The Lorentz logarithmic barrier has parameter two

Continuity of the Euclidean norm makes $K$ closed, and its triangle inequality makes $K$ convex and closed under addition and nonnegative scaling. If $(t,u)$ and its negative are both in $K$, then $t\ge \mathrm{\parallel }u\mathrm{\parallel }$ and $-t\ge \mathrm{\parallel }u\mathrm{\parallel }$, forcing $t=0,u=0$; hence $K$ is pointed. The point $(1,0)$ is interior, so $K$ is proper. If $(a,v)\in K$ and $(t,u)\in K$, then $at+v^{T}u\ge at-\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_2\ge 0$, so $K\subseteq K^{\ast }$ by [F2]. Conversely, suppose $(a,v)\in K^{\ast }$. If $v=0$ and $a<\mathrm{\parallel }v{\mathrm{\parallel }}_2=0$, then pairing with $(1,0)\in K$ gives $a<0$, a contradiction. If $v\mathrm{\ne }0$ and $a<\mathrm{\parallel }v{\mathrm{\parallel }}_2$, pairing with $(1,-v\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_2)\in K$ gives $a-\mathrm{\parallel }v{\mathrm{\parallel }}_2<0$, also a contradiction. Thus $a\ge \mathrm{\parallel }v{\mathrm{\parallel }}_2$ and $K^{\ast }=K$.

Write $\rho =t^2-\mathrm{\parallel }u{\mathrm{\parallel }}_2^2>0$. An orthogonal change of spatial coordinates followed by the Lorentz boost ${\rho }^{-1\mathrm{/}2}\left(\begin{array}{cc}t& -\mathrm{\parallel }u\mathrm{\parallel }\\ -\mathrm{\parallel }u\mathrm{\parallel }& t\end{array}\right)$ on the time and first spatial coordinate preserves $\rho$ and maps $(t,u)$ to $(\sqrt{\rho },0)$. Since $F$ is invariant under this invertible linear cone map, the chain rule transports its Hessian, third derivative, and local norms; the Newton decrement is invariant as well. At the canonical point $\mathrm{\nabla }F=(-2\mathrm{/}\sqrt{\rho },0)$ and $D^{2}F=(2\mathrm{/}\rho )I$. For directions $h=(a,p)$ and $k=(b,v)$, direct differentiation gives $D^{3}F[h,k,k]=-\frac{4\{a(b^2+\mathrm{\parallel }v{\mathrm{\parallel }}_2^2)+2b{p}^{T}v\}}{{\rho }^{3\mathrm{/}2}}$. These formulas also show the Hessian is positive definite.

By Cauchy–Schwarz and $2\mathrm{\mid }b\mathrm{\mid }\mathrm{\parallel }v{\mathrm{\parallel }}_2\le b^2+\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$, the numerator in step 1.2 has absolute value at most $4\sqrt{2}\sqrt{a^2+\mathrm{\parallel }p{\mathrm{\parallel }}_2^2}(b^2+\mathrm{\parallel }v{\mathrm{\parallel }}_2^2)\mathrm{/}{\rho }^{3\mathrm{/}2}=2\mathrm{\parallel }h{\mathrm{\parallel }}_x\mathrm{\parallel }k{\mathrm{\parallel }}_x^2$. If $\mathrm{\parallel }h{\mathrm{\parallel }}_x<1$, set $A=\mathrm{\mid }a\mathrm{\mid }\mathrm{/}\sqrt{\rho }$ and $B=\mathrm{\parallel }p{\mathrm{\parallel }}_2\mathrm{/}\sqrt{\rho }$. Then $A^2+B^2<1\mathrm{/}2$, so $1+a\mathrm{/}\sqrt{\rho }\ge 1-A>0$ and $(1+a\mathrm{/}\sqrt{\rho }{)}^2-B^2\ge (1-A{)}^2-B^2>2(A-1\mathrm{/}2{)}^2\ge 0$. Thus $\sqrt{\rho }+a>\mathrm{\parallel }p{\mathrm{\parallel }}_2$ and the Dikin ball stays inside $D$. The Lorentz transformation carries this inclusion to every $x$. Approaching a finite boundary point makes $\rho \downarrow 0$, hence $F\to +\mathrm{\infty }$. Finally, $F(\alpha x)=F(x)-2\log \alpha$, and step 1.2 gives $\mathrm{\parallel }DF(x){\mathrm{\parallel }}_{x,\ast }^2=2$.