Combination Theorem for Bounded Real-Valued Functions/Absolute Value Rule

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Theorem

Let $S$ be a set.

Let $\R$ denote the real number line.

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

Let $\size f : S \to \R$ denote the absolute value of $f$, that is, $\size f$ denotes the mapping defined by:

$\forall s \in S : \map {\size f} s = \size{\map f s}$

Then:

$\size f$ is a bounded real-valued function

Proof

By definition of bounded real-valued function

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

We have:

\(\ds \forall s \in S: \, \)

\(\ds \bigsize{\map {\size f} s}\)

\(=\)

\(\ds \bigsize{\size{\map f s} }\)

Definition of $\size f$

\(\ds \)

\(=\)

\(\ds \size{\map f s}\)

Definition of Absolute Value

\(\ds \)

\(\le\)

\(\ds M\)

Definition of Bounded Real-Valued Function

It follows that $\size f$ is a bounded real-valued function by definition.

$\blacksquare$