Supremum Metric on Differentiability Class is Metric

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

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

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 $d: A \times A \to \R$ 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:


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

where $f$ and $g$ are continuous functions on $\left[{a \,.\,.\, b}\right]$ which are of differentiability class $r$.

Proof of $M1$
So axiom $M1$ holds for $d$.

Proof of $M2$
Let $f, g, h \in A$.

Let $c \in \left[{a \,.\,.\, b}\right]$.

Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an upper bound for:
 * $S := \left\{ {\sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{f^{\left({i}\right)} \left({c}\right) - g^{\left({i}\right)} \left({c}\right)}\right\vert: c \in \left[{a \,.\,.\, b}\right]}\right\}$

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

So axiom $M2$ holds for $d$.

Proof of $M3$
So axiom $M3$ holds for $d$.

Proof of $M4$
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
 * $\forall f, g \in A: d \left({f, g}\right) \ge 0$

Suppose $f, g \in A: d \left({f, g}\right) = 0$.

Then:

So axiom $M4$ holds for $d$.