Supremum Metric on Differentiability Class is Metric

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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:

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

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


Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map d {f, f}\) \(=\) \(\ds \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {f^{\paren i} } x}\) Definition of $d$
\(\ds \) \(=\) \(\ds \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size 0\)
\(\ds \) \(=\) \(\ds 0\)

So Metric Space Axiom $(\text M 1)$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

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

Let $c \in \closedint a b$.

\(\ds c\) \(\in\) \(\ds \closedint a b\)
\(\ds \leadsto \ \ \) \(\ds \sup_{i \mathop \in \set {0, 1, 2, \ldots, r} } \size {\map {f^{\paren i} } c - \map {h^{\paren i} } c}\) \(\le\) \(\ds \sup_{i \mathop \in \set {0, 1, 2, \ldots, r} } \size {\map {f^{\paren i} } c - \map {g^{\paren i} } c} + \sup_{i \mathop \in \set {0, 1, 2, \ldots, r} } \size {\map {g^{\paren i} } c - \map {h^{\paren i} } c}\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x} + \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {g^{\paren i} } x - \map {h^{\paren i} } x}\) Definition of Supremum of Real-Valued Function
\(\ds \) \(=\) \(\ds \map d {f, g} + \map d {g, h}\) Definition of $d$

Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:

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

So:

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

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

\(\ds \map d {f, g}\) \(=\) \(\ds \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x}\) Definition of $d$
\(\ds \) \(=\) \(\ds \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {g^{\paren i} } x - \map {f^{\paren i} } x}\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \map d {g, f}\) Definition of $d$

So Metric Space Axiom $(\text M 3)$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

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:

\(\ds \map d {f, g}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {f^{\paren i} } x}\) \(=\) \(\ds 0\) Definition of $d$
\(\ds \leadsto \ \ \) \(\ds \forall x \in \closedint a b: \, \) \(\ds \map f x\) \(=\) \(\ds \map g x\) Definition of Absolute Value
\(\ds \leadsto \ \ \) \(\ds f\) \(=\) \(\ds g\) Equality of Mappings

So Metric Space Axiom $(\text M 4)$ holds for $d$.

$\blacksquare$


Sources

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