Unit-speed confinement before an artificial outer boundary

The base projection $x$ is continuous by the compressed-ray convention. At a nonreflection time, the time-normalized Hamilton rule gives a spatial derivative of norm one: in the interior this is the Euclidean velocity, and at glancing the characteristic covector is tangent and its tangential Hamilton velocity has norm one. This also applies at infinite-order glancing. At an isolated reflection time, the ray is interior on each adjacent interval; its one-sided phase limits give one-sided unit spatial velocities. Consequently, for each unit vector $e\in {\mathbb{R}}^d$, the continuous real function $f_e(t)=e\cdot x(t)$ has a right derivative at every $t<T$, with $\mathrm{\mid }(f_e{)}_{+}^{\mathrm{\prime }}(t)\mathrm{\mid }\le 1$. The endpoint right derivative uses the same local rule.

Under the additional obstacle hypotheses, $D=B_L(0)\setminus \mathrm{\Theta }$ is open and bounded because $\mathrm{\Theta }$ is closed. Its boundary is $\mathrm{\partial }\mathrm{\Theta }\cup \mathrm{\partial }{B}_L(0)$: near $\mathrm{\partial }\mathrm{\Theta }$ it agrees with the exterior since $\mathrm{\Theta }\subset B_R(0)$ and $R<L$, while near $\mathrm{\partial }{B}_L(0)$ the obstacle is absent. At a point on neither of these boundaries, a sufficiently small neighborhood lies entirely on a fixed side of each set, so no further boundary points occur. The two boundary components are disjoint and smooth; the obstacle charts are the original one-sided charts with the exterior side chosen, and the sphere has smooth one-sided graph charts. Thus $D$ has the required smooth boundary.

First obtain absolute continuity without assuming that a compact time interval has only finitely many reflections. Fix $0\le a<b\le T$ and put $f=f_e$. If $f(b)-f(a)>2(b-a)$, then the continuous function $g(u)=f(u)-2u$ satisfies $g(b)>g(a)$. Its attained minimum on $[a,b]$ therefore occurs at some $c<b$. All sufficiently small positive right difference quotients of $g$ at $c$ are nonnegative, whereas $g_{+}^{\mathrm{\prime }}(c)=f_{+}^{\mathrm{\prime }}(c)-2\le -1$, a contradiction. Apply the same argument to $-f$ to obtain $\mathrm{\mid }f(b)-f(a)\mathrm{\mid }\le 2(b-a)$. For any finite family of intervals with disjoint interiors, ${\sum }_j\mathrm{\mid }f(v_j)-f(u_j)\mathrm{\mid }\le 2{\sum }_j(v_j-u_j)$; choosing $\delta =\epsilon \mathrm{/}2$ proves scalar absolute continuity. On a singleton interval the increments vanish.

Apply the absolute-continuity fundamental theorem to $f_e$ on $[s,t]$ when $s<t$. Wherever its ordinary derivative exists, it equals its right derivative, so $\mathrm{\mid }{f}_e^{\mathrm{\prime }}\mathrm{\mid }\le 1$ almost everywhere. Hence $\mathrm{\mid }e\cdot (x(t)-x(s))\mathrm{\mid }\le {\int }_s^t\mathrm{\mid }{f}_e^{\mathrm{\prime }}(r)\mathrm{\mid }dr\le t-s$. If $x(t)\mathrm{\ne }x(s)$ choose $e=(x(t)-x(s))\mathrm{/}\mathrm{\mid }x(t)-x(s)\mathrm{\mid }$ to obtain $\mathrm{\mid }x(t)-x(s)\mathrm{\mid }\le t-s$; otherwise the inequality is immediate. Interchanging the endpoints handles $t<s$. The triangle inequality now gives $\mathrm{\mid }x(t)\mathrm{\mid }\le \mathrm{\mid }x(0)\mathrm{\mid }+t\le R+t$.

By step 3.1, $\mathrm{\mid }x(t)\mathrm{\mid }\le R+t\le R+T<L$. At every point visited by the segment, choose a neighborhood lying inside $B_L(0)$; on this neighborhood $D$ and $\mathrm{\Omega }$ agree. The Euclidean wave symbol, the local compression map, and all interior, reflected and glancing rules are therefore identical in these two domains. Keep the original compressed curve in these identical local charts. It satisfies the generalized-ray rules in $D$ on the entire segment and cannot meet $\mathrm{\partial }{B}_L(0)$. This is an identification of a prescribed curve, so neither a global continuation theorem nor uniqueness is used.