# Supremum Metric on Bounded Real Functions on Closed Interval is Metric

## Theorem

Let $\closedint a b \subseteq \R$ be a closed real interval.

Let $A$ be the set of all bounded real functions $f: \closedint a b \to \R$.

Let $d: A \times A \to \R$ be the supremum metric on $A$.

Then $d$ is a metric.

## Proof 1

The interval is an instance of a set.

Hence Supremum Metric on Bounded Real-Valued Functions is Metric can be directly applied.

## Proof 2

We have that the supremum metric on $A \times A$ is defined as:

$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$

where $f$ and $g$ are bounded real functions.

So:

$\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$

for all $x \in \closedint a b$.

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 $\text M 1$

 $\ds \map d {f, f}$ $=$ $\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map f x}$ Definition of $d$ $\ds$ $=$ $\ds \sup_{x \mathop \in \closedint a b} \size 0$ $\ds$ $=$ $\ds 0$

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 $c \in \closedint a b$.

 $\ds c$ $\in$ $\ds \closedint a b$ $\ds \leadsto \ \$ $\ds \size {\map f c - \map h c}$ $\le$ $\ds \size {\map f c - \map g c} + \size {\map g c - \map h c}$ Triangle Inequality for Real Numbers $\ds$ $\le$ $\ds \sup_{x \mathop \in \closedint a b} \size {\map f c - \map g c} + \sup_{x \mathop \in \closedint a b} \size {\map g c - \map h c}$ 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:

$S := \set {\size {\map f c - \map h c}: c \in \closedint a b}$

So:

$\map d {f, g} + \map d {g, h} \ge \sup S = \map d {f, h}$

So Metric Space Axiom $\text M 2$ holds for $d$.

$\Box$

### Proof of Metric Space Axiom $\text M 3$

 $\ds \map d {f, g}$ $=$ $\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$ Definition of $d$ $\ds$ $=$ $\ds \sup_{x \mathop \in \closedint a b} \size {\map g x - \map f x}$ Definition of Absolute Value $\ds$ $=$ $\ds \map d {g, f}$ 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 \ \$ $\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$ $=$ $\ds 0$ Definition of $d$ $\ds \leadsto \ \$ $\, \ds \forall x \in \closedint a b: \,$ $\ds \map f x$ $=$ $\ds \map g x$ Definition of Absolute Value $\ds \leadsto \ \$ $\ds f$ $=$ $\ds g$ Equality of Mappings

So Metric Space Axiom $\text M 4$ holds for $d$.

$\blacksquare$