Equivalence of Definitions of Bounded Real-Valued Function

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:
 * $\forall x \in S: L \le f \left({x}\right)$
 * $\forall x \in S: f \left({x}\right) \le H$

By Equivalence of Definitions of Bounded Subset of Real Numbers:
 * $\exists K \in \R_{\le 0}: \forall x \in S: \left|{f \left({x}\right)}\right| \le K$

So $f$ is bounded according to definition 2.

Definition 2 implies Definition 1
Let $f$ be bounded according to definition 2:
 * $\exists K \in \R_{\le 0}: \forall x \in S: \left|{f \left({x}\right)}\right| \le K$

Then by Equivalence of Definitions of Bounded Subset of Real Numbers:
 * $\exists K \in \R_{\le 0}: \forall x \in S: -K \le f \left({x}\right)$

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

and
 * $\exists K \in \R_{\le 0}: \forall x \in S: f \left({x}\right) \le K$

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