself at a point where we write every such vector field in the Dirac integral way \begin{equation} \textbf{F}(\textbf{r})=\int_{V}\textbf{F}(\textbf{r}')\delta(\textbf{r}-\textbf{r}')d\textbf{r}' \end{equation} and then use the identity \begin{equation} \delta(\textbf{r}-\textbf{r}')=-\frac{1}{4\pi}\nabla{}^2\left( \frac{1}{|\textbf{r}-\textbf{r}'|}\right) \end{equation} in order to get: \begin{equation} \textbf{F}(\textbf{r})=\int_{V}\textbf{F}(\textbf{r}')\left[-\frac{1}{4\pi}\nabla{}^2\left( \frac{1}{|\textbf{r}-\textbf{r}'|}\right) \right]d\textbf{r}'=-\int_{V}\textbf{F}(\textbf{r}')\left[\frac{1}{4\pi}\nabla{}^2\left( \frac{1}{|\textbf{r}-\textbf{r}'|}\right) \right]d\textbf{r}' \end{equation} The question is the following: What are the conditions that allow the Laplace operator $\nabla{}^2$ to come out of the integral? It has to do something with uniform convergence but I am not sure.
Thanks!
Suppose $\psi=\psi(\vec{r}-\vec{r}')$, then $$ \frac{\partial\psi}{\partial x_i}=\frac{\partial\psi}{\partial (x_i-x_i')}=-\frac{\partial\psi}{\partial x'_i} $$ This means (let $\nabla'$ be nabla in $\vec{r}'$) $$ \nabla'^2\frac{1}{|\vec{r}-\vec{r}'|}=-\nabla'\cdot \nabla\frac{1}{|\vec{r}-\vec{r}'|}=\nabla^2 \frac{1}{|\vec{r}-\vec{r}'|} $$ Now in your integral (by the way in the integral you should have $\nabla'^2$ not $\nabla^2$) you can replace $\nabla'^2$ by $\nabla^2$. This allows you to bring $\nabla^2$ out of the integral since it does not involve any differentiation with respect to the integration variable $\vec{r}'$.