text:
\section{Preliminaries on the analysis of the solution}\label{cans}

We now take the initial data on the $v$ axis to be the data on the event horizon of a subextremal Reissner-Nordstr\"{o}m
solution with mass $M>0$. So, we
choose initial data as in~\eqref{iu}$-$\eqref{iv} with $\zeta_0(0)=0$.
Moreover, we assume $\zeta_0$ to be continuous.
Since in this case the function $\varpi_0$ is constant equal to $M$, we also denote $M$ by $\varpi_0$.
In particular, when $\Lambda<0$, which corresponds to the Reissner-Nordstr\"{o}m anti-de Sitter solution, and when $\Lambda=0$, which corresponds to the Reissner-Nordstr\"{o}m solution, we assume that
$$
r\mapsto (1-\mu)(r,\varpi_0)=1-\frac{2\varpi_0}{r}+\frac{e^2}{r^2}-\frac\Lambda 3r^2
$$
has two zeros $r_-(\varpi_0)=r_-<r_+=r_+(\varpi_0)$. When $\Lambda>0$, which corresponds to the Reissner-Nordstr\"{o}m de Sitter solution, we assume that
$r\mapsto (1-\mu)(r,\varpi_0)$ has three zeros $r_-(\varpi_0)=r_-<r_+=r_+(\varpi_0)<r_c=r_c(\varpi_0)$.

\begin{center}
\begin{psfrags}
\psfrag{a}{{\tiny $\Lambda=0$}}
\psfrag{b}{{\tiny $\Lambda<0$}}
\psfrag{d}{{\tiny $\Lambda>0$}}
\psfrag{z}{{\tiny $r_0$}}
\psfrag{y}{{\tiny $\ $}}
\psfrag{r}{{\tiny $r$}}
\psfrag{n}{{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$1-\mu(r,\varpi_0)$}}
\psfrag{p}{{\tiny $r_+$}}
\psfrag{m}{{\tiny \!\!\!\!\!\!$r_-$}}
\psfrag{c}{{\tiny $r_c$}}
\includegraphics[scale=.6]{fig6.eps}\\
\includegraphics[scale=.6]{fig8.eps}\ \ \ \
\includegraphics[scale=.6]{fig10.eps}
\end{psfrags}
\end{center}

We define $\eta$ to be the function
$$
\eta=\vlinha.
$$
The functions $(r,\varpi)\mapsto\eta(r,\varpi)$ and $(r,\varpi)\mapsto (1-\mu)(r,\varpi)$
are related by
\begin{equation}\label{eta}
\eta=-\,\frac{r^2}{2}\partial_r(1-\mu).
\end{equation}
We define the function $\eta_0:\R^+\to\R$ by
$$
\eta_0(r)=\vlinhazz.
$$
We will repeatedly use the fact that $\eta(r,\varpi)\leq\eta_0(r)$ (see Lemma~\ref{sign}).
If $\Lambda\leq 0$, then $\eta_0'<0$. So $\eta_0$ is strictly decreasing and has precisely one zero.
The zero is located between $r_-$ and $r_+$. If $\Lambda>0$, then $\eta_0''$ is positive, so $\eta_0$ is strictly convex
and has precisely two zeros: one zero is located between $r_-$ and $r_+$ and the other zero is located between $r_+$ and $r_c$.
We denote by $r_0$ the zero of $\eta_0$ between $r_-$ and $r_+$ in both cases.

\begin{center}
\begin{psfrags}
\psfrag{a}{{\tiny $\Lambda=0$}}
\psfrag{b}{{\tiny $\Lambda<0$}}
\psfrag{d}{{\tiny $\Lambda>0$}}
\psfrag{e}{{\tiny \!$\eta_0$}}
\psfrag{z}{{\tiny \!\!\!\!\!$r_0$}}
\psfrag{y}{{\tiny $\ $}}
\psfrag{r}{{\tiny $r$}}
\psfrag{p}{{\tiny $r_+$}}
\psfrag{m}{{\tiny \!\!\!\!\!\!$r_-$}}
\psfrag{c}{{\tiny $r_c$}}
\includegraphics[scale=.6]{fig7.eps}\\
\includegraphics[scale=.6]{fig9.eps}\ \ \ \
\includegraphics[scale=.6]{fig11.eps}
\end{psfrags}
\end{center}

According to~\eqref{lambda_u}, we have $\partial_u\lambda(0,0)=-\partial_r(1-\mu)(r_+,\varpi_0)<0$. Since $\lambda(0,0)=0$, we may choose $U$ small enough so that $\lambda(u,0)$ is negative for $u\in\,]0,U]$.
Again we denote by ${\cal P}$ the maximal past set where the solution of the characteristic initial value problem is defined.
In Part~1 we saw that
$\lambda$ is negative on ${\cal P}\setminus\{0\}\times[0,\infty[$, and so, as $\kappa$ is positive (from \eqref{kappa_u} and \eqref{iv}), then $1-\mu$ is negative on ${\cal P}\setminus\{0\}\times[0,\infty[$.

Using the above, we can thus particularize the result of Part 1 on signs and monotonicities to the case where the initial data is\/~\eqref{iu} and\/~\eqref{iv} as follows.
\begin{Lem}[Sign and monotonicity]\label{sign}
Suppose that $(r,\nu,\lambda,\omega,\theta,\zeta,\kappa)$ is the maximal solution of the characteristic initial value problem\/~\eqref{r_u}$-$\eqref{kappa_at_u}, with initial
conditions\/~\eqref{iu} and\/~\eqref{iv}.
Then:
\begin{itemize}
\item $\kappa$ is positive;
\item $\nu$ is negative;
\item $\lambda$ is negative on ${\cal P}\setminus\{0\}\times[0,\infty[$;
\item $1 - \mu$ is negative on ${\cal P}\setminus\{0\}\times[0,\infty[$;
\item $r$ is decreasing with both $u$ and $v$;
\item $\omega$ is nondecreasing with both $u$ and $v$.
\end{itemize}
\end{Lem}

Using~\eqref{lambda_u} and~\eqref{theta_u}, we obtain
 \begin{equation}\label{theta_lambda}
  \partial_u\frac\theta\lambda=
	-\,\frac\zeta r-\frac\theta\lambda\frac{\nu}{1-\mu}\partial_r(1-\mu),
 \end{equation}
and analogously, using~\eqref{nu_v} and~\eqref{zeta_v},
 \begin{equation}\label{zeta_nu}
  \partial_v\frac\zeta\nu=
	-\,\frac\theta r-\frac\zeta\nu\frac{\lambda}{1-\mu}\partial_r(1-\mu).
 \end{equation}

Given $0 < \ckr < r_+$, let us denote by $\cg_{\ckr}$ the level set of the radius function $$\cg_{\ckr}:=\{(u,v)\in{\cal P}:r(u,v)=\ckr\}.$$
If nonempty, $\cg_{\ckr}$ is a connected $C^1$ spacelike curve, since both $\nu$ and $\lambda$ are negative on ${\cal P}\setminus\{0\}\times[0,\infty[$.
Using the Implicit Function Theorem, the facts that $r(0,v)=r_+$,
$r(u,0)=r_+-u$, the signs of $\nu$ and $\lambda$, and the
breakdown criterion given in Theorem~\ref{bdown},
one can show that $\cg_{\ckr}$ can be parametrized by a $C^1$ function $$v\mapsto (\uckr(v),v),$$
whose domain is $[0,\infty[$ if $\ckr\geq r_+-U$, or an interval of the form $[\vckr(U),\infty[$, for some $\vckr(U)>0$, if $\ckr<r_+-U$\,. Alternatively, $\cg_{\ckr}$ can also be parametrized by a $C^1$ function
$$u\mapsto (u,\vckr(u)),$$ whose domain is always an interval of the form $\,]\uckr(\infty),\min\{r_+-\ckr,U\}]$, for some $\uckr(\infty) \geq 0$\,.
We prove below that if $\ckr>r_-$, then $\uckr(\infty)=0$.

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{a}{{\tiny $\vckr(U)$}}
\psfrag{y}{{\tiny $(\ug(v),\tilde v)$}}
\psfrag{d}{{\tiny \!\!$u$}}
\psfrag{u}{{\tiny $v$}}
\psfrag{x}{{\tiny $\ug(v)$}}
\psfrag{f}{{\tiny \!\!\!\!$U$}}
\psfrag{j}{{\tiny \!\!\!\!$\cg_{\ckr}$}}
\psfrag{h}{{\tiny \!\!$\cg_{\ckrp}$}}
\psfrag{v}{{\tiny $u$}}
\psfrag{b}{{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$\uckr(\infty)$}}
\psfrag{c}{{\tiny $v$}}
\psfrag{s}{{\tiny \ \ $(u,\vckr(u))$}}
\psfrag{t}{{\tiny \ \ $(\uckr(v),v)$}}
\includegraphics[scale=.8]{fig14b.eps}
\end{psfrags}
\end{turn}
\end{center}

To analyze the solution we partition the domain into four regions (see figure below). We start by choosing $\ckrm$ and $\ckrp$ such that
 $r_-<\ckrm<r_0<\ckrp<r_+$. In Section~\ref{red-shift} we treat the region $\ckrp\leq r\leq r_+$.
In Section~\ref{medium} we consider the region $\ckrm\leq r\leq\ckrp$. In Section~\ref{large} we treat the region where $(u,v)$ is such that
$$
\vgr(u)\leq v\leq (1+\beta)\,\vgr(u),
$$
with $\beta>0$ appropriately chosen (we will denote the curve $v=(1+\beta)\,\vgr(u)$ by $\gam$). Finally, in Section~\ref{huge} we consider the region where $(u,v)$ is such that
$$
v\geq (1+\beta)\,\vgr(u).
$$
The reader should regard $\ckrm$, $\ckrp$ and $\beta$ as fixed. Later, they will have to be carefully chosen for our arguments to go through.

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{u}{{\tiny $u$}}
\psfrag{v}{{\tiny $v$}}
\psfrag{x}{{\tiny $\!\!\!\cg_{\ckrp}$}}
\psfrag{y}{{\tiny $\!\!\cg_{\ckrm}$}}
\psfrag{z}{{\tiny $\gam$}}
\psfrag{e}{{\tiny $\!\!\!\!\!U$}}
\includegraphics[scale=.7]{fig22b.eps}
\end{psfrags}
\end{turn}
\end{center}

The crucial step consists in estimating the fields $\frac\theta\lambda$ and $\frac\zeta\nu$. Once this is done, the other estimates follow easily.
By integrating~\eqref{theta_lambda} and~\eqref{zeta_nu}, we obtain
\begin{eqnarray}
\frac\theta\lambda(u,v)&=&\frac\theta\lambda(\uckr(v),v)e^{-\int_{\uckr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\nonumber\\
&&-\int_{\uckr(v)}^{u}\frac\zeta r(\tilde u,v)e^{-\int_{\tilde u}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\bar{u},v)\,d\bar{u}}\,d\tilde u,\label{field_6}\\
\frac\zeta\nu(u,v)&=&\frac\zeta\nu(u,\vckr(u))e^{-\int_{\vckr(u)}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\nonumber\\
&&-\int_{\vckr(u)}^{v}\frac\theta r(u,\tilde v)e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\bar{v})\,d\bar{v}}\,d\tilde v.\label{field_7}
\end{eqnarray}

Formula~\eqref{field_6} is valid provided that $\uckr(v)$ is defined and $\uckr(v)\leq u$, since the domain ${\cal P}$ is a past set; it also holds if we replace $\uckr(v)$ by $0$.
Similarly, formula~\eqref{field_7} is valid provided that $\vckr(u)$ is defined and $\vckr(u)\leq v$; again it holds if we replace $\vckr(u)$ by $0$.

\section{The region $J^-(\cg_{\ckrp})$}\label{red-shift}

Recall that $r_0<\ckrp<r_+$.
In this section, we treat the region $\ckrp\leq r\leq r_+$, that is, $J^-(\cg_{\ckrp})$.\footnote{Throughout this paper we follow the usual notations for the causal structure of the quotient Lorentzian manifold with coordinates $(u,v)$ and time orientation such that $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$ are future-pointing.}
Our first goal is to estimate~\eqref{z_00} for $\frac\zeta\nu$.
This will allow us to obtain the lower bound~\eqref{k_min} for $\kappa$,
which will then be used to improve estimate~\eqref{z_00} to~\eqref{z_0_bis}.
Finally, we successively bound $\frac\theta\lambda$, $\theta$, $\varpi$, and use this to prove that the domain of
$\vgrp(\,\cdot\,)$ is $\,]0,\min\{r_+-\ckrp,U\}]$.

In this region, the solution with general $\zeta_0$ can then be considered as a small perturbation of the
Reissner-Nordstr\"{o}m solution~\eqref{primeira}$-$\eqref{ultima}: $\omega$ is close to $\omega_0$, $\kappa$ is close to $1$ and $\zeta, \theta$ are
close to $0$. Besides, the smaller $U$ is, the closer the approximation.

Substituting~\eqref{field_6} in~\eqref{field_7}  (with both $\uckr(v)$ and $\vckr(u)$ replaced by 0), we get
\begin{eqnarray}
\frac\zeta\nu(u,v)&=&
\frac\zeta\nu(u,0)e^{-\int_{0}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\label{double_zero}\\
&&+\int_{0}^{v}\frac\theta\lambda(0,\tilde v)e^{-\int_{0}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,\tilde v)\,d\tilde u}\times\nonumber\\
&&\qquad\qquad\times
\Bigl[\frac{(-\lambda)}{r}\Bigr](u,\tilde v)e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\bar{v})\,d\bar{v}}\,d\tilde v\nonumber\\
&&+\int_{0}^{v}\left(\int_{0}^{u}\Bigl[\frac\zeta\nu\frac{(-\nu)}r\Bigr](\tilde u,\tilde v)
e^{-\int_{\tilde u}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\bar{u},\tilde v)\,d\bar{u}}\,d\tilde u\right)\times\nonumber\\
&&\qquad\qquad\times\Bigl[\frac{(-\lambda)}{r}\Bigr](u,\tilde v)
e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\bar{v})\,d\bar{v}}\,d\tilde v.\nonumber
\end{eqnarray}
We make the change of coordinates
\begin{equation}
(u,v)\mapsto(r(u,v),v)\quad \Leftrightarrow\quad (r,v)\mapsto (\ug(v),v).
\label{coordinates}
\end{equation}
The coordinates $(r,v)$ are called Bondi coordinates.
We denote by $\hatz$ the function $\frac\zeta\nu$ written in these new coordinates, so that
$$
\frac\zeta\nu(u,v)=\hatz(r(u,v),v)
\quad \Leftrightarrow\quad
\hatz(r,v)=\frac\zeta\nu(\ug(v),v).
$$
The same notation will be used for other functions.
In the new coordinates, \eqref{double_zero} may be written
\begin{eqnarray}
\hatz(r,v)&=&
\frac\zeta\nu(\ug(v),0)e^{-\int_{0}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\tilde v)\,d\tilde v}\label{triple_zero}\\
&&+\int_{0}^{v}\widehat{\frac\theta\lambda}(r_+,\tilde v)
e^{\int_{r(\ug(v),\tilde v)}^{r_+}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\tilde{s},\tilde v)\,d\tilde{s}}\times\nonumber\\
&&\qquad\qquad\times
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\bar{v})\,d\bar{v}}\,d\tilde v\nonumber\\
&&+\int_{0}^{v}\left(\int_{r(\ug(v),\tilde v)}^{r_+}\frac{1}{\tilde s}\Bigl[\hatz\Bigr](\tilde s,\tilde v)
e^{\int_{r(\ug(v),\tilde v)}^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},\tilde v)\,d\bar{s}}\,d\tilde s\right)\times\nonumber\\
&&\qquad\qquad\times
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]
e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\bar{v})\,d\bar{v}}\,d\tilde v.\nonumber
\end{eqnarray}

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{a}{{\tiny $\tilde v$}}
\psfrag{y}{{\tiny $(\ug(v),\tilde v)$}}
\psfrag{u}{{\tiny $v$}}
\psfrag{z}{{\tiny \!\!\!\!$r_+-s_1$}}
\psfrag{x}{{\tiny $\ug(v)$}}
\psfrag{f}{{\tiny $U$}}
\psfrag{g}{{\tiny $\cg_{s_2}$}}
\psfrag{h}{{\tiny $\cg_{s_1}$}}
\psfrag{v}{{\tiny $u$}}
\psfrag{b}{{\tiny $v$}}
\psfrag{c}{{\tiny $0$}}
\psfrag{p}{{\tiny $(\ug(v),v)$}}
\includegraphics[scale=.8]{fig12b.eps}
\end{psfrags}
\end{turn}
\end{center}

We have $\theta(0,v)=0$ and, from~\eqref{theta_u}, $\partial_u\theta(0,v)=0$, whereas $\lambda(0,v)=0$ and, from~\eqref{lambda_u}, $\partial_u\lambda(0,v)<0$. Writing
\[
\frac{\theta}{\lambda} (u,v) = \frac{\int_0^u \partial_u\theta(\tilde{u},v) d\tilde{u}}{\int_0^u \partial_u\lambda(\tilde{u},v) d\tilde{u}},
\]
it is easy to show that the function $\frac\theta\lambda$ can be extended as a continuous function to $\{0\}\times[0,\infty[$, with $\frac\theta\lambda(0,v)=0$.
Substituting this into~\eqref{field_6} (again with $\uckr(v)$ replaced by 0) yields
\begin{eqnarray*}
\frac\theta\lambda(u,v)
&=&-\int_{0}^{u}\frac\zeta r(\tilde u,v)e^{-\int_{\tilde u}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\bar{u},v)\,d\bar{u}}\,d\tilde u.
\end{eqnarray*}
We can rewrite this in the new coordinates as
\begin{eqnarray}
\widehat{\frac\theta\lambda}(r,v)&=&\int_{r}^{r_+}\frac 1 {\tilde s}\Bigl[\hatz\Bigr](\tilde s,v)
e^{\int_r^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},v)\,d\bar{s}}\,d\tilde s.\label{t_z}
\end{eqnarray}

A key point is to bound the exponentials that appear in~\eqref{triple_zero} and~\eqref{t_z}.
As we go on, this will be done several times in different ways.

\begin{Lem}\label{run}
 Assume that there exists $\alpha \geq 0$ such that, for $0 \leq \tilde{v} \leq v$, the following bounds hold:
$$
e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\bar{v})\,d\bar{v}}\leq e^{-\alpha(v-\tilde v)}
$$
and
$$
e^{\int_{r(\ug(v),\tilde v)}^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},\tilde v)\,d\bar{s}}\leq 1.
$$
Then~\eqref{triple_zero} implies
\begin{equation}\label{z_0}
\Bigl|\hatz\Bigr|(r,v)\leq e^{\frac{(r_+-r)^2}{rr_+}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.
\end{equation}
\end{Lem}
\begin{proof}
Combining~\eqref{triple_zero} with $\widehat{\frac\theta\lambda}(r_+,v)\equiv 0$ and the bounds on the exponentials, we have
\begin{eqnarray}
\Bigl|\hatz\Bigr|(r,v)&\leq&|\zeta_0|(\ug(v))e^{-\alpha v}\label{four_zero}\\
&&+\int_{0}^{v}\int_{r(\ug(v),\tilde v)}^{r_+}\frac{1}{\tilde s}\Bigl[\hatz\Bigr](\tilde s,\tilde v)
\,d\tilde s\times\nonumber\\
&&\qquad\qquad\qquad\times
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]
e^{-\alpha(v-\tilde v)}\,d\tilde v.\nonumber
\end{eqnarray}
For $r\leq s< r_+$, define
$$
{\cal Z}_{(r,v)}^\alpha(s)=\left\{
\begin{array}{ll}
\max_{\tilde v\in[0,v]}\left\{e^{\alpha\tilde v}\Bigl|\hatz\Bigr|(s,\tilde v)\right\}&{\rm if}\ r_+-\ug(v)\leq s<r_+,\\
\max_{\tilde v\in[\vgs(\ug(v)),v]}\left\{e^{\alpha\tilde v}\Bigl|\hatz\Bigr|(s,\tilde v)\right\}&{\rm if}\ r\leq s\leq r_+-\ug(v).
\end{array}
\right.
$$
Here the maximum is taken over the projection of $J^-(\ug(v),v)\cap\cg_s$ on the $v$-axis (see the figure below).

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{a}{{\tiny $\vgsd(\ug(v))$}}
\psfrag{y}{{\tiny $(\ug(v),\vgsd(\ug(v)))$}}
\psfrag{u}{{\tiny $v$}}
\psfrag{z}{{\tiny $r_+-s_1$}}
\psfrag{x}{{\tiny $\ug(v)$}}
\psfrag{f}{{\tiny $U$}}
\psfrag{g}{{\tiny $\cg_{s_2}$}}
\psfrag{h}{{\tiny $\cg_{s_1}$}}
\psfrag{v}{{\tiny $u$}}
\psfrag{b}{{\tiny $v$}}
\psfrag{c}{{\tiny $0$}}
\psfrag{p}{{\tiny $(\ug(v),v)$}}
\includegraphics[scale=.8]{fig3b.eps}
\end{psfrags}
\end{turn}
\end{center}

Note that ${\cal Z}_{(r,v)}^\alpha(r)=e^{\alpha v}\bigl|\hatz\bigr|(r,v)$. Inequality~\eqref{four_zero} implies
\begin{eqnarray}
{\cal Z}_{(r,v)}^\alpha(r)&\leq& |\zeta_0|(\ug(v))\nonumber\\
&&+\int_{0}^{v}\int_{r(\ug(v),\tilde v)}^{r_+}{\cal Z}_{(r,v)}^\alpha(\tilde s)
\,d\tilde s
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{[r(\ug(v),\tilde v)]^2}\Bigr]
\,d\tilde v\nonumber\\
&\leq& \max_{s\in[r,r_+]}|\zeta_0|(\ugs(v))\nonumber\\
&&+\int_{r}^{r_+}{\cal Z}_{(r,v)}^\alpha(\tilde s)
\,d\tilde s\int_{0}^{v}
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{[r(\ug(v),\tilde v)]^2}\Bigr]
\,d\tilde v\nonumber\\
&\leq& \max_{u\in[0,\ug(v)]}|\zeta_0|(u)
+
\Bigl(\frac 1r-\frac 1{r_+-\ug(v)}\Bigr)
\int_{r}^{r_+}{\cal Z}_{(r,v)}^\alpha(\tilde s)
\,d\tilde s.\nonumber
\end{eqnarray}
We still consider $r\leq s <r_+$.
Let $\tilde v\in[0,v]$ if $r_+-\ug(v)\leq s<r_+$,
and $\tilde v\in [\vgs(\ug(v)),v]$ if $r\leq s\leq r_+-\ug(v)$.
In this way $(\ugs(\tilde v),\tilde v)\in J^-(\ug(v),v)$.
In the same way one can show that
$$
e^{\alpha\tilde v}\Bigl|\hatz\Bigr|(s,\tilde v)\leq \max_{u\in[0,\ugs(v)]}|\zeta_0|(u)
+
\Bigl(\frac 1s-\frac 1{r_+-\ugs(v)}\Bigr)
\int_{s}^{r_+}{\cal Z}_{(r,v)}^\alpha(\tilde s)
\,d\tilde s
$$
because $J^-(\ugs(\tilde v),\tilde v)\cap \cg_{\tilde s}\subset J^-(\ug(v),v)\cap \cg_{\tilde s}$, for $s\leq\tilde s<r_+$, and so
${\cal Z}_{(s,\tilde v)}^\alpha(\tilde s)\leq {\cal Z}_{(r,v)}^\alpha(\tilde s)$.
Since $ \ugs(v)\leq \ug(v)$ for $r\leq s<r_+$, we have
\begin{eqnarray*}
{\cal Z}_{(r,v)}^\alpha(s)
&\leq& \max_{u\in[0,\ug(v)]}|\zeta_0|(u)
+
\Bigl(\frac 1r-\frac 1{r_+}\Bigr)
\int_{s}^{r_+}{\cal Z}_{(r,v)}^\alpha(\tilde s)
\,d\tilde s.
\end{eqnarray*}
Using Gronwall's inequality, we get
$$
{\cal Z}_{(r,v)}^\alpha(r)\leq e^{\frac{(r_+-r)^2}{rr_+}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u).
$$
This establishes~\eqref{z_0}.
\end{proof}

\begin{Lem}\label{lso}
 Let $r_0\leq r<r_+$ and $v>0$. Then
\begin{equation}\label{z_00}
\Bigl|\hatz\Bigr|(r,v)\leq e^{\frac{(r_+-r)^2}{rr_+}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u).
\end{equation}
\end{Lem}
\begin{proof}
We bound the exponentials in~\eqref{triple_zero}.
From~\eqref{eta}, the definition of $\eta$ and $\varpi\geq\varpi_0$,
$$
-\partial_r(1-\mu)=\frac{2\eta}{r^2}\leq\frac{2\eta_0}{r^2}=-\partial_r(1-\mu)(r,\varpi_0)\leq 0.
$$
Therefore in the region $J^-(\cg_{r_0})$ the exponentials are bounded by 1.
Applying Lemma~\ref{run} with $\alpha=0$ we obtain~\eqref{z_0} with $\alpha=0$, which is precisely~\eqref{z_00}.
\end{proof}

According to~\eqref{z_00}, the function $\frac\zeta\nu$ is bounded in the region $J^-(\cg_{r_0})$, say by $\hat\delta$.
From~\eqref{kappa_u},
\begin{eqnarray}
\kappa(u,v)&=&e^{\int_0^u\bigl(\frac{\zeta^2}{r\nu}\bigr)(\tilde u,v)\,d\tilde u}\nonumber\\
&\geq&e^{\hat\delta^2\int_0^u(\frac{\nu}{r})(\tilde u,v)\,d\tilde u}\nonumber\\
&\geq&\left(\frac{r_0}{r_+}\right)^{\hat\delta^2}.\label{k_min}
\end{eqnarray}

We recall from Part 1 that equations (\ref{r_v}), (\ref{nu_v}), (\ref{omega_v}) and (\ref{kappa_at_u}) imply
\begin{equation}\label{Ray}
\partial_v\left(\frac{1-\mu}{\nu}\right)=-\,\frac{\theta^2}{\nu r\kappa},
\end{equation}
which is the Raychaudhuri equation in the $v$ direction. We also recall that the integrated form of~\eqref{omega_u} is
\begin{eqnarray}
\omega(u,v)=\omega_0(v)e^{-\int_0^u\bigl(\frac{\zeta^2}{r\nu}\bigr)(u',v)\,du'}\qquad\qquad\qquad\qquad\qquad\qquad\qquad&&\nonumber\\ \qquad\qquad\ \  +
\int_0^ue^{-\int_s^u\frac{\zeta^2}{r\nu}(u',v)\,du'}\left(\frac{1}{2}\left(1+\frac{e^2}{r^2}
-\frac{\Lambda}{3}r^2\right)\frac{\zeta^2}{\nu}\right)(s,v)\,ds.&&\label{omega_final}
\end{eqnarray}
These will be used in the proof of the following result.

\begin{Prop}\label{psd}
  Let $r_0<\ckrp\leq r<r_+$ and $v>0$. Then there exists $\alpha>0$ (given by~\eqref{alp_new} below) such that
\begin{equation}\label{z_0_bis}
\Bigl|\hatz\Bigr|(r,v)\leq e^{\frac{(r_+-r)^2}{rr_+}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v},
\end{equation}
\begin{equation}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)
\leq \hat C_r\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v},\label{t_0}
\end{equation}
\begin{equation}
|\hat\theta|(r,v)\leq C\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.\label{theta_f}
\end{equation}
For $(u,v)\in J^-(\cg_{\ckrp})$, and $U$ sufficiently small, we have
\begin{equation}\label{mass_2}
\varpi_0\leq\varpi(u,v)\leq \varpi_0+
C\left(\sup_{\tilde u\in[0,u]}|\zeta_0|(\tilde u)\right)^2.
\end{equation}
Moreover, the curve $\cg_{\ckrp}$ intersects every line of constant $u$ provided that $0 < u \leq \min\{r_+-\ckrp,U\}$. Therefore, $\ugrp(\infty)=0$.
\end{Prop}
\begin{proof}
In $J^-(\cg_{\ckrp})$, we have
$\partial_r(1-\mu)(r,\varpi_0)
\geq\min_{r\in[\ckrp,r_+]}\partial_r(1-\mu)(r,\varpi_0)>0$ and
\begin{eqnarray}
-\,\frac{\lambda}{1-\mu}\partial_r(1-\mu)&\leq&-\kappa\, \partial_r(1-\mu)(r,\varpi_0)\nonumber\\
&\leq &-\inf_{J^-(\cg_{\ckrp})}\kappa\times\partial_r(1-\mu)(r,\varpi_0)\nonumber\\
&\leq &
-\Bigl(\frac{\ckrp}{r_+}\Bigr)^{\hat\delta^2}\min_{r\in[\ckrp,r_+]}\partial_r(1-\mu)(r,\varpi_0)\nonumber\\
&=:&-\alpha<0,\label{alp_new}
\end{eqnarray}
where we have used~\eqref{k_min} (with $\ckrp$ instead of $r_0$).
Thus, we can improve the bounds on the exponentials in~\eqref{triple_zero} that involve integrals in $v$ as follows:
$$
e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\bar{v})\,d\bar{v}}\leq e^{-\alpha(v-\tilde v)}.
$$
Since
$$
\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\leq 0,
$$
as before, we have
$$
e^{\int_{r(\ug(v),\tilde v)}^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},\tilde v)\,d\bar{s}}\leq 1.
$$
We apply Lemma~\ref{run} again, this time with a positive $\alpha$, to get~\eqref{z_0_bis}.

Now we may use~\eqref{t_z} and~\eqref{z_0_bis} to obtain
\begin{eqnarray*}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)&\leq& e^{\frac{(r_+-r)^2}{rr_+}}\ln\Bigl(\frac{r_+}r\Bigr)\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}\nonumber\\
&=&\hat C_r\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.
\end{eqnarray*}

In order to bound $\varpi$ in $J^-(\cg_{\ckrp})$,
we note that
$$
-\int_0^{u}\left(\frac{\zeta^2}{r\nu}\right)(\tilde u,v)\,d\tilde u\leq C^2\left(\sup_{\tilde u\in[0,u]}|\zeta_0|(\tilde u)\right)^2\ln\left(\frac{r_+}{\ckrp}\right),
$$
$$
\left|1+\frac{e^2}{r^2}-\frac\Lambda 3r^2\right|\leq 1+\frac{e^2}{\ckrp^2}+\frac{|\Lambda|}{3}r_+^2
$$
and
$$
-\int_0^{u}\nu(\tilde u,v)\,d\tilde u=r_+-r(u,v)\leq r_+-\ckrp.
$$
From~\eqref{omega_final}, we conclude that
$$
\varpi(u,v)\leq \varpi_0e^{C\left(\sup_{\tilde u\in[0,u]}|\zeta_0|(\tilde u)\right)^2}+
C\left(\sup_{\tilde u\in[0,u]}|\zeta_0|(\tilde u)\right)^2.
$$
Inequality~\eqref{mass_2} follows from $e^x\leq 1+2x$, for small $x$, since
$\zeta_0$ is continuous and $\zeta_0(0)=0$.

Given that $\kappa\leq 1$, we have $(1-\mu)\leq\lambda$. Moreover, since $\varpi$ is bounded
in the region $J^-(\cg_{\ckrp})$, $1-\mu$ is bounded from below, and so $\lambda$ is also bounded from below.
Hence~\eqref{t_0} implies~\eqref{theta_f}.

Let $0<u\leq\min\{r_+-\ckrp,U\}$. We claim that
\begin{equation}\label{stop}
\sup\left\{v\in[0,\infty[:(u,v)\in J^-(\cg_{\ckrp})\right\}<\infty.
\end{equation}
To see this, first note that~\eqref{nu_v} shows that $v\mapsto\nu(u,v)$ is decreasing in $J^-(\cg_{\ckrp})$, as $\partial_r(1-\mu)\geq 0$ for $r_0 \leq r \leq r_+$ (recall that $\eta(r,\omega) \leq \eta_0(r)$).
Then~\eqref{Ray} shows $v\mapsto(1-\mu)(u,v)$ is also decreasing in $J^-(\cg_{\ckrp})$. Thus, as long as $v$
is such that $(u,v)\in J^-(\cg_{\ckrp})$, we have
$(1-\mu)(u,v)\leq (1-\mu)(u,0)<0$.
Combining the previous inequalities with~\eqref{k_min}, we get
$$
\lambda(u,v)\leq \left(\frac{\ckrp}{r_+}\right)^{\hat\delta^2}(1-\mu)(u,0)<0.
$$
Finally, if \eqref{stop} did not hold for a given $u$, we would have
\begin{eqnarray*}
r(u,v)&=&r(u,0)+\int_0^v\lambda(u,v')\,dv'\\ &\leq& r(u,0)+\left(\frac{\ckrp}{r_+}\right)^{\hat\delta^2}(1-\mu)(u,0)\,v\to-\infty,
\end{eqnarray*}
as $v \to \infty$, which is a contradiction. This establishes the claim.
\end{proof}

\section{The region $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$}\label{medium}

In this section, we treat the region $\ckrm\leq r\leq\ckrp$. Recall that we assume that $r_-<\ckrm<r_0<\ckrp<r_+$. By decreasing $\ckrm$, if necessary, we will also assume that
\begin{equation}\label{assume}
-(1-\mu)(\ckrm,\varpi_0) \leq - (1-\mu)(\ckrp,\varpi_0).
\end{equation}

In Subsection~\eqref{medium-1}, we obtain estimates~\eqref{z-0} and~\eqref{t-0-a} for $\frac\zeta\nu$ and
$\frac\theta\lambda$, which will allow us to obtain the lower bound~\eqref{k_min_2} for $\kappa$, the upper bound~\eqref{mass_2_bis} for $\varpi$,
and to prove that the domain of
$\vgr(\,\cdot\,)$ is $\,]0,\min\{r_+-\ckrm,U\}]$.
In Subsection~\eqref{nu-lambda}, we obtain upper and lower bounds for $\lambda$ and $\nu$, as well as more information about the region $\ckrm\leq r\leq\ckrp$.
In Subsection~\eqref{medium-3}, we use the results from the previous subsection
to improve the estimates on $\frac\zeta\nu$ and $\frac\theta\lambda$ to~\eqref{z_f} and~\eqref{t-f}.
We also obtain the bound~\eqref{theta_ff} for $\theta$.

As in the previous section, the solution with general $\zeta_0$ is qualitatively still a small perturbation of the Reissner-Nordstr\"{o}m solution~\eqref{primeira}$-$\eqref{ultima}: $\omega$, $\kappa$, $\zeta$ and $\theta$ remain close to $\omega_0$, $1$ and $0$, respectively. Moreover, $\lambda$ is bounded from below by a negative constant, and away from zero by a constant depending on $\ckrp$ and $\ckrm$, as is also the case in the Reissner-Nordstr\"{o}m solution (see equation~\eqref{primeira}). Likewise, $\nu$ has a similar behavior to its Reissner-Nordstr\"{o}m counterpart (see equation~\eqref{segunda}): when multiplied by $u$, $\nu$ behaves essentially like $\lambda$.

\subsection{First estimates}\label{medium-1}
By reducing $U>0$, if necessary, we can assume  $U\leq r_+-\ckrp$.
We turn our attention to the region $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$.
Substituting~\eqref{field_6} in~\eqref{field_7} with $\ckr=\ckrp$, we get
\begin{eqnarray}
\frac\zeta\nu(u,v)&=&
\frac\zeta\nu(u,\vgrp(u))e^{-\int_{\vgrp(u)}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\label{double}\\
&&+\int_{\vgrp(u)}^{v}\frac\theta\lambda(\ugrp(\tilde v),\tilde v)e^{-\int_{\ugrp(\tilde v)}^{u}
\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,\tilde v)\,d\tilde u}\times\nonumber\\
&&\qquad\qquad\times
\Bigl[\frac{(-\lambda)}{r}\Bigr](u,\tilde v)e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\bar{v})\,d\bar{v}}\,d\tilde v\nonumber\\
&&+\int_{\vgrp(u)}^{v}\left(\int_{\ugrp(\tilde v)}^{u}\Bigl[\frac\zeta\nu\frac{(-\nu)}r\Bigr](\tilde u,\tilde v)
e^{-\int_{\tilde u}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\bar{u},\tilde v)\,d\bar{u}}\,d\tilde u\right)\times\nonumber\\
&&\qquad\qquad\times\Bigl[\frac{(-\lambda)}{r}\Bigr](u,\tilde v)
e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\bar{v})\,d\bar{v}}\,d\tilde v.\nonumber
\end{eqnarray}
We make the change of coordinates~\eqref{coordinates}.
Then, \eqref{double} may be written
\begin{eqnarray}
\hatz(r,v)&=&
\hatz(\ckrp,\vgrp(\ug(v)))e^{-\int_{\vgrp(\ug(v))}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\tilde v)\,d\tilde v}\label{triple}\\
&&+\int_{\vgrp(\ug(v))}^{v}\widehat{\frac\theta\lambda}(\ckrp,\tilde v)
e^{\int_{r(\ug(v),\tilde v)}^{\ckrp}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\tilde{s},\tilde v)\,d\tilde{s}}\times\nonumber\\
&&\qquad\qquad\times
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\bar{v})\,d\bar{v}}\,d\tilde v\nonumber\\
&&+\int_{\vgrp(\ug(v))}^{v}\left(\int_{r(\ug(v),\tilde v)}^{\ckrp}\Bigl[\hatz\frac{1}{\tilde s}\Bigr](\tilde s,\tilde v)
e^{\int_{r(\ug(v),\tilde v)}^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},\tilde v)\,d\bar{s}}\,d\tilde s\right)\times\nonumber\\
&&\qquad\qquad\times
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]
e^{-\int_{\tilde v}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](\ug(v),\bar{v})\,d\bar{v}}\,d\tilde v.\nonumber
\end{eqnarray}
For $(r,v)$ such that $(\ug(v),v)\in J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$, $\vgrp(\ug(v))$ is well defined because $U\leq r_+-\ckrp$.

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{a}{{\tiny $\tilde v$}}
\psfrag{y}{{\tiny $(\ug(v),\tilde v)$}}
\psfrag{d}{{\tiny $\vgrp(\ug(v))$}}
\psfrag{u}{{\tiny $v$}}
\psfrag{z}{{\tiny \!\!\!\!$r_+-s_1$}}
\psfrag{x}{{\tiny $\ug(v)$}}
\psfrag{f}{{\tiny $U=r_+-\ckrp$}}
\psfrag{g}{{\tiny \!\!$\cg_{\ckr}$}}
\psfrag{j}{{\tiny \!\!$\cg_{s}$}}
\psfrag{h}{{\tiny \!\!$\cg_{\ckrp}$}}
\psfrag{v}{{\tiny $u$}}
\psfrag{b}{{\tiny $v$}}
\psfrag{c}{{\tiny $0$}}
\psfrag{p}{{\tiny $(\ug(v),v)$}}
\includegraphics[scale=.8]{fig13c.eps}
\end{psfrags}
\end{turn}
\end{center}

\begin{Lem}\label{sto}
Let $\ckrm\leq r\leq\ckrp$. Then
 \begin{equation}\label{z-0}
\Bigl|\hatz\Bigr|(r,v)\leq \tilde C \max_{u\in[0,\ug(v)]}|\zeta_0|(u),
\end{equation}
 \begin{equation}\label{t-0-a}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)\leq C \max_{u\in[0,\ug(v)]}|\zeta_0|(u).
\end{equation}
\end{Lem}
\begin{proof}
From~\eqref{assume} we have
\begin{eqnarray}
 (1-\mu)(r,\varpi)&\leq& (1-\mu)(r,\varpi_0)\nonumber\\
 &\leq& \max\left\{(1-\mu)(\ckrm,\varpi_0),(1-\mu)(\ckrp,\varpi_0)\right\}\label{umm}\\
 &=&(1-\mu)(\ckrm,\varpi_0)\nonumber
\end{eqnarray}
and
\begin{eqnarray*}
\frac{\partial_r(1-\mu)}{1-\mu}&=&\frac{2\eta/r^2}{-(1-\mu)}\ \leq\ \frac{2\eta_0/r^2}{-(1-\mu)}\ \leq\  \frac{2\eta_0(\ckrm)/\ckrm^2}{-(1-\mu)}\\
&\leq& \frac{2\eta_0(\ckrm)/\ckrm^2}{-(1-\mu)(r,\varpi_0)}\ \leq\  \frac{2\eta_0(\ckrm)/\ckrm^2}{-(1-\mu)(\ckrm,\varpi_0)}\ =:\ c_{\ckrm}.
\end{eqnarray*}
(For the second inequality, see the graph of $\eta_0$ in Section~\ref{cans}.)
Each of the five exponentials in~\eqref{triple} is bounded by
\begin{equation}\label{c}
e^{c_{\ckrm}(\ckrp-\ckrm)}=:C.
\end{equation}
Hence, for $(r,v)$ such that $(\ug(v),v)\in J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$, we have from~\eqref{triple}
\begin{eqnarray}
\Bigl|\hatz\Bigr|(r,v)&\leq&C\Bigl|\hatz\Bigr|(\ckrp,\vgrp(\ug(v)))\label{four}\\
&&+C^2\int_{\vgrp(\ug(v))}^{v}\Bigl|\widehat{\frac\theta\lambda}\Bigr|(\ckrp,\tilde v)
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]\,d\tilde v\nonumber\\
&&+C^2\int_{\vgrp(\ug(v))}^{v}\int_{r(\ug(v),\tilde v)}^{\ckrp}\Bigl|\hatz\Bigr|(\tilde s,\tilde v)
\,d\tilde s
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{[r(\ug(v),\tilde v)]^2}\Bigr]
\,d\tilde v.\nonumber
\end{eqnarray}
For $r\leq s\leq \ckrp$, define
\begin{equation}\label{z}
 {\cal Z}_{(r,v)}(s)=\max_{\tilde v\in[\vgs(\ug(v)),v]}\Bigl|\hatz\Bigr|(s,\tilde v)
\end{equation}
and
\begin{equation}\label{t}
{\cal T}_{(r,v)}(\ckrp)=\max_{\tilde v\in[\vgrp(\ug(v)),v]}\Bigl|\widehat{\frac\theta\lambda}\Bigr|(\ckrp,\tilde v).
\end{equation}

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{a}{{\tiny $\vgs(\ug(v))$}}
\psfrag{y}{{\tiny $(\ug(v),\vgs(\ug(v)))$}}
\psfrag{u}{{\tiny $v$}}
\psfrag{z}{{\tiny \!\!\!\!$r_+-s_1$}}
\psfrag{x}{{\tiny $\ug(v)$}}
\psfrag{f}{{\tiny $U=r_+-\ckrp$}}
\psfrag{g}{{\tiny \!\!$\cg_{\ckr}$}}
\psfrag{j}{{\tiny \!\!$\cg_{s}$}}
\psfrag{h}{{\tiny \!\!$\cg_{\ckrp}$}}
\psfrag{v}{{\tiny $u$}}
\psfrag{b}{{\tiny $v$}}
\psfrag{c}{{\tiny $0$}}
\psfrag{p}{{\tiny $(\ug(v),v)$}}
\includegraphics[scale=.8]{fig4c.eps}
\end{psfrags}
\end{turn}
\end{center}
\noindent Recall that $[\vgs(\ug(v)),v]$ is the projection of $J^-(\ug(v),v)\cap\cg_s$ on the $v$-axis.

Note that ${\cal Z}_{(r,v)}(r)=\bigl|\hatz\bigr|(r,v)$. Inequality~\eqref{four} implies
\begin{eqnarray}
{\cal Z}_{(r,v)}(r)&\leq& C{\cal Z}_{(r,v)}(\ckrp)\nonumber\\
&&+C^2\int_{\vgrp(\ug(v))}^{v}{\cal T}_{(r,v)}(\ckrp)
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{r(\ug(v),\tilde v)}\Bigr]\,d\tilde v\nonumber\\
&&+C^2\int_{\vgrp(\ug(v))}^{v}\int_{r(\ug(v),\tilde v)}^{\ckrp}{\cal Z}_{(r,v)}(\tilde s)
\,d\tilde s
\Bigl[\frac{(-\lambda)(\ug(v),\tilde v)}{[r(\ug(v),\tilde v)]^2}\Bigr]
\,d\tilde v\nonumber\\
&\leq& C{\cal Z}_{(r,v)}(\ckrp)
+C^2
\ln\Bigl(\frac{\ckrp}{r}\Bigr)
{\cal T}_{(r,v)}(\ckrp)
+C^2
\Bigl(\frac 1r-\frac 1\ckrp\Bigr)
\int_{r}^{\ckrp}{\cal Z}_{(r,v)}(\tilde s)
\,d\tilde s.\nonumber
\end{eqnarray}
Again consider $r\leq s\leq\ckrp$ and let $\tilde v\in [\vgs(\ug(v)),v]$, so that $(\ugs(\tilde v),\tilde v)\in J^-(\ug(v),v)\cap J^+(\cg_{\ckrp})$.
In the same way one can show that
$$
\Bigl|\hatz\Bigr|(s,\tilde v)\leq C{\cal Z}_{(r,v)}(\ckrp)
+C^2
\ln\Bigl(\frac{\ckrp}{s}\Bigr)
{\cal T}_{(r,v)}(\ckrp)
+C^2
\Bigl(\frac 1s-\frac 1\ckrp\Bigr)
\int_{s}^{\ckrp}{\cal Z}_{(r,v)}(\tilde s)
\,d\tilde s,
$$
because $J^-(\ugs(\tilde v),\tilde v)\cap \cg_{\ckrp}\subset J^-(\ug(v),v)\cap \cg_{\ckrp}$.
Therefore,
$$
{\cal Z}_{(r,v)}(s)\leq C{\cal Z}_{(r,v)}(\ckrp)
+C^2
\ln\Bigl(\frac{\ckrp}{r}\Bigr)
{\cal T}_{(r,v)}(\ckrp)
+C^2
\Bigl(\frac 1r-\frac 1\ckrp\Bigr)
\int_{s}^{\ckrp}{\cal Z}_{(r,v)}(\tilde s)
\,d\tilde s.
$$
Using Gronwall's inequality, we get
\begin{equation}\label{zf}
{\cal Z}_{(r,v)}(r)\leq C\Bigl[{\cal Z}_{(r,v)}(\ckrp)+C\ln\Bigl(\frac{\ckrp}{r}\Bigr){\cal T}_{(r,v)}(\ckrp)\Bigr]e^{\frac{C^2(\ckrp-r)^2}{r\ckrp}}.
\end{equation}

To bound ${\cal Z}_{(r,v)}$ and ${\cal T}_{(r,v)}$, it is convenient at this point to use~\eqref{z_00} and
\begin{eqnarray}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)&\leq
&\hat C_r \max_{u\in[0,\ug(v)]}|\zeta_0|(u)\label{t_00}
\end{eqnarray}
(valid for $\ckrp \leq r < r_+$), in spite of having the better estimates~\eqref{z_0_bis} and~\eqref{t_0}.
Indeed, if these better estimates are used, the improvement is just $e^{-\alpha \vgrp(\ug(v))}$ (that is, an exponential factor computed over $\cg_{\ckrp}$ for the same value of $u$); to turn this into an exponential decay in $v$ we must first obtain a more accurate control of the various quantities in the region $\ckrm\leq r\leq\ckrp$.
Applying first the definition~\eqref{z} and then~\eqref{z_00}, we have
\begin{eqnarray}
{\cal Z}_{(r,v)}(\ckrp)&=&\max_{\tilde v\in[\vgrp(\ug(v)),v]}\Bigl|\hatz\Bigr|(\ckrp,\tilde v)\nonumber\\
&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\max_{\tilde v\in[\vgrp(\ug(v)),v]}\ \ \max_{u\in[0,\ugrp(\tilde v)]}|\zeta_0|(u)\nonumber\\
&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u),\label{part-1a}
\end{eqnarray}
because $\ugrp(\tilde v)\leq \ug(v)$.
 Applying first the definition~\eqref{t} and then~\eqref{t_00}, we have
\begin{eqnarray}
{\cal T}_{(r,v)}(\ckrp)&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\ln\Bigl(\frac{r_+}{\ckrp}\Bigr)\max_{\tilde v\in[\vgrp(\ug(v)),v]}\ \
\max_{u\in[0,\ugrp(\tilde v)]}|\zeta_0|(u)\nonumber\\
&\leq&\hat C_{\ckrp}\max_{u\in[0,\ug(v)]}|\zeta_0|(u).\label{part-2a}
\end{eqnarray}
We use~\eqref{part-1a} and~\eqref{part-2a} in~\eqref{zf}. This yields~\eqref{z-0}.

Finally, writing~\eqref{field_6} in the $(r,v)$ coordinates (with $\ckr = \ckrp$) gives
\begin{eqnarray*}
\widehat{\frac\theta\lambda}(r,v)&=&\widehat{\frac\theta\lambda}(\ckrp,v)
e^{\int_r^{\ckrp}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\tilde{s},v)\,d\tilde{s}}\\
&&+\int_{r}^{\ckrp}\Bigl[\hatz\frac 1 {\tilde s}\Bigr](\tilde s,v)
e^{\int_r^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},v)\,d\bar{s}}\,d\tilde s.
\end{eqnarray*}
The exponentials are bounded by the constant $C$ in~\eqref{c}. We use the estimates~\eqref{t_00} and~\eqref{z-0} to obtain
\begin{eqnarray*}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)&\leq&C\Bigl|\widehat{\frac\theta\lambda}\Bigr|(\ckrp,v)
+C\int_{r}^{\ckrp}\Bigl[\Bigl|\hatz\Bigr|\frac 1 {\tilde s}\Bigr](\tilde s,v)\,d\tilde s\\
&\leq&C\hat C_{\ckrp}\max_{u\in[0,u_{\ckrp}(v)]}|\zeta_0|(u)\\ &&+C
\tilde C\ln\Bigl(\frac{\ckrp}{r}\Bigr) \max_{u\in[0,\ug(v)]}|\zeta_0|(u)\\
&=&C\max_{u\in[0,\ug(v)]}|\zeta_0|(u),
\end{eqnarray*}
which is~\eqref{t-0-a}.
\end{proof}

According to~\eqref{z_00} and~\eqref{z-0}, the function $\frac\zeta\nu$ is bounded in the region $J^-(\cg_{\ckrm})$, let us say by $\hat\delta$.
Arguing as in the deduction of~\eqref{k_min}, we obtain
\begin{equation}
\kappa(u,v)
\geq\left(\frac{\ckrm}{r_+}\right)^{\hat\delta^2}.\label{k_min_2}
\end{equation}

\begin{Lem}
For $(u,v)\in J^-(\cg_{\ckrm})$, and $U \leq r_+ - \ckrp$ sufficiently small, we have
\begin{equation}\label{mass_2_bis}
\varpi_0\leq\varpi(u,v)\leq \varpi_0+
C\left(\sup_{\tilde u\in[0,u]}|\zeta_0|(\tilde u)\right)^2.
\end{equation}
The curve $\cg_{\ckrm}$ intersects every line of constant $u$. Therefore, $\ugr(\infty)=0$.
\end{Lem}
\begin{proof}
The proof of~\eqref{mass_2_bis} is identical to the proof of~\eqref{mass_2}.

Because $\varpi$ is bounded, the function $1-\mu$ is bounded below in $J^-(\cg_{\ckrm})$.
Also, by~\eqref{umm}, the function $1-\mu$ is bounded above in $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$ by $(1-\mu)(\ckrm,\varpi_0)$.

 We claim that for each $0<u\leq U$
\begin{equation}\label{stop2}
\sup\left\{v\in\left[0,\infty\right[:(u,v)\in J^-(\cg_{\ckrm})\right\}<\infty.
\end{equation}
The proof is similar to the proof of~\eqref{stop}: since $\kappa$ is bounded below by a positive constant
and $1-\mu$ is bounded above by a negative constant, $\lambda$ is bounded above by a negative constant in $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$, say $-c_\lambda$.
Then, as long as $(u,v)$ belongs to $J^-(\cg_{\ckrm})$, we have the upper bound for $r(u,v)$ given by
$$
r(u,v)\leq r_+-u-c_\lambda v,
$$
since $0<u\leq U\leq r_+-\ckrp$). Finally, if \eqref{stop2} did not hold for a given $u$, we would have $r(u,v)\to -\infty$
as $v \to \infty$, which is a contradiction. This proves the claim.
\end{proof}

\subsection{Estimates for $\nu$, $\lambda$ and the region $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$}\label{nu-lambda}

\begin{Lem}
In the region $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$,
we have the following estimates from above and from below on $\lambda$ and $\nu$:
\begin{equation}\label{similar1}
-\tilde C\leq\lambda\leq -\tilde c
\end{equation}
and
\begin{equation}\label{similar2}
-\frac{\tilde C}{u}\leq\nu\leq-\frac{\tilde c}{u},
\end{equation}
where the constants $\tilde c$ and $\tilde C$ depend on $\ckrp$ and $\ckrm$.

Furthermore, if $0 < \delta < r_+-r_0$ and $(u,v)\in\cg_{r_+ - \delta}$ then
\begin{equation}\label{bound_lambda}
-C\,\partial_r(1-\mu)(r_+,\varpi_0)\,\delta\leq\lambda(u,v)\leq -c\,\partial_r(1-\mu)(r_+,\varpi_0)\,\delta
\end{equation}
and
\begin{equation}\label{bound_nu}
-C\frac{\delta}{u}\leq\nu(u,v)\leq-c\frac{\delta}{u},
\end{equation}
where the constants $0<c<1<C$ may be chosen independently of $\delta$.
Given $\eps>0$ then $1-\eps<c<1$ and $1<C<1+\eps$ for small enough $\delta$.
\end{Lem}
\begin{proof}
From~\eqref{kappa_u} we obtain
(the Raychaudhuri equation)
$$
\partial_u\left(\frac{\lambda}{1-\mu}\right)=\frac{\lambda}{1-\mu}\left(\frac{\zeta}{\nu}\right)^2\frac{\nu}{r},
$$
and from~\eqref{Ray} we obtain (the Raychaudhuri equation)
\begin{equation}\label{ray_v_bis}
\partial_v\left(\frac{\nu}{1-\mu}\right)=\frac{\nu}{1-\mu}\left(\frac{\theta}{\lambda}\right)^2\frac{\lambda}{r}.
\end{equation}
Let $\hat\delta>0$. By decreasing $U$, if necessary, using~\eqref{z_0_bis}, \eqref{t_0}, \eqref{z-0} and~\eqref{t-0-a}, we have
$\bigl|\frac{\theta}{\lambda}\bigr|<\hat\delta$ and
$\bigl|\frac{\zeta}{\nu}\bigr|<\hat\delta$ in $J^-(\cg_{\ckrm})$.
Since
$\int_0^u\frac\nu r(\tilde u,v)\,d\tilde{u}=\ln\Bigl(\frac{r(u,v)}{r_+}\Bigr)$,
$\int_0^v\frac\lambda r(u,\tilde v)\,d\tilde{v}=\ln\Bigl(\frac{r(u,v)}{r_+-u}\Bigr)$,
$\frac{\ckrm}{r_+}\leq\frac{r(u,v)}{r_+}\leq 1$ and $\frac{\ckrm}{r_+}\leq\frac{r(u,v)}{r_+-u}\leq 1$,
for $(u,v)\in J^-(\cg_{\ckrm})$ we have
\begin{equation}\label{close_one}
\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}\leq e^{\int_0^u\bigl((\frac{\zeta}{\nu})^2\frac{\nu}{r}\bigr)(\tilde u,v)\,d\tilde u}\leq 1,
\end{equation}
\begin{equation}\label{perto_um}
\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}\leq e^{\int_0^v\bigl((\frac{\theta}{\lambda})^2\frac{\lambda}{r}\bigr)(u,\tilde v)\,d\tilde v}\leq 1.
\end{equation}
So,
integrating the Raychaudhuri equations, we get
\begin{equation}\label{lambda_mu}
\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}=
\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}\frac{\lambda}{1-\mu}(0,v)\leq\frac{\lambda}{1-\mu}(u,v)\leq\frac{\lambda}{1-\mu}(0,v)=1
\end{equation}
(as $\kappa(0,v)=1$), and
\begin{equation}\label{nu_mu}
-\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}\frac{1}{1-\mu}(u,0)
\leq
\frac{\nu}{1-\mu}(u,v)\leq\frac{\nu}{1-\mu}(u,0)=-\,\frac{1}{1-\mu}(u,0).
\end{equation}
To bound $(1-\mu)(u,0)$, using~\eqref{lambda_u} and~\eqref{kappa_u}, we compute
\begin{equation}\label{mu_u}
 \partial_u(1-\mu)=\partial_u\Bigl(\frac\lambda\kappa\Bigr)
 =\nu\partial_r(1-\mu)-(1-\mu)\frac{\nu}{r}\Bigl(\frac\zeta\nu\Bigr)^2.
\end{equation}
At the point $(u,v)=(0,0)$ this yields
$$
\partial_u(1-\mu)(0,0)=-\partial_r(1-\mu)(r_+,\varpi_0).
$$
Fix $0<\eps<1$. Since the function
$u\mapsto(1-\mu)(u,0)$ is $C^1$, by decreasing $U$ if necessary, we have
$$
-\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1-\eps}<\partial_u(1-\mu)(u,0)<-\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1+\eps}
$$
for $0\leq u\leq U$, and so
$$
-\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1-\eps}\,u<(1-\mu)(u,0)<-\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1+\eps}\,u.
$$
Using these inequalities in~\eqref{nu_mu} immediately gives
\begin{equation}\label{nu_mu_bis}
\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}
\frac{1-\eps}{\partial_r(1-\mu)(r_+,\varpi_0)\,u}\leq\frac{\nu}{1-\mu}(u,v)\leq\frac{1+\eps}{\partial_r(1-\mu)(r_+,\varpi_0)\,u}.
\end{equation}

To obtain bounds on $\lambda$ and $\nu$ from~\eqref{lambda_mu} and~\eqref{nu_mu_bis},
recall that, in accordance with~\eqref{umm}, in the region $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$ the function $1-\mu$ is bounded
above by a negative constant.
On the other hand, the bounds we obtained earlier on $\varpi$ in
$J^-(\cg_{\ckrm})$ imply that $1-\mu$ is bounded below in $J^-(\cg_{\ckrm})$.
In summary, there exist $\overline{c}$ and $\overline{C}$ such that
$$
-\overline{C}\leq 1-\mu\leq -\overline{c}.
$$
Therefore, from~\eqref{lambda_mu} and~\eqref{nu_mu_bis}, in the region $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$,
we get~\eqref{similar1} and~\eqref{similar2}:
$$
-\overline{C}\leq\lambda\leq-\overline{c}\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2},
$$
$$
-\overline{C}
\frac{1+\eps}{\partial_r(1-\mu)(r_+,\varpi_0)}\frac 1u
\leq\nu\leq
-\overline{c}\Bigl(\frac{\ckrm}{r_+}\Bigr)^{\hat\delta^2}\frac{1-\eps}{\partial_r(1-\mu)(r_+,\varpi_0)}
\frac1 u.
$$
By decreasing $\overline c$ and increasing $\overline C$, if necessary, we can guarantee~\eqref{similar1} and~\eqref{similar2}
hold, without having to further decrease $U$.

Now suppose that $(u,v)\in\cg_{r_+ - \delta}$. Then
\begin{eqnarray}
(1-\mu)(u,v)&=&(1-\mu)(r_+ - \delta,\varpi)\nonumber\\
&\leq&(1-\mu)(r_+ - \delta,\varpi_0)\nonumber\\
&\leq&-\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1+\eps}\,\delta,\label{negative}
\end{eqnarray}
where $\eps$ is any fixed positive number, provided that $\delta$ is sufficiently small. If $\delta$ is not small, then~\eqref{negative} also holds
but with $1+\eps$ replaced by a larger constant.

Using again~\eqref{mu_u},
$$
(1-\mu)(u,v)=-\int_0^ue^{-\int_{\tilde u}^u\bigl(\frac{\nu}{r}\bigl(\frac\zeta\nu\bigr)^2\bigr)(\bar u,v)\,d\bar u}
\Bigl(\frac{2\nu}{r^2}\eta\Bigr)(\tilde u,v)\,d\tilde u.
$$
We take into account that
$$
e^{-\int_{\tilde u}^u\bigl(\frac{\nu}{r}\bigl(\frac\zeta\nu\bigr)^2\bigr)(\bar u,v)\,d\bar u}\leq\Bigl(\frac{r_+}{r_+-\delta}\Bigr)^{\hat{\delta}^2}
$$
and
\begin{eqnarray*}
-\,\frac{2\nu}{r^2}\eta&\geq& 2\nu\left(-\,\frac{e^2}{r^3}-\frac\Lambda 3r+\frac{\varpi_0}{r^2}\right)+2\nu\frac{\tilde{\delta}}{r^2}\\
&=&\nu\partial_r(1-\mu)(r,\varpi_0)+2\nu\frac{\tilde{\delta}}{r^2}
\end{eqnarray*}
provided $U$ is chosen small enough so that $\varpi\leq\varpi_0+\tilde{\delta}$ in $J^-(\cg_{r_+-\delta})$. We get
\begin{eqnarray}
(1-\mu)(u,v)
&\geq&\Bigl(\frac{r_+}{r_+-\delta}\Bigr)^{\hat{\delta}^2}
\left((1-\mu)(r_+-\delta,\varpi_0)-\,\frac{2\tilde{\delta}\delta}{r_+(r_+-\delta)}\right)\nonumber\\
&\geq&-\,\Bigl(\frac{r_+}{r_+-\delta}\Bigr)^{\hat{\delta}^2}
\left(\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1-\eps}+\frac{4\tilde{\delta}}{r_+^2}\right)\,\delta,\label{negative_below}
\end{eqnarray}
where $0<\eps<1$,
provided $\delta$ is sufficiently small.
We notice that in
the case under consideration the integration is done between $r_+$ and $r_+-\delta$ and so
the left hand sides of~\eqref{close_one} and~\eqref{perto_um}
can be improved to
$\bigl(\frac{r_+-\delta}{r_+}\bigr)^{\hat\delta^2}$.
Estimates
\eqref{lambda_mu}, \eqref{negative} and~\eqref{negative_below} yield, for $\hat\delta\leq 1$,
\begin{eqnarray}
-\,\Bigl(\frac{r_+}{r_+-\delta}\Bigr)
\left(\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1-\eps}+\frac{4\tilde{\delta}}{r_+^2}\right)\,\delta\leq\lambda\qquad\qquad\qquad\qquad&&\label{bound_lambda2}\\
\leq
-\Bigl(\frac{r_+-\delta}{r_+}\Bigr)\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{1+\eps}\,\delta,&&\nonumber
\end{eqnarray}
 whereas estimates
\eqref{nu_mu_bis}, \eqref{negative} and~\eqref{negative_below} yield, again for $\hat\delta\leq 1$,
\begin{eqnarray}
-\,\Bigl(\frac{r_+}{r_+-\delta}\Bigr)
\left(\frac{1+\eps}{1-\eps}+\frac{4\tilde{\delta}(1+\eps)}{r_+^2\partial_r(1-\mu)(r_+,\varpi_0)}\right)\,\frac\delta u\leq\nu
\qquad\qquad\ \ &&\label{bound_nu2}\\
\leq
-\Bigl(\frac{r_+-\delta}{r_+}\Bigr)\frac{1-\eps}{1+\eps}\,\frac\delta u.&&\nonumber
\end{eqnarray}
Estimates~\eqref{bound_lambda}
and~\eqref{bound_nu} are established. Note that $u \leq \delta$
when $(u,v)\in\cg_{r_+-\delta}$. Since
\[
c = c(\delta, \eps, \tilde \delta) = c(\delta, \eps(U,\delta), \tilde \delta(U)) = c(\delta, \eps(U(\delta),\delta), \tilde \delta(U(\delta))),
\]
and analogously for $C$, we see that $c$ and $C$ can be chosen arbitrarily close to one, provided that $\delta$
is sufficiently small.
\end{proof}

\begin{Lem} Let $\eps>0$. If $\delta$ is sufficiently small, then
for any point $(u,v)\in \cg_{r_+-\delta}$ we have
\begin{equation}\label{region_delta}
\delta\, e^{-[\partial_r(1-\mu)(r_+,\varpi_0)+\eps]\,v}
\leq u \leq
\delta\, e^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\eps]\,v}.
\end{equation}
For any point $(u,v)\in J^-(\cg_{\ckrm})\cap J^+(\cg_{r_+-\delta})$ we have
\begin{equation}\label{region}
\delta\, e^{-[\partial_r(1-\mu)(r_+,\varpi_0)+\eps]\,v}
\leq u\leq\  \delta\, e^{\frac{r_+}{\tilde c}}e^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\eps]\,v}.
\end{equation}
\end{Lem}
\begin{proof}
Obviously, we have
$$
r(\udz(v),v)=r_+-\delta.
$$
Since $r$ is $C^1$ and $\nu$ does not vanish, $v\mapsto\udz(v)$ is $C^1$.
Differentiating both sides of the last equality with respect to $v$ we obtain
$$
\udp(v)=-\,\frac{\lambda(\udz(v),v)}{\nu(\udz(v),v)}.
$$
Using~\eqref{bound_lambda} and~\eqref{bound_nu}, we have
$$
-\,\frac{C}{c}\partial_r(1-\mu)(r_+,\varpi_0)\udz(v)\leq
\udp(v)\leq-\,\frac{c}{C}\partial_r(1-\mu)(r_+,\varpi_0)\udz(v).
$$
Integrating the last inequalities between $0$ and $v$, as $\udz(0)=\delta$, we have
$$\delta e^{-\,\frac{C}{c}\partial_r(1-\mu)(r_+,\varpi_0)v}\leq
\udz(v)\leq \delta e^{-\,\frac{c}{C}\partial_r(1-\mu)(r_+,\varpi_0)v}.
$$
This proves~\eqref{region_delta}.

Let $(u,v)\in J^-(\cg_{\ckrm})\cap J^+(\cg_{r_+ - \delta})$.
Integrating~\eqref{similar2} between $\udz(v)$ and $u$, we get
$$
1\leq\frac{u}{\udz(v)}\leq e^{\frac{r_+}{\tilde c}}.
$$
Combining $\vdz(u)\leq v$ with the first inequality in~\eqref{region_delta} applied at the point $(u,\vdz(u))$,
\begin{eqnarray*}
u&\geq& \delta\, e^{-[\partial_r(1-\mu)(r_+,\varpi_0)+\eps]\,\vdz(u)}\\
&\geq& \delta\, e^{-[\partial_r(1-\mu)(r_+,\varpi_0)+\eps]\,v},
\end{eqnarray*}
and combining $u\leq e^{\frac{r_+}{\tilde c}}\udz(v)$ with the second inequality in~\eqref{region_delta} applied at the point $(\udz(v),v)$,
\begin{eqnarray*}
u&\leq& e^{\frac{r_+}{\tilde c}}\udz(v)\ \leq\  \delta\, e^{\frac{r_+}{\tilde c}}e^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\eps]\,v}.
\end{eqnarray*}
\end{proof}

\subsection{Improved estimates}\label{medium-3}

\begin{Lem}
 Let $\ckrm\leq r\leq\ckrp$. Then
\begin{equation}\label{z_f}
 \Bigl|\hatz\Bigr|(r,v)\leq \tilde C_{\ckrm} \max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.
\end{equation}
\end{Lem}
\begin{proof}
Applying first the definition~\eqref{z} and then~\eqref{z_0_bis},
\begin{eqnarray}
{\cal Z}_{(r,v)}(\ckrp)&=&\max_{\tilde v\in[\vgrp(\ug(v)),v]}\Bigl|\hatz\Bigr|(\ckrp,\tilde v)\nonumber\\
&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\max_{\tilde v\in[\vgrp(\ug(v)),v]}\ \ \max_{u\in[0,\ugrp(\tilde v)]}|\zeta_0|(u)e^{-\alpha \tilde v}\nonumber\\
&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\max_{u\in[0,\ugrp(\vgrp(\ug(v)))]}|\zeta_0|(u)e^{-\alpha \vgrp(\ug(v))}\nonumber\\
&=&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha \vgrp(\ug(v))}\label{zr}
\end{eqnarray}
because $\ugrp(\tilde v)\geq \ugrp(v)$.
Integrating~\eqref{similar1} between $\vgrp(u_r(v))$ and $v$, we get
\begin{equation}\label{diff}
v-\vgrp(\ug(v))\leq{\textstyle\frac{\ckrp-r}{\tilde c}}=:
 c_{r,\ckrp}\leq c_{\ckrm,\ckrp}.
\end{equation}
This allows us to continue the estimate~\eqref{zr}, to obtain
\begin{eqnarray}
{\cal Z}_{(r,v)}(\ckrp)\leq e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}e^{\alpha c_{\ckrm,\ckrp}}\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.\label{part-1}
\end{eqnarray}
 Applying first the definition~\eqref{t} and then~\eqref{t_0}, and repeating the computations that lead to~\eqref{zr} and~\eqref{diff},
\begin{eqnarray}
{\cal T}_{(r,v)}(\ckrp)&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}\ln\Bigl(\frac{r_+}{\ckrp}\Bigr)\max_{\tilde v\in[\vgrp(\ug(v)),v]}\ \
\max_{u\in[0,\ugrp(\tilde v)]}|\zeta_0|(u)e^{-\alpha \tilde v}\nonumber\\
&\leq&e^{\frac{(r_+-\ckrp)^2}{\ckrp r_+}}e^{\alpha c_{\ckrm,\ckrp}}\ln\Bigl(\frac{r_+}{\ckrp}\Bigr)\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}\label{part-2}
\end{eqnarray}
We use~\eqref{part-1} and~\eqref{part-2} in~\eqref{zf}. This yields~\eqref{z_f}.
\end{proof}

\begin{Lem}
 Let $\ckrm\leq r\leq\ckrp$. Then
 \begin{equation}\label{t-f}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)\leq\overline{C}\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v},
\end{equation}
\begin{equation}
|\hat\theta|(r,v)\leq\check C\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.\label{theta_ff}
\end{equation}
\end{Lem}
\begin{proof}
Just like inequality~\eqref{t-0-a} was obtained from~\eqref{t_00} (that is, \eqref{t_0} with $\alpha=0$) and~\eqref{z-0}, inequality~\eqref{t-f} will be obtained from~\eqref{t_0} and~\eqref{z_f}.
Writing~\eqref{field_6} in the $(r,v)$ coordinates,
\begin{eqnarray*}
\widehat{\frac\theta\lambda}(r,v)&=&\widehat{\frac\theta\lambda}(\ckrp,v)
e^{\int_r^{\ckrp}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\tilde{s},v)\,d\tilde{s}}\\
&&+\int_{r}^{\ckrp}\Bigl[\hatz\frac 1 {\tilde s}\Bigr](\tilde s,v)
e^{\int_r^{\tilde s}\bigl[\frac{1}{\widehat{1-\mu}}\widehat{\partial_r(1-\mu)}\bigr](\bar{s},v)\,d\bar{s}}\,d\tilde s.
\end{eqnarray*}
The exponentials are bounded by the constant $C$ in~\eqref{c}. We use the estimates~\eqref{t_0} and~\eqref{z_f} to obtain
\begin{eqnarray*}
\Bigl|\widehat{\frac\theta\lambda}\Bigr|(r,v)&\leq&C\Bigl|\widehat{\frac\theta\lambda}\Bigr|(\ckrp,v)
+C\int_{r}^{\ckrp}\Bigl[\Bigl|\hatz\Bigr|\frac 1 {\tilde s}\Bigr](\tilde s,v)\,d\tilde s\\
&\leq&C\hat C_{\ckrp}\max_{u\in[0,\ugrp(v)]}|\zeta_0|(u)e^{-\alpha v}\\ &&+C
\tilde C_{\ckrm}\ln\Bigl(\frac{\ckrp}{r}\Bigr) \max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}\\
&=&\overline{C}\max_{u\in[0,\ug(v)]}|\zeta_0|(u)e^{-\alpha v}.
\end{eqnarray*}

Using~\eqref{similar1}, the function $\lambda$ is bounded from below in $J^-(\cg_{\ckrm})\cap J^+(\cg_{\ckrp})$. Hence~\eqref{t-f} implies~\eqref{theta_ff}.
\end{proof}

\begin{Rmk}
For use in\/ {\rm Part~3}, we observe that~\eqref{t_0} and~\eqref{t-f} imply
\begin{equation}\label{canto-part}
\lim_{\stackrel{(u,v)\to(0,\infty)}{{\mbox{\tiny{$(u,v)\in J^-(\Gamma_{\ckrm})$}}}}}\,\Bigl|\frac\theta\lambda\Bigr|(u,v)=0.
\end{equation}
\end{Rmk}

\section{The region $J^-(\gam)\cap J^+(\cg_{\ckrm})$}\label{large}

In this section, we define a curve $\gam$ to the future of $\cg_{\ckrm}$.
Our first aim is to obtain the bounds in Corollary~\ref{cor-8}, $r(u,v)\geq r_--\frac\eps 2$ and $\varpi(u,v)\leq\varpi_0+\frac\eps 2$,
for $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$ with $u\leq U_\eps$. In the process, we will bound
$\int_{\ugr(v)}^u\bigl[\bigl|\frac\zeta\nu\bigr||\zeta|\bigr](\tilde u,v)\,d\tilde u$ (this is inequality~\eqref{K}).
Then we will obtain a lower bound on $\kappa$, as well as upper and lower bounds on $\lambda$ and $\nu$. Therefore
this region, where $r$ may already be below $r_-$, is still a small perturbation of the Reissner-Nordstr\"{o}m solution.

We choose a positive number\footnote{We always have $-\partial_r(1-\mu)(r_-,\varpi_0)>\partial_r(1-\mu)(r_+,\varpi_0)$ (see Appendix~A of Part~3). So,
in particular, we may choose $\beta=-\,\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{\partial_r(1-\mu)(r_-,\varpi_0)}$.}
\begin{equation}\label{beta}
0<\beta<{\textstyle\frac 12\left(\sqrt{1-8\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{\partial_r(1-\mu)(r_-,\varpi_0)}}-1\right)},
\end{equation}
and define $\gam=\gamma_{\ckrm,\beta}$ to be the curve parametrized by
\begin{equation}\label{small_gamma}
u\mapsto\big(u,(1+\beta)\,\vgr(u)),
\end{equation}
for $u\in[0,U]$.
Since the curve $\cg_{\ckrm}$ is spacelike, so is $\gam$ ($u\mapsto\vgr(u)$ is strictly decreasing).
\begin{Lem}\label{boot_r}
For each $\beta$ satisfying~\eqref{beta} there exist $r_-<\overline{\ckrm}<r_0$ and $0<\eps_0<r_-$ for which, whenever\/ $\ckrm$ and $\eps$ are chosen satisfying $r_-<\ckrm\leq\overline{\ckrm}$ and $0<\eps\leq\eps_0$, the following holds:
there exists $U_\eps$ (depending on $\ckrm$ and $\eps$) such that if $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$,
with $0<u\leq U_\eps$, and
\begin{equation}\label{r_omega}
r(u,v)\geq r_--\eps,
\end{equation}
then
\begin{equation}\label{r_omega_2}
r(u,v)\geq r_--{\textstyle\frac\eps 2}\quad{\rm and}\quad\varpi(u,v)\leq\varpi_0+{\textstyle\frac\eps 2}.
\end{equation}
\end{Lem}
\begin{Cor}\label{cor-8}
Suppose that $\beta$ is given satisfying~\eqref{beta}, and let $\overline{\ckrm}$ and $\eps_0$ be as in the previous lemma. Fix $r_-<\ckrm\leq\overline{\ckrm}$ and $0<\eps<\eps_0$. If $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$ with $0<u\leq U_\eps$, then
\begin{equation}
r(u,v)\geq r_--{\textstyle\frac\eps 2}\quad{\rm and}\quad\varpi(u,v)\leq\varpi_0+{\textstyle\frac\eps 2}.\label{r-baixo}
\end{equation}
\end{Cor}
\begin{proof}
On $\cg_{\ckrm}$ we have $r=\ckrm>r_->r_--{\textstyle\frac\eps 2}$.
Suppose that there exists a point $(u,v)\in J^-(\gam)\cap
J^+(\cg_{\ckrm})$, with $0<u\leq U_\eps$, such that $r(u,v)<r_--\frac\eps 2$.
Then there exists a point $(\tilde u,v)$, with $0<\tilde u<u\leq U_\eps$, such that $r_--\eps\leq r(\tilde u,v)<r_--{\frac\eps 2}$.
The point $(\tilde u,v)$ belongs to $J^-(\gam)\cap
J^+(\cg_{\ckrm})$.
Applying Lemma~\ref{boot_r} at the point $(\tilde u,v)$, we reach a contradiction.
The rest of the argument is immediate.
\end{proof}
\begin{proof}[Proof of\/ {\rm Lemma~\ref{boot_r}}]
Let $(u,v)\in J^-(\gam)\cap
J^+(\cg_{\ckrm})$ such that~\eqref{r_omega} holds.
Because of the monotonicity properties of $r$,
$$
\min_{J^-(u,v)\cap
J^+(\cg_{\ckrm})}r\geq r_--\eps.
$$
According to Proposition~13.2 of~\cite{Dafermos2} (this result depends only on equations \eqref{theta_u} and \eqref{zeta_v}, and so does not depend on the presence of $\Lambda$), there exists a constant $\underline C$ (depending on $r_--\eps_0$) such that
\begin{eqnarray}
&&\int_{\vgr(u)}^v|\theta|(u,\tilde v)\,d\tilde v+\int_{\ugr(v)}^u|\zeta|(\tilde u,v)\,d\tilde u\label{Big}\\
&&\qquad\leq\underline C\left(
\int_{\vgr(u)}^v|\theta|(\ugr(v),\tilde v)\,d\tilde v+\int_{\ugr(v)}^u|\zeta|(\tilde u,\vgr(u))\,d\tilde u
\right).\nonumber
\end{eqnarray}

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{a}{{\tiny $\vgr(u)$}}
\psfrag{y}{{\tiny $(\ug(v),\vgsd(\ug(v)))$}}
\psfrag{u}{{\tiny $v$}}
\psfrag{j}{{\tiny $\cg_{\ckrm}$}}
\psfrag{x}{{\tiny $\ug(v)$}}
\psfrag{f}{{\tiny \!\!\!\!$U$}}
\psfrag{d}{{\tiny \!\!$u$}}
\psfrag{v}{{\tiny $u$}}
\psfrag{b}{{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$\ugr(v)$}}
\psfrag{c}{{\tiny $v$}}
\psfrag{p}{{\tiny $(u,v)$}}
\includegraphics[scale=.8]{fig5b.eps}
\end{psfrags}
\end{turn}
\end{center}

The first integral on the right hand side of~\eqref{Big} can be estimated using~\eqref{theta_f}, \eqref{theta_ff} and~\eqref{small_gamma}:
\begin{eqnarray*}
\int_{\vgr(u)}^v|\theta|(\ugr(v),\tilde v)\,d\tilde v
&\leq& C\sup_{[0,u]}|\zeta_0|\int_{\vgr(u)}^ve^{-\alpha\tilde v}\,d\tilde v\\
&\leq&C\sup_{[0,u]}|\zeta_0|e^{-\alpha\vgr(u)}\beta\vgr(u)\\
&\leq&C\sup_{[0,u]}|\zeta_0|e^{-\,\frac\alpha{1+\beta}v}\beta v\\
&\leq&\tilde C\sup_{[0,u]}|\zeta_0|e^{-\,\frac\alpha{1+\beta^+}v},
\end{eqnarray*}
where we have used $\vgr(u)=\frac{\vgam(u)}{1+\beta}\geq\frac{v}{1+\beta}$ and $\vgr(u)\leq v$,
and denoted by $\beta^+$ a fixed number strictly greater than $\beta$.
The second integral on the right hand side of~\eqref{Big} can be estimated using~\eqref{z_0_bis}, \eqref{z_f} and~\eqref{small_gamma}:
\begin{eqnarray*}
\int_{\ugr(v)}^u|\zeta|(\tilde u,\vgr(u))\,d\tilde u&=&\int_{\ugr(v)}^u\Bigl|\frac\zeta\nu\Bigr|(-\nu)(\tilde u,\vgr(u))\,d\tilde u\nonumber\\
&\leq&C(r_+-\ckrm)\sup_{[0,u]}|\zeta_0|e^{-\alpha \vgr(u)}\nonumber\\
&\leq&\tilde C\sup_{[0,u]}|\zeta_0|e^{-\,\frac\alpha{1+\beta}v}.\nonumber
\end{eqnarray*}
These lead to the following estimate for the left hand side of~\eqref{Big}:
\begin{eqnarray}
\int_{\vgr(u)}^v|\theta|(u,\tilde v)\,d\tilde v+\int_{\ugr(v)}^u|\zeta|(\tilde u,v)\,d\tilde u\leq
C\sup_{[0,u]}|\zeta_0|e^{-\,\frac\alpha{1+\beta^+}v}.\label{int_theta_zeta}
\end{eqnarray}
In order to use~\eqref{field_7}, note that, using $\eta(r,\varpi) \leq \eta_0(r)$,
\begin{eqnarray*}
e^{-\int_{\vgr(u)}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}&\leq&
e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\vgr(u)}\\ &\leq& e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta v}.
\end{eqnarray*}
Thus,
\begin{eqnarray}
\nonumber
\Bigl|\frac\zeta\nu\Bigr|(u,v)&\leq&\Bigl|\frac\zeta\nu\Bigr|(u,\vgr(u))
e^{-\int_{\vgr(u)}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}
\\
\nonumber
&&+\int_{\vgr(u)}^{v}\frac{|\theta|} r(u,\bar v)e^{-\int_{\bar v}^{v}\bigl[\frac{\lambda}{1-\mu}
\partial_r(1-\mu)\bigr](u,\tilde{v})\,d\tilde{v}}\,d\bar v
\\
\nonumber
&\leq&C\sup_{[0,u]}|\zeta_0|e^{-\alpha \vgr(u)}e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta v}
\\
\nonumber
&&+\frac{e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta v}}{r_--\eps_0}\int_{\vgr(u)}^{v}|\theta|(u,\bar v)\,d\bar v
\\
\nonumber
&\leq&C\sup_{[0,u]}|\zeta_0|e^{-\frac\alpha{1+\beta} v}e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta v}
\\
\nonumber
&&+\frac{e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta v}}{r_--\eps_0}C\sup_{[0,u]}|\zeta_0|e^{-\,\frac\alpha{1+\beta^+}v}
\\
&\leq&C\sup_{[0,u]}|\zeta_0|e^{-\bigl(\frac\alpha{1+\beta^+}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\bigr) v}.\label{zetaNuGamma}
\end{eqnarray}
Clearly, the right hand side of the last inequality also bounds $\max_{\tilde u\in[\ugr(v),u]}\bigl|\frac\zeta\nu\bigr|(\tilde u,v)$.
In order to bound $\varpi(u,v)$, note that
\begin{eqnarray}
&&\int_{\ugr(v)}^u\Bigl[\Bigl|\frac\zeta\nu\Bigr||\zeta|\Bigr](\tilde u,v)\,d\tilde u\nonumber\\
&&\ \ \ \ \leq
C\sup_{[0,u]}|\zeta_0|e^{-\Bigl(\frac\alpha{1+\beta^+}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\Bigr) v}
\int_{\ugr(v)}^u|\zeta|(\tilde u,v)\,d\tilde u\nonumber\\
&&\ \ \ \ \leq C\bigl(\sup_{[0,u]}|\zeta_0|\bigr)^2e^{-\Bigl(\frac{2\alpha}{1+\beta^+}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\Bigr) v}.\label{K}
\end{eqnarray}
Using~\eqref{omega_final} and the last estimate, we get
\begin{eqnarray*}
\varpi(u,v)&\leq&
\varpi(\ugr(v),v)e^{\frac{1}{r_--\eps_0}\int_{\ugr(v)}^u\bigl[\bigl|\frac\zeta\nu\bigr||\zeta|\bigr](\tilde u,v)\,d\tilde u}\\
&&+C
\int_{\ugr(v)}^ue^{\frac{1}{r_--\eps_0}\int_{s}^u\bigl[\bigl|\frac\zeta\nu\bigr||\zeta|\bigr](\tilde u,v)\,d\tilde u}
\Bigl[\Bigl|\frac\zeta\nu\Bigr||\zeta|\Bigr](s,v)\,ds\\
&\leq&\varpi(\ugr(v),v)e^{C\bigl(\sup_{[0,u]}|\zeta_0|\bigr)^2e^{-\Bigl(\frac{2\alpha}{1+\beta^+}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\Bigr) v}}\\
&&+Ce^{C\bigl(\sup_{[0,u]}|\zeta_0|\bigr)^2e^{-\Bigl(\frac{2\alpha}{1+\beta^+}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\Bigr) v}}\times\\
&&\qquad\qquad\qquad \times
\bigl(\sup_{[0,u]}|\zeta_0|\bigr)^2e^{-\Bigl(\frac{2\alpha}{1+\beta^+}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta\Bigr) v}.
\end{eqnarray*}
Let $\delta>0$.
Using the definition of $\alpha$ in~\eqref{alp_new}, the constant in the exponent
\begin{equation}\label{exponent}
\frac{2\alpha}{1+\beta+\delta}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta
\end{equation}
is positive for
\begin{equation}\label{success}
\textstyle
\beta<\frac 12\left(
\sqrt{(1+\delta)^2-8
\frac{\mbox{\tiny $(\frac{\ckrp}{r_+})^{\hat\delta^2}$}\min_{r\in[\ckrp,r_+]}\partial_r(1-\mu)(r,\varpi_0)}{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}}-(1+\delta)
\right).
\end{equation}
Now, the right hand side tends to
$$
{\textstyle\frac 12\left(\sqrt{1-8\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{\partial_r(1-\mu)(r_-,\varpi_0)}}-1\right)}
$$
as $(\ckrp,\eps_0,\delta)\to(r_+,0,0)$. So, if $\beta$ satisfies~\eqref{beta}, we may choose $\ckrp$, $\eps_0$ and $\delta$ such that~\eqref{success} holds.
Having done this, equations~\eqref{mass_2_bis} and~\eqref{region} now imply that for each $0 <\bar\eps < \eps_0$ there exists $\bar U_{\bar\eps} > 0$ such that
\begin{eqnarray*}
\varpi(u,v)
&\leq&\varpi_0+{\textstyle\frac{\bar\eps} 2},
\end{eqnarray*}
provided that  $u\leq\bar U_{\bar\eps}$.
Since $1-\mu$
is nonpositive and $1-\mu=(1-\mu)(r,\varpi_0)-\frac{2(\varpi-\varpi_0)}{r}$, we have
$$\textstyle
(1-\mu)(r(u,v),\varpi_0)\leq\frac{2(\varpi(u,v)-\varpi_0)}{r}\leq\frac{\bar\eps}{r_--\eps_0}.
$$
Hence, by inspection of the graph of $(1-\mu)(r,\varpi_0)$, there exists $\bar\eps_0$ such that for $0<\bar\eps\leq\bar\eps_0$, we have $r(u,v)>r_--\frac\eps 2$ provided that  $u\leq\bar U_{\bar\eps}$.
For $0<u\leq U_\eps:=\min\{\bar U_{\bar\eps_0},\bar U_\eps\}$, both inequalities~\eqref{r_omega_2} hold.
\end{proof}
\begin{Rmk}
 Given $\eps>0$, we may choose $U$ sufficiently small so that if $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$, then
\begin{equation}\label{H}
 \kappa(u,v)\geq 1-\eps.
\end{equation}
This is a consequence of~\eqref{k_min_2} and~\eqref{K}, since $r$ is bounded away from zero.
\end{Rmk}

Consider the reference subextremal Reissner-Nordstr\"{o}m black hole with renormalized mass $\varpi_0$, charge parameter $e$ and cosmological constant $\Lambda$.
The next remark will turn out to be crucial in Part~3.
\begin{Rmk}
Suppose that there exist positive constants $C$ and $s$ such that
$|\zeta_0(u)|\leq Cu^s$. Then, instead of choosing $\beta$ according to~\eqref{beta}, in\/ {\rm Lemma~\ref{boot_r}} we may choose
\begin{equation}\label{beta-s}
0<\beta<{\textstyle\frac 12\left(\sqrt{1-8\frac{(1+s)\partial_r(1-\mu)(r_+,\varpi_0)}{\partial_r(1-\mu)(r_-,\varpi_0)}}-1\right)}.
\end{equation}
\end{Rmk}
\begin{proof}
Let $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$.
According to~\eqref{region}, we have
\begin{eqnarray}
u&\leq&Ce^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\eps]\vgr(u)}\nonumber\\
&\leq&Ce^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\eps]\frac v{1+\beta}}\nonumber\\
&\leq&Ce^{-\partial_r(1-\mu)(r_+,\varpi_0)\frac v{1+\beta^+}}.\label{region_bis}
\end{eqnarray}
Thus, the exponent in the upper bound for $\varpi$ in~\eqref{exponent} may be replaced by
$$
\frac{2s\partial_r(1-\mu)(r_+,\varpi_0)}{1+\beta+\delta}+
\frac{2\alpha}{1+\beta+\delta}+\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\beta.
$$
This is positive for
\begin{equation}\label{success-s}
\textstyle
\beta<\frac 12\left(
\sqrt{(1+\delta)^2-8
\frac{\mbox{\tiny $[(\frac{\ckrp}{r_+})^{\hat\delta^2}$}+s]\min_{r\in[\ckrp,r_+]}\partial_r(1-\mu)(r,\varpi_0)}{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}}-(1+\delta)
\right).
\end{equation}
Given $\beta$ satisfying~\eqref{beta-s}, we can guarantee that it satisfies the condition above by choosing $(\ckrp,\eps_0,\delta)$ sufficiently close to $(r_+,0,0)$.
\end{proof}
\begin{Cor}
If, for example, $|\zeta_0|(u)\leq e^{-1/{u^2}}$, then instead of choosing $\beta$ according to~\eqref{beta}, in\/ {\rm Lemma~\ref{boot_r}}
we may choose any positive $\beta$.
\end{Cor}

\begin{Lem}\label{lambda-and-nu}
Suppose that $\beta$ is given satisfying~\eqref{beta}.
Choose $\ckrm$ and $\eps_0$ as in the statement of\/ {\rm Lemma~\ref{boot_r}}.
Let $\gam$ be the curve parametrized by~\eqref{small_gamma}.
Let also $\delta>0$, $\beta^-<\beta$ and $\beta^+>\beta$.
There exist constants, $\tilde c$, $\tilde C$, $\overline c$ and $\overline C$, such that
 for  $(u,v)\in\gam$, with $0<u\leq U_{\eps_0}$, we have
 \begin{eqnarray}
 &&
 \tilde ce^{(1+\delta)\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\frac\beta{1+\beta}\,v}\label{lambda-below}\\
 &&\qquad\qquad\qquad\qquad\leq
  -\lambda(u,v)\leq\nonumber\\
  &&\qquad\qquad\qquad\qquad\qquad\qquad
  \tilde Ce^{(1-\delta)\partial_r(1-\mu)(\ckrm,\varpi_0)\frac\beta{1+\beta}\,v}\label{lambda-above}
 \end{eqnarray}
 and
  \begin{eqnarray}
 &&
 \overline
 cu^{\mbox{\tiny$-\,\frac{1+\beta^+}{1+\beta^-}\frac{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta$}\,-1}
 \nonumber\\
 &&\qquad\qquad\qquad\qquad\leq
  -\nu(u,v)\leq\nonumber\\
  &&\qquad\qquad\qquad\qquad\qquad\qquad
  \overline Cu^{\mbox{\tiny$-\,\frac{1+\beta^-}{1+\beta^+}\frac{\partial_r(1-\mu)(\ckrm,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta$}\,-1}.
  \label{nu-above}
  \end{eqnarray}
\end{Lem}
\begin{proof}
Let us first outline the proof.
According to~\eqref{lambda_u} and~\eqref{nu_v},
\begin{eqnarray}
-\lambda(u,v)&=&-\lambda(\ugr(v),v)e^{\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u},\label{l}\\
-\nu(u,v)&=&-\nu(u,\vgr(u))e^{\int_{\vgr(u)}^v\bigl[\kappa\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}.\label{n}
\end{eqnarray}
In this region we cannot proceed as was done in the previous section because we cannot guarantee $1-\mu$ is bounded away from zero.
The idea now is to use these two equations to estimate $\lambda$ and $\nu$. For this we need to obtain lower and upper bounds for
\begin{equation}\label{int-1}
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u
\end{equation}
and
\begin{equation}\label{int-2}
\int_{\vgr(u)}^v\kappa(u,\tilde v)\,d\tilde v,
\end{equation}
when $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$.
The estimates for~\eqref{int-2}, and thus for $\nu$, are easy to obtain.
We estimate~\eqref{int-1} by comparing it with
\begin{equation}\label{int-3}
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,\vgr(\tilde u))\,d\tilde u.
\end{equation}
Using~\eqref{ray_v_bis}, we see that~\eqref{int-1} is bounded above by~\eqref{int-3}. We can also bound \eqref{int-1} from below by \eqref{int-3}, divided by $1+\eps$, once we show that
$$
e^{\int_{\vgr(\bar u)}^v\bigl[\big|\frac\theta\lambda\bigr|\frac{|\theta|}{r}\bigr](\bar u,\tilde v)\,d\tilde v}\leq 1+\eps.
$$
The estimates for $\frac\theta\lambda$ are obtained via~\eqref{field_6} and via upper estimates for~\eqref{int-1}.
To bound~\eqref{int-3} we use the fact that the integrals of $\nu$ and $\lambda$ along $\cg_{\ckrm}$ coincide.

We start the proof by differentiating the equation
$$
r(\tilde u,\vgr(\tilde u))={\ckrm}
$$
with respect to $\tilde u$, obtaining
\begin{equation}
\nu(\tilde u,\vgr(\tilde u))+\lambda(\tilde u,\vgr(\tilde u))\vgr'(\tilde u)=0.\label{derivada}
\end{equation}
For $(u,v)\in J^+(\cg_{\ckrm})$,
integrating~\eqref{derivada} between $\ugr(v)$ and $u$, we get
$$
\int_{\ugr(v)}^u\nu(\tilde u,\vgr(\tilde u))\,d\tilde u+\int_{\ugr(v)}^u\lambda(\tilde u,\vgr(\tilde u))\vgr'(\tilde u)\,d\tilde u =0.
$$
By making the change of variables $\tilde v=\vgr(\tilde u)$, this last equation can be rewritten as
\begin{equation}
 \int_{\ugr(v)}^u\nu(\tilde u,\vgr(\tilde u))\,d\tilde u-\int_{\vgr(u)}^v\lambda(\ugr(\tilde v),\tilde v)\,d\tilde v=0,
\label{integral}
\end{equation}
as $\vgr(\ugr(v))=v$ and $\frac{d\tilde v}{d\tilde u}=\vgr'(\tilde u)$.

We may bound the integral of $\lambda$ along $\cg_{\ckrm}$ in terms of the integral of $\kappa$ along $\cg_{\ckrm}$ in the following way:
\begin{eqnarray}
&&-\max_{\cg_{\ckrm}}(1-\mu)
\int_{\vgr(u)}^v\kappa(\ugr(\tilde v),\tilde v)\,d\tilde v\label{F}\\
&&\qquad\qquad\leq -\int_{\vgr(u)}^v\lambda(\ugr(\tilde v),\tilde v)\,d\tilde v\leq\nonumber \\
&&\qquad\qquad\qquad\qquad-\min_{\cg_{\ckrm}}(1-\mu)
\int_{\vgr(u)}^v\kappa(\ugr(\tilde v),\tilde v)\,d\tilde v.\label{A}
\end{eqnarray}
Analogously, we may bound the integral of $\nu$ along $\cg_{\ckrm}$ in terms of the integral of $\frac{\nu}{1-\mu}$ along $\cg_{\ckrm}$ in the following way:
\begin{eqnarray}
&&-\max_{\cg_{\ckrm}}(1-\mu)
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,\vgr(\tilde u))\,d\tilde u\label{B}\\
&&\qquad\ \ \leq-\int_{\ugr(v)}^u\nu(\tilde u,\vgr(\tilde u))\,d\tilde u\leq\nonumber \\
&&\qquad\qquad\qquad -\min_{\cg_{\ckrm}}(1-\mu)
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,\vgr(\tilde u))\,d\tilde u.\label{D}
\end{eqnarray}

Let now $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$. Using successively~\eqref{ray_v_bis},
\eqref{B}, \eqref{integral} and~\eqref{A}, we get
\begin{eqnarray}
&&\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u\nonumber\\
&&\qquad\qquad\leq\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,\vgr(\tilde u))\,d\tilde u\nonumber\\
&&\qquad\qquad\leq \mbox{\tiny$\frac{1}{-\max_{\cg_{\ckrm}}(1-\mu)}$}\int_{\ugr(v)}^u-\nu(\tilde u,\vgr(\tilde u))\,d\tilde u\nonumber\\
&&\qquad\qquad= \mbox{\tiny$\frac{1}{-\max_{\cg_{\ckrm}}(1-\mu)}$}\int_{\vgr(u)}^v-\lambda(\ugr(\tilde v),\tilde v)\,d\tilde v\nonumber\\
&&\qquad\qquad\leq \mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\int_{\vgr(u)}^v\kappa(\ugr(\tilde v),\tilde v)\,d\tilde v\label{argue3}\\
&&\qquad\qquad\leq \mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\beta \vgr(u)\label{G}\\
&&\qquad\qquad \leq \mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\,\beta v.\nonumber
\end{eqnarray}

We can now bound the field $\frac\theta\lambda$ for $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$.
Using~\eqref{field_6},
\begin{eqnarray}
\Bigl|\frac\theta\lambda\Bigr|(u,v)&\leq &\Bigl|\frac\theta\lambda\Bigr|(\ugr(v),v)
e^{-\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\nonumber\\
&&+\int_{\ugr(v)}^{u}\frac{|\zeta|} r(\bar u,v)e^{-\int_{\bar u}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde{u},v)\,d\tilde{u}}\,d\bar u.\label{C}
\end{eqnarray}
We can bound the exponentials in~\eqref{C} by
\begin{eqnarray*}
&&e^{-\int_{\bar u}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad\leq e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\int_{\bar u}^{u}\bigl[\frac{\nu}{1-\mu}\bigr](\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad\leq e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\bigr](\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad\leq e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\beta v}.
\end{eqnarray*}
Combining this inequality with~\eqref{t-f}, \eqref{r-baixo} and~\eqref{int_theta_zeta},
leads to
\begin{eqnarray}
\nonumber
&&\Bigl|\frac\theta\lambda\Bigr|(u,v)
\leq \left(C\sup_{[0,u]}|\zeta_0|e^{-\alpha v}+C\sup_{[0,u]}|\zeta_0|e^{-\frac\alpha{1+\beta^+} v}\right)\times
\\
\nonumber
&&\qquad\qquad\qquad\qquad\qquad\qquad\times
e^{-\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\beta v}\\
&&\qquad\qquad\leq C\sup_{[0,u]}|\zeta_0|e^{-\left(\frac\alpha{1+\beta^+}+
\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\beta\right)v}.\label{thetaLambdaGamma}
\end{eqnarray}
We consider the two possible cases. Suppose first that the exponent in~\eqref{thetaLambdaGamma} is nonpositive.
Then, from~\eqref{int_theta_zeta} we get
\begin{eqnarray*}
\int_{\vgr(u)}^v\Bigl[\Big|\frac\theta\lambda\Bigr||\theta|\Bigr](u,\tilde v)\,d\tilde v
&\leq& C \int_{\vgr(u)}^v|\theta|(u,\tilde v)\,d\tilde v
\\ &\leq&
C\sup_{[0,u]}|\zeta_0|e^{-\,\frac{\alpha}{1+\beta^+}v}.
\end{eqnarray*}
Suppose now the exponent in~\eqref{thetaLambdaGamma} is positive.
Using~\eqref{thetaLambdaGamma} and~\eqref{int_theta_zeta} again,
\begin{eqnarray*}
&&\int_{\vgr(u)}^v\Bigl[\Big|\frac\theta\lambda\Bigr||\theta|\Bigr](u,\tilde v)\,d\tilde v\\
&&\ \leq C\sup_{[0,u]}|\zeta_0|e^{-\left(\frac\alpha{1+\beta^+}+
\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\beta\right)v}\int_{\vgr(u)}^v|\theta|(u,\tilde v)\,d\tilde v
\\
&&\ \leq
C(\sup_{[0,u]}|\zeta_0|)^2e^{-\left(\frac{2\alpha}{1+\beta^+}+
\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\beta\right)v}.
\end{eqnarray*}
Therefore, in either case, given $\eps>0$ we may choose $U$ sufficiently small so that if $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$, then
\begin{equation}\label{integrating_factor}
e^{\frac 1{r_--\eps_0}\int_{\vgr(\bar u)}^v\bigl[\big|\frac\theta\lambda\bigr||\theta|\bigr](\bar u,\tilde v)\,d\tilde v}\leq 1+\eps,
\end{equation}
for $\bar u\in[\ugr(v),u]$.

Next we use~\eqref{ray_v_bis}, \eqref{B}, \eqref{D} and~\eqref{integrating_factor}.
We may bound the integral of $\nu$ along $\cg_{\ckrm}$ in terms of the integral of $\frac{\nu}{1-\mu}$ on the segment
$\bigl[\ugr(v),u\bigr]\times\{v\}$ in the following way:
\begin{eqnarray}
&&-\max_{\cg_{\ckrm}}(1-\mu)
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u\label{Q}\\
&&\leq-\max_{\cg_{\ckrm}}(1-\mu)
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,\vgr(\tilde u))\,d\tilde u\nonumber\\
&&\qquad\ \ \leq-\int_{\ugr(v)}^u\nu(\tilde u,\vgr(\tilde u))\,d\tilde u\leq\nonumber \\
&&\qquad\qquad\qquad-\min_{\cg_{\ckrm}}(1-\mu)
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,\vgr(\tilde u))\,d\tilde u\leq\nonumber\\
&&\qquad\qquad\qquad-(1+\eps)\min_{\cg_{\ckrm}}(1-\mu)
\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u.\label{E}
\end{eqnarray}

Now we consider $(u,v)\in\gam$.
In~\eqref{G} we obtained an upper bound for $\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u$. Now we use~\eqref{E} to obtain a lower bound for this quantity.
Applying successively~\eqref{E}, \eqref{integral}, \eqref{F}, and~\eqref{H},
\begin{eqnarray}
&&\int_{\ugr(v)}^u\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u\nonumber\\
&&\qquad\qquad\geq \mbox{\tiny$\frac{1}{-(1+\eps)\min_{\cg_{\ckrm}}(1-\mu)}$}\int_{\ugr(v)}^u-\nu(\tilde u,\vgr(\tilde u))\,d\tilde u\nonumber\\
&&\qquad\qquad= \mbox{\tiny$\frac{1}{-(1+\eps)\min_{\cg_{\ckrm}}(1-\mu)}$}\int_{\vgr(u)}^v-\lambda(\ugr(\tilde v),\tilde v)\,d\tilde v\nonumber\\
&&\qquad\qquad\geq \mbox{\tiny$\frac{\max_{\cg_{\ckrm}}(1-\mu)}{(1+\eps)\min_{\cg_{\ckrm}}(1-\mu)}$}\int_{\vgr(u)}^v\kappa(\ugr(\tilde v),\tilde v)\,d\tilde v\label{argue}\\
&&\qquad\qquad\geq \mbox{\tiny$\frac{(1-\eps)\max_{\cg_{\ckrm}}(1-\mu)}{(1+\eps)\min_{\cg_{\ckrm}}(1-\mu)}$}\beta \vgr(u)\nonumber\\
&&\qquad\qquad = \mbox{\tiny$\frac{(1-\eps)}{(1+\eps)}\frac{\max_{\cg_{\ckrm}}(1-\mu)}{\min_{\cg_{\ckrm}}(1-\mu)}$}\,{\textstyle\frac\beta{1+\beta}} v.\nonumber
\end{eqnarray}
Thus,
\begin{eqnarray}
&&e^{\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\nonumber\\
&&\qquad\qquad\leq
e^{\bigl[\max_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\partial_r(1-\mu)\bigr]
\int_{\ugr(v)}^{u}\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u}\nonumber\\
&&\qquad\qquad\leq
e^{\bigl[\max_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\partial_r(1-\mu)\bigr]
\mbox{\tiny$\frac{(1-\eps)}{(1+\eps)}\frac{\max_{\cg_{\ckrm}}(1-\mu)}{\min_{\cg_{\ckrm}}(1-\mu)}$}\, \frac\beta{1+\beta} v}\nonumber\\
&&\qquad\qquad\leq
e^{\bigl[\partial_r(1-\mu)(\ckrm,\varpi_0)+\max_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\mbox{\tiny$\frac{2(\varpi-\varpi_0)}{r^2}$}\bigr]
\mbox{\tiny$\frac{(1-\eps)}{(1+\eps)}\frac{\max_{\cg_{\ckrm}}(1-\mu)}{\min_{\cg_{\ckrm}}(1-\mu)}$}\,\frac\beta{1+\beta} v}\nonumber\\
&&\qquad\qquad\leq
e^{\bigl[\partial_r(1-\mu)(\ckrm,\varpi_0)+\mbox{\tiny$\frac\eps{(r_--\eps_0)^2}$}\bigr]
\mbox{\tiny$\frac{(1-\eps)}{(1+\eps)}\frac{\max_{\cg_{\ckrm}}(1-\mu)}{\min_{\cg_{\ckrm}}(1-\mu)}$}\, \frac{\beta}{1+\beta} v}.\label{I}
\end{eqnarray}
On the other hand, using~\eqref{G},
\begin{eqnarray}
&&e^{\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\nonumber\\
&&\qquad\qquad\geq
e^{\bigl[\min_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\partial_r(1-\mu)\bigr]
\int_{\ugr(v)}^{u}\frac\nu{1-\mu}(\tilde u,v)\,d\tilde u}\nonumber\\
&&\qquad\qquad\geq
e^{\bigl[\min_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\partial_r(1-\mu)\bigr]
\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\, \frac\beta{1+\beta}v}\nonumber\\
&&\qquad\qquad\geq
e^{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)
\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}$}\, \frac\beta{1+\beta}v}.\label{J}
\end{eqnarray}

We continue assuming $(u,v)\in\gam$. Taking into account~\eqref{similar1},
estimate \eqref{I} allows us to obtain an upper bound for $-\lambda(u,v)$,
\begin{eqnarray*}
-\lambda(u,v)&=&-\lambda(\ugr(v),v)e^{\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\\
&\leq& C
e^{\mbox{\tiny$\frac{(1-\eps)}{(1+\eps)}\frac{\max_{\cg_{\ckrm}}(1-\mu)}{\min_{\cg_{\ckrm}}(1-\mu)}$}
\bigl[\partial_r(1-\mu)(\ckrm,\varpi_0)+\mbox{\tiny$\frac\eps{(r_--\eps_0)^2}$}\bigr]
\,\frac\beta{1+\beta} v}\\
&\leq&\tilde Ce^{(1-\delta)\partial_r(1-\mu)(\ckrm,\varpi_0)\frac\beta{1+\beta}\,v},
\end{eqnarray*}
and estimate~\eqref{J} allows us to obtain a lower bound for $-\lambda(u,v)$,
\begin{eqnarray*}
-\lambda(u,v)&=&-\lambda(\ugr(v),v)e^{\int_{\ugr(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\\
&\geq &c
e^{\mbox{\tiny$\frac{\min_{\cg_{\ckrm}}(1-\mu)}{\max_{\cg_{\ckrm}}(1-\mu)}\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\frac\beta{1+\beta}$}
\, v}\\
&\geq&\tilde ce^{(1+\delta)\partial_r(1-\mu)(r_--\eps_0,\varpi_0)\frac\beta{1+\beta}\,v}.
\end{eqnarray*}
Next, we turn to the estimates on $\nu$. Let, again, $(u,v)\in \gam$. Using~\eqref{H},
\begin{eqnarray*}
(1-\eps){\textstyle\frac\beta{1+\beta}}\,v\leq\int_{\vgr(u)}^v\kappa(u,\tilde v)\,d\tilde v\leq{\textstyle\frac\beta{1+\beta}}\,v.
\end{eqnarray*}
These two inequalities imply
\begin{eqnarray}
&&e^{\int_{\vgr(u)}^v[\kappa\partial_r(1-\mu)](u,\tilde v)\,d\tilde v}\nonumber\\
&&\qquad\qquad\leq
e^{\bigl[\max_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\partial_r(1-\mu)\bigr]
\int_{\vgr(u)}^v\kappa(u,\tilde v)\,d\tilde v}\nonumber\\
&&\qquad\qquad\leq
e^{(1-\eps)\bigl[\partial_r(1-\mu)(\ckrm,\varpi_0)+\mbox{\tiny$\frac\eps{(r_--\eps_0)^2}$}\bigr]
\mbox{\tiny$\frac\beta{1+\beta}$}\, v}\label{L}
\end{eqnarray}
and
\begin{eqnarray}
&&e^{\int_{\vgr(u)}^v[\kappa\partial_r(1-\mu)](u,\tilde v)\,d\tilde v}\nonumber\\
&&\qquad\qquad\geq
e^{\bigl[\min_{J^-(\gam)\cap J^+(\cg_{\ckrm})}\partial_r(1-\mu)\bigr]
\int_{\vgr(u)}^v\kappa(u,\tilde v)\,d\tilde v}\nonumber\\
&&\qquad\qquad\geq
e^{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)
\mbox{\tiny$\frac\beta{1+\beta}$}\, v}.\label{M}
\end{eqnarray}
We note that according to~\eqref{region} we have
\begin{eqnarray}
&&ce^{-\partial_r(1-\mu)(r_+,\varpi_0)\frac{v}{1+\beta^-}}\label{O}\\
&&\leq ce^{-[\partial_r(1-\mu)(r_+,\varpi_0)+\tilde\eps]\frac{v}{1+\beta}}\nonumber\\
&&= ce^{-[\partial_r(1-\mu)(r_+,\varpi_0)+\tilde\eps]\vgr(u)}\nonumber\\
&&\qquad\qquad\leq u\leq \nonumber\\
&&\qquad\qquad\qquad\qquad Ce^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\tilde\eps]\vgr(u)}= \nonumber\\
&&\qquad\qquad\qquad\qquad Ce^{-[\partial_r(1-\mu)(r_+,\varpi_0)-\tilde\eps]\frac{v}{1+\beta}}\leq \nonumber\\
&&\qquad\qquad\qquad\qquad Ce^{-\partial_r(1-\mu)(r_+,\varpi_0)\frac{v}{1+\beta^+}},\label{P}
\end{eqnarray}
as $(u,v)\in \gam$. (The bound~\eqref{P} is actually valid in $J^-(\gam)\cap J^+(\cg_{\ckrm})$, see~\eqref{region_bis}.) Recalling~\eqref{n} and \eqref{similar2}, and using~\eqref{L} and~\eqref{O},
\begin{eqnarray*}
-\nu(u,v)&=&-\nu(u,\vgr(u))e^{\int_{\vgr(u)}^v[\kappa\partial_r(1-\mu)](u,\tilde v)\,d\tilde v}\\
&\leq&\frac{C}{u}e^{(1-\eps)\bigl[\partial_r(1-\mu)(\ckrm,\varpi_0)+\mbox{\tiny$\frac\eps{(r_--\eps_0)^2}$}\bigr]
\mbox{\tiny$\frac\beta{1+\beta}$}\, v}\\
&\leq&\frac{C}{u}e^{\partial_r(1-\mu)(\ckrm,\varpi_0)
\mbox{\tiny$\frac\beta{1+\beta^+}$}\, v}\\
&\leq&\overline Cu^{\mbox{\tiny$-\,\frac{1+\beta^-}{1+\beta^+}\frac{\partial_r(1-\mu)(\ckrm,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta$}\,-1},
\end{eqnarray*}
whereas
using~\eqref{M} and~\eqref{P},
\begin{eqnarray*}
-\nu(u,v)&=&-\nu(u,\vgr(u))e^{\int_{\vgr(u)}^v[\kappa\partial_r(1-\mu)](u,\tilde v)\,d\tilde v}\\
&\geq&\frac{c}{u}e^{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)
\mbox{\tiny$\frac\beta{1+\beta}$}\, v}\\
&\geq&\overline cu^{\mbox{\tiny$-\,\frac{1+\beta^+}{1+\beta^-}\frac{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta$}\,-1}.
\end{eqnarray*}
\end{proof}

\begin{Rmk}
Since $-\partial_r(1-\mu)(r_-,\varpi_0)>\partial_r(1-\mu)(r_+,\varpi_0)$ (see\/ {\rm Appendix~A} of\/ {\rm Part~3}), we can make our choice of $\beta$
and other parameters ($\ckrm$, $\eps_0$, $U$)
so that
$$\textstyle-\,\frac{1+\beta^-}{1+\beta^+}\frac{\partial_r(1-\mu)(\ckrm,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta\,-1>0$$
and
$$\textstyle-\,\frac{1+\beta^+}{1+\beta^-}\frac{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta\,-1>0.$$
Having done so, for $(u,v)$ on the curve $\gam$, we obtain
$$cu^{s_2}\leq-\nu(u,v)\leq Cu^{s_1},$$
with $0<s_1<s_2$.
\end{Rmk}

\section{The region $J^+(\gam)$}\label{huge}

Using~\eqref{l} and~\eqref{n}, we wish to obtain upper bounds for $-\lambda$ and for $-\nu$ in the future of $\gam$ while $r$ is greater than or equal to $r_--\eps$.
To do so, we partition this set into two regions, one where the mass is close to $\varpi_0$ and another one where
the mass is not close to $\varpi_0$. In the former case $\partial_r(1-\mu)<0$ and in the latter case $\frac{\partial_r(1-\mu)}{1-\mu}$ is bounded.
This information is used to bound the exponentials that appear in~\eqref{l} and~\eqref{n}.

Here the solution with general $\zeta_0$ departs qualitatively from the Reissner-Nordstr\"{o}m solution~\eqref{primeira}$-$\eqref{ultima}, but the radius function remains bounded away from zero, and approaches $r_-$ as $u \to 0$. This shows that the existence of a Cauchy horizon is a stable property when $\zeta_0$ is perturbed away from zero.
\begin{Lem}\label{ln-above}
Let $0<\eps_0<r_-$. There exists $0<\eps\leq\eps_0$ such that for $(u,v)\in \{r>r_--\varepsilon\}\cap J^+(\gam)$
we have
\begin{eqnarray}
-\lambda(u,v)&\leq&Ce^{(1-\delta)\partial_r(1-\mu)(\ckrm,\varpi_0)\frac\beta{1+\beta}\,v},\label{up-l}\\
-\nu(u,v)
&\leq&Cu^{\mbox{\tiny$-\,\frac{1+\beta^-}{1+\beta^+}\frac{\partial_r(1-\mu)(\ckrm,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta$}\,-1}.\label{up-n}
\end{eqnarray}
\end{Lem}
\begin{proof}
We recall that on $\gam$ the function $r$ is bounded above by $\ckrm$ and
that
$$
\eta=\eta_0+\varpi_0-\varpi.
$$
The minimum of $\eta_0$ in the interval $[r_--\eps_0,\ckrm]$ is positive, since $\eta_0(\ckrm)>0$.
If
\begin{equation}\label{varpi-m}
\varpi<\varpi_0+\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)
\end{equation}
then clearly
\begin{equation}\label{eta-positive}
\eta>0.
\end{equation}
On the other hand, if
\begin{equation}\label{varpi-p}
\varpi\geq\varpi_0+\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)
\end{equation}
then, for $r\in [r_--\eps,\ckrm]$,
$$
(1-\mu)(r,\varpi)\leq (1-\mu)(r,\varpi_0)-\,\frac{2\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)}{\ckrm},
$$
where we used
$$
(1-\mu)(r,\varpi)=(1-\mu)(r,\varpi_0)+\frac{2(\varpi_0-\varpi)}{r}.
$$
Choosing $0<\eps\leq\eps_0$ such that
$$
\max_{r\in[r_--\eps,\ckrm]}(1-\mu)(r,\varpi_0)\leq \frac{\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)}{\ckrm}
$$
we have
\begin{equation}\label{mu-negative}
(1-\mu)(r,\varpi)\leq -\,\frac{\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)}{\ckrm}<0.
\end{equation}
In case~\eqref{varpi-m} we have (recall~\eqref{eta})
$$
\frac{\nu}{1-\mu}\partial_r(1-\mu)<0\qquad{\rm and}\qquad \frac{\lambda}{1-\mu}\partial_r(1-\mu)<0.
$$
In case~\eqref{varpi-p}, the absolute value of
$$
-\,\frac{1}{1-\mu}\partial_r(1-\mu)
$$
is bounded, say by $C$. Indeed, this is a consequence of two facts: (i) the denominators $1-\mu$ and $r$ are bounded away from zero (we recall
$\eta$ also has a denominator equal to $r$);
(ii) the equality
\begin{equation}\label{omega_bigb}
\lim_{\omega\rightarrow+\infty}-\,\frac{1}{1-\mu}\partial_r(1-\mu)=\frac 1r.
\end{equation}
We define
\begin{eqnarray*}
\Pi_v&=&
\biggl\{u\in\,]0,U]:(u,v)\in \{r>r_--\varepsilon\}\cap J^+(\gam)\ \\
&&\qquad\qquad\qquad\qquad{\rm and}\ \varpi(u,v)<\varpi_0+\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)\biggr\},
\end{eqnarray*}
\begin{eqnarray*}
\Pi^v&=&
\biggl\{u\in\,]0,U]:(u,v)\in \{r>r_--\varepsilon\}\cap J^+(\gam)\ \\
&&\qquad\qquad\qquad\qquad{\rm and}\ \varpi(u,v)\geq\varpi_0+\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)\biggr\},
\end{eqnarray*}
\begin{eqnarray*}
\tilde\Pi_u&=&
\biggl\{v\in\,]0,\infty[\,:(u,v)\in \{r>r_--\varepsilon\}\cap J^+(\gam)\ \\
&&\qquad\qquad\qquad\qquad{\rm and}\ \varpi(u,v)<\varpi_0+\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)\biggr\}
\end{eqnarray*}
and
\begin{eqnarray*}
\tilde\Pi^u&=&
\biggl\{v\in\,]0,\infty[\,:(u,v)\in \{r>r_--\varepsilon\}\cap J^+(\gam)\ \\
&&\qquad\qquad\qquad\qquad{\rm and}\ \varpi(u,v)\geq\varpi_0+\min_{r\in [r_--\eps_0,\ckrm]}\eta_0(r)\biggr\}.
\end{eqnarray*}
In order to estimate $\lambda$, we observe that
\begin{eqnarray*}
&&e^{\int_{\ugam(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad= e^{\int_{\tilde u\in[\ugam(v),u]\cap\Pi_v}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad\ \ \ \,\times e^{\int_{\tilde u\in[\ugam(v),u]\cap\Pi^v}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad\leq 1\times e^{C\int_{\tilde u\in[\ugam(v),u]\cap\Pi^v}(-\nu)(\tilde u,v)\,d\tilde u}\\
&&\qquad\qquad\leq 1\times e^{C(\ckrm-(r_--\eps))}=:\hat C.
\end{eqnarray*}
Similarly, to estimate $\nu$ we note that
\begin{eqnarray*}
&&e^{\int_{\vgam(u)}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\\
&&\qquad\qquad= e^{\int_{\tilde v\in[\vgam(u),v]\cap\tilde\Pi_u}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\\
&&\qquad\qquad\ \ \ \,\times e^{\int_{\tilde v\in[\vgam(u),v]\cap\tilde\Pi^u}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\\
&&\qquad\qquad\leq 1\times e^{C\int_{\tilde v\in[\vgam(u),v]\cap\tilde\Pi^u}(-\lambda)(u,\tilde v)\,d\tilde v}\\
&&\qquad\qquad\leq 1\times e^{C(\ckrm-(r_--\eps))}=\hat C.
\end{eqnarray*}
In conclusion, let $(u,v)\in \{r>r_--\varepsilon\}\cap J^+(\cg_{\ckrm})$. Using~\eqref{l} and~\eqref{lambda-above}, we have
\begin{eqnarray}
-\lambda(u,v)&=&
-\lambda(\ugam(v),v)e^{\int_{\ugam(v)}^{u}\bigl[\frac{\nu}{1-\mu}\partial_r(1-\mu)\bigr](\tilde u,v)\,d\tilde u}\label{lambda_above}\\
&\leq&\hat C
\tilde Ce^{(1-\delta)\partial_r(1-\mu)(\ckrm,\varpi_0)\frac\beta{1+\beta}\,v}.\nonumber
\end{eqnarray}
Similarly, using~\eqref{n} and~\eqref{nu-above}, we have
\begin{eqnarray*}
-\nu(u,v)&=&
-\nu(u,\vgam(u))e^{\int_{\vgam(u)}^{v}\bigl[\frac{\lambda}{1-\mu}\partial_r(1-\mu)\bigr](u,\tilde v)\,d\tilde v}\\
&\leq&\hat C
\overline Cu^{\mbox{\tiny$-\,\frac{1+\beta^-}{1+\beta^+}\frac{\partial_r(1-\mu)(\ckrm,\varpi_0)}{\partial_r(1-\mu)(r_+,\varpi_0)}\beta$}\,-1}.
\end{eqnarray*}
\end{proof}

\begin{Lem}\label{infty}
 Let $\delta>0$. There exists $\tilde U_\delta$ such that for $(u,v)\in J^+(\gam)$ with $u<\tilde U_\delta$, we have
 $$
 r(u,v)>r_--\delta.
 $$
 \end{Lem}
\begin{proof}
We denote by $\overline\eps$ the value of $\eps$ that is provided in Lemma~\ref{ln-above}.
Let $\delta>0$. Without loss of generality, we assume that $\delta$ is less than or equal to $\overline\eps$.
Choose the value of $\eps$ in Corollary~\ref{cor-8} equal to $\delta$. This determines an $U_\eps$ as in the statement of that corollary.
Let $(u,v)\in J^+(\gam)$ with $u\leq U_\eps$.
Then
 $$
 r(u,\vgam(u))\geq r_--\frac\delta 2\qquad{\rm and}\qquad r(\ugam(v),v)\geq r_--\frac\delta 2
 $$
because $\ugam(v)\leq u$.
Here
$$
u\mapsto (u,\vgam(u))\quad {\rm and}\quad v\mapsto (\ugam(v),v)
$$
are parameterizations of the spacelike curve $\gam$.
Integrating~\eqref{up-n}, we obtain
\begin{equation}\label{ok}
-\int_{\ugam(v)}^u\frac{\partial r}{\partial u}(s,v)\,ds\leq
\int_{\ugam(v)}^uCs^{p-1}\,ds,
\end{equation}
for a positive $p$. This estimate is valid for $(u,v)\in \{r>r_--\overline\eps\}\cap J^+(\gam)$. It yields
\begin{eqnarray}
r(u,v)&\geq& r(\ugam(v),v)-\frac Cp(u^p-(\ugam(v))^p)\nonumber\\
&\geq&r_--\frac\delta 2-\frac Cpu^p\ >\ r_--\delta,\label{r-below}
\end{eqnarray}
provided $u<\min\Bigl\{U_\eps,\sqrt[p]{\frac{\delta p}{2C}}\Bigr\}=:\tilde U_\delta$.
Since $\delta$ is less than or equal to $\overline\eps$ and $\gam\subset \{r>r_--\overline\eps\}$, if $(u,v)\in J^+(\gam)$ and $u<\tilde U_\delta$, then
$(u,v)\in \{r>r_--\overline\eps\}$ and
the estimate~\eqref{ok} does indeed apply.

Alternatively, we can obtain~\eqref{r-below} integrating~\eqref{up-l}:
$$
-\int_{\vgam(u)}^v\frac{\partial r}{\partial v}(u,s)\,ds\leq
\int_{\vgam(u)}^vCe^{-qs}\,ds,
$$
for a positive $q$. This yields
\begin{eqnarray*}
r(u,v)&\geq& r(u,\vgam(u))-\frac Cq\Bigl(e^{-q\vgam(u)}-e^{-qv}\Bigr)\\
&\geq&r_--\frac\delta 2-\frac Cqe^{-q\vgam(u)}\\ &\geq&\ r_--\frac\delta 2-{\tilde C}u^{\tilde q},
\end{eqnarray*}
for a positive $\tilde q$, according to~\eqref{O}. For $u<\min\Bigl\{U_\eps,\sqrt[\tilde q]{\frac{\delta}{2\tilde C}}\Bigr\}$
we obtain, once more,
$$
r(u,v)>r_--\delta.
$$
\end{proof}

\begin{Cor}
If $\delta<r_-$ then ${\cal P}$ contains $[0,\tilde U_\delta]\times[0,\infty[$. Moreover, estimates~\eqref{up-l} and~\eqref{up-n} hold on~$J^+(\gam)$.
\end{Cor}
Due to the monotonicity of $r(u,\,\cdot\,)$ for each fixed $u$, we may define
$$
r(u,\infty)=\lim_{v\to\infty}r(u,v).
$$
As $r(u_2,v)<r(u_1,v)$ for $u_2>u_1$, we have that $r(\,\cdot\,,\infty)$ is nonincreasing.
\begin{Cor}\label{Gana}
We have
\begin{equation}\label{r-menos}
\lim_{u\searrow 0}r(u,\infty)=r_-.
\end{equation}
\end{Cor}

The previous two corollaries prove Theorem~\ref{r-stability}.
The argument in~\cite[Section~11]{Dafermos2}, shows that, as in the case when $\Lambda=0$, the spacetime is then extendible across the Cauchy horizon with $C^0$ metric.

\section{Two effects of any nonzero field}

This section contains two results concerning the structure of the solutions with general $\zeta_0$.
Theorem~\ref{rmenos} asserts that only in the case of the Reissner-Nordstr\"{o}m solution does the curve $\cg_{r_-}$ coincide with
the Cauchy horizon: if the field $\zeta_0$ is not identically zero, then the curve $\cg_{r_-}$ is contained in ${\cal P}$.

Lemma~\ref{l-kappa} states that, in contrast with what happens with the Reissner-Nordstr\"{o}m solution,
and perhaps unexpectedly,
the presence of a nonzero field immediately causes the integral $\int_0^\infty \kappa(u,v)\,dv$ to be finite for any $u>0$.
This implies that the affine parameter of any outgoing null geodesic inside the event horizon is finite at the Cauchy horizon.

For each $u>0$, we define
\begin{eqnarray*}
\varpi(u,\infty)&=&\lim_{v\nearrow+\infty}\varpi(u,v).
\end{eqnarray*}
This limit exists, and $u\mapsto\varpi(u,\infty)$ is an increasing function.

\begin{Thm}\label{rmenos}
 Suppose that there exists a positive sequence $(u_n)$ converging to $0$ such that $\zeta_0(u_n)\neq 0$.
 Then $r(u,\infty)<r_-$ for all $u \in \left] 0, U \right]$.
\end{Thm}

\begin{proof}
 The proof is by contradiction. Assume that $r(\bar{u},\infty) = r_-$ for some $\bar{u} \in \left]0,U\right]$.
Then $r(u,\infty) = r_-$ for all $u \in \left]0, \bar{u}\right]$. Let $0<\delta<u\leq \bar{u}$. Clearly,
$$
r(u,v)=r(\delta,v)+\int_\delta^u\nu(s,v)\,ds.
$$
Fatou's Lemma implies that $$\liminf_{v\to\infty}\int_\delta^u-\nu(s,v)\,ds\geq \int_\delta^u\liminf_{v\to\infty}-\nu(s,v)\,ds.$$
So,
\begin{eqnarray}
r_-=\lim_{v\to\infty}r(u,v)&=&\lim_{v\to\infty}r(\delta,v)-\lim_{v\to\infty}\int_\delta^u-\nu(s,v)\,ds\nonumber\\
&=&r_--\liminf_{v\to\infty}\int_\delta^u-\nu(s,v)\,ds\nonumber\\
&\leq&r_--\int_\delta^u\liminf_{v\to\infty}-\nu(s,v)\,ds.\label{fatou}
\end{eqnarray}
Since $\delta$ is arbitrary, this inequality implies that $\liminf_{v\to\infty}-\nu(u,v)$ is equal to zero for almost all $u \in \left]0, \bar{u}\right]$.
However, we will now show that, under the hypothesis on $\zeta_0$, $\liminf_{v\to\infty}-\nu(u,v)$ cannot be zero for any positive $u$ if $r(u,\infty)\equiv r_-$.

First, assume that $\varpi(u,\infty)=\infty$ for a certain $u$.
Then, using~\eqref{omega_bigb},
$$\lim_{v\rightarrow\infty}\frac{\partial_r(1-\mu)}{1-\mu}(u,v)=-\frac{1}{r_-}<0.$$
We may choose $V=V(u)>0$ such that $\frac{\partial_r(1-\mu)}{1-\mu}(u,v)<0$ for $v>V$. Integrating~\eqref{nu_v}, for $v>V$,
 \begin{eqnarray*}
  -\nu(u,v)
  &=&
  -\nu(u,V)e^{\int_V^v\left[\frac{\partial_r(1-\mu)}{1-\mu}\lambda\right](u,\tilde v)\,d\tilde v}
  \\
  &\geq&
  -\nu(u,V)>0.
 \end{eqnarray*}
Thus, for such a $u$, it is impossible for $\liminf_{v\to\infty}-\nu(u,v)$ to be equal to zero.

 Now assume $\varpi(u,\infty)<\infty$. The hypothesis on $\zeta_0$ and~\eqref{omega_u} imply that
$\varpi(u,0)>\varpi_0$ for each $u>0$, and so
  $\varpi(u,\infty)>\varpi_0$ for each $u>0$.  Then,
 $$(1-\mu)(u,\infty)=(1-\mu)(r_-,\varpi(u,\infty))< (1-\mu)(r_-,\varpi_0)=0.$$
We may choose $V=V(u)>0$ such that $-(1-\mu)(u,v)\geq C(u)>0$ for $v>V$.
Hence, integrating~\eqref{omega_v}, for $v>V$,
\begin{eqnarray*}
\varpi(u,v)&=&\varpi(u,V)+\frac 12\int_V^{v}\Bigl[-(1-\mu)\frac{\theta^2}{-\lambda}\Bigr](u,v)\,dv\\
&\geq&\varpi(u,V)+\frac {C(u)}2\int_V^{v}\Bigl[\frac{\theta^2}{-\lambda}\Bigr](u,v)\,dv
\end{eqnarray*}
Since $\varpi(u,\infty)<\infty$, letting $v$ tend to $+\infty$, we conclude
$$\int_V^{\infty}\Bigl[\frac{\theta^2}{-\lambda}\Bigr](u,v)\,dv<\infty.$$
 Finally, integrating~\eqref{ray_v_bis} starting from $V$, we see that $\frac{\nu(u,\infty)}{(1-\mu)(u,\infty)}>0$. Since $(1-\mu)(u,\infty)<0$,
once again we conclude that $\liminf_{v\to\infty}-\nu(u,v)=-\nu(u,\infty)>0$.
 \end{proof}

\begin{Lem}\label{l-kappa}
 Suppose that there exists a positive sequence $(u_n)$ converging to $0$ such that $\zeta_0(u_n)\neq 0$.
Then
\begin{equation}\label{integral-k}
\int_{0}^\infty\kappa(u,v)\,dv<\infty\ {\rm for\ all}\ u>0.
\end{equation}
\end{Lem}
\begin{proof}
We claim that for some decreasing sequence $(u_n)$ converging to 0, $$(1-\mu)(u_n,\infty)<0.$$
To prove our claim, we consider three cases.

{\em Case 1}. If $\varpi(u,\infty)=\infty$ for each $u>0$ then $(1-\mu)(u,\infty)=-\infty$.

{\em Case 2}. If $\lim_{u\searrow 0}\varpi(u,\infty)>\varpi_0$ then, using Corollary~\ref{Gana},
$$
\lim_{u\searrow 0}(1-\mu)(u,\infty)=(1-\mu)(r_-,\lim_{u\searrow 0}\varpi(u,\infty))<(1-\mu)(r_-,\varpi_0)=0.
$$

{\em Case 3}.
Suppose that
 $\lim_{u\searrow 0}\varpi(u,\infty)=\varpi_0$.
For sufficiently small $u$ and $(u,v)\in J^+(\cg_{\ckrm})$, we have
$$
\eta(u,v)\geq 0
$$
(see~\eqref{eta-positive}).
So, we may define
$\nu(u,\infty)=\lim_{v\nearrow+\infty}\nu(u,v)$. By Lebesgue's Monotone Convergence Theorem, we have
\begin{equation}\label{lebesgue-2}
r(u,\infty)=r(\delta,\infty)+\int_\delta^u\nu(s,\infty)\,ds.
\end{equation}
Note that different convergence theorems have to be used in~\eqref{fatou} and~\eqref{lebesgue-2}.
If $\nu(u,\infty)$ were zero almost everywhere, then $r(u,\infty)$ would be a constant. If the constant were $r_-$ we would be contradicting Theorem~\ref{rmenos}.
If the constant were smaller than $r_-$ we would be contradicting Lemma~\ref{infty}.
We conclude there must exist a sequence $u_n\searrow 0$ such that
$\nu(u_n,\infty)<0$.
Integrating~\eqref{ray_v_bis}, we get
$$
\frac{\nu(u,\infty)}{(1-\mu)(u,\infty)}\leq\frac{\nu(u,0)}{(1-\mu)(u,0)}<\infty.
$$
Therefore, $(1-\mu)(u_n,\infty)<0$. This proves our claim.

For any fixed index $n$, there exists a $v_n$ such that
$$(1-\mu)(u_n,v)<\frac 12(1-\mu)(u_n,\infty)=:-\,\frac 1{c_n},$$ for $v\geq v_n$.
It follows that
$$
\kappa(u_n,v)\leq c_n(-\lambda(u_n,v)),\ {\rm for}\ v\geq v_n.
$$
Using the estimate~\eqref{up-l} for $-\lambda$, we have
$$
\int_{v_n}^\infty\kappa(u_n,v)\,dv<\infty.
$$
Hence
$
\int_{0}^\infty\kappa(u_n,v)\,dv<\infty
$.
Recalling that $u\mapsto\kappa(u,v)$ is nonincreasing, we get~\eqref{integral-k}.
\end{proof}

\begin{Cor}\label{affine}
Let $u>0$. Consider an outgoing null geodesic $t\mapsto(u,v(t))$ for $({\cal M},g)$, with $g$ given by
$$
g=-\Omega^2(u,v)\,dudv+r^2(u,v)\,\sigma_{{\mathbb S}^2}.
$$
Then $v^{-1}(\infty)<\infty$, i.e.\
the affine parameter is finite at the Cauchy horizon.
\end{Cor}
\begin{proof}
The function $v(\,\cdot\,)$ satisfies
\begin{equation}\label{geodesic}
\ddot{v}+\Gamma^v_{vv}(u,v)\,\dot{v}^2=0,
\end{equation}
where the Christoffel symbol $\Gamma^v_{vv}$ is given by
$$\Gamma^v_{vv}=\partial_v\ln\Omega^2.$$
So, we may rewrite~\eqref{geodesic} as
$$
\frac{\ddot{v}}{\dot{v}}=-\partial_t(\ln\Omega^2)(u,v).
$$
We integrate both sides of this equation to obtain
$$
\ln \dot v+\ln c=-\ln\Omega^2(u,v),
$$
with $c>0$, or
$$
\frac{dt}{dv}=c\,\Omega^2(u,v).
$$
Integrating both sides of the previous equation once again, the affine parameter $t$ is given by
$$
t=v^{-1}(0)+c\int_0^v\Omega^2(u,\bar v)\,d\bar v=v^{-1}(0)-4c\int_0^v(\nu\kappa)(u,\bar v)\,d\bar v.
$$

If $\zeta_0$ vanishes in a neighborhood of the origin, the solution corresponds to the Reissner-Nordstr\"{o}m solution. The function
$\kappa$ is identically 1 and, using~\eqref{ray_v_bis}, $\frac{\nu}{1-\mu}=C(u)$, with $C(u)$ a positive function of $u$.
Thus, $\nu=C(u)(1-\mu)=C(u)\lambda$ and
$$
\int_0^\infty\Omega^2(u,\bar v)\,d\bar v=-4cC(u)\int_0^\infty\lambda(u,\bar v)\,d\bar v=4cC(u)(r_+-u-r_-)<\infty.
$$

On the other hand, suppose that there exists a positive sequence $(u_n)$ converging to $0$ such that $\zeta_0(u_n)\neq 0$.
Then, since $\nu$ is continuous, it satisfies the bound~\eqref{up-n} for large $v$, and~\eqref{integral-k} holds. So we also
have
$$
\int_0^\infty \Omega^2(u,\bar v)\,d\bar v<\infty.
$$
\end{proof}


TITLE: Proving that a set $A$ is dense in $M$ iff $A^c$ has empty interior
QUESTION [3 upvotes]: Prove that a set $A$ is dense in a metric space $(M,d)$ iff $A^c$ has empty interior.

Attempt:
I think I proved the converse correctly, but I'm not sure how to start the forward direction.
$(\Leftarrow)$
Assume $A^c$ has empty interior. Show that $M=\operatorname{cl}(A)$, where $\operatorname{cl}(A)$ denotes the closure of $A$. We also note that $\operatorname{cl}(A)=(\operatorname{int}(A^c))^c=\emptyset^c=M$, where $\operatorname{int}(A^c)$ denotes the set of interior points of $A^c$.
$\blacksquare$
Any help would be appreciated. Thanks.

REPLY [2 votes]: Just recall that in any topological space $X$ the closure of the complement is equal to the complement of the interior. Formally $(E^\circ)^c=\overline{(E^c)}$ for any $E\subseteq X$. Take $E^\circ=\emptyset$ for one implication. Assume $E^c$ to be dense to get the other implication.