Supremum Metric on Bounded Real-Valued Functions is Metric

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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 $d$.

$\blacksquare$


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