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.

$\blacksquare$