Lindquist Identity:$\big(|b|^{p-2}b-|a|^{p-2}a\big)(b-a)=\frac{|b|^{p-2}+|a|^{p-2}}{2}|b-a|^2+\frac{1}{2}\big(|b|^{p-2}-|a|^{p-2}\big)(|b|^2-|a|^2)$

Question

When investigating the properties of the one-dimensional "French tower" function $$ g: \mathbb{R} \to \mathbb{R}, \ z \mapsto \begin{cases} | z |^{p - 2} z, & \text{if } z \ne 0, \\ 0, & \text{else.} \end{cases} $$ for $p \in (1, \infty)$ these lecture notes (in German, Beispiel 13.5) mention a certain Lindquist identity, stating $$ \big( | b |^{p - 2} b - | a |^{p - 2} a\big) (b - a) = \frac{| b |^{p - 2} + | a |^{p - 2}}{2} | b - a |^2 + \frac{1}{2}\big(| b |^{p - 2} - | a |^{p - 2}\big)(| b |^2 - | a |^2). $$ Where does this come from and is this related to the Swedish mathematician Anders Lindquist?

Answer 1

The RHS becomes $$ | b |^{p} - | b |^{p - 2} a \cdot b - | a |^{p - 2} a \cdot b + | a |^{p}. $$ The LHS is \begin{align} & \frac{1}{2} | b |^{p} - | b |^{p - 2} a \cdot b + \frac{1}{2} | b |^{p - 2} | a |^2 + \frac{1}{2} | a |^{p - 2} | b |^{2} - | a |^{p - 2} a \cdot b + \frac{1}{2} | a |^p \\ + & \frac{1}{2} | b |^{p} - \frac{1}{2} | b |^{p - 2} | a |^2 - \frac{1}{2} | a |^{p - 2} | b |^2 + \frac{1}{2} | a |^p. \end{align} Now group similar terms to see the equality.