Supremum Norm is Norm

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.

$\blacksquare$