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 $M1$

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

So axiom $M1$ holds for $d$.

$\Box$

Proof of $M2$

Let $f, g, h \in A$.

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

 $\displaystyle c$ $\in$ $\displaystyle \left[{a \,.\,.\, b}\right]$ $\displaystyle \implies \ \$ $\displaystyle \sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{f^{\left({i}\right)} \left({c}\right) - h^{\left({i}\right)} \left({c}\right)}\right\vert$ $\le$ $\displaystyle \sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{f^{\left({i}\right)} \left({c}\right) - g^{\left({i}\right)} \left({c}\right)}\right\vert + \sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{g^{\left({i}\right)} \left({c}\right) - h^{\left({i}\right)} \left({c}\right)}\right\vert$ Triangle Inequality for Real Numbers $\displaystyle$ $\le$ $\displaystyle \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert + \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{g^{\left({i}\right)} \left({x}\right) - h^{\left({i}\right)} \left({x}\right)}\right\vert$ Definition of Supremum $\displaystyle$ $=$ $\displaystyle d \left({f, g}\right) + d \left({g, h}\right)$ Definition of $d$

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 $d$.

$\Box$

Proof of $M3$

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

$\forall f, g \in A: d \left({f, g}\right) \ge 0$

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

Then:

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

So axiom $M4$ holds for $d$.

$\blacksquare$