Supremum Metric is Metric

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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 $\text M 2$

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 $\text M 2$
\(\ds \) \(\le\) \(\ds \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x} + \sup_{x \mathop \in S} \map {d'} {\map g x, \map h x}\) 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 $\text M 2$ 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

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