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 $\closedint a b$.

Let $d: S \times S \to \R$ be the $L^1$ metric on $\closedint a b$:

$\ds \forall f, g \in S: \map d {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$


Then $d$ is a metric.


Proof

Metric Space Axiom $(\text M 1)$

\(\ds \map d {f, f}\) \(=\) \(\ds \int_a^b \size {\map f t - \map f t} \rd t\) Definition of $d$
\(\ds \) \(=\) \(\ds \int_a^b 0 \rd t\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds 0\) Definite Integral of Constant

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

$\Box$


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

\(\ds \size {\map f t - \map g t} + \size {\map g t - \map h t}\) \(\ge\) \(\ds \size {\map f t - \map h t}\) Triangle Inequality for Real Numbers
\(\ds \leadsto \ \ \) \(\ds \int_a^b \size {\map f t - \map g t} \rd t + \int_a^b \size {\map g t - \map h t} \rd t\) \(\ge\) \(\ds \int_a^b \size {\map f t - \map h t} \rd t\) Relative Sizes of Definite Integrals
\(\ds \leadsto \ \ \) \(\ds \map d {f, g} + \map d {g, h}\) \(\ge\) \(\ds \map d {f, h}\) Definition of $d$

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

$\Box$


Metric Space Axiom $(\text M 3)$

\(\ds \map d {f, g}\) \(=\) \(\ds \int_a^b \size {\map f t - \map g t} \rd t\) Definition of $d$
\(\ds \) \(=\) \(\ds \int_a^b \size {\map g t - \map f t} \rd t\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \map d {g, f}\) Definition of $d$

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

$\Box$


Metric Space Axiom $(\text M 4)$

\(\ds \forall t \in \closedint a b: \, \) \(\ds \size {\map f t - \map g t}\) \(\ge\) \(\ds 0\) Definition of Absolute Value
\(\ds \leadsto \ \ \) \(\ds \int_a^b \size {\map f t - \map g t} \rd t\) \(\ge\) \(\ds 0\) Sign of Function Matches Sign of Definite Integral

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

$\map d {f, g} = 0 \implies f = g$

on $\closedint a b$.

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

$\blacksquare$


Sources

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