How can I prove the following? $$\vec{\nabla}\cdot(\vec{\nabla}^2\vec{F}) = \vec{\nabla}^2(\vec{\nabla}\cdot\vec{F})$$ $$\vec{F}:\Bbb{R^3}\mapsto\Bbb{R^3}$$
I am confused because on the left part I use $\vec{\nabla}^2$ on $\vec{F}$ but on the right part I first do the inner product of $\vec{\nabla}\cdot\vec{F}$ which should result in a number and then $\vec{\nabla}^2$ of a number.
On
$$\nabla\cdot(\nabla^2\vec{F}) = \nabla^2( \nabla \cdot\vec{F})$$ I'll use Cartesian coordinates.
LHS: $$\nabla\cdot(\nabla^2\vec{F}) = \nabla \cdot (\nabla^2 F_x, \nabla^2 F_y, \nabla^2 F_z) \\ =\partial_x \nabla^2 F_x + \partial_y \nabla^2 F_y + \partial_z \nabla^2 F_z$$
RHS: $$\nabla^2( \nabla \cdot\vec{F}) = \nabla^2( \partial_x F_x + \partial_y F_y + \partial_z F_z)\\ =\nabla^2 \partial_x F_x + \nabla^2 \partial_y F_y + \nabla^2 \partial_z F_z $$ Equal by switching order of partial derivatives.
On
$\def\p{\partial}\def\n{\nabla}\def\c{\cdot}$Assuming $\n^2\equiv\n\c\n,\,$ the formula is simply a statement about the commutativity and associativity of partial derivatives $$\eqalign{ \n\c(\n\c\n F) &= (\n\c\n)\,(\n\c F) \\ \p_k(\p_j\p_j F_k) &= (\p_j\p_j)\,(\p_k F_k) \\ \p_j\p_k &\equiv \p_k\p_j \\ }$$ where a repeated index implies summation (aka the Einstein convention) and $\,\p_k\phi = \large\frac{\p\phi}{\p x_k}$
Let $\textbf{F} = (F_1,F_2,F_3)^T$. Then the RHS is : $$ \nabla^2\ (\nabla \cdot F) = \nabla^2 \left( \partial_x F_1 + \partial_y F_2 + \partial_z F_3 \right) = (\partial_{xxx} F_1 + \partial_{yxx} F_2 + \partial_{zxx} F_3) + (\partial_{xyy} F_1 + \partial_{yyy} F_2 + \partial_{zyy} F_3 ) + (\partial_{xzz} F_1 + \partial_{yzz} F_2 + \partial_{zzz} F_3 ) $$ where $\partial_{abc}$ means first derivate w.r.t $a$ then w.r.t $b$ then w.r.t $c$. I also put parenthesis to make things a bit clearer but they are of course not needed ! I'll let you do the LHS since you didn't mention problem with it :)