# Supremum Metric on Continuous Real Functions is Metric

## Theorem

Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a closed real interval.

Let $\mathscr C \left[{a \,.\,.\, b}\right]$ be the set of all continuous functions $f: \left[{a \,.\,.\, b}\right] \to \R$.

Let $d$ be the supremum metric on $\mathscr C \left[{a \,.\,.\, b}\right]$.

Then $d$ is a metric.

## Proof 1

Let $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ be the set of all bounded real functions $f: \left[{a \,.\,.\, b}\right] \to \R$.

From Supremum Metric on Continuous Real Functions is Subspace of Bounded, $\left({\mathscr C \left[{a \,.\,.\, b}\right], d_{\mathscr C} }\right)$ is a (metric) subspace of $\left({\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right), d}\right)$.

The result follows from Subspace of Metric Space is Metric Space.

$\blacksquare$

## Proof 2

Let $A := \mathscr D^r \left[{a \,.\,.\, b}\right]$ be the set of all continuous functions $f: \left[{a \,.\,.\, b}\right] \to \R$ which are of differentiability class $r$.

Let $d_r: A \times A \to \R$ be the supremum metric on $A$, defined as:

- $\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \mathop \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert$

where:

- $f$ and $g$ are continuous functions on $\left[{a \,.\,.\, b}\right]$ which are $r$ times differentiable
- $r \in \N$ is a natural number.

From Supremum Metric on Differentiability Class is Metric, $d_r$ is a metric on $A$.

By definition, real functions which are continuous on $\left[{a \,.\,.\, b}\right]$ are of differentiability class $0$.

Thus setting $r = 0$ it follows that $\mathscr C \left[{a \,.\,.\, b}\right]$ is $\mathscr D^0 \left[{a \,.\,.\, b}\right]$.

The result follows.

$\blacksquare$