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

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.
