# Supremum Metric on Bounded Real-Valued Functions is Metric

## Theorem

Let $X$ be a set.

Let $A$ be the set of all bounded real-valued functions $f: X \to \R$.

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

Then $d$ is a metric.

## Proof 1

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

$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in X} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$

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

From Real Number Line is Metric Space, the real numbers $\R$ together with the absolute value function form a metric space.

The result follows by Supremum Metric is Metric.

$\blacksquare$

## Proof 2

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

$\displaystyle \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in X} \size {\map f x - \map g x}$

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

So:

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

for all $x \in X$.

First note that we have:

 $\displaystyle \size {\map f x - \map g x}$ $=$ $\displaystyle \size {\map f x + \paren {-\map g x} }$ $\displaystyle$ $\le$ $\displaystyle \size {\map f x} + \size {\paren {-\map g x} }$ Triangle Inequality for Real Numbers $\displaystyle$ $=$ $\displaystyle \size {\map f x} + \size {\map g x}$ Definition of Absolute Value $\displaystyle$ $\le$ $\displaystyle K + L$

and so the right hand side exists.

### Proof of $M1$

 $\displaystyle \map d {f, f}$ $=$ $\displaystyle \sup_{x \mathop \in X} \size {\map f x - \map f x}$ Definition of $d$ $\displaystyle$ $=$ $\displaystyle \sup_{x \mathop \in X} \size 0$ $\displaystyle$ $=$ $\displaystyle 0$

So axiom $M1$ holds for $d$.

$\Box$

### Proof of $M2$

Let $f, g, h \in A$.

Let $c \in X$.

 $\displaystyle c$ $\in$ $\displaystyle X$ $\displaystyle \leadsto \ \$ $\displaystyle \size {\map f c - \map h c}$ $\le$ $\displaystyle \size {\map f c - \map g c} + \size {\map g c - \map h c}$ Triangle Inequality for Real Numbers $\displaystyle$ $\le$ $\displaystyle \sup_{x \mathop \in X} \size {\map f c - \map g c} + \sup_{x \mathop \in X} \size {\map g c - \map h c}$ Definition of Supremum of Real-Valued Function $\displaystyle$ $=$ $\displaystyle \map d {f, g} + \map d {g, h}$ Definition of $d$

Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:

$S := \set {\size {\map f c - \map h c}: c \in X}$

So:

$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup S = d \left({f, h}\right)$

So axiom $M2$ holds for $d$.

$\Box$

### Proof of $M3$

 $\displaystyle \map d {f, g}$ $=$ $\displaystyle \sup_{x \mathop \in X} \size {\map f x - \map g x}$ Definition of $d$ $\displaystyle$ $=$ $\displaystyle \sup_{x \mathop \in X} \size {\map g x - \map f x}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle \map d {g, f}$ Definition of $d$

So axiom $M3$ holds for $d$.

$\Box$

### Proof of $M4$

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:

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

So axiom $M4$ holds for $d$.

$\blacksquare$