Cone bumps: for every point of a Cauchy level there is a smooth bump with past timelike gradient on its positive region below that level

(A truncated cone inside a small convex normal neighborhood.) Since $b>a$, the point $p\in S_b$ satisfies $t(p)=b>a$, so $p\in I^{+}(S_a)$ by [L1]. Let $\delta >0$ be prescribed. Choose a Euclidean chart ball $E_{{\rho }_0}$ around $p$ with compact closure in $O\cap I^{+}(S_a)$, let $C_0$ be the comparison constant of [L7] on its closure, and choose $0<\rho \le {\rho }_0$ with $2\sqrt{C_0}\rho <\delta$. By [L2] there is a sufficiently small strictly convex normal neighborhood $C$ of $p$ with $\bar{C}$ compact and $\bar{C}\subseteq E_{\rho }\subseteq I^{+}(S_a)$; then ${\mathrm{diam}}_h(C)\le {\mathrm{diam}}_h(E_{\rho })\le 2\sqrt{C_0}\rho <\delta$ by [L7]. We claim that some $p^{\mathrm{\prime }}\ll p$ satisfies $J^{+}(p^{\mathrm{\prime }})\cap J^{-}(S_b)\subseteq C$. Suppose not: choose a sequence $p_k^{\mathrm{\prime }}\to p$ along a future timelike curve with $p_k^{\mathrm{\prime }}\ll p$ and $p_k^{\mathrm{\prime }}\in C$, and points $r_k\in (J^{+}(p_k^{\mathrm{\prime }})\cap J^{-}(S_b))\setminus C$. For each $k$ the causal segment from $p_k^{\mathrm{\prime }}$ to $r_k$ leaves $C$, so it has a first exit point $q_k\in \mathrm{\partial }C\cap J^{+}(p_k^{\mathrm{\prime }})\cap J^{-}(S_b)$. Fix $r\ll p$ with $r\le p_k^{\mathrm{\prime }}$ for all large $k$ (take $r$ before the sequence on the timelike curve through $p$); then $q_k\in J^{+}(r)\cap J^{-}(S_b)$, which is compact by [L1], so a subsequence $q_k\to q\in J^{+}(r)\cap J^{-}(S_b)\cap \mathrm{\partial }C$. Since $p_k^{\mathrm{\prime }}\to p$ and $p_k^{\mathrm{\prime }}\le q_k$, closedness of the causal relation [L1] gives $p\le q$. Also $q\in J^{-}(S_b)$ gives some $z\in S_b$ with $q\le z$. Monotonicity of $t$ [L1] gives $b=t(p)\le t(q)\le t(z)=b$, so $t(q)=b$, that is, $q\in S_b$; acausality of $S_b$ [L1] then forces $q=z$, and $p\le q$ with $p,q\in S_b$ likewise forces $q=p$. Hence $p=q\in \mathrm{\partial }C$, contradicting $p\in C$ with $C$ open. This proves the claim.

(The future-branch cone function.) On $C$ let $v(q^{\mathrm{\prime }})={\exp }_{p^{\mathrm{\prime }}}^{-1}(q^{\mathrm{\prime }})$ and $Q(q^{\mathrm{\prime }})=-g_{p^{\mathrm{\prime }}}(v(q^{\mathrm{\prime }}),v(q^{\mathrm{\prime }}))$, smooth by [L2]. Fix a future unit tangent $n$ at $p^{\mathrm{\prime }}$ and set $a(q^{\mathrm{\prime }})=-g_{p^{\mathrm{\prime }}}(v(q^{\mathrm{\prime }}),n)$. Define $f:C\to [0,\mathrm{\infty })$ by $f(q^{\mathrm{\prime }})=\phi (Q(q^{\mathrm{\prime }}))$ when $a(q^{\mathrm{\prime }})>0$, and $f(q^{\mathrm{\prime }})=0$ when $a(q^{\mathrm{\prime }})\le 0$. The set $\{Q>0,a>0\}$ is exactly $I_C^{+}(p^{\mathrm{\prime }})$ by [L2], so $f>0$ exactly there and $f(p)>0$. At any nonvertex point with $a=0$, the rest-space positivity gives $Q<0$ and f is locally zero. Elsewhere smoothness follows from composition and flatness across $Q=0$. At the vertex, derivatives of $\phi (Q)$ are polynomially bounded factors times flat derivatives, tending to zero faster than any power of the inverse-vector norm; the zero extension across the separating hyperplane is therefore smooth. Only smoothness on $C$ is claimed; extension across the boundary of $C$ will be performed after multiplication by the compact cutoff.

(The gradient is past timelike.) Where $f>0$ one has $\mathrm{\nabla }f={\phi }^{\mathrm{\prime }}(Q)\mathrm{\nabla }Q$ with ${\phi }^{\mathrm{\prime }}(Q)>0$ [L4], hence $\mathrm{\nabla }f=-2{\phi }^{\mathrm{\prime }}(Q)P(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ by [L3]; now $q^{\mathrm{\prime }}\in I^{+}(p^{\mathrm{\prime }})$ means $v$ is future timelike, so $P(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ is future timelike with the same causal character [L2, L3], and a positive multiple of its negative is past timelike. Hence $\mathrm{\nabla }f$ is past timelike at every point where $f>0$.

(The cutoff and the bump.) Let $K=J^{+}(p^{\mathrm{\prime }})\cap J^{-}(S_b)$, a compact subset of the open set $C$ by step 1.1 and [L1]. By [L5] choose a smooth cutoff $\chi :M\to [0,1]$ with $\chi =1$ on a neighborhood $U_0$ of $K$ and $\mathrm{supp}\chi \subseteq C$, and define $h_p{\mathrm{\mid }}_C:=\frac{\chi f}{f(p)},h_p:=0\text{ outside }C\mathrm{.}$ On $C$ use the displayed product and define $h_p=0$ outside $C$. Since $\mathrm{supp}\chi$ is compactly contained in $C$, this extension is smooth across the boundary of $C$; thus $h_p$ is a smooth function with compact support in $C\subseteq I^{+}(S_a)=\{t>a\}$ and $h_p\ge 0$; this is (a). Since $p\in K\subseteq U_0$, near $p$ one has $\chi =1$ and $h_p=f\mathrm{/}f(p)$, so $h_p(p)=1$ and, by continuity, $\{h_p>1\mathrm{/}2\}$ contains a neighborhood of $p$: this is (b). For (c): let $q^{\mathrm{\prime }}\in \{h_p>0\}\cap J^{-}(S_b)$. Since $h_p>0$ forces $\chi >0$ and $f>0$, we have $q^{\mathrm{\prime }}\in \{f>0\}=I_C^{+}(p^{\mathrm{\prime }})\subseteq J^{+}(p^{\mathrm{\prime }})$, so $q^{\mathrm{\prime }}\in J^{+}(p^{\mathrm{\prime }})\cap J^{-}(S_b)=K\subseteq U_0$; on the neighborhood $U_0$ of $K$ the cutoff equals one, so on a neighborhood of $q^{\mathrm{\prime }}$ contained in $U_0$ one has $h_p=f\mathrm{/}f(p)$ and $\mathrm{\nabla }{h}_p=\mathrm{\nabla }f\mathrm{/}f(p)$, which is past timelike by step 2.1. At a point of the future null boundary or its vertex, all derivatives of f vanish; outside the future cone f is locally zero. On the cutoff plateau containing K, these facts imply that a nonzero gradient at a point of $J^{-}(S_b)$ forces f>0. This proves the nonzero-gradient assertion in (d). Statement (c) follows above; the remaining part of (d) is the identity $\{h_p>0\}\cap J^{-}(S_b)=\{f>0\}\cap J^{-}(S_b)$ established in the same computation, together with $\chi =1$ on $U_0\supseteq K\supseteq \{f>0\}\cap J^{-}(S_b)$.