L1 Metric on Closed Real Interval is Metric

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Theorem

Let $S$ be the set of all real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

Let $d: S \times S \to \R$ be the $L^1$ metric on $\left[{a \,.\,.\, b}\right]$:

$\displaystyle \forall f, g \in S: d \left({f, g}\right) := \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t$


Then $d$ is a metric.


Proof

Proof of $M1$

\(\displaystyle d \left({f, f}\right)\) \(=\) \(\displaystyle \int_a^b \left\vert{f \left({t}\right) - f \left({t}\right)}\right\vert \ \mathrm d t\) Definition of $d$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b 0 \ \mathrm d t\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle 0\) Definite Integral of Constant

So axiom $M1$ holds for $d$.

$\Box$


Proof of $M2$

\(\displaystyle \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert + \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert\) \(\ge\) \(\displaystyle \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert\) Triangle Inequality for Real Numbers
\(\displaystyle \implies \ \ \) \(\displaystyle \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t + \int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert \ \mathrm d t\) \(\ge\) \(\displaystyle \int_a^b \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert \ \mathrm d t\) Relative Sizes of Definite Integrals
\(\displaystyle \implies \ \ \) \(\displaystyle d \left({f, g}\right) + d \left({g, h}\right)\) \(\ge\) \(\displaystyle d \left({f, h}\right)\) Definition of $d$

So axiom $M2$ holds for $d$.

$\Box$


Proof of $M3$

\(\displaystyle d \left({f, g}\right)\) \(=\) \(\displaystyle \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t\) Definition of $d$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \left\vert{g \left({t}\right) - f \left({t}\right)}\right\vert \ \mathrm d t\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle d \left({g, f}\right)\) Definition of $d$

So axiom $M3$ holds for $d$.

$\Box$


Proof of $M4$

\(\, \displaystyle \forall t \in \left[{a \,.\,.\, b}\right]: \, \) \(\displaystyle \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert\) \(\ge\) \(\displaystyle 0\) Definition of Absolute Value
\(\displaystyle \implies \ \ \) \(\displaystyle \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t\) \(\ge\) \(\displaystyle 0\) Sign of Function Matches Sign of Definite Integral

From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:

$d \left({f, g}\right) = 0 \implies f = g$

on $\left[{a \,.\,.\, b}\right]$.

So axiom $M4$ holds for $d$.

$\blacksquare$


Sources