Definition:Bounded Mapping

Definition
Let $\left({T, \preceq}\right)$ be a poset.

Let $f: S \to T$ be a mapping.

Let the codomain of $f$ be bounded.

Then $f$ is defined as being bounded.

That is, $f$ is bounded if it is both bounded above and bounded below.

Real-valued Function
A real-valued function $f: S \to \R$ is bounded if there is a number $K \ge 0$ such that $\left|{f \left({x}\right)}\right| \le K$ for all $x \in S$.

See Bounded Set of Real Numbers‎ for a demonstration that this definition is compatible with boundedness on an ordered set.

Function Attaining its Bounds
If a real-valued function $f: S \to \R$ is bounded, then $f \left({S}\right)$ is by definition a bounded subset of $\R$, and hence has a supremum and infimum.

These are the bounds on $f \left({S}\right)$, which may or may not be in $f \left({S}\right)$.

If $\inf \left({f \left({S}\right)}\right) \in f \left({S}\right)$ and $\sup \left({f \left({S}\right)}\right) \in f \left({S}\right)$, then $f$ attains its bounds on $S$.

Complex-Valued Function
A complex-valued function $f: S \to \C$ is called bounded if the real-valued function $\left|{f}\right|: S \to \R$ is bounded, where $\left|{f}\right|$ is the modulus of $f$.

That is, $f$ is bounded if there is a constant $K \ge 0$ such that $\left|{f \left({z}\right)}\right| \le K$ for all $z \in S$.

This coincides with the definition of a bounded mapping into a metric space, using the standard metric on $\C$.

See Complex Plane is Metric Space.