Can $\nabla \cdot (f \nabla g) = f \nabla^{2} g - g \nabla^{2} f$?

Question

Is this possible to show? I know that $\nabla \cdot (f \vec{\nabla} g) = f \nabla^{2} g + \vec{\nabla} f \cdot \vec{\nabla} g$ but can the other be proven? (Assume that $ f \vec{\nabla} g$ is a vector field denoted by $\textbf{F}$)

Answer 1

Concrete counterexample for title, $f(x) = |x|^2 = g(x), x\in\mathbb R^d$, then $$\mathrm{LHS} = \sum_{i=1}^d \partial_i 2 (x_i |x|^2 ) = \sum_{i=1}^d 2|x|^2 + 2 x_i (2x_i) = (2d+4)|x|^2 \neq 0 = \mathrm{RHS}$$

The example is not necessary to understand the issue - you can see that there is at most one derivative falling on $f$, so you have no chance of an equality without a method of transferring derivatives over from $g$. This method could be integration by parts (c.f. comment above)

Answer 2

If $~\mu(x,y,z)~$ is a scalar point function and $~\vec f(x,y,z)~$be a vector point function, then $$\vec\nabla \cdot(\mu \vec f)=(\vec\nabla \mu)\cdot \vec f + \mu(\vec\nabla\cdot \vec f)\tag1$$ Putting $~\mu=f~$ and $~\vec f=\vec\nabla g~$ in $(1)$, $$\vec\nabla \cdot(f~ \vec \nabla g)=(\vec\nabla f)\cdot (\vec\nabla g) + f(\vec\nabla\cdot \vec\nabla g)$$ $$=(\vec\nabla f)\cdot (\vec\nabla g) + f~\vec\nabla~^2 g$$ So $$\vec\nabla \cdot (f~ \vec\nabla g)\ne f ~\vec\nabla^{2} g - g~ \vec\nabla^{2} f$$ as $~(\vec\nabla f)\cdot (\vec\nabla g)\ne - g~ \vec\nabla^{2} f~$