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 $\BB$.
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Continuous on Closed Interval Real-Valued Function
Let $I = \closedint a b$ be a closed interval.
Let $\struct {\map C I, +, \, \cdot \,}_\R$ be the vector space of real-valued functions, continuous on $I$.
Let $\map x t \in \map C I$ 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 $\struct {\map C I, +, \, \cdot \,}_\R$.
Space of Bounded Sequences
The supremum norm on the vector 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.
$\blacksquare$