Supremum Metric on Differentiability Class is Metric

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

Let $r \in \N$ be a natural number.

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

Then $d$ is a metric.

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


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

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

Proof of $\text M 1$
So axiom $\text M 1$ holds for $d$.

Proof of $\text M 2$
Let $f, g, h \in A$.

Let $c \in \closedint a b$.

Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
 * $S := \set {\sup_{i \mathop \in \set {0, 1, 2, \ldots, r} } \size {\map {f^{\paren i} } c - \map {g^{\paren i} } c}: c \in \closedint a b}$

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

So axiom $\text M 2$ holds for $d$.

Proof of $\text M 3$
So axiom $\text M 3$ holds for $d$.

Proof of $\text M 4$
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:

So axiom $\text M 4$ holds for $d$.