Supremum Metric on Bounded Real Functions on Closed Interval is Metric

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $A$ be the set of all bounded real functions $f: \closedint a b \to \R$.

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


Then $d$ is a metric.


Proof 1

The interval is an instance of a set.

Hence Supremum Metric on Bounded Real-Valued Functions is Metric can be directly applied.


Proof 2

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

$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$

where $f$ and $g$ are bounded real functions.

So:

$\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$

for all $x \in \closedint a b$.


First note that we have:

\(\ds \size {\map f x - \map g x}\) \(=\) \(\ds \size {\map f x + \paren {-\map g x} }\)
\(\ds \) \(\le\) \(\ds \size {\map f x} + \size {\paren {-\map g x} }\) Triangle Inequality for Real Numbers
\(\ds \) \(=\) \(\ds \size {\map f x} + \size {\map g x}\) Definition of Absolute Value
\(\ds \) \(\le\) \(\ds K + L\)

and so the right hand side exists.


Proof of Metric Space Axiom $\text M 1$

\(\ds \map d {f, f}\) \(=\) \(\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map f x}\) Definition of $d$
\(\ds \) \(=\) \(\ds \sup_{x \mathop \in \closedint a b} \size 0\)
\(\ds \) \(=\) \(\ds 0\)

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

$\Box$


Proof of Metric Space Axiom $\text M 2$

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

Let $c \in \closedint a b$.

\(\ds c\) \(\in\) \(\ds \closedint a b\)
\(\ds \leadsto \ \ \) \(\ds \size {\map f c - \map h c}\) \(\le\) \(\ds \size {\map f c - \map g c} + \size {\map g c - \map h c}\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \sup_{x \mathop \in \closedint a b} \size {\map f c - \map g c} + \sup_{x \mathop \in \closedint a b} \size {\map g c - \map h c}\) 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:

$S := \set {\size {\map f c - \map h c}: c \in \closedint a b}$

So:

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

So Metric Space Axiom $\text M 2$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $\text M 3$

\(\ds \map d {f, g}\) \(=\) \(\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}\) Definition of $d$
\(\ds \) \(=\) \(\ds \sup_{x \mathop \in \closedint a b} \size {\map g x - \map f x}\) 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:

\(\ds \map d {f, g}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}\) \(=\) \(\ds 0\) 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