Combination Theorem for Bounded Real-Valued Functions/Negation Rule
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Theorem
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f :S \to \R$ be bounded real-valued function.
Let $-f : S \to \R$ denote the pointwise negation of $f$, that is, $-f$ denotes the mapping defined by:
- $\forall s \in S : \map {\paren{-f} } s = - \map f s$
Then:
- $-f$ 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:
\(\ds \forall s \in S: \, \) | \(\ds \size{\map {\paren{-f} } s}\) | \(=\) | \(\ds \size{-\map f s}\) | Definition of Absolute Value of Real-Valued Function | ||||||||||
\(\ds \) | \(=\) | \(\ds \size{\map f s}\) | Absolute Value of Negative | |||||||||||
\(\ds \) | \(\le\) | \(\ds M\) | Definition of Bounded Real-Valued Function |
It follows that $-f$ is a bounded real-valued function by definition.
$\blacksquare$