Combination Theorem for Bounded Real-Valued Functions/Product 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 $f g : S \to \R$ be the pointwise product of $f$ and $g$, that is, $f g$ is the mappping defined by:

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

Then:

$f g$ 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

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

Let $M = M_f M_g$.

We have:

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

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

\(=\)

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

Definition of Pointwise Product

\(\ds \)

\(=\)

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

Absolute Value Function is Completely Multiplicative

\(\ds \)

\(\le\)

\(\ds M_f M_g\)

Definition of Bounded Real-Valued Function

\(\ds \)

\(=\)

\(\ds M\)

Definition of $M$

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

$\blacksquare$