Supremum Metric is Metric
Theorem
Let $S$ be a set.
Let $M = \struct {A', d'}$ be a metric space.
Let $A$ be the set of all bounded mappings $f: S \to M$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
Proof
We have that the supremum metric on $A \times A$ is defined as:
- $\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}$
where $f$ and $g$ are bounded mappings.
First note that we have:
| \(\ds \size {\map f x - \map g x}\) | \(=\) | \(\ds \size {\map f x + \paren {-\map g x} }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \size {\map f x} + \size {\paren {-\map g x} }\) | Triangle Inequality for Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size {\map f x} + \size {\map g x}\) | Definition of Absolute Value | |||||||||||
| \(\ds \) | \(\le\) | \(\ds K + L\) |
and so the right hand side exists.
Proof of Metric Space Axiom $(\text M 1)$
| \(\ds \map d {f, f}\) | \(=\) | \(\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map f x}\) | Definition of $d$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x \mathop \in S} \size 0\) | as $d'$ fulfils Metric Space Axiom $(\text M 1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
So Metric Space Axiom $(\text M 1)$ holds for $d$.
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
Let $f, g, h \in A$.
Let $x, y \in S$.
| \(\ds c\) | \(\in\) | \(\ds S\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {d'} {\map f c, \map h c}\) | \(\le\) | \(\ds \map {d'} {\map f c, \map g c} + \map {d'} {\map g c, \map h c}\) | as $d'$ fulfils Metric Space Axiom $(\text M 2)$: Triangle Inequality | ||||||||||
| \(\ds \) | \(\le\) | \(\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x} + \sup_{x \mathop \in S} \map {d'} {\map g x, \map h x}\) | Definition of Supremum of Real-Valued Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {f, g} + \map d {g, h}\) | Definition of $d$ |
Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
- $X := \set {\map {d'} {\map f c, \map g c}: c \in X}$
So:
- $\map d {f, g} + \map d {g, h} \ge \sup X = \map d {f, h}$
So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d$.
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
| \(\ds \map d {f, g}\) | \(=\) | \(\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}\) | Definition of $d$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x \mathop \in S} \map {d'} {\map g x, \map f x}\) | as $d'$ fulfils Metric Space Axiom $(\text M 3)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {g, f}\) | Definition of $d$ |
So Metric Space Axiom $(\text M 3)$ holds for $d$.
$\Box$
Proof of Metric Space Axiom $(\text M 4)$
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
- $\forall f, g \in A: \map d {f, g} \ge 0$
Suppose $f, g \in A: \map d {f, g} = 0$.
Then:
| \(\ds \map d {f, g}\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}\) | \(=\) | \(\ds 0\) | Definition of $d$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall x \in S: \, \) | \(\ds \map {d'} {\map f x, \map g x}\) | \(=\) | \(\ds 0\) | as $d'$ fulfils Metric Space Axiom $(\text M 1)$ and Metric Space Axiom $(\text M 4)$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds f\) | \(=\) | \(\ds g\) | as $d'$ fulfils Metric Space Axiom $(\text M 4)$ |
So Metric Space Axiom $(\text M 4)$ holds for $d$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$