Definition:Bounded Mapping

Definition

Let $S$ be a set.

Let $\struct {T, \preceq}$ be an ordered set.

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

Let the image of $f$ be bounded.

Then $f$ is bounded.

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

Real-Valued Function

$f$ is bounded on $S$ if and only if:

$f$ is bounded above on $S$

and also:

$f$ is bounded below on $S$.

Complex-Valued Function

Let $f: S \to \C$ be a complex-valued function.

Then $f$ is bounded if and only if the real-valued function $\cmod f: S \to \R$ is bounded, where $\cmod f$ is the modulus of $f$.

That is, $f$ is bounded if there is a constant $K \ge 0$ such that $\cmod {\map f z} \le K$ for all $z \in S$.

Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.

Let $f: S \to R$ be a mapping from $S$ into $R$.

Then $f$ is bounded if and only if the real-valued function $\norm {\,\cdot\,} \circ f: S \to \R$ is bounded, where $\norm {\,\cdot\,} \circ f$ is the composite of $\norm {\,\cdot\,}$ and $f$.

That is, $f$ is bounded if there is a constant $K \in \R_{\ge 0}$ such that $\norm{f \paren {s}} \le K$ for all $s \in S$.

Metric Space

Let $M$ be a metric space.

Let $f: X \to M$ be a mapping from any set $X$ into $M$.

Then $f$ is a bounded mapping if and only if $f \sqbrk X$ is bounded in $M$.

Unbounded Mapping

$f$ is unbounded if and only if it is neither bounded above or bounded below.

Also see

• Results about bounded mappings can be found here.