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.
$\blacksquare$