Smooth causal-cone collar barrier with screen directions

In RN, $d\tau =dv-dr$ and the inverse $v,r$ block of [F1] gives $g^{-1}(d\tau ,d\tau )=f-2,$ which equals $-2$ at $r_H$. After shrinking a radial collar it remains negative, and $d\tau (N)=2-f\mathrm{/}2>0$. In KN, direct inversion of [F2] gives $g^{-1}(d\tau ,d\tau )=\frac{a^2{\sin }^2\theta -2R+\mathrm{\Delta }}{\mathrm{\Sigma }}\mathrm{.}$ At $r_H$, $\mathrm{\Delta }=0$, $R=r_H^2+a^2$, and the numerator satisfies $a^2{\sin }^2\theta -2R\le -2r_H^2-a^2<0$. Compactness of $S^2$ and continuity therefore give a uniform radial collar where this norm stays negative. There $d\tau (N)=1+(1+f_0^2)\mathrm{/}2>0$. Thus $\mathrm{\nabla }\tau$ is past timelike in either case by [F3-F4]. In a local future orthonormal tetrad, write $-\mathrm{\nabla }\tau =(T^0,\mathbf{T})$ and a future causal $Z=(Z^0,\mathbf{Z})$. Then $T^0>\mathrm{\parallel }\mathbf{T}\mathrm{\parallel }$ and $Z^0\ge \mathrm{\parallel }\mathbf{Z}\mathrm{\parallel }$, with $Z^0>0$, by the signature and selected cone. By [F9], $g(-\mathrm{\nabla }\tau ,Z)=-T^0{Z}^0+\mathbf{T}\cdot \mathbf{Z}\le -(T^0-\mathrm{\parallel }\mathbf{T}\mathrm{\parallel })Z^0<0.$ Therefore $d\tau (Z)=g(\mathrm{\nabla }\tau ,Z)>0$.

At an RN root, $\xi ={\mathrm{\partial }}_v$ satisfies ${\xi }^{\mathrm{♭}}=-fdv+dr=dr$, and $g(\xi ,N)=-1<0$ there, so it is future null by [F5]. At a KN root put $R_H=r_H^2+a^2$ and $\chi ={\mathrm{\partial }}_v+\frac{a}{R_H}\mathrm{\Psi }\mathrm{.}$ Substitution in the metric of [F2] gives ${\chi }^{\mathrm{♭}}=({\mathrm{\Sigma }}_H\mathrm{/}{R}_H)dr$, where ${\mathrm{\Sigma }}_H=r_H^2+a^2{\cos }^2\theta >0$. Also $dr(N)=-(1+f_0^2)\mathrm{/}2<0$, so $g(\chi ,N)={\chi }^{\mathrm{♭}}(N)<0$ and $\chi$ is future null. In both cases $dr$ on the root is a positive multiple of the covector of a future null generator. In an orthonormal frame, if future null $K=(K^0,\mathbf{K})$ and future causal $Z=(Z^0,\mathbf{Z})$, then $K^0=\mathrm{\parallel }\mathbf{K}\mathrm{\parallel }$, $Z^0\ge \mathrm{\parallel }\mathbf{Z}\mathrm{\parallel }$, and $g(K,Z)=-K^0{Z}^0+\mathbf{K}\cdot \mathbf{Z}\le -K^0{Z}^0+\mathrm{\parallel }\mathbf{K}\mathrm{\parallel }\mathrm{\parallel }\mathbf{Z}\mathrm{\parallel }\le 0.$ Consequently $dr(Z)\le 0$ on $r=r_H$.

Put $u:=-g^{-1}(d\tau ,d\tau )>0$ on the collar by step 1.1. The unique positive root $\sqrt{u}$ is smooth: at every $u_0>0$ the map $s\mapsto s^2$ has derivative $2\sqrt{u_0}\mathrm{\ne }0$, so [F7] gives a smooth local inverse; uniqueness makes these local inverses agree. Hence $n_{\tau }:=-\mathrm{\nabla }\tau \mathrm{/}\sqrt{u}$ is a smooth future unit timelike field. At $v=0$, cover the compact sphere section at $r=r_H$ by finitely many relatively compact neighborhoods on which [F6] supplies smooth future orthonormal tetrads $(E_0,E_1,E_2,E_3)$. Shrink $\delta$ so the closed radial collar over each smaller patch stays in its tetrad neighborhood. Extend these tetrads to all $v$ by the isometric translations of [F1-F2]; both metrics and $d\tau$ are $v$-translation invariant. On each neighborhood set $P_i:=E_i+g(E_i,n_{\tau })n_{\tau }$ for $i=1,2,3$. Then $d\tau (P_i)=0$. If ${\sum }_i{c}_i{P}_i=0$, then ${\sum }_i{c}_i{E}_i$ is a multiple of $n_{\tau }$; its squared norm is nonnegative, while a nonzero multiple of $n_{\tau }$ has negative squared norm. Hence all $c_i=0$. The kernel of $d\tau$ is positive definite, so finite Gram-Schmidt gives a smooth orthonormal screen $(e_1,e_2,e_3)$ after shrinking each neighborhood. Its normalization roots are smooth by the same inverse argument. Every future causal vector with $d\tau (Z)=1$ has $Z=\frac{n_{\tau }+b_1{e}_1+b_2{e}_2+b_3{e}_3}{d\tau (n_{\tau })},b_1^2+b_2^2+b_3^2\le 1.$ Indeed, write $Z=s{n}_{\tau }+W$ with $W\perp n_{\tau }$. Since $n_{\tau }$ is future timelike and $Z$ is future causal, [F5] gives $g(n_{\tau },Z)<0$, so $s=-g(n_{\tau },Z)>0$. Causality gives $g(W,W)\le s^2$, and $d\tau (Z)=s\sqrt{u}=1$.

On each finite coordinate patch of step 2.1, the function $G(r,\omega ,b):=dr(Z)$ for the displayed normalized vector is smooth in the radial variable, the compact sphere-coordinate subset and the closed unit ball $b\in {\mathbb{R}}^3$. Their products with the closed radial interval and unit ball are closed bounded Euclidean sets, hence compact by [F8]. Applying the extreme-value theorem [F8] to $\mathrm{\mid }{\mathrm{\partial }}_{r}G\mathrm{\mid }$ on each product and taking the largest of finitely many bounds gives one finite $C\ge 0$, uniform in $v$. At $r=r_H$, step 1.2 gives $G(r_H,\omega ,b)\le 0$. The one-variable mean-value theorem [F10] therefore gives, throughout the collar, $G(r,\omega ,b)\le G(r_H,\omega ,b)+C\mathrm{\mid }r-r_H\mathrm{\mid }\le C\mathrm{\mid }r-r_H\mathrm{\mid }\mathrm{.}$ For any nonzero future causal $Z$, step 1.1 gives $d\tau (Z)>0$; apply the normalized estimate to $Z\mathrm{/}d\tau (Z)$ and multiply back. The zero vector is trivial. This proves the asserted cone bound including all angular screen directions.