Let $H$ be a real Hilbert space with inner product $\langle\cdot, \cdot \rangle: H \times H \rightarrow \mathbb{R}$, and induced norm $\left\| \cdot \right\|: H \rightarrow \mathbb{R}_{\geq 0}$.
Let $X \subset H$ be a closed, bounded, convex set and $P_X: H \rightarrow X$ the metric projection onto $X$, i.e., $P_X(x) := \arg \min_{y \in X} \left\| x-y\right\|$.
Let $f: H \rightarrow H$ be a pseudocontraction, i.e.,
$$ \left\| f(x) – f(y) \right\|^2 \leq \left\| x-y\right\|^2 + \left\| f(x) – f(y) – (x-y) \right\|^2 $$
for all $x,y \in H$.
I am looking for additional assumptions on $f$ and $X$ such that the mapping $P_X(f(\cdot))$ is a pseudocontraction as well.
Observations and comments:
$P_X$ is firmly nonexpansive, i.e.,
$$ \left\| P_X(x) – P_X(y) \right\|^2 \leq \langle x-y, P_X(x) – P_X(y) \rangle $$
$f$ is a pseudocontraction if and only if $x \mapsto x – f(x)$ is accretive. A map $g: H \rightarrow H$ is accretive if $ \langle g(x) – g(y), z \rangle \geq 0$ for all $x,y \in H$, $z \in J(x-y)$, where $J$ is the normalized duality mapping, i.e., $J(x) = \{ z \in X^* \mid \left\| z \right\|^2 = \left\| x \right\|^2 = \langle x, z \rangle \}$.
$f$ is a pseudocontraction if and only if
$$ \langle f(x) – f(y), x-y \rangle \leq \left\| x-y\right\|^2 $$
for all $x,y \in H$.
$f$ is a pseudocontraction if and only if $I – f$ is monotone; $g: H \rightarrow H$ is monotone if $\langle g(x) – g(y), x-y\rangle \geq 0$ for all $x,y \in H$.
The composition of a firmly nonexpansive mapping with a pseudocontraction is not always a pseudocontraction.