Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 339 / 2 / b / Solution

If $T(w)=w$, the defining equation becomes $M(w)+F(w)=M(w)$, so $\boxed{F(w)=0}$.

A map is firmly nonexpansive in the $M$-inner product whenPut $p=T(v)$ and $q=T(w)$. The two implicit equations giveBecause $F$ is a monotone operator,Thereforewhich is precisely firm nonexpansiveness. In particular, the preconditioned proximal point algorithm map $T=(M+F{)}^{-1}M$ is nonexpansive in the norm induced by the positive-definite matrix $M$.