 Solution

A firmly nonexpansive mapping satisfiesWriting $d=z-w$ and $a=Tz-Tw$, this is equivalent toIt is also equivalent to $\Vert 2a-d{\Vert }_2\le \Vert d{\Vert }_2$, so $R=2T-I$ is a nonexpansive mapping. Consequently firm nonexpansiveness is exactly $1/2$-averagedness. A proximal map of a proper lower semicontinuous convex function is a basic example. A $1/2$-averaged operator is also $\alpha$-averaged for every $\alpha \in [1/2,1)$, by combining its nonexpansive part with the identity; the equivalence with firm nonexpansiveness specifically uses the parameter $1/2$.