A ray trapped between two convex obstacles

Equip ${\mathbb{R}}^n$ with the Euclidean metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$. Apply the second assertion in [F4] to the nonempty compact disjoint sets ${\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2$ to obtain $a\in {\mathrm{\Theta }}_1$ with $d(a,{\mathrm{\Theta }}_2)=d({\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2)>0$. Apply its first assertion to the point $a$ and the nonempty compact set ${\mathrm{\Theta }}_2$ to obtain $b\in {\mathrm{\Theta }}_2$ with $\mathrm{\mid }a-b\mathrm{\mid }=d(a,{\mathrm{\Theta }}_2)$. Thus the pair minimizes the distance between the bodies, and $\mathrm{\mid }a-b\mathrm{\mid }>0$. Both points lie on their respective boundaries: otherwise moving an interior point a short distance toward the other point would decrease the distance. The open segment from $a$ to $b$ meets neither body, since any point of it in either body would give a strictly shorter distance to the other endpoint.

For each of the two endpoint pairs $(p,q)=(a,b)$ and $(p,q)=(b,a)$, use a smooth boundary chart from [F1], writing its rigid motion as $Q(z)=Uz+c$ with $U$ orthogonal and $Q(p)=(y_0,h(y_0))$. For $v\in {\mathbb{R}}^{n-1}$ the curve ${\gamma }_v(s)=Q^{-1}(y_0+sv,h(y_0+sv))$ lies in that boundary for all sufficiently small $\mathrm{\mid }s\mathrm{\mid }$, since the chart domains are open. It has ${\gamma }_v(0)=p$ and velocity ${\gamma }_v^{\mathrm{\prime }}(0)=U^{\mathsf{T}}(v,Dh(y_0)v)$ by [F8]. These velocities give exactly the boundary tangent hyperplane: any differentiable boundary curve through $p$ has chart coordinates $(y(s),h(y(s)))$, whose derivative has this form by [F8], and each such vector is realized by ${\gamma }_v$. For $f_q(z)=\mathrm{\mid }z-q{\mathrm{\mid }}^2$, the identity $f_q(p+w)-f_q(p)=2(p-q)\cdot w+\mathrm{\mid }w{\mathrm{\mid }}^2$ gives $D{f}_q(p)w=2(p-q)\cdot w$. By step 1.1, $f_q\circ {\gamma }_v$ has a minimum at the interior parameter $0$. Thus [F7] gives $(p-q)\cdot U^{\mathsf{T}}(v,Dh(y_0)v)=0$ for every $v$. Consequently $b-a$ is perpendicular to both boundary tangent hyperplanes.

Put $\mathrm{\ell }=\mathrm{\mid }b-a\mathrm{\mid }$ and $e=(b-a)\mathrm{/}\mathrm{\ell }$. For every integer $k\in \mathbb{Z}$, define $x(t)=a+(t-2k\mathrm{\ell })e$ on $[2k\mathrm{\ell },(2k+1)\mathrm{\ell }]$ and $x(t)=b-(t-(2k+1)\mathrm{\ell })e$ on $[(2k+1)\mathrm{\ell },(2k+2)\mathrm{\ell }]$. These formulas agree at their endpoints and define a continuous $2\mathrm{\ell }$-periodic path on all of $\mathbb{R}$. On the open segments use temporal covector $-1$ and spatial covector $e$ or $-e$, respectively. Incidence at each endpoint is normal and nonzero, so [F2] reverses the normal covector and preserves the zero tangential covector. Compression identifies the incoming and outgoing endpoints. The reflection times form a discrete subset of $\mathbb{R}$; away from them the characteristic is a straight unit-speed segment. Thus this is a complete admissible periodic ray in exactly the supplier convention, with image $[a,b]$. Both endpoints lie in $B_R(0)$; by [F6] this ball is convex, and by [F5] the whole segment $[a,b]$ therefore lies there. The box $K=[-R,R{]}^n$ is compact by [F9] and contains $B_R(0)$, since $\mathrm{\mid }{z}_j\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }<R$ for every coordinate of $z\in B_R(0)$. Hence the whole ray lies in $K$. The ray never meets $\omega$, so GCC fails for every $T>0$ by [F2]; it never leaves that compact set, so non-trapping fails by [F3].