Supremum Norm is Norm

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Theorem

Let $S$ be a set.

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $K \in \set {\R, \C}$.

Let $\BB$ be the set of bounded mappings $S \to X$.

Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\BB$.


Then $\norm {\, \cdot \,}_\infty$ is a norm on $\BB$.



Continuous on Closed Interval Real-Valued Function

Let $I = \closedint a b$ be a closed interval.

Let $\struct {\map \CC I, +, \, \cdot \,}_\R$ be the vector space of real-valued functions, continuous on $I$.

Let $\map x t \in \map \CC I$ be a continuous real function.

Let $\size {\, \cdot \,}$ be the absolute value.

Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on real-valued functions, continuous on $I$.


Then $\norm {\, \cdot \,}_\infty$ is a norm over $\struct {\map \CC I, +, \, \cdot \,}_\R$.


Space of Bounded Sequences

The supremum norm on the space of bounded sequences is a norm.


Proof

First:

\(\ds \norm f_\infty\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sup_{x \mathop \in S} \norm {\map f x}\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\, \ds \forall x \in S: \, \) \(\ds \norm {\map f x}\) \(=\) \(\ds 0\) since $\norm {\, \cdot \,}$ is a norm, and hence non-negative
\(\ds \leadstoandfrom \ \ \) \(\, \ds \forall x \in S: \, \) \(\ds \map f x\) \(=\) \(\ds 0\) since $\norm x = 0 \iff x = 0$
\(\ds \leadstoandfrom \ \ \) \(\ds f\) \(=\) \(\ds 0\)


Now let $\lambda \in K, f \in \BB$

We have:

\(\ds \norm {\lambda f}_\infty\) \(=\) \(\ds \sup_{x \mathop \in S} \norm {\lambda \map f x}\)
\(\ds \) \(=\) \(\ds \size \lambda_K \sup_{x \mathop \in S} \norm {\map f x}\) Multiple of Supremum, and because $\norm {\, \cdot \,}$ is a norm
\(\ds \) \(=\) \(\ds \size \lambda_K \, \norm f_\infty\)


Finally let $f, g \in \BB$.

We have:

\(\ds \norm {f + g}_\infty\) \(=\) \(\ds \sup_{x \mathop \in S} \norm {\map f x + \map g x}\)
\(\ds \) \(\le\) \(\ds \sup_{x \mathop \in S} \paren {\norm {\lambda \map f x} + \norm {\lambda \map g x} }\) because $\norm {\, \cdot \,}$ is a norm
\(\ds \) \(\le\) \(\ds \sup_{x \mathop \in S} \norm {\lambda \map f x} + \sup_{x \mathop \in S} \norm {\lambda \map g x}\) Supremum of Sum
\(\ds \) \(=\) \(\ds \norm f_\infty + \norm g_\infty\)


Thus $\norm {\, \cdot \,}_\infty$ has the defining properties of a norm.

$\blacksquare$