Double-commutator $[f,[f, - \Delta]] = -2 |\nabla f|^2.$

Question

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.

Answer 1

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}