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An operator is $\alpha$-averaged for $0<\alpha <1$ when $T=(1-\alpha )I+\alpha R$ for a nonexpansive mapping $R$. Averaging controls the update residual and enables fixed-point iteration. A firmly nonexpansive mapping is exactly a $1/2$-averaged operator.

The Browder convergence theorem for averaged operators states: if $T:{\mathbb{R}}^d\to {\mathbb{R}}^d$ is an averaged operator and its fixed-point set $\mathrm{Fix}T=\{z:Tz=z\}$ is nonempty, then for every starting point, $z^{k+1}=T{z}^k$ converges in norm to a fixed point of $T$. The theorem assumes a fixed point exists; averaging by itself does not create one.

The finite-dimensional mechanism is worth making explicit. For any fixed point $p$, the averaged-operator inequality givesThis is Fejér monotonicity with respect to the fixed-point set. It makes the sequence bounded and, by summation, makes ${\sum }_k\Vert z^{k+1}-z^k{\Vert }_2^2$ finite. Hence its residual tends to zero. Choose a convergent subsequence $z^{k_j}\to \bar{z}$; continuity of the nonexpansive mapping $T$ implies $T\bar{z}=\bar{z}$. Apply Fejér monotonicity with $p=\bar{z}$. The nonincreasing distances $\Vert z^k-\bar{z}{\Vert }_2$ have a subsequence tending to zero, so the whole sequence converges to $\bar{z}$. This last compactness step uses finite dimension; a corresponding general Hilbert space statement usually gives weak convergence.

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 $1/2$.

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.