The reversed triangle inequality and convexity of the future timelike cone

Fix an event $o\in A$ and an inertial frame $(o;e_0,e_1,e_2,e_3)$ with its coordinates; by [F1] every vector $u$ is written uniquely as $u=u^0{e}_0+u_s$ with $u_s\in S$, the pairing is $\eta (u,u^{\mathrm{\prime }})=-u^0{u}^{\mathrm{\prime }0}+\delta (u_s,u_s^{\mathrm{\prime }})$, and $u\in C^{+}⟺u^0>\mathrm{\mid }{u}_s\mathrm{\mid }$.

Let $v,w$ be future timelike. Then $(v+w{)}^0=v^0+w^0>\mathrm{\mid }{v}_s\mathrm{\mid }+\mathrm{\mid }{w}_s\mathrm{\mid }\ge \mathrm{\mid }{v}_s+w_s\mathrm{\mid }=\mathrm{\mid }(v+w{)}_s\mathrm{\mid }$: the first inequality adds the two strict inequalities $v^0>\mathrm{\mid }{v}_s\mathrm{\mid }$ and $w^0>\mathrm{\mid }{w}_s\mathrm{\mid }$ that [F1] gives from future timelikeness, the second is the triangle inequality of the spatial norm, and the displayed equalities are additivity of coordinates; hence $v+w$ is future timelike.

Let $a:=\sqrt{-\eta (v,v)}$, $b:=\sqrt{-\eta (w,w)}$ and $C:=\sqrt{-\eta (v+w,v+w)}$, which is a positive real by step 2.1; bilinearity and symmetry of $\eta$ give $C^2=-\eta (v+w,v+w)=-\eta (v,v)-2\eta (v,w)-\eta (w,w)=a^2+b^2+2(-\eta (v,w))$, and $-\eta (v,w)\ge ab$ by the reversed Cauchy-Schwarz inequality, so multiplying by the positive number $2$ and adding the constant $a^2+b^2$ yields $C^2\ge a^2+b^2+2ab=(a+b{)}^2$.

For $t\in (0,1)$ the vectors $tv$ and $(1-t)w$ are future timelike: $(tv{)}^0=t{v}^0>t\mathrm{\mid }{v}_s\mathrm{\mid }=\mathrm{\mid }t{v}_s\mathrm{\mid }$ because $t>0$ preserves the strict inequality and $\mathrm{\mid }t{v}_s\mathrm{\mid }=t\mathrm{\mid }{v}_s\mathrm{\mid }$ by the coordinate formula $\mathrm{\mid }z\mathrm{\mid }=\sqrt{(z^1{)}^2+(z^2{)}^2+(z^3{)}^2}$ of [F2] together with uniqueness of the nonnegative root, and symmetrically $(1-t)w^0>(1-t)\mathrm{\mid }{w}_s\mathrm{\mid }=\mathrm{\mid }(1-t)w_s\mathrm{\mid }$; for $t=0$ and $t=1$ the vector $tv+(1-t)w$ equals $w$ respectively $v$, which are future timelike.

For the base case $k=1$ of (c): a future timelike vector $v_1$ is its own sum, $\sqrt{-\eta (v_1,v_1)}\ge \sqrt{-\eta (v_1,v_1)}$ holds with equality, and $v_1=1\cdot v_1$ exhibits $v_1$ as a positive multiple of the single future timelike vector $u:=v_1$, so the equality characterization of (c) is satisfied for $k=1$.

Since $C\ge 0$ and $a+b\ge 0$ with $C^2\ge (a+b{)}^2$, monotonicity of squaring on the nonnegatives [L4] gives $C\ge a+b$: this is the inequality of (a).

For $t\in (0,1)$, (a) applied to the future timelike vectors $tv$ and $(1-t)w$ of step 3.2 gives $tv+(1-t)w$ future timelike and $\sqrt{-\eta (tv+(1-t)w,tv+(1-t)w)}\ge \sqrt{-\eta (tv,tv)}+\sqrt{-\eta ((1-t)w,(1-t)w)}=t\sqrt{-\eta (v,v)}+(1-t)\sqrt{-\eta (w,w)}$, the two displayed equalities using bilinearity and uniqueness of the nonnegative root; this is the inequality of (b) on the open interval.

Equality holds in (a) if and only if $C=a+b$, which by monotonicity of squaring on the nonnegatives [L4] is equivalent to $C^2=(a+b{)}^2$, hence, subtracting the common term $a^2+b^2$ from the two displays of step 3.1 and multiplying by $1\mathrm{/}2$, to $-\eta (v,w)=ab$, and by the equality case of the reversed Cauchy-Schwarz inequality [F3] to $v$ and $w$ being positive proportional; this proves (a).

Induction step, membership and inequality: assume $k\ge 1$ and that the claim of (c) holds for the integer $k$; let $v_1,\dots ,v_{k+1}$ be future timelike and put $s:=v_1+\cdots +v_k$, which is future timelike by the induction hypothesis; applying (a) of steps 2.1 and 4.1 to the future timelike vectors $s$ and $v_{k+1}$ gives $s+v_{k+1}=v_1+\cdots +v_{k+1}$ future timelike and $\sqrt{-\eta (s+v_{k+1},s+v_{k+1})}\ge \sqrt{-\eta (s,s)}+\sqrt{-\eta (v_{k+1},v_{k+1})}\ge {\sum }_{i=1}^k\sqrt{-\eta (v_i,v_i)}+\sqrt{-\eta (v_{k+1},v_{k+1})}$, the second inequality being the induction hypothesis.

Equality in the interior inequality of (b) for $0<t<1$ holds if and only if $tv$ and $(1-t)w$ are positive proportional, by the equality case of (a) established in step 5.1 applied to the interior inequality of (b) of step 4.2; since $t>0$ and $1-t>0$, this is equivalent to $v$ and $w$ being positive proportional; at $t=0$ and $t=1$ the inequality reduces to an identity $b\ge b$ or $a\ge a$, and the statement makes no interior equality claim there, so (b) is proved.

Induction step, equality characterization: with the notation of step 5.2, if equality holds for the $k+1$ vectors, then both inequalities in the chain of step 5.2 are equalities; the second equality means that the equality case of the induction hypothesis applies to $v_1,\dots ,v_k$, so there are a future timelike $u$ and reals $c_1,\dots ,c_k>0$ with $v_i=c_{i}u$ and $s=({\sum }_{i=1}^k{c}_i)u$; the first equality is the equality case of (a) for $s$ and $v_{k+1}$, so $s$ and $v_{k+1}$ are positive proportional, and since $s\mathrm{\ne }0$ this gives $v_{k+1}=c_{k+1}u$ for some real $c_{k+1}>0$, so all $k+1$ vectors are positive multiples of the single future timelike vector $u$; conversely, if $v_i=c_{i}u$ with $c_i>0$ and $u$ future timelike for $i=1,\dots ,k+1$, then $v_1+\cdots +v_{k+1}=({\sum }_i{c}_i)u$ and bilinearity with uniqueness of the nonnegative root give $\sqrt{-\eta ({\sum }_i{v}_i,{\sum }_i{v}_i)}=({\sum }_i{c}_i)\sqrt{-\eta (u,u)}={\sum }_i\sqrt{-\eta (v_i,v_i)}$, so equality holds.

By [L6], applied to the property "$v_1,\dots ,v_k$ future timelike implies the conclusions of (c) for these vectors", the base case of step 3.3 and the induction steps 5.2 and 6.2 prove (c) for every integer $k\ge 1$: membership and inequality from step 5.2, and the equality characterization from step 6.2.