Combination Theorem for Bounded Real-Valued Functions/Maximum Rule
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Theorem
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded real-valued functions.
Let $f \vee g$ denote the pointwise maximum of $f$ and $g$, that is, $f \vee g$ is the mapping defined by:
- $\forall s \in S : \map {\paren{f \vee g} } s = \max \set{\map f s, \map g s}$
Then:
- $f \vee g$ is a bounded real-valued function
Proof
By definition of bounded real-valued function
- $\exists M_f \in \R_{\ge 0} : \forall s \in S : \size{\map f s} \le M_f$
and
- $\exists M_g \in \R_{\ge 0} : \forall s \in S : \size{\map g s} \le M_g$
From Negative of Absolute Value:
- $\forall s \in S : \map f s \le \size{\map f s}$
and
- $\forall s \in S : \map g s \le \size{\map g s}$
Let $M = \max \set{M_f, M_g}$.
We have:
\(\ds \forall s \in S: \, \) | \(\ds \bigsize{\map {\paren{f \vee g} } s}\) | \(=\) | \(\ds \bigsize{\max\set {\map f s, \map g s} }\) | Definition of Pointwise Maximum of Real-Valued Functions | ||||||||||
\(\ds \) | \(=\) | \(\ds \bigsize{\max\set {\size{\map f s}, \size{\map g s} } }\) | Max Operation Preserves Total Ordering | |||||||||||
\(\ds \) | \(\le\) | \(\ds \bigsize{\max\set {M_f, M_g} }\) | Max Operation Preserves Total Ordering | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigsize M\) | Definition of $M$ | |||||||||||
\(\ds \) | \(=\) | \(\ds M\) | as $M \ge 0$ |
It follows that $f \vee g$ is a bounded real-valued function by definition.
$\blacksquare$