A map $T$ is a <firmly nonexpansive mapping> when
$$
\|Tx-Ty\|^2\leq(Tx-Ty)^T(x-y)
$$
for all $x,y$. Put $p=\operatorname{prox}_f(x)$ and $q=\operatorname{prox}_f(y)$. Then $x-p\in\partial f(p)$ and $y-q\in\partial f(q)$. Monotonicity of the <subdifferential> gives
$$
[(x-p)-(y-q)]^T(p-q)\geq0,
$$
which rearranges to
$$
\boxed{\|p-q\|^2\leq(p-q)^T(x-y).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 339 / 3 / a / 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 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> $M$.

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

{wiki=Nonexpansive_mapping}

A mapping $T$ on an <inner-product space> is firmly nonexpansive when
$$
\|T(v)-T(w)\|^2
\leq\langle T(v)-T(w),v-w\rangle.
$$
Every firmly nonexpansive mapping is <nonexpansive>, and the resolvent of a maximal <monotone operator> is firmly nonexpansive.

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