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.
$\blacksquare$