Double-commutator $[f,[f, - \Delta]] = -2 |\nabla f|^2.$ This book (proof of Theorem 3.2) in chapter 3.1 claims click me
that an easy computation shows that $$[f,[f, - \Delta]] = -2 |\nabla f|^2.$$
where $[.,.]$ denotes the commutator. Unfortunately, I really don't see how one gets this easily out of it. Besides, the equation below this one is fairly direct to me.
Does anybody have an idea.
 A: Note that the commutator acts in a way similar to the derivative, therefore, e.g.
\begin{align}[f,[\partial_x^2,f]]&=[f,\partial_x\circ[\partial_x,f]+[\partial_x,f]\circ\partial_x]=
[f,\partial_x\circ f_x+f_x\circ\partial_x]=\\
&=\underbrace{[f,\partial_x]}_{=-f_x}\circ f_x+\partial_x\circ\underbrace{[f,f_x]}_{=0}+\underbrace{[f,f_x]}_{=0}\circ \partial_x+f_x\underbrace{\circ[f,\partial_x]}_{=-f_x}=-2f_x^2.
\end{align}
A: It's a good idea to be careful and act a commutator on a dummy function to figure out which it does when you are encountering it for the first time. The commutator may just act like a function being multiplied, e.g. $| \nabla f|^2 $, but it also may act like an operator, e.g. $f \Delta $. You probably won't be able to tell which it is intuitively, but your bases are covered if you always act the operator on a dummy function. You can see this by acting some simple example commutators, $[\partial,f]$ and $[\partial^2,f]$, on dummy function $g$:
\begin{align*}
[\partial,f]g = & \, (\partial f - f \partial)g &[\partial^2,f]g = & \, (\partial^2 f - f \partial^2)g \\
=& \, \partial (fg) - f \partial g & =& \, \partial^2 (fg) - f \partial^2 g\\
=& \, f \partial g + g \, \partial f -f\partial g & =& \, f \partial^2 g + 2 (\partial f)(\partial g) +g \, \partial^2 f -f\partial^2 g\\
=& \, g \, \partial f & =& \, 2 (\partial f)(\partial g) +g \, \partial^2 f\\
\therefore  [\partial,f]=& \, \partial f &\therefore [\partial^2,f] = & \, \partial^2 f + 2 (\partial f)\partial  
\end{align*}
So $[\partial,f]$ can be written as a function being multiplied, but $[\partial^2,f]$ cannot. Now we can repeat this with the commutator the OP described.
\begin{align*}
[f,[f,-\Delta]]g
=& \, (f[f,-\Delta]-[f,-\Delta]f)g\\
=& \, (-f^2 \Delta+2f\Delta f -\Delta f^2)g\\
=& \, -f^2 \Delta g +2f\Delta(fg)-\Delta(f^2 g)\\
=& \, -f^2 \Delta g +2f\Delta(fg)-(f\Delta(fg)+2\nabla f\cdot\nabla (fg)+fg \, \Delta f)\\ 
=& \, -f^2 \Delta g +f\Delta(fg)-2\nabla f\cdot\nabla (fg)-fg \, \Delta f\\
=& \, -f^2 \Delta g +f(f\Delta g+2\nabla f\cdot\nabla g+g \, \Delta f)-2\nabla f\cdot(f\nabla g+ g \nabla f)-fg \, \Delta f\\
= & -2 \nabla f \cdot g \nabla f\\
\therefore [f,[f,-\Delta]] =& -2 |\nabla f|^2
\end{align*}
