L2 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 $L^2$ metric on $\closedint a b$:

$\displaystyle \forall f, g \in S: \map {d_2} {f, g} := \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}$


Then $d_2$ is a metric.


Proof

Proof of $M1$

\(\displaystyle \map {d_2} {f, f}\) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map f t - \map f t}^2 \rd t}^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\int_a^b 0^2 \rd t}^{\frac 1 2}\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle 0\) Definite Integral of Constant

So axiom $M1$ holds for $d_2$.

$\Box$


Proof of $M2$

It is required to be shown:

$\map {d_2} {f, g} + \map {d_2} {g, h} \ge d_2 \map {d_2} {f, h}$

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


Let:

$(1): \quad t \in \closedint a b$
$(2): \quad \map f t - \map g t = \map r t$
$(3): \quad \map g t - \map h t = \map s t$

Thus we need to show that:

$\displaystyle \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2} \ge \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}$


We have:

\(\displaystyle \map {d_2} {f, g} + \map {d_2} {g, h}\) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map r t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map s t}^2 \rd t}^{\frac 1 2}\)
\(\displaystyle \) \(\ge\) \(\displaystyle \paren {\int_a^b \paren {\map r t + \map s t}^2 \rd t}^{\frac 1 2}\) Minkowski's Inequality for Integrals
\(\displaystyle \) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map f t - \map g t + \map g t - \map h t}^2 \rd t}^{\frac 1 2}\) Definition of $\map r t$ and $\map s t$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {d_2} {f, h}\) Definition of $d_2$

So axiom $M2$ holds for $d_2$.

$\Box$


Proof of $M3$

\(\displaystyle \map {d_2} {f, g}\) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\int_a^b \paren {\map g t - \map f t}^2 \rd t}^{\frac 1 2}\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle \map {d_2} {g, f}\) Definition of $d_2$

So axiom $M3$ holds for $d_2$.

$\Box$


Proof of $M4$

\(\, \displaystyle \forall t \in \closedint a b: \, \) \(\displaystyle \map f t - \map g t\) \(\ne\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\map f t - \map g t}^2\) \(\ge\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}\) \(\ge\) \(\displaystyle 0\) Sign of Function Matches Sign of Definite Integral

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

$\map {d_2} {f, g} = 0 \implies f = g$

on $\closedint a b$.

So axiom $M4$ holds for $d_2$.

$\blacksquare$


Sources