# Equivalence of Definitions of Bounded Real-Valued Function

## Theorem

The following definitions of the concept of Bounded Real-Valued Function are equivalent:

### Definition 1

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

$f$ is bounded above on $S$

and also:

$f$ is bounded below on $S$.

### Definition 2

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

$\exists K \in \R_{\ge 0}: \forall x \in S: \size {\map f x} \le K$

where $\size {\map f x}$ denotes the absolute value of $\map f x$.

### Definition 3

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

$\exists a, b \in \R_{\ge 0}: \forall x \in S: \map f x \in \closedint a b$

where $\closedint a b$ denotes the (closed) real interval from $a$ to $b$.

## Proof

Let $S$ be a set.

Let $f: S \to \R$ be a real-valued function.

### Definition 1 implies Definition 2

Let $f$ be bounded according to definition 1:

bounded above in $S$ by $H \in \R$
bounded below in $S$ by $L \in \R$

Thus by definition:

$\forall x \in S: L \le \map f x$
$\forall x \in S: \map f x \le H$
$\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$

So $f$ is bounded according to definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $f$ be bounded according to definition 2:

$\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$
$\exists K \in \R_{\le 0}: \forall x \in S: -K \le \map f x$

and so $f$ is bounded below in $S$ by $-K \in \R$

and

$\exists K \in \R_{\le 0}: \forall x \in S: \map f x \le K$

and so $f$ is bounded above in $S$ by $K \in \R$.

So $f$ is bounded according to definition 2.

$\Box$

### Definition 1 implies Definition 3

Let $f$ be bounded according to definition 1:

bounded above in $S$ by $H \in \R$
bounded below in $S$ by $L \in \R$

Thus by definition:

$\forall x \in S: L \le \map f x$
$\forall x \in S: \map f x \le H$

That is:

$\map f x \in \closedint L H$

Thus $f$ is bounded according to definition 3.

$\Box$

### Definition 3 implies Definition 1

Let $f$ be bounded according to definition 3:

$\exists a, b \in \R_{\le 0}: \forall x \in S: \map f x \in \closedint a b$

Thus by definition:

$\forall x \in S: a \le \map f x$

and so $f$ is bounded below in $S$ by $a \in \R$

$\forall x \in S: \map f x \le b$

and so $f$ is bounded above in $S$ by $b \in \R$.

So $f$ is bounded according to definition 1.

$\blacksquare$