= Solution
For $0<\alpha<1$, an <averaged operator> has the form
$$
\boxed{T=(1-\alpha)I+\alpha R}
$$
for some <nonexpansive mapping> $R$, meaning $\|Rz-Rw\|_2\leq\|z-w\|_2$ for all $z,w$. The word averaging refers to a <convex combination> with the identity. It does not mean averaging function values over a neighborhood.
An equivalent and useful <averaged-operator inequality> is
$$
\boxed{\|Tz-Tw\|_2^2\leq\|z-w\|_2^2-\frac{1-\alpha}{\alpha}\|(I-T)z-(I-T)w\|_2^2}.
$$
To see it, put $d=z-w$ and $r=Rz-Rw$ and expand
$$
\|(1-\alpha)d+\alpha r\|_2^2
=(1-\alpha)\|d\|_2^2+\alpha\|r\|_2^2-\alpha(1-\alpha)\|d-r\|_2^2.
$$
Use $\|r\|_2\leq\|d\|_2$ and $(I-T)z-(I-T)w=\alpha(d-r)$. Conversely, substituting $R=[T-(1-\alpha)I]/\alpha$ into the same expansion proves its <nonexpansiveness> from the displayed inequality. The strict range $0<\alpha<1$ matters for convergence of unrelaxed iterations: a general <nonexpansive mapping> such as $T=-I$ can have nonconvergent alternating iterates.
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