Let $P,Q:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be two $1/2$-averaged operators, which means that both $P$ and $Q$ are firmly nonexpansive. Also, let $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a general nonexpansive operator. I have two questions:
What can we say about the composition $C= P \circ T \circ Q$ in general? I believe it is at least nonexpansive, but can we guarantee that $C$ is $\alpha$-averaged for some $\alpha \in (0,1)$?
If we give $T$ a little more structure, say $T=2H-\text{Id}$, where $H$ is a $1/2$-averaged operator, can we affirm that $C$ is averaged somehow?
No. For counterexample, let $P=Q=\mathrm{Id}$ and $T$ be nonexpansive but not averaged. Then $C=T$, which is nonexpansive but not averaged.
No. For counterexample, let $P=Q=\mathrm{Id}$ and $H = 1/2 A + 1/2 \mathrm{Id}$, where $A$ is nonexpansive but not averaged. Then, $T = C = A$, which is nonexpansive but not averaged.