Lp Metric on Closed Real Interval is Metric

From ProofWiki
Jump to: navigation, search

Theorem

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

Let $p \in \R_{\ge 1}$.

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

$\displaystyle \forall f, g \in S: d \left({f, g}\right) := \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}$


Then $d_p$ is a metric.


Proof

Proof of $M1$

\(\displaystyle d_p \left({f, f}\right)\) \(=\) \(\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - f \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int_a^b 0^p \ \mathrm d t}\right)^{\frac 1 p}\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle 0\) Definite Integral of Constant

So axiom $M1$ holds for $d_p$.

$\Box$


Proof of $M2$

It is required to be shown:

$d_p \left({f, g}\right) + d_p \left({g, h}\right) \ge d_p \left({f, h}\right)$

for all $f, g, h \in S$.


Let:

$(1): \quad t \in \left[{a \,.\,.\, b}\right]$
$(2): \quad \left\vert{f \left({t}\right) - g \left({t}\right) = r \left({t}\right)}\right\vert$
$(3): \quad \left\vert{g \left({t}\right) - h \left({t}\right) = s \left({t}\right)}\right\vert$

Thus we need to show that:

$\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p} + \left({\int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p} \ge \left({\int_a^b \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}$


We have:

\(\displaystyle d_p \left({f, g}\right) + d_p \left({g, h}\right)\) \(=\) \(\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p} + \left({\int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int_a^b \left({r \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p} + \left({\int_a^b \left({s \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p}\)
\(\displaystyle \) \(\ge\) \(\displaystyle \left({\int_a^b \left({r \left({t}\right) + s \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p}\) Minkowski's Inequality for Integrals
\(\displaystyle \) \(=\) \(\displaystyle \left({\int_a^b \left({\left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert + \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}\) Definition of $r \left({t}\right)$ and $s \left({t}\right)$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - h \left({t}\right\vert}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle d_p \left({f, h}\right)\) Definition of $d_p$

So axiom $M2$ holds for $d_p$.

$\Box$


Proof of $M3$

\(\displaystyle d_p \left({f, g}\right)\) \(=\) \(\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int_a^b \left\vert{g \left({t}\right) - f \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle d_p \left({g, f}\right)\) Definition of $d_p$

So axiom $M3$ holds for $d_p$.

$\Box$


Proof of $M4$

\(\, \displaystyle \forall t \in \left[{a \,.\,.\, b}\right]: \, \) \(\displaystyle f \left({t}\right)\) \(\ne\) \(\displaystyle g \left({t}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{f \left({t}\right) - g \left({t}\right\vert}\right)^p\) \(\ge\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}\) \(\ge\) \(\displaystyle 0\) Sign of Function Matches Sign of Definite Integral

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

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

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

So axiom $M4$ holds for $d_p$.

$\blacksquare$


Sources