So I would like to prove the following identity for a scalar $f$ and a vector $v$
$\left[\nabla^2\left(v\cdot \nabla\right)-\left(v\cdot \nabla\right)\nabla^2\right]f=0$
for the special case where
$\nabla\cdot v=0$ and $\nabla\times v=0$
and lets say we're in 3 dimensions.
I can re-write the identity I want to show as
$\nabla^2\left(A\cdot B\right)=\left( A\cdot \nabla\right)\left(\nabla\cdot B\right)$.
where
$B:=\nabla f$
and the same conditions hold now on $A$ as they did on $v$ and we now have te additional identity $\nabla\times B=0$.
It feels like this should be true, and indeed even seems trivial but I just cant get it to work.
I have tried many things but always end up with some cross terms involving first derivatives of $A$ in a product with first derivatives of $B$. These terms are of the form
$\left(\partial_x A_y\right)\left(\partial_y B_x\right)$ and $\left(\partial_x A_x\right)\left(\partial_y B_y\right)$ etc.
I'm starting to wonder whether it might not be true.
Am I going crazy?