Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 325 / 2 / i / Solution

For $0<\alpha <1$, an averaged operator has the formfor some nonexpansive mapping $R$, meaning $\Vert Rz-Rw{\Vert }_2\le \Vert z-w{\Vert }_2$ 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 isTo see it, put $d=z-w$ and $r=Rz-Rw$ and expandUse $\Vert r{\Vert }_2\le \Vert d{\Vert }_2$ 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.