L2 Metric on Closed Real Interval is Metric

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 $L^2$ metric on $\left[{a \,.\,.\, b}\right]$:
 * $\displaystyle \forall f, g \in S: d_2 \left({f, g}\right) := \left({\int_a^b \left({f \left({t}\right) - g \left({t}\right)}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}$

Then $d$ is a metric.

Proof of $M1$
So axiom $M1$ holds for $d_2$.

Proof of $M2$
It is required to be shown:
 * $d \left({f, g}\right) + d \left({g, h}\right) \ge d \left({f, h}\right)$

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

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

Thus we need to show that:
 * $\displaystyle \left({\int_a^b \left({f \left({t}\right) - g \left({t}\right)}\right)^2 \ \mathrm d t}\right)^{\frac 1 2} + \left({\int_a^b \left({g \left({t}\right) - h \left({t}\right)}\right)^2 \ \mathrm d t}\right)^{\frac 1 2} \ge \left({\int_a^b \left({f \left({t}\right) - h \left({t}\right)}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}$

We have:

So axiom $M2$ holds for $d_2$.

Proof of $M3$
So axiom $M3$ holds for $d_2$.

Proof of $M4$
From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:
 * $d_2 \left({f, g}\right) = 0 \implies f = g$

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

So axiom $M4$ holds for $d_2$.