Supremum Metric on Continuous Real Functions is Metric

Theorem

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

Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.

Let $d$ be the supremum metric on $\mathscr C \closedint a b$.

Then $d$ is a metric.

Proof 1

Let $\map {\mathscr B} {\closedint a b, \R}$ be the set of all bounded real functions $f: \closedint a b \to \R$.

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

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

$\blacksquare$

Proof 2

Let $A := \mathscr D^r \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$ which are of differentiability class $r$.

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

$\ds \forall f, g \in A: \map d {f, g} := \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x}$

where:

$f$ and $g$ are continuous functions on $\closedint a b$ 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 $\closedint a b$ are of differentiability class $0$.

Thus setting $r = 0$ it follows that $\mathscr C \closedint a b$ is $\mathscr D^0 \closedint a b$.

The result follows.

$\blacksquare$