Second derivative growth under an explicit horizon decay hypothesis

In a local sphere chart with angular metric components ${\gamma }_{AB}$, the inverse of the ingoing $(v,r)$ metric block is $\left(\begin{array}{cc}0& 1\\ 1& D\end{array}\right)$ and the full density is $w=r^2\sqrt{\det ({\gamma }_{AB})}$. For $J^a=g^{ab}{\mathrm{\partial }}_b\tilde{\psi }$, spherical symmetry gives $J^v={\psi }_r$, $J^r={\psi }_v+D{\psi }_r$, and $J^A=0$. This vector field is $C^2$, so [F3] applies. By [F1], [F4] and metric compatibility, ${\mathrm{\square }}_g\tilde{\psi }={\mathrm{\nabla }}_a{J}^a$. Applying the density formula and cancelling the angular density, which is independent of $v,r$, yields $0={\mathrm{\square }}_g\tilde{\psi }=\frac{1}{r^2}[{\mathrm{\partial }}_v(r^2{\psi }_r)+{\mathrm{\partial }}_r\text{\hspace{0.17em}⁣}(r^2({\psi }_v+D{\psi }_r))]\mathrm{.}$ Expanding and using equality of the continuous mixed partials gives $2r^2{\psi }_{vr}+2r{\psi }_v+{\mathrm{\partial }}_r(r^{2}D{\psi }_r)=0$. With $\varphi =r\psi$, the identities $2r^2{\psi }_{vr}+2r{\psi }_v=2r{\varphi }_{vr}$ and $r^{2}D{\psi }_r=D(r{\varphi }_r-\varphi )$ reduce this to $2{\varphi }_{vr}+D{\varphi }_{rr}+D^{\mathrm{\prime }}{\varphi }_r-\frac{D^{\mathrm{\prime }}}{r}\varphi =0.$

Let $f(v):=\varphi (v,m)$, which is continuous and tends to zero by [A2], and define $F(v):={\int }_{v_0}^{v}f(s)ds$ for any fixed $v_0\in \mathbb{R}$. By [F5], $F$ exists and $F^{\mathrm{\prime }}=f$. Given $\epsilon >0$, choose $S\ge v_0$ so that $\mathrm{\mid }f(s)\mathrm{\mid }<\epsilon$ for every $s\ge S$. For each $v>\max \{S,0\}$, the mean value theorem [F6] on $[S,v]$ gives a $c\in (S,v)$ with $F(v)-F(S)=f(c)(v-S)$; hence $\mid \frac{F(v)}{v}\mid \le \frac{\mathrm{\mid }F(S)\mathrm{\mid }}{v}+\epsilon \frac{v-S}{v}\mathrm{.}$ Taking $v\to +\mathrm{\infty }$ gives $\mathrm{lim\hspace{0.17em}sup}\mathrm{\mid }F(v)\mathrm{/}v\mathrm{\mid }\le \epsilon$. Since $\epsilon$ was arbitrary, $F(v)\mathrm{/}v\to 0$.

Differentiate the equation in step 1.1 once in $r$, which is permitted because $\varphi$ is $C^3$. The result is $2{\varphi }_{vrr}+D{\varphi }_{rrr}+2D^{\mathrm{\prime }}{\varphi }_{rr}+(D^{\mathrm{\prime }\mathrm{\prime }}-\frac{D^{\mathrm{\prime }}}{r}){\varphi }_r-(\frac{D^{\mathrm{\prime }\mathrm{\prime }}}{r}-\frac{D^{\mathrm{\prime }}}{r^2})\varphi =0.$ Direct differentiation gives $D(m)=D^{\mathrm{\prime }}(m)=0$ and $D^{\mathrm{\prime }\mathrm{\prime }}(m)=2\mathrm{/}{m}^2$. Evaluating at $r=m$ and using [F2] therefore yields the exact horizon transport identity ${\mathrm{\partial }}_v{\varphi }_{rr}(v,m)=-\frac{H}{m^2}+\frac{\varphi (v,m)}{m^3}\mathrm{.}$

Define $P(v):=-H(v-v_0)\mathrm{/}{m}^2+F(v)\mathrm{/}{m}^3$ and $G(v):={\varphi }_{rr}(v,m)$. By [F5], $F^{\mathrm{\prime }}(v)=\varphi (v,m)$, so $P^{\mathrm{\prime }}(v)=-H\mathrm{/}{m}^2+\varphi (v,m)\mathrm{/}{m}^3=G^{\mathrm{\prime }}(v)$ by step 2.1. Both $P$ and $G$ are continuous on $[v_0,\mathrm{\infty })$ and differentiable on its interior; [F7] makes $G-P$ constant there. Since $P(v_0)=0$, this gives ${\varphi }_{rr}(v,m)-{\varphi }_{rr}(v_0,m)=-\frac{H}{m^2}(v-v_0)+\frac{F(v)}{m^3}\mathrm{.}$ Divide by $v$ and use step 1.2; the fixed terms ${\varphi }_{rr}(v_0,m)\mathrm{/}v$ and $H{v}_0\mathrm{/}(m^{2}v)$ tend to zero, while $F(v)\mathrm{/}(m^{3}v)\to 0$. Therefore ${\varphi }_{rr}(v,m)\mathrm{/}v\to -H\mathrm{/}{m}^2$. Since $H\mathrm{\ne }0$, the limit is nonzero and its absolute value is $\mathrm{\mid }H\mathrm{\mid }\mathrm{/}{m}^2$, proving the stated linear growth.