Supremum Metric on Bounded Real Functions on Closed Interval is Metric/Proof 2

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Theorem

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

Let $A$ be the set of all bounded real functions $f: \left[{a \,.\,.\, b}\right] \to \R$.

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:

$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$

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

So:

$\exists K, L \in \R: \left\vert{f \left({x}\right)}\right\vert \le K, \left\vert{g \left({x}\right)}\right\vert \le L$

for all $x \in \left[{a \,.\,.\, b}\right]$.


First note that we have:

\(\displaystyle \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert\) \(=\) \(\displaystyle \left\vert{f \left({x}\right) + \left({- g \left({x}\right)}\right)}\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{f \left({x}\right)}\right\vert + \left\vert{\left({- g \left({x}\right)}\right)}\right\vert\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{f \left({x}\right)}\right\vert + \left\vert{g \left({x}\right)}\right\vert\) Definition of Absolute Value
\(\displaystyle \) \(\le\) \(\displaystyle K + L\)

and so the right hand side exists.


Proof of $M1$

\(\displaystyle d \left({f, f}\right)\) \(=\) \(\displaystyle \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - f \left({x}\right)}\right\vert\) Definition of $d$
\(\displaystyle \) \(=\) \(\displaystyle \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{0}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So axiom $M1$ holds for $d$.

$\Box$


Proof of $M2$

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

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

\(\displaystyle c\) \(\in\) \(\displaystyle \left[{a \,.\,.\, b}\right]\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{f \left({c}\right) - h \left({c}\right)}\right\vert\) \(\le\) \(\displaystyle \left\vert{f \left({c}\right) - g \left({c}\right)}\right\vert + \left\vert{g \left({c}\right) - h \left({c}\right)}\right\vert\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(\le\) \(\displaystyle \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({c}\right) - g \left({c}\right)}\right\vert + \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{g \left({c}\right) - h \left({c}\right)}\right\vert\) Definition of Supremum of Real-Valued Function
\(\displaystyle \) \(=\) \(\displaystyle d \left({f, g}\right) + d \left({g, h}\right)\) Definition of $d$

Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an upper bound for:

$S := \left\{ {\left\vert{f \left({c}\right) - h \left({c}\right)}\right\vert: c \in \left[{a \,.\,.\, b}\right]}\right\}$

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 $M3$

\(\displaystyle d \left({f, g}\right)\) \(=\) \(\displaystyle \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert\) Definition of $d$
\(\displaystyle \) \(=\) \(\displaystyle \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{g \left({x}\right) - f \left({x}\right)}\right\vert\) 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:

\(\displaystyle d \left({f, g}\right)\) \(=\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert\) \(=\) \(\displaystyle 0\) 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 $d$.

$\blacksquare$


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