=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_p \left({f, f}\right)       | r = \left({\int_a^b \left\vert{f \left({t}\right) - f \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\int_a^b 0^p \ \mathrm d t}\right)^{\frac 1 p}       | c = {{Defof|Absolute Value}} }} {{eqn | r = 0       | c = Definite Integral of Constant }} {{end-eqn}} So axiom $M1$ holds for $d_p$. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = d_p \left({f, g}\right) + d_p \left({g, h}\right)       | r = \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}       | c = Definition of $d_p$ }} {{eqn | r = \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}       | c =  }} {{eqn | o = \ge       | r = \left({\int_a^b \left({r \left({t}\right) + s \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p}       | c = Minkowski's Inequality for Integrals }} {{eqn | r = \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}       | c = Definition of $r \left({t}\right)$ and $s \left({t}\right)$ }} {{eqn | r = \left({\int_a^b \left\vert{f \left({t}\right) - h \left({t}\right\vert}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}       | c = Triangle Inequality for Real Numbers }} {{eqn | r = d_p \left({f, h}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M2$ holds for $d_p$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_p \left({f, g}\right)       | r =  \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r =  \left({\int_a^b \left\vert{g \left({t}\right) - f \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}       | c = {{Defof|Absolute Value}} }} {{eqn | r = d_p \left({g, f}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M3$ holds for $d_p$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | lo= \forall t \in \left[{a \,.\,.\, b}\right]:       | l = f \left({t}\right)       | o = \ne       | r = g \left({t}\right)       | c =  }} {{eqn | ll= \implies       | l = \left\vert{f \left({t}\right) - g \left({t}\right\vert}\right)^p       | o = \ge       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}       | o = \ge       | r = 0       | c = Sign of Function Matches Sign of Definite Integral }} {{end-eqn}} 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 $d_p$. {{qed}}