Nonexpansive map

Remarks

Equivalently, $f$ is Lipschitz with Lipschitz constant at most $1$. In operator theory, $X = Y = C$ is often a subset of a Banach space with the norm distance, and the condition reads $\norm{f(x) - f(x')} \le \norm{x - x'}$.

Example: let $H$ be a real Hilbert space, $C \subseteq H$ a nonempty closed convex set, and $P_C(x)$ the nearest point of $C$ to $x$. The variational inequality $\ip{x - P_C x}{c - P_C x} \le 0$ for $c \in C$ gives $\norm{P_C x - P_C y}^2 \le \ip{x - y}{P_C x - P_C y} \le \norm{x - y}\norm{P_C x - P_C y}$, so $P_C$ is nonexpansive.

Non-example: $f(x) = 2x$ on $\R$ gives $\abs{f(1) - f(0)} = 2 > 1 = \abs{1 - 0}$.

Every contraction mapping is nonexpansive, but not conversely. The Browder–Göhde–Kirk theorem gives fixed points: a nonexpansive map of a nonempty bounded closed convex subset of a uniformly convex Banach space into itself has a fixed point.