= Solution
A <firmly nonexpansive mapping> satisfies
$$
\boxed{\|Tz-Tw\|_2^2\leq\langle Tz-Tw,z-w\rangle\quad\text{for all }z,w}.
$$
Writing $d=z-w$ and $a=Tz-Tw$, this is equivalent to
$$
\|a\|_2^2+\|d-a\|_2^2\leq\|d\|_2^2.
$$
It is also equivalent to $\|2a-d\|_2\leq\|d\|_2$, so $R=2T-I$ is a <nonexpansive mapping>. Consequently \b[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 $1/2$.
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