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

A map is <firmly nonexpansive mapping>[firmly nonexpansive] in the $M$-inner product when
$$
\|T(v)-T(w)\|_M^2
\leq\langle T(v)-T(w),v-w\rangle_M.
$$
Put $p=T(v)$ and $q=T(w)$. The two implicit equations give
$$
F(p)=M(v-p),\qquad F(q)=M(w-q).
$$
Because $F$ is a <monotone operator>,
$$
0\leq\langle F(p)-F(q),p-q\rangle
=\langle(v-w)-(p-q),p-q\rangle_M.
$$
Therefore
$$
\boxed{
\|p-q\|_M^2
\leq\langle p-q,v-w\rangle_M,}
$$
which 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> $M$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-339/2/b/solution