Combination Theorem for Bounded Real-Valued Functions/Multiple Rule

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

Theorem

Let $\struct{S, \tau_S}$ be a topological space.

Let $\R$ denote the real number line.

Let $f :S \to \R$ be bounded real-valued function.

Let $\lambda \in \R$.

Let $\lambda f : S \to \R$ be the pointwise scalar multiplication of $f$ by $\lambda$, that is, $\lambda f$ is the mappping defined by:

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

Then:

$\lambda f$ is a bounded real-valued function

Proof

By definition of bounded real-valued function

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

Let $M = \size{\lambda} M_f$.

We have:

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

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

\(=\)

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

Definition of Pointwise Scalar Multiplication of Real-Valued Function

\(\ds \)

\(=\)

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

Absolute Value Function is Completely Multiplicative

\(\ds \)

\(\le\)

\(\ds \size{\lambda} M_f\)

Definition of Bounded Real-Valued Function

\(\ds \)

\(=\)

\(\ds M\)

definition of $M$

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

$\blacksquare$