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 $\closedint a b$.

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

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:

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

on $\closedint a b$.

So axiom $M4$ holds for $d_2$.