The spacelike restart surface at an exposed boundary point: maximizers on the boundary, the level hypersurface and its radial tube

(Maximum and its boundary location.) The compact set $K$ is nonempty because it contains $p$; ${\tau }_q$ is finite and upper semicontinuous on $K$, so its maximum is attained and ${\tau }_0\ge {\tau }_q(p)>0$ by [L1]. A maximizer $r$ belongs to $I^{-}(q)\cap U^c\cap J^{+}(\mathrm{\Sigma })$, hence to $W$ by the assumed containment, and to $I^{+}(\mathrm{\Sigma })$ since $r\mathrm{\notin }\mathrm{\Sigma }\subset U$. If $r$ were interior to $U^c$, the timelike radial geodesic from $r$ to $q$ could be continued a little to the past within $W\cap \mathrm{int}(U^c)\cap I^{+}(\mathrm{\Sigma })$ by [L2]. Its extended positive proper length exceeds ${\tau }_0$, contradicting maximality on $K$. Thus every maximizer lies in $W\cap \mathrm{\partial }U\subset C_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}}$. If $x\in W$ has ${\tau }_q(x)>{\tau }_0$ and is outside $U$, then $x\in K$ because $W\subset I^{+}(\mathrm{\Sigma })$, a contradiction. The same boundary-location argument applies to each $x\in W\cap U^c\cap J^{+}(\mathrm{\Sigma })$ with ${\tau }_q(x)={\tau }_0$. This proves (a).

(The complete level surface and compact ambient closure.) On $W\cap I^{-}(q)$ define $R(x)=\sqrt{-g_q(\xi (x),\xi (x))}$ with $\xi ={\exp }_q^{-1}x$; by [L1], $R={\tau }_q$ there. The Gauss differential [L2] gives $\mathrm{\nabla }R=-P(q,x)\mathrm{/}R$, so $g(\mathrm{\nabla }R,\mathrm{\nabla }R)=-1$ and $\mathrm{\nabla }R$ is future-directed; its negative $X$ is unit past-directed. The level ${\tau }_q={\tau }_0>0$ is therefore a smooth spacelike hypersurface in $W$. Its connected components are open and closed in this regular level, so the specified component $S$ is embedded without boundary and closed relative to $W$. It contains the chosen boundary maximizer, and every point of $S\setminus U$ is a boundary maximizer by step 1.1. Hence (b) holds. Let $A={\bar{S}}^M\subset \bar{W}$; this is compact, although $S$ itself need not be compact. In the normal chart of $C$, every limit point of $S$ still has past timelike inverse vector of norm squared $-{\tau }_0^2$. Thus $R$ is smooth on an open neighbourhood of $A$ in $C$, and $A$ lies in its regular level $R={\tau }_0$.

(Quantitative normal flow over the ambient closure.) Extend $X=-\mathrm{\nabla }R$ to its smooth domain near $A$. The radial formula shows directly that its flow is $F(s,t)={\exp }_q((1+t\mathrm{/}{\tau }_0)\xi (s))$ while defined, so $R(F(s,t))={\tau }_0+t$ and its lines are unit affinely parametrized past timelike geodesics by [L2]. Compactness of $A$ and openness of the smooth flow domain give $a>0$, an open neighbourhood $B$ of $A$ in the extended level, and a compact ambient set containing every $F(s,t)$ for $s\in B$, $\mathrm{\mid }t\mathrm{\mid }\le a$ after shrinking $B,a$. Smoothness of $dF$ and the positive metric on the extended level permit a further uniform shrinking with $\frac{1}{2}{q}_0\le q_t\le 2q_0$, where $q_t(V,V)=g(d{F}_{t}V,d{F}_{t}V)$ for $V$ tangent to that level. To see uniformity, the $h$-unit tangent bundle over compact $A$ is compact, $q_0$ has a positive minimum there and $q_t-q_0$ tends uniformly to zero; the bounds then persist near $A$. Likewise choose constants $c,C>0$ with $q_0\ge ch{\mathrm{\mid }}_{TS}$ and $\mathrm{\mid }X{\mathrm{\mid }}_h\le C$ throughout this compact flow range. The flow speed gives $d_h(F(s,t),s)\le C\mathrm{\mid }t\mathrm{\mid }$. These are bounds on a compact extended level neighbourhood, not an assumption that $S$ is compact.

(Smooth thin width with an actual derivative estimate.) Let $O\subset W$ be any prescribed open neighbourhood of $S$, and let $b:S\to (0,\mathrm{\infty })$ be any prescribed continuous width ceiling, taking $O=W$, $b=1$ if none is specified. Set $\beta (s)=\min \{a\mathrm{/}2,{\tau }_0\mathrm{/}8,b(s),d_h(s,M\setminus O)\mathrm{/}(2C)\}>0$. By [L4] choose a countable locally finite smooth partition $({\rho }_i)$ with compact supports in $S$, indexed by a subset of the positive integers. Put $M_i={\sup }_S\mathrm{\mid }d{\rho }_i{\mathrm{\mid }}_{q_0}<\mathrm{\infty }$. Choose positive constants $a_i$ with $a_i\le \frac{1}{2}{\inf }_{\mathrm{supp}{\rho }_i}\beta$ and $a_i{M}_i\le 2^{-i}\mathrm{/}(8\sqrt{2})$, omitting any identically zero partition member. Countable Choice suffices. Define $\delta ={\sum }_i{a}_i{\rho }_i$. Local finiteness makes it smooth and positive; the partition identity gives $\delta \le \beta \mathrm{/}2$, while $\mathrm{\mid }d\delta {\mathrm{\mid }}_{q_0}\le {\sum }_i{a}_i\mathrm{\mid }d{\rho }_i{\mathrm{\mid }}_{q_0}\le 1\mathrm{/}(8\sqrt{2})$. The speed estimate and distance bound make $F(s,t)\in O$ for $\mathrm{\mid }t\mathrm{\mid }<\delta (s)$. Also $\delta (s)\to 0$ whenever $s\to e\in A\setminus S$, since such $e$ lies in $\mathrm{\partial }W$ and $\delta \le d_h(s,M\setminus W)\mathrm{/}(4C)$. This constructs the required smooth width from quantified partition bounds; scaling an arbitrary continuous function is not used.

(Injectivity, the open tube and its metric.) Restrict the smooth flow to ${\mathcal{D}}_{\delta }=\{(s,t):s\in S,\mathrm{\mid }t\mathrm{\mid }<\delta (s)\}$. It is defined by step 4.1, and ${\tau }_q(F(s,t))={\tau }_0+t$ because its image lies in $W$. If $F(s,t)=F(s^{\mathrm{\prime }},t^{\mathrm{\prime }})$, this identity gives $t=t^{\mathrm{\prime }}$ and flow uniqueness gives $s=s^{\mathrm{\prime }}$. Its differential is invertible at the zero section because $X$ is transverse to $S$, and everywhere by the flow group law; hence [L2] makes it a diffeomorphism onto an open $N\subset O$. Vectors $d{F}_{t}V$ are tangent to the $R$-level and orthogonal to $X$, so $F^{\ast }g=-d{t}^2+q_t$. All construction claims in (c), including its quantitative inequalities, now hold.

(Causal graphs and finite base length.) For a future causal curve $\gamma (\lambda )=F(s(\lambda ),t(\lambda ))$ in $N$, [L5] and step 5.1 give $q_t(s^{\mathrm{\prime }},s^{\mathrm{\prime }})\le (t^{\mathrm{\prime }}{)}^2$ almost everywhere. Since $\mathrm{\nabla }R$ is future timelike, $t^{\mathrm{\prime }}=dR({\gamma }^{\mathrm{\prime }})<0$ almost everywhere, and the fundamental theorem makes $t$ strictly decreasing. The lower comparison bound gives $\mathrm{\mid }{s}^{\mathrm{\prime }}{\mathrm{\mid }}_{q_0}\le \sqrt{2}\mathrm{\mid }{t}^{\mathrm{\prime }}\mathrm{\mid }$ and hence the total $q_0$-length is at most $\sqrt{2}\mathrm{var}(t)\le 2\sqrt{2}{\sup }_S\delta <\mathrm{\infty }$. Thus the image is a graph in the time coordinate; integrated length on subintervals gives a locally Lipschitz graph in any base chart, without assuming that the inverse of the original time parametrization is a $C^1$ map. It crosses $t=0$ at most once. At either time end, tail $q_0$-length tends to zero; $q_0\ge ch$ from step 3.1 makes $s(\lambda )$ Cauchy in $d_h$, with a unique limit in compact $A$. Past curves are handled by reversal. This proves (c)1 and supplies the endpoint control used below.

(Exact radial interior.) Let $y=F(s,t_y)\in N\cap U$. For $t_y>0$, all radial points with $0<u\le t_y$ lie in $U$ by step 1.1 because their time separation exceeds ${\tau }_0$; the base may lie on $\mathrm{\partial }U$ and is excluded from this assertion. For $t_y<0$, $s$ is to the chronological past of $y$. Concatenate a timelike curve from $\mathrm{\Sigma }$ to $s$ (since $s\in W\subset I^{+}(\mathrm{\Sigma })$) with the radial segment from $s$ to $y$. Its endpoint is $y\in U\cap I^{+}(\mathrm{\Sigma })$, so [L3]'s past-convexity clause puts the whole concatenation, including $s$ and the radial segment, in $U$. If $t_y=0$, $s=y\in U$ directly. If $s\in U$ and $t_y>0$, causal convexity with endpoints $y,s\in U$ also puts the inclusive segment in $U$. This proves precisely the radial assertions in (c)3.

(Past endpoint of a hypothetical avoiding curve.) Suppose a future causal curve $\gamma$ inextendible in $U\cup N$ avoids $U$. Then it lies in $N$ and has $t\le 0$ by step 1.1; equality may occur and is not excluded without proof. At its past end the decreasing time has supremum limit $H\le 0$, and step 6.1 gives $s\to e\in A$. If $e\in S$, continuity and the tube inequalities imply $H\ge -\delta (e)$. If $H>-\delta (e)$, then also $H<\delta (e)$ since $\delta (e)>0$, so the smooth extended flow gives endpoint $F(e,H)\in N$, contradicting inextendibility. The only remaining possibility is $H=-\delta (e)$. Along the curve set $r=t+\delta (s)>0$. The actual derivative bound yields $r^{\mathrm{\prime }}=t^{\mathrm{\prime }}+d\delta (s^{\mathrm{\prime }})\le -(1-1\mathrm{/}8)\mathrm{\mid }{t}^{\mathrm{\prime }}\mathrm{\mid }$, hence for any two ordered parameter values ${\lambda }_1<{\lambda }_2$, $r({\lambda }_1)-r({\lambda }_2)\ge (7\mathrm{/}8)(t({\lambda }_1)-t({\lambda }_2))>0$. A past-end limit $r\to H+\delta (e)=0$ is impossible because it is at least the positive value of $r$ at each fixed later parameter. This excludes the signed lower face with the correct direction of monotonicity; no analysis of a second endpoint is needed.

(Noncompact surface ends and the Cauchy conclusion.) If instead $e\in A\setminus S$, relative closedness gives $e\in \mathrm{\partial }W$. Step 4.1 gives $\delta (s)\to 0$, so the tube inequality forces $H=0$ and the smooth extended flow gives $\gamma \to e$. The point $e\in \bar{W}\subset C\subset I^{+}(\mathrm{\Sigma })$ has past timelike inverse exponential of radial norm ${\tau }_0$, so $e\in I^{-}(q)$ by [L2]. If $e\mathrm{\notin }U$, the assumed containment $J^{-}(q)\cap U^c\cap J^{+}(\mathrm{\Sigma })\subset W$ would give $e\in W$, contradicting $e\in \mathrm{\partial }W$. Therefore $e\in U$, again an endpoint in $U\cup N$. With step 7.1 this proves there is no $U$-avoiding inextendible curve. For any inextendible curve meeting $U$, [L5] makes each maximal $U$-portion inextendible in $U$, so it meets $\mathrm{\Sigma }$ once. If two parameter visits to $\mathrm{\Sigma }$ existed, [L3]'s causal convexity puts the intervening segment in $U$, making both visits belong to the same maximal $U$-portion, a contradiction. This proves (c)2 for the entire noncompact surface.

(The map comparison and admissible thinner tubes.) For the conditional local isometry $\mathrm{\Theta }$ of (c)3, its differential agrees with $d\mathrm{\Phi }$ on $TS$ and, since both are future isometries at $S$, on the future unit normal by [L3], [L6]. They therefore agree on $X$, the past unit normal. The punctured radial segment of step 6.2 lies in $U$, so $\phi$ maps it to an affinely parametrized geodesic by [L6], with limiting value and velocity at the base supplied by its smooth extension $\mathrm{\Phi }$. The $\mathrm{\Theta }$-image is a geodesic with the identical base value and velocity. Geodesic uniqueness [L2] gives $\mathrm{\Theta }(y)=\phi (y)$. Finally any smaller smooth positive width with the derivative bound in (c)4 inherits the flow domain, metric comparisons, speed bounds and vanishing at $A\setminus S$. The same proof uses $\mathrm{\mid }d\tilde{\delta }(s^{\mathrm{\prime }})\mathrm{\mid }\le \frac{1}{4}\mathrm{\mid }{t}^{\mathrm{\prime }}\mathrm{\mid }$ in step 7.1, still giving strict signed-face monotonicity; all other steps are unchanged. Step 4.1 already allows any prescribed open $O$ and ceiling $b$. Thus (c)3-4 and all conclusions hold.