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

Let $d_p: S \times S \to \R$ be the $L^p$ metric on $\closedint a b$:
 * $\ds \forall f, g \in S: \map {d_p} {f, g} := \paren {\int_a^b \size {\map f t - \map g t}^p \rd t}^{\frac 1 p}$

Then $d_p$ is a metric.

Proof of
So holds for $d_p$.

Proof of
It is required to be shown:
 * $\map {d_p} {f, g} + \map {d_p} {g, h} \ge \map {d_p} {f, h}$

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

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

Thus we need to show that:
 * $\ds \paren {\int_a^b \size {\map f t - \map g t}^p \rd t}^{\frac 1 p} + \paren {\int_a^b \size {\map g t - \map h t}^p \rd t}^{\frac 1 p} \ge \paren {\int_a^b \size {\map f t - \map h t}^p \rd t}^{\frac 1 p}$

We have:

So holds for $d_p$.

Proof of
So holds for $d_p$.

Proof of
From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:
 * $\map {d_p} {f, g} = 0 \leadsto f = g$

on $\closedint a b$.

So holds for $d_p$.