$\int_{0}^{+\infty}\overline{h}^{2}\,dr$ is bounded by a continuous function
of $u$, then (7.10) and (7.11) tell us that, for almost all $u$, $\xi^{*}$
belongs to the Sobolev space $H^{1}(0,+\infty)$. Therefore, by the Sobolev
embedding theorem, for almost all $u$, $\xi^{*}(u,.)\in C^{1/2}[0,+\infty)$
and $\xi^{*}(u,s)\to 0$ as $s\to+\infty$. Consequently, for almost all $u$,
$\xi(u,r)$ is continuous with respect to $r$ and $\xi(u,r)\to 0$ as $r\to 0$.

Now, for any $\delta>0$, the function $\xi$ is expressed as

$\xi(u,r)=\xi(u,\delta)+\int_{\delta}^{r}\Big{(}\overline{g}\frac{(h-\overline{h})}{s}-c^{2}s\overline{h}\,e^{\nu+\lambda}\Big{)}\,ds,$
(7.12)

and taking $\delta\to 0$ in this equation, the result of the lemma follows. ∎

###### Proof of Proposition 7.2.

From (7.8), we have that $\frac{\xi}{\sqrt{e^{\nu+\lambda}}}(.,\delta)\in
L^{2}(0,u_{0})$. It follows, from (7.12), that so does
$\frac{\xi}{\sqrt{e^{\nu+\lambda}}}(.,r)$ for all $r>0$. Moreover, it is also
true by (7.8) that, for any $(u_{1},r_{1})\in\mathcal{O}$, the restriction of
$\xi$ to the characteristic $\chi(\cdot;u_{1},r_{1})$ belongs to
$L^{2}(0,u_{1})$.

Now, let $(s,\,\xi^{*},\,e^{\nu^{*}+\lambda^{*}})$ be defined as in (7.9). For
every $s_{0}\in[0,+\infty)$, by the Sobolev inequality we get

$\sup_{s\geq s_{0}}\xi^{*2}(s)\leq
c_{*}\int_{s_{0}}^{+\infty}\Big{(}\xi^{*2}+\big{(}\partial_{s}\xi^{*}\big{)}^{2}\Big{)}\,ds,$
(7.13)

where $c_{*}$ is a constant, independent of $s_{0}$. Since
$e^{\nu^{*}+\lambda^{*}}$ is a non-increasing function of $s$, it follows that

$\Big{(}\frac{\xi^{*2}}{e^{\nu^{*}+\lambda^{*}}}\Big{)}(s_{0})\leq\frac{c_{*}}{e^{(\nu^{*}+\lambda^{*})(s_{0})}}\int_{s_{0}}^{+\infty}\Big{(}\xi^{*2}+\big{(}\partial_{s}\xi^{*}\big{)}^{2}\Big{)}\,ds\leq
c_{*}\int_{s_{0}}^{+\infty}\frac{1}{e^{\nu^{*}+\lambda^{*}}}\Big{(}\xi^{*2}+\big{(}\partial_{s}\xi^{*}\big{)}^{2}\Big{)}\,ds.$

Therefore, we have

$\sup_{s\geq 0}\Big{(}\frac{\xi^{*2}}{e^{\nu^{*}+\lambda^{*}}}\Big{)}(s)\leq
c_{*}\int_{0}^{+\infty}\frac{1}{e^{\nu^{*}+\lambda^{*}}}\Big{(}\xi^{*2}+\big{(}\partial_{s}\xi^{*}\big{)}^{2}\Big{)}\,ds.$
(7.14)

Taking into account (7.10) and the second inequality of (7.11), from (7.14) we
deduce that

$\displaystyle\sup_{r\in[0,r_{0}]}\Big{(}\frac{\xi^{2}}{e^{\nu+\lambda}}\Big{)}(r)$
$\displaystyle\leq
c_{*}\Big{(}\int_{0}^{r_{0}}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr+c^{4}r_{0}^{3}\int_{0}^{r_{0}}\overline{h}^{2}\,dr+\frac{1}{2\pi}\big{(}e^{(\nu+\lambda)(r_{0})}-e^{(\nu+\lambda)(0)}\big{)}\Big{)}$
(7.15) $\displaystyle\leq
c_{*}\Big{(}\int_{0}^{r_{0}}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr+c^{4}r_{0}^{3}F(u)+\frac{1}{2\pi}\big{(}e^{(\nu+\lambda)(r_{0})}-e^{(\nu+\lambda)(0)}\big{)}\Big{)},$

where $F(u)$ is a continuous function of $u$ and we applied (2.7). Since
$e^{\nu+\lambda}$ is continuous on $[0,u_{0}]\times[0,+\infty)$, taking
$r_{0}\to 0$ in (7.15), we derive from (7.8) that, at almost all $u$,
$\frac{\xi^{2}}{e^{\nu+\lambda}}\to 0$ as $r\to 0$, as claimed.

Now, by integrating (7.15) with respect to $u$ we obtain

$\displaystyle\int_{0}^{u_{0}}\sup_{r\in[0,r_{0}]}\Big{(}\frac{\xi^{2}}{e^{\nu+\lambda}}\Big{)}(u,r)\,du$
(7.16) $\displaystyle\leq
c_{*}\Big{(}\iint_{\mathcal{O}(r_{0})}\frac{\xi^{2}}{re^{\nu+\lambda}}\,drdu+c^{4}r_{0}^{3}\int_{0}^{u_{0}}F(u)\,du+\frac{1}{2\pi}\int_{0}^{u_{0}}\big{(}e^{(\nu+\lambda)(r_{0})}-e^{(\nu+\lambda)(0)}\big{)}\,du\Big{)}.$

It then follows from (7.8) that

$\sup_{r\in[0,1]}\Big{|}\xi\,e^{-(\nu+\lambda)/2}\Big{|}\in L^{2}(0,u_{0}).$
(7.17)

On the other hand, since

$\xi(u,r)=\xi(u,1)+\int_{1}^{r}\Big{(}\overline{g}\frac{(h-\overline{h})}{s}-c^{2}s\overline{h}\,e^{\nu+\lambda}\Big{)}\,ds,$

we have, for all $r>1$,

$\displaystyle\Big{(}\frac{\xi^{2}}{r^{2}e^{\nu+\lambda}}\Big{)}(u,r)$
$\displaystyle\leq
2\Big{(}\frac{\xi^{2}}{e^{\nu+\lambda}}\Big{)}(u,1)+\frac{2}{r^{2}e^{\nu+\lambda}}\Big{(}\int_{1}^{r}\Big{(}\overline{g}\frac{(h-\overline{h})}{s}-c^{2}s\overline{h}\,e^{\nu+\lambda}\Big{)}\,ds\Big{)}^{2}$
$\displaystyle\leq
2\Big{(}\frac{\xi^{2}}{e^{\nu+\lambda}}\Big{)}(u,1)+\frac{2}{r}\int_{1}^{r}\Big{(}(h-\overline{h})^{2}+c^{4}s^{2}\overline{h}^{2}\Big{)}\,ds\leq
2\Big{(}\frac{\xi^{2}}{e^{\nu+\lambda}}\Big{)}(u,1)+F_{1}(u),$

where $F_{1}(u)$ is a continuous function of $u$ and we applied (2.7).
Combining this result with (7.17), gives the first statement in (7.6), as
claimed.

Finally, since by dominated convergence,

$\int_{0}^{u_{0}}\big{(}e^{(\nu+\lambda)(r_{0})}-e^{(\nu+\lambda)(0)}\big{)}\,du\to
0,\quad\text{as $r\to 0$},$

taking $r_{0}\to 0$ in (7.16), we complete the proof with
$\int_{0}^{u_{0}}\Big{(}\frac{\xi^{2}}{e^{\nu+\lambda}}\Big{)}(u,r)\,du\to
0,\quad\text{as $r\to 0$}.$ ∎

###  7.2 Compactness arguments based on the mass equation

It is convenient to introduce the function
$\gamma(u,r):=-\frac{\nu+\lambda}{4\pi}=\int_{r}^{+\infty}\frac{(h-\overline{h})^{2}}{s}\,ds.$
Before we start studying in more detail the terms that appear in the evolution
equation for the Hawking mass, we recall that (cf. (5.11))

$\partial_{u}\bigg{(}\int_{r}^{+\infty}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{s+\epsilon}\,ds\bigg{)}=\int_{r}^{+\infty}\partial_{u}\bigg{(}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{s+\epsilon}\bigg{)}\,ds,$

so that, by straightforward calculations,

$\displaystyle\partial_{u}\bigg{(}\int_{r}^{+\infty}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{s+\epsilon}\,ds\bigg{)}=\int_{r}^{+\infty}\frac{2(h_{\epsilon}-\overline{h}_{\epsilon})}{s+\epsilon}\Big{(}Dh_{\epsilon}+\frac{1}{2}\overline{g}_{\epsilon}\partial_{r}h_{\epsilon}-D\overline{h}_{\epsilon}-\frac{1}{2}\overline{g}_{\epsilon}\partial_{r}\,\overline{h}_{\epsilon}\Big{)}\,ds$
$\displaystyle=\int_{r}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{s+\epsilon}\Bigg{(}\frac{(g_{\epsilon}-\overline{g}_{\epsilon})(h_{\epsilon}-\overline{h}_{\epsilon})}{s+\epsilon}-c^{2}(s+\epsilon)\,\overline{h}_{\epsilon}e^{\nu_{\epsilon}+\lambda_{\epsilon}}+\overline{g}_{\epsilon}\partial_{r}h_{\epsilon}-\frac{\xi_{\epsilon}}{s+\epsilon}-\frac{\overline{g}_{\epsilon}}{s+\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})\Bigg{)}\,ds.$

Combining this equation with

$\frac{1}{r+\epsilon}\partial_{r}\overline{g}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}+\frac{\overline{g}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})}{r+\epsilon}\partial_{r}h_{\epsilon}-\frac{\overline{g}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{(r+\epsilon)^{2}}=\frac{g_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{2(r+\epsilon)^{2}}+\partial_{r}\Big{(}\overline{g}_{\epsilon}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{2(r+\epsilon)}\Big{)},$

we obtain

$\displaystyle\partial_{u}\bigg{(}\int_{r}^{+\infty}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{s+\epsilon}\,ds\bigg{)}$
(7.18)
$\displaystyle=-\frac{1}{2(r+\epsilon)}\overline{g}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}+\int_{r}^{+\infty}\Big{(}\frac{1}{2}g_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})-\xi_{\epsilon}-c^{2}(s+\epsilon)^{2}\overline{h}_{\epsilon}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\Big{)}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\,ds.$

So, for any smooth function $\varphi$ whose support is compact and contained
in the interior of $\mathcal{O}$, we have

$\displaystyle\iint_{\mathcal{O}}\bigg{(}-\frac{\overline{g}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{2(r+\epsilon)}+\int_{r}^{+\infty}\Big{(}\frac{g_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})}{2}-\xi_{\epsilon}-c^{2}(s+\epsilon)^{2}\overline{h}_{\epsilon}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\Big{)}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\,ds\bigg{)}\varphi\,dudr$
(7.19)
$\displaystyle=-\iint_{\mathcal{O}}\bigg{(}\int_{r}^{+\infty}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{s+\epsilon}\,ds\bigg{)}(\partial_{u}\varphi)dudr,$

and, by (6.10),

$\iint_{\mathcal{O}}\bigg{(}\int_{r}^{+\infty}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{s+\epsilon}\,ds\bigg{)}\partial_{u}\varphi\,dudr\to\iint_{\mathcal{O}}\gamma\partial_{u}\varphi\,dudr.$
(7.20)

We now study the left-hand side of (7.19). First of all, thanks to (2) in
(2.6) and Hölder’s inequality, we have

$\displaystyle\bigg{|}\int_{r}^{+\infty}g_{\epsilon}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{(s+\epsilon)^{2}}\,ds\bigg{|}$
$\displaystyle\leq\frac{1+4\pi
c^{2}\mathcal{A}^{2}_{\epsilon}(u,r)}{(r+\epsilon)^{2}}\int_{r}^{+\infty}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}ds,$
(7.21)
$\displaystyle\bigg{|}\int_{r}^{+\infty}\overline{h}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})e^{\nu_{\epsilon}+\lambda_{\epsilon}}\,ds\bigg{|}$
$\displaystyle\leq\frac{1}{r+\epsilon}\bigg{|}\int_{r}^{+\infty}(r+\epsilon)\,\overline{h}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})e^{\nu_{\epsilon}+\lambda_{\epsilon}}\,ds\bigg{|}$
$\displaystyle\leq\frac{1}{r+\epsilon}\bigg{(}\int_{r}^{+\infty}\big{(}r+\epsilon)^{2}\overline{h}_{\epsilon}^{2}ds\bigg{)}^{1/2}\bigg{(}\int_{r}^{+\infty}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}ds\bigg{)}^{1/2}.$

By the fourth and fifth assertions of Theorem 2.1,
$\int_{r}^{+\infty}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}ds$ and
$\int_{r}^{+\infty}\big{(}s+\epsilon)^{2}\overline{h}_{\epsilon}^{2}ds$ are
bounded by a continuous function of $u$ and, consequently, (7.21) implies (as
$r\to+\infty$)

$\int_{r}^{+\infty}g_{\epsilon}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{(s+\epsilon)^{2}}\,ds\to
0,\quad\qquad\int_{r}^{+\infty}\overline{h}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})e^{\nu_{\epsilon}+\lambda_{\epsilon}}\,ds\to
0.$

Combining this result with the property that (for any $r_{0}>\delta>0$)
$h_{\epsilon},\,\overline{h}_{\epsilon},\,g_{\epsilon}$ converge uniformly on
$\mathcal{O}_{\delta,r_{0}}$ to $h,\,\overline{h},\,g$, respectively, and
proceeding similarly as in (6.10) we deduce that

$\displaystyle\int_{r}^{+\infty}g_{\epsilon}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{(s+\epsilon)^{2}}\,ds$
$\displaystyle\to\int_{r}^{+\infty}g_{h}\frac{(h-\overline{h})^{2}}{s^{2}}\,ds\qquad\text{uniformly
on $\mathcal{O}_{\delta,r_{0}}$,}$ (7.22)
$\displaystyle\int_{r}^{+\infty}\overline{h}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})e^{\nu_{\epsilon}+\lambda_{\epsilon}}\,ds$
$\displaystyle\to\int_{r}^{+\infty}\overline{h}(h-\overline{h})e^{\nu+\lambda}\,ds\qquad\text{uniformly
on $\mathcal{O}_{\delta,r_{0}}$. }$

Next, by integration by parts we find

$\displaystyle\int_{r}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\xi_{\epsilon}\,ds$
(7.23)
$\displaystyle=\xi_{\epsilon}\int_{r}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\,ds+\int_{r}^{+\infty}\bigg{(}\overline{g}_{\epsilon}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})}{s+\epsilon}-c^{2}(s+\epsilon)\,\overline{h}_{\epsilon}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\bigg{)}\bigg{(}\int_{s}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s_{1}+\epsilon)^{2}}\,ds_{1}\bigg{)}\,ds.$

Applying Hölder’s inequality we have

$\displaystyle\int_{r}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\,ds$
$\displaystyle\leq\bigg{(}\frac{1}{3(r+\epsilon)^{3}}\bigg{)}^{1/2}\bigg{(}\int_{r}^{+\infty}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}ds\bigg{)}^{1/2}\leq\frac{1}{(r+\epsilon)^{3/2}}F_{2}(u),$

where $F_{2}(u)$ is a continuous function of $u$, by Theorem 2.1. Similarly to
(7.22), for any $r_{0}\geq\delta>0$ the two expressions

$\int_{r}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\,ds\quad\text{and}\quad\int_{r}^{+\infty}\bigg{(}\overline{g}_{\epsilon}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{s+\epsilon}-c^{2}(s+\epsilon)\,\overline{h}_{\epsilon}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\bigg{)}\bigg{(}\int_{s}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s_{1}+\epsilon)^{2}}\,ds_{1}\bigg{)}\,ds$

converge uniformly on $\mathcal{O}_{\delta,r_{0}}$ to, respectively,

$\int_{r}^{+\infty}\frac{h-\overline{h}}{s^{2}}\,ds\quad\text{and}\quad\int_{r}^{+\infty}\bigg{(}\overline{g}\frac{h-\overline{h}}{s}-c^{2}s\overline{h}\,e^{\nu+\lambda}\bigg{)}\bigg{(}\int_{s}^{+\infty}\frac{h-\overline{h}}{s_{1}^{2}}\,ds_{1}\bigg{)}\,ds.$

Taking these expressions as well as (7.2) into account in (7.23), we obtain

$\iint_{\mathcal{O}}\bigg{(}\int_{r}^{+\infty}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{(s+\epsilon)^{2}}\xi_{\epsilon}\,ds\bigg{)}\varphi\,dudr\to\iint_{\mathcal{O}}\bigg{(}\int_{r}^{+\infty}\frac{h-\overline{h}}{s^{2}}\xi
ds\bigg{)}\varphi\,dudr,$ (7.24)

where, similarly to (7.23), we used integration by parts. Therefore, we
conclude from (7.22) and (7.24) that the left-hand side of (7.19) converges to

$\iint_{\mathcal{O}}\bigg{(}-\frac{1}{2r}\overline{g}(h-\overline{h})^{2}+\int_{r}^{+\infty}\Big{(}\frac{1}{2}g(h-\overline{h})-\xi-c^{2}s^{2}\overline{h}\,e^{\nu+\lambda}\Big{)}\frac{(h-\overline{h})}{s^{2}}\,ds\bigg{)}\varphi\,dudr.$

In view of this result together with (7.21), we infer from (7.19) that
$\gamma$ is weakly differentiable with respect to $u$ in $\mathcal{O}$ and

$\displaystyle
D\gamma=\int_{r}^{+\infty}\Big{(}\frac{1}{2}g(h-\overline{h})-\xi-c^{2}s^{2}\overline{h}\,e^{\nu+\lambda}\Big{)}\frac{(h-\overline{h})}{s^{2}}\,ds.$
(7.25)

Consequently, $e^{\nu+\lambda}$ is weakly differentiable with respect to $u$
in $\mathcal{O}$ and $D\big{(}e^{\nu+\lambda}\big{)}=-4\pi
e^{\nu+\lambda}D\gamma.$ We next compute the evolution equation for
$\frac{1}{r}\int_{\delta}^{r}g\,ds$, with $\delta>0$. We compute

$\displaystyle
D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}=\frac{\overline{g}}{2r}\Big{(}\frac{1}{r}\int_{\delta}^{r}(1-4\pi
c^{2}s^{2}\overline{h}^{2})e^{\nu+\lambda}\,ds-(1-4\pi
c^{2}r^{2}\overline{h}^{2})e^{\nu+\lambda}\Big{)}$
$\displaystyle\quad+\frac{1}{r}\int_{0}^{r}\Big{(}D\big{(}e^{\nu+\lambda}\big{)}+\frac{1}{2}\overline{g}\partial_{r}\big{(}e^{\nu+\lambda}\big{)}-8\pi
c^{2}s^{2}e^{\nu+\lambda}\overline{h}\partial_{u}\overline{h}-4\pi
c^{2}s^{2}\overline{h}^{2}\partial_{u}\big{(}e^{\nu+\lambda}\big{)}\Big{)}\,ds$
$\displaystyle=\frac{\overline{g}}{2r^{2}}(r\overline{g}-\delta\overline{g}(\delta))-\frac{1}{2r}\overline{g}g-\frac{4\pi}{r}\int_{\delta}^{r}\Big{(}gD\gamma-\frac{1}{2}g\overline{g}\frac{(h-\overline{h})^{2}}{s}+c^{2}s\overline{h}(h-\overline{h})\overline{g}e^{\nu+\lambda}+2\,c^{2}s^{2}\overline{h}D\overline{h}\,e^{\nu+\lambda}\Big{)}\,ds$

and, from (7.25),

$\displaystyle\int_{\delta}^{r}gD\gamma
ds=r\overline{g}D\gamma-\delta\overline{g}(\delta)D\gamma(\delta)-\int_{\delta}^{r}s\overline{g}\partial_{r}D\gamma
ds,$
$\displaystyle\int_{\delta}^{r}\overline{g}\Big{(}s\partial_{r}D\gamma+\frac{g}{2}\frac{(h-\overline{h})^{2}}{s}-c^{2}s\overline{h}(h-\overline{h})e^{\nu+\lambda}\Big{)}\,ds=\int_{\delta}^{r}\frac{(h-\overline{h})}{s}\overline{g}\xi
ds,$

so that, in view of (7.4) and (7.7) we arrive at the identity

$\displaystyle D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}$
$\displaystyle=\frac{\overline{g}}{2r^{2}}(r\overline{g}-\delta\overline{g}(\delta))-\frac{1}{2r}\overline{g}g-4\pi\overline{g}D\gamma+\frac{4\pi\,\delta}{r}\overline{g}(\delta)D\gamma(\delta)+\frac{2\,c^{2}\pi}{r}\int_{\delta}^{r}\xi\partial_{r}\xi
ds$ (7.26)
$\displaystyle=f_{\delta}-\frac{1}{2r}\overline{g}(g-\overline{g})-4\pi\overline{g}D\gamma+\frac{2\pi}{r}\xi^{2},$

where

$f_{\delta}(u,\delta):=-\frac{\overline{g}}{2r^{2}}\delta\overline{g}(u,\delta)-\frac{c^{2}\pi}{r}\xi^{2}(u,\delta)+\frac{4\pi\,\delta}{r}\overline{g}(u,\delta)D\gamma(u,\delta).$

The first term of $f_{\delta}$ converges to $0$, as $\delta\to 0$, and by
Proposition 7.2, so does the second term for almost all $u$. Therefore, in
order to conclude that $f_{\delta}\to 0$, at almost all $u$, as $\delta\to 0$,
it suffices to check the lemma stated next.

###  7.3 Final argument

######  Lemma 7.3.

For almost all $u$, one has $\delta D\gamma(u,\delta)\to 0$ as $\delta\to 0$.

###### Proof.

In view of (7.25) we have

$\displaystyle|D\gamma|$
$\displaystyle=\Big{|}\int_{\delta}^{+\infty}\Big{(}\frac{1}{2}g(h-\overline{h})-\xi-c^{2}r^{2}\overline{h}\,e^{\nu+\lambda}\Big{)}\frac{(h-\overline{h})}{r^{2}}\,dr\Big{|}$
(7.27)
$\displaystyle\leq\frac{1}{2}\int_{\delta}^{+\infty}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r^{2}}\,dr+c^{4}\pi\,\int_{\delta}^{+\infty}\overline{h}^{2}(h-\overline{h})^{2}e^{\nu+\lambda}\,dr$
$\displaystyle\quad+c^{2}\int_{\delta}^{+\infty}|\overline{h}(h-\overline{h})|e^{\nu+\lambda}\,dr+\int_{\delta}^{+\infty}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr.$

By a direct computation, we have

$\displaystyle\delta\int_{\delta}^{+\infty}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r^{2}}\,dr=\delta\int_{\delta}^{\sqrt{\delta}}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r^{2}}\,dr+\delta\int_{\sqrt{\delta}}^{+\infty}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r^{2}}\,dr$
$\displaystyle\leq\int_{\delta}^{\sqrt{\delta}}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r}\,dr+\sqrt{\delta}\int_{\sqrt{\delta}}^{+\infty}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r}\,dr\leq\frac{1}{4\pi}\Big{(}e^{(\nu+\lambda)(\sqrt{\delta})}-e^{(\nu+\lambda)(\delta)}+\sqrt{\delta}(1-e^{(\nu+\lambda)(\sqrt{\delta})})\Big{)}$

and, since the right-hand side of the previous inequality converges to $0$, as
$\delta\to 0$, we obtain

$\delta\int_{\delta}^{+\infty}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r^{2}}\,dr\to
0\quad\text{as $\delta\to 0$}.$

Combining this result with the fact that $r\,\overline{h}$ is bounded by a
continuous function of $u$ (cf. (2.6)), we get

$\delta\int_{\delta}^{+\infty}e^{\nu+\lambda}\overline{h}^{2}(h-\overline{h})^{2}\,dr=\delta\int_{\delta}^{+\infty}\big{(}r\,\overline{h}\big{)}^{2}\Big{(}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r^{2}}\Big{)}\,dr\to
0\quad\text{as $\delta\to 0$}.$

Next, by Hölder’s inequality and (iii) in Section 6.2 we deduce that

$\delta\int_{\delta}^{+\infty}|\overline{h}(h-\overline{h})|e^{\nu+\lambda}\,dr\leq\delta\Big{(}\int_{\delta}^{+\infty}|\overline{h}|^{2}\,dr\Big{)}^{1/2}\Big{(}\int_{\delta}^{+\infty}(h-\overline{h})^{2}\,dr\Big{)}^{1/2}\to
0\quad\text{as $\delta\to 0$,}$

and the last integral in (7.27) can be rewritten as

$\int_{\delta}^{+\infty}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr=\int_{\delta}^{\sqrt{\delta}}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr+\int_{\sqrt{\delta}}^{1}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr+\int_{1}^{+\infty}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr.$

By Hölder’s inequality, we also obtain

$\displaystyle\delta\int_{\delta}^{\sqrt{\delta}}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr\leq\int_{\delta}^{\sqrt{\delta}}|\xi|\frac{|h-\overline{h}|}{r}\,dr$
$\displaystyle\leq\Big{(}\int_{\delta}^{\sqrt{\delta}}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr\Big{)}^{1/2}\Big{(}\int_{\delta}^{\sqrt{\delta}}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r}\,dr\Big{)}^{1/2}$
$\displaystyle=\Big{(}\int_{\delta}^{\sqrt{\delta}}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr\Big{)}^{1/2}\Big{(}\frac{1}{4\pi}(e^{(\nu+\lambda)(\sqrt{\delta})}-e^{(\nu+\lambda)(\delta)})\Big{)}^{1/2}.$

Since, at almost all $u$, $\frac{\xi}{\sqrt{re^{\nu+\lambda}}}\in
L^{2}(0,r_{0})$ for all $r_{0}>0$, then
$\int_{\delta}^{\sqrt{\delta}}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr\to 0$ at
almost all $u$ as $\delta\to 0$. Hence, by the previous inequality

$\delta\int_{\delta}^{\sqrt{\delta}}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr\to
0\quad\text{as $\delta\to 0$}.$

Similarly, we obtain

$\displaystyle\delta\int_{\sqrt{\delta}}^{1}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr\leq\sqrt{\delta}\int_{0}^{1}|\xi|\frac{|h-\overline{h}|}{r}\,dr$
$\displaystyle\leq\sqrt{\delta}\Big{(}\int_{0}^{1}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr\Big{)}^{1/2}\Big{(}\int_{0}^{1}e^{\nu+\lambda}\frac{(h-\overline{h})^{2}}{r}\,dr\Big{)}^{1/2}$
$\displaystyle\leq\Big{(}\frac{\delta}{4\pi}\Big{)}^{1/2}\Big{(}\int_{0}^{1}\frac{\xi^{2}}{re^{\nu+\lambda}}\,dr\Big{)}^{1/2}\to
0\quad\text{as $\delta\to 0$}.$

To complete the proof of the lemma, it only remains to check that at almost
all $u$

$\delta\int_{1}^{+\infty}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr\to
0\quad\text{as $\delta\to 0$}.$ (7.28)

In fact, by Hölder’s inequality

$\displaystyle\int_{1}^{+\infty}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr$
$\displaystyle\leq\Big{(}\int_{1}^{+\infty}\frac{\xi^{2}}{r^{4}}\,dr\Big{)}^{1/2}\Big{(}\int_{1}^{+\infty}(h-\overline{h})^{2}\,dr\Big{)}^{1/2}$
$\displaystyle\leq\sup_{r\geq
1}\Big{(}\frac{|\xi|}{r}\Big{)}\Big{(}\int_{1}^{+\infty}\frac{1}{r^{2}}\,dr\Big{)}^{1/2}\Big{(}\int_{1}^{+\infty}(h-\overline{h})^{2}\,dr\Big{)}^{1/2}.$

Observe that, thanks to (2.7), $\int_{1}^{+\infty}(h-\overline{h})^{2}\,dr$ is
bounded by a continuous function of $u$. Furthermore, by Proposition 7.2,
$\sup_{r\geq 1}\frac{|\xi|}{r}$ is finite for almost all $u$. Therefore,
$\int_{1}^{+\infty}|\xi|\frac{|h-\overline{h}|}{r^{2}}\,dr$ is finite for
almost all $u$, hence (7.28) holds for almost all $u$, as desired. ∎

We have established that, for almost all $u$ and all $r>0$, $f_{\delta}\to 0$
as $\delta\to 0$. It then follows from (7.26) that

$\lim_{\delta\to
0}D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}=-\frac{1}{2r}\overline{g}(g-\overline{g})-4\pi\overline{g}D\gamma+\frac{2\pi}{r}\xi^{2}.$
(7.29)

Furthermore, for any smooth function $\varphi$ whose support is a compact set
contained in the interior of $\mathcal{O}$, we have

$-\iint_{\mathcal{O}}\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}D\varphi\,dudr=\iint_{\mathcal{O}}D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}\varphi\,dudr.$

Since $\frac{1}{r}\int_{\delta}^{r}g\,ds\to\overline{g}$ for $\delta\to 0$
uniformly in the support of $\varphi$, it follows that

$-\iint_{\mathcal{O}}\overline{g}D\varphi\,dudr=\lim_{\delta\to
0}\iint_{\mathcal{O}}D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}\varphi\,dudr.$
(7.30)

On the other hand, from Proposition 7.2 and the proof of Lemma 7.3 we deduce
that $\varphi f_{\delta}$ is dominated by an integrable function and converges
to $0$ almost everywhere as $\delta\to 0$. Therefore, by the dominated
convergence theorem we have

$\lim_{\delta\to
0}\iint_{\mathcal{O}}D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}\varphi\,dudr=\iint_{\mathcal{O}}\lim_{\delta\to
0}D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}\varphi\,dudr.$ (7.31)

Hence, we conclude from (7.26), Lemma 7.3 and (7.29) –(7.31) that
$\overline{g}$ is weakly differentiable in $\mathcal{O}$ and

$D\overline{g}=\lim_{\delta\to
0}D\Big{(}\frac{1}{r}\int_{\delta}^{r}g\,ds\Big{)}=-\frac{1}{2r}\overline{g}(g-\overline{g})-4\pi\overline{g}D\gamma+\frac{2\pi}{r}\xi^{2}.$

This means that the Hawking mass $m$ is weakly differentiable in $\mathcal{O}$
and

$\displaystyle Dm$
$\displaystyle=-\frac{r}{2e^{\nu+\lambda}}\Big{(}D\overline{g}-\frac{\overline{g}}{e^{\nu+\lambda}}D\big{(}e^{\nu+\lambda}\big{)}\Big{)}-\frac{\overline{g}(g-\overline{g})}{4e^{\nu+\lambda}}-c^{2}\pi
r^{2}\overline{h}^{2}\overline{g}=-\frac{\pi}{re^{\nu+\lambda}}\xi^{2}-c^{2}\pi
r^{2}\overline{h}^{2}\overline{g}.$

Finally, we also check similarly that
$\int_{0}^{r}\Big{(}\frac{e^{\nu+\lambda}}{\overline{g}}\,ds\Big{)}$ is weakly
differentiable in $\mathcal{O}$ and

$D\Big{(}\int_{0}^{r}\frac{e^{\nu+\lambda}}{\overline{g}}\,dr\Big{)}=-2\pi\,\int_{0}^{r}\Big{(}\frac{e^{\nu+\lambda}}{\overline{g}^{2}}\frac{\xi^{2}}{r}\Big{)}\,dr-\frac{1}{2}e^{(\nu+\lambda)(u,0)}.$

Integrating this equation along the characteristics $\chi(u;u_{1},r_{1})$, we
obtain

$\displaystyle\int_{0}^{r_{1}}\Big{(}\frac{e^{\nu+\lambda}}{\overline{g}}\Big{)}(u_{1},r)\,dr+2\pi\,\iint_{\mathcal{O}(u_{1},r_{1})}\Big{(}\frac{e^{\nu+\lambda}}{\overline{g}^{2}}\frac{\xi^{2}}{(r+\epsilon)}\Big{)}\,drdu+\frac{1}{2}\int_{0}^{u_{1}}e^{(\nu+\lambda)(u,0)}\,du=\int_{0}^{r_{0}}\Big{(}\frac{e^{\nu+\lambda}}{\overline{g}}\Big{)}(0,r)\,dr,$

with
$\mathcal{O}(u_{1},r_{1})=\big{\\{}(u,r)\,/\,0<u<u_{1},\,0<r<\chi(u;u_{1},r_{1})\big{\\}}.$
Consequently, the generalized solution exists globally and this concludes the
proof of Theorem 1.2.

#### Acknowledgments.

This work was done in part when PLF was a visiting research fellow at the
Courant Institute for Mathematical Sciences, New York University, and a
visiting professor at the School of Mathematical Sciences, Fudan University,
Shanghai. The second author was supported by FCT sabbatical grant
SFRH/BSAB/130242/2017, CMAT, Univ. Minho, through FCT/Portugal projects
UIDB/00013/2020 and FEDER Funds COMPETE as well as CAMGSD, IST-ID, through
projects UIDB/04459/2020 and UIDP/04459/2020. The third author (TCN) was a
postdoc fellow of the Fondation des Sciences Mathématiques de Paris (FSMP).

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##  Appendix A Continuity and compactness of the solution mapping

#### Continuity of the mapping $\Psi_{b}$.

Assume that $(t_{i},h_{i})\to(t,h)$ in $[0,1]\times X_{b}$ (that is, in the
$C^{1}$ norm). Let $k$ and $k_{i}$ be solutions of the transport equation
(3.7) associated with $(t,h)$ and $(t_{i},h_{i})$, respectively. Let us prove
that $k_{i}$ converges to $k$ in the $C^{1}$ norm. Since the set of all smooth
functions in $X_{b}$ is dense, we can pick up a sequence of smooth functions
$\\{z_{i}\\}\subset X_{b}$ such that $\|h_{i}-z_{i}\|_{C^{1}}\to 0$. We then
define $w_{i}$ to be the unique solution of (3.7) associated with
$(t_{i},\,z_{i})$. Obviously, we have

$\|k_{i}-k\|_{C^{1}}\leq\|k_{i}-w_{i}\|_{C^{1}}+\|w_{i}-k\|_{C^{1}}$ (A.1)

and we analyze first the term $\|w_{i}-k\|_{C^{1}}$. By definition, we have

$\displaystyle
D_{a,\,t_{i},z_{i}}w_{i}-\frac{t_{i}}{2(r+\epsilon)}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})w_{i}$
$\displaystyle=-\frac{1}{2(r+\epsilon)}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})\overline{w}_{i}-\frac{c^{2}(r+\epsilon)}{2}e^{\nu_{z_{i}}+\lambda_{z_{i}}}\overline{w}_{i},$
(A.2) $\displaystyle
D_{a,t,h}k-\frac{t}{2(r+\epsilon)}(g_{a,t,h}-\overline{g}_{a,t,h})k$
$\displaystyle=-\frac{1}{2(r+\epsilon)}(g_{a,t,h}-\overline{g}_{a,t,h})\,\overline{h}-\frac{c^{2}(r+\epsilon)}{2}e^{\nu_{h}+\lambda_{h}}\overline{h}$

and, therefore,

$D_{a,t,h}(w_{i}-k)-\frac{t}{2r}(g_{a,t,h}-\overline{g}_{a,t,h})(w_{i}-k)=P(t,h,\,t_{i},\,z_{i}),$
(A.3)

with

$\displaystyle P(t,h,\,t_{i},\,z_{i})$
$\displaystyle:=\frac{1}{2(r+\epsilon)}(g_{a,t,h}-\overline{g}_{a,t,h})\,\overline{h}-\frac{1}{2(r+\epsilon)}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})\overline{w}_{i}$
(A.4)
$\displaystyle\quad+\frac{c^{2}(r+\epsilon)}{2}e^{\nu_{h}+\lambda_{h}}\overline{h}-\frac{c^{2}(r+\epsilon)}{2}e^{\nu_{z_{i}}+\lambda_{z_{i}}}\overline{w}_{i}+\frac{1}{2}\big{(}\rho(\overline{g}_{a,\,t_{i},z_{i}})-\rho(\overline{g}_{a,t,h})\big{)}\partial_{r}w_{i}$
$\displaystyle\quad+\frac{1}{2(r+\epsilon)}\big{(}t_{i}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})-t(g_{a,t,h}-\overline{g}_{a,t,h})\big{)}w_{i}.$

Since $\|h-z_{i}\|_{C^{1}}$ and $|t_{i}-t|$ tend to $0$, it follows from the
continuity of $\Psi_{b}$ and the $C^{2}$ bound on $k$ that

$P(t,h,\,t_{i},\,z_{i})\to 0\qquad\mbox{in $C^{0}$}.$ (A.5)

On the other hand, since $w_{i}(0,r)=k(0,r)=h_{0}(r)$, along the
characteristics $\chi_{h,u_{1}}=\chi_{a,t,h}(\cdot;u_{1},r_{1})$ we can write
the transport equation (A.3) in the form

$(w_{i}-k)(u_{1},r(u_{1}))=\int_{0}^{u_{1}}[P(t,h,\,t_{i},\,z_{i})]_{\chi_{h,u_{1}}}e^{\int_{u_{2}}^{u_{1}}\frac{t}{2(r_{3}+\epsilon)}(g_{a,t,h}-\overline{g}_{a,t,h})\big{|}_{\chi_{h,u_{1}}}\,du_{3}}\,du_{2}.$
(A.6)

Hence, from (A.5) and (A.6) we deduce that

$w_{i}\to k$ in $C^{0}$. (A.7)

Differentiating now the equations (A.2) with respect to $r$, we obtain

$\displaystyle
D_{a,t,h}(\partial_{r}k)-\frac{1}{2(r+\epsilon)}\Big{(}(g_{a,t,h}-\overline{g}_{a,t,h})\rho^{\prime}(\overline{g}_{a,t,h})+t(g_{a,t,h}-\overline{g}_{a,t,h})\Big{)}\partial_{r}k=f_{1}(h,k,t)$
$\displaystyle
D_{a,\,t_{i},z_{i}}(\partial_{r}w_{i})-\frac{1}{2(r+\epsilon)}\Big{(}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})\rho^{\prime}(\overline{g}_{a,\,t_{i},z_{i}})+t_{i}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})\Big{)}\partial_{r}w_{i}=f_{1}(z_{i},w_{i},t_{i}),$

where $f_{1}$ is given as in (3.14). Comparing the previous two equations we
find

$\displaystyle D_{a,t,h}(\partial_{r}w_{i}-\partial_{r}k)-$
$\displaystyle\frac{1}{2(r+\epsilon)}\Big{(}(g_{a,t,h}-\overline{g}_{a,t,h})\rho^{\prime}(\overline{g}_{a,t,h})+t(g_{a,t,h}-\overline{g}_{a,t,h})\Big{)}\,(\partial_{r}w_{i}-\partial_{r}k)=\widetilde{P}_{i}(t,h,\,t_{i},\,z_{i}),$
(A.8)

with

$\displaystyle\widetilde{P}(t,h,\,t_{i},\,z_{i})$
$\displaystyle:=f_{1}(h,k,t)-f_{1}(z_{i},w_{i},t_{i})+\frac{1}{2}(\rho(\overline{g}_{a,\,t_{i},z_{i}})-\rho(\overline{g}_{a,t,h}))\partial_{rr}w_{i}$
$\displaystyle\quad+\frac{1}{2(r+\epsilon)}\Big{(}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})\rho^{\prime}(\overline{g}_{a,\,t_{i},z_{i}})-(g_{a,t,h}-\overline{g}_{a,t,h})\rho^{\prime}(\overline{g}_{a,t,h})\Big{)}\partial_{r}w_{i}$
$\displaystyle\quad+\frac{1}{2(r+\epsilon)}\Big{(}t_{i}(g_{a,\,t_{i},z_{i}}-\overline{g}_{a,\,t_{i},z_{i}})-t(g_{a,t,h}-\overline{g}_{a,t,h})\Big{)}\partial_{r}w_{i}.$

Since $\partial_{r}w_{i}(0,r)=\partial_{r}k(0,r)=\partial_{r}h_{0}$, then
solving (A.8) along the characteristic $\chi_{h,u_{1}}$ yields us

$\displaystyle\partial_{r}(w_{i}-k)(u_{1},r(u_{1}))=\int_{0}^{u_{1}}\widetilde{P}(t,h,\,t_{i},\,z_{i})\big{|}_{\chi_{h,u_{1}}}e^{\int_{u_{2}}^{u_{1}}\frac{1}{2(r_{3}+\epsilon)}\big{(}(g_{a,t,h}-\overline{g}_{a,t,h})\rho^{\prime}(\overline{g}_{a,t,h})+t(g_{a,t,h}-\overline{g}_{a,t,h})\big{)}\big{|}_{\chi_{h,u_{1}}}\,du_{3}}\,du_{2}.$
(A.9)

From the basic continuity properties in Section 3.2, we see that
$\|\partial_{rr}w_{i}\|_{C^{0}}$ and $\|\partial_{r}w_{i}\|_{C^{0}}$ are
uniformly bounded. We deduce from (A.7) and (LABEL:formula_integration_C1)
that $\widetilde{P}(t,h,t_{i},z_{i})\to 0$, hence, by
(LABEL:formula_integration_C1), $\partial_{r}w_{i}$ converges to
$\partial_{r}k$ in $C^{0}$. Combining this result with (A.7), we find that
$\|w_{i}-k\|_{C^{1}}\to 0$ and, similarly, $\|w_{i}-k_{i}\|_{C^{1}}\to 0$.
Therefore, returning to (A.1) we conclude that $k_{i}$ converges to $k$ in the
$C^{1}$ norm, as claimed.

#### Compactness of the mapping $\Psi_{b}$.

Given any bounded sequence $(t_{i},h_{i})$ in $[0,1]\times X_{b}$, let us
denote by $k_{i}$ the unique solution of the transport equation (3.7)
associated with $(t_{i},h_{i})$. We need to prove that there exists a
subsequence (still denoted by $\\{k_{i}\\}$) such that $\\{k_{i}\\}$ converges
in the space $X_{b}$. Without loss of generality we assume that $h_{i}$ is
smooth. Since $\\{h_{i}\\}$ is bounded in $C^{1}$, a standard compact
embedding theorem tells us that there exists a subsequence (still denoted by
$\\{h_{i}\\}$) such that $\\{h_{i}\\}$ converges in the $C^{0}$ norm. In
particular, given any $\alpha>0$ we can guarantee that

$|h_{i}-h_{j}|\leq\alpha\quad\text{ for all sufficiently large }i,j.$ (A.10)

After passing to a subsequence, we may assume that $|t_{i}-t_{j}|\leq\alpha$.
An analysis similar to the one used to prove the continuity of $\Psi_{b}$
shows that, along the characteristics
$\chi_{h_{j},u_{1}}=\chi_{a,t_{j},h_{j}}(\cdot;u_{1},r_{1})$,

$\displaystyle(k_{i}-k_{j})(u_{1},r(u_{1}))$
$\displaystyle=\int_{0}^{u_{1}}P(t_{i},h_{i},\,t_{j},h_{j})\big{|}_{\chi_{h_{j},u_{1}}}e^{\int_{u_{2}}^{u_{1}}\frac{t_{j}}{2(r_{3}+\epsilon)}(g_{a,\,t_{j},h_{j}}-\rho(\overline{g}_{a,\,t_{j},h_{j}}))\big{|}_{\chi_{h_{j},u_{1}}}\,du_{3}}\,du_{2},$
$\displaystyle\partial_{r}(k_{i}-k_{j})(u_{1},r(u_{1}))$
$\displaystyle=\int_{0}^{u_{1}}\widetilde{P}(t_{i},h_{i},\,t_{j},h_{j})\big{|}_{\chi_{h_{j},u_{1}}}e^{\int_{u_{2}}^{u_{1}}\frac{1}{2(r_{3}+\epsilon)}\big{(}(g_{a,\,t_{j},h_{j}}-\overline{g}_{a,\,t_{j},h_{j}})\rho^{\prime}(\overline{g}_{a,\,t_{j},h_{j}})+t_{j}(g_{a,\,t_{j},h_{j}}-\overline{g}_{a,\,t_{j},h_{j}})\big{)}\big{|}_{\chi_{h_{j},u_{1}}}du_{3}}\,du_{2},$

where $P$ and $\widetilde{P}$ are defined as before. Thanks to the continuity
of $\Psi_{b}$ and the $C^{2}$ bound available on $k$, this implies that there
exists a constant $a_{7}>0$ depending on $\|h_{i}\|_{C^{1}}$ and
$(h_{0},a,b,\epsilon)$ such that $\|k_{i}-k_{j}\|_{C^{1}}\leq a_{7}\alpha$. We
conclude that $\\{k_{i}\\}$ converges in $C^{1}$, and this completes the proof
of Proposition 3.1.

##  Appendix B Derivation of energy estimates

We give here a proof of Proposition 4.5. We start from the identity

$\partial_{r}(D\overline{h}_{\epsilon,b})+\frac{1}{r}D\overline{h}_{\epsilon,b}=\frac{1}{2r^{2}}e^{\nu_{\epsilon,b}-\lambda_{\epsilon,b}}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})-\frac{c^{2}}{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\overline{h}_{\epsilon,b},$

which by integration gives us

$D_{\epsilon}\overline{h}_{\epsilon,b}=\frac{1}{2(r+\epsilon)}\int_{0}^{r}\left(\overline{g}_{\epsilon,b}\frac{h_{\epsilon,b}-\overline{h}_{\epsilon,b}}{r+\epsilon}-c^{2}(r+\epsilon)\,\overline{h}_{\epsilon,b}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\right)dr.$
(B.1)

Thanks to Lemma 4.2, on the spacetime domain $[0,c_{1}]\times[0,b]$ we have

$\partial_{r}\overline{g}_{\epsilon,b}+(r+\epsilon)\partial_{r}\big{(}\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}\big{)}-4\pi\,\partial_{u}\big{(}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\big{)}-8\pi(r+\epsilon)\big{(}\partial_{r}\,\overline{h}_{\epsilon,b}D_{\epsilon}\overline{h}_{\epsilon,b}-\frac{c^{2}}{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\overline{h}_{\epsilon,b}^{2}\big{)}=0.$

Integrating this identity and using (B.1) we obtain

$\displaystyle
4\pi\,\partial_{u}\bigg{(}\int_{0}^{b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\,dr\bigg{)}$
$\displaystyle=\int_{0}^{b}\bigg{(}\partial_{r}\overline{g}_{\epsilon,b}+(r+\epsilon)\partial_{r}\big{(}\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}\big{)}\bigg{)}dr-8\pi\,\int_{0}^{b}(r+\epsilon)\big{(}\partial_{r}\overline{h}_{\epsilon,b}D_{\epsilon}\overline{h}_{\epsilon,b}-\frac{c^{2}}{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\overline{h}_{\epsilon,b}^{2}\big{)}dr$
(B.2) $\displaystyle\leq
1+\bigg{(}(r+\epsilon)\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}-\int_{0}^{r}\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}\,ds\bigg{)}(b)+4\pi
c^{2}\int_{0}^{b}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,dr$
$\displaystyle\quad-4\pi\,\int_{0}^{b}\partial_{r}\,\overline{h}_{\epsilon,b}\bigg{(}\int_{0}^{r}\overline{g}_{\epsilon,b}\partial_{r}\,\overline{h}_{\epsilon,b}\,ds-c^{2}\int_{0}^{r}(s+\epsilon)\,\overline{h}_{\epsilon,b}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,ds\bigg{)}dr.$

We now need to estimate each of the terms above. Observe that by the first
equation in (4.8)

$\displaystyle(b+\epsilon)\big{(}\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}\big{)}(b)$
$\displaystyle=\bigg{(}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{2}+2\pi\,\overline{g}_{\epsilon,b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\bigg{)}(b)$
$\displaystyle\int_{0}^{b}\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}\,ds$
$\displaystyle=\int_{0}^{b}\bigg{(}\frac{1}{2}\partial_{r}\overline{g}_{\epsilon,b}+2\pi\,\overline{g}_{\epsilon,b}\frac{(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}}{s+\epsilon}\bigg{)}\,ds\geq
0,$

hence

$\displaystyle\bigg{(}(r+\epsilon)\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}-\int_{0}^{r}\overline{g}_{\epsilon,b}\partial_{r}\nu_{\epsilon,b}\,ds\bigg{)}(b)$
$\displaystyle\leq\bigg{(}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{2}+2\pi\,\overline{g}_{\epsilon,b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\bigg{)}(b).$
(B.3)

Next, we will use (4.13) for estimating
$\int_{0}^{+\infty}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,dr$.
Indeed, we note first that if there exists
$(u_{\epsilon},r_{\epsilon})\in[0,c_{1}]\times[0,3]$ such that

$(r_{\epsilon}+\epsilon)\big{(}|\overline{h}_{\epsilon,b}|e^{\frac{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}{2}}\big{)}(u_{\epsilon},r_{\epsilon})\geq\frac{2}{c}\sqrt{\frac{m_{\epsilon,b}(u,b)}{2\pi}},$

thanks to (4.13), we may take
$r_{\epsilon}^{\prime}\in(r_{\epsilon},r_{\epsilon}+1]$ such that

$(r_{\epsilon}^{\prime}+\epsilon)\big{(}|\overline{h}_{\epsilon,b}|e^{\frac{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}{2}}\big{)}(u_{\epsilon},r_{\epsilon}^{\prime})\leq\frac{1}{c}\sqrt{\frac{m_{\epsilon,b}(u,b)}{2\pi}}.$

Consequently, we have
$|\overline{h}_{\epsilon,b}|(u_{\epsilon},r_{\epsilon})\geq
2|\overline{h}_{\epsilon,b}|(u_{\epsilon},r_{\epsilon}^{\prime}),$ and thus

$|\overline{h}_{\epsilon,b}(u_{\epsilon},r_{\epsilon})-\overline{h}_{\epsilon,b}(u_{\epsilon},r_{\epsilon}^{\prime})|\geq\frac{1}{2}|\overline{h}_{\epsilon,b}(u_{\epsilon},r_{\epsilon})|.$
(B.4)

Now, assume that there exists $r<1-\epsilon$ satisfying

$\sqrt{r+\epsilon}\big{(}\overline{h}_{\epsilon,b}e^{\frac{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}{2}}\big{)}(u,r)\geq\frac{4}{c^{2}}\sqrt{\frac{m_{\epsilon,b}(u,b)}{2\pi}}.$
(B.5)

If
$|\overline{h}_{\epsilon,b}(u,r)-\overline{h}_{\epsilon,b}(u,1-\epsilon)|\leq\frac{1}{2}|\overline{h}_{\epsilon,b}(u,r)|,$
we obtain

$\big{(}|\overline{h}_{\epsilon,b}|e^{\frac{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}{2}}\big{)}(u,1-\epsilon)\geq\frac{\sqrt{r+\epsilon}}{2}\big{(}|\overline{h}_{\epsilon,b}|e^{\frac{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}{2}}\big{)}(u,r)\geq\frac{2}{c^{2}}\sqrt{\frac{m_{\epsilon,b}(u,b)}{2\pi}},$

hence, by (B.4), there exists $r^{\prime}\in(1-\epsilon,2-\epsilon]$
satisfying

$|\overline{h}_{\epsilon,b}(u,1-\epsilon)-\overline{h}_{\epsilon,b}(u,r^{\prime})|\geq\frac{1}{2}|\overline{h}_{\epsilon,b}(u,1-\epsilon)|.$

This tells us that if such $r$ exists (i.e. $r<1-\epsilon$ and satisfies
(B.5)), then there is $r_{*}\in(r,2-\epsilon]$ such that

$|\overline{h}_{\epsilon,b}(u,r)-\overline{h}_{\epsilon,b}(u,r_{*})|\geq\frac{1}{4}|\overline{h}_{\epsilon,b}(u,r)|.$
(B.6)

Combined with the fact that (cf. also (4.16))

$4\pi\int_{r}^{b}(s+\epsilon)(\partial_{r}\,\overline{h}_{\epsilon,b})^{2}ds\geq
4\sqrt{\pi}\bigg{|}\int_{r}^{r_{*}}\partial_{r}\,\overline{h}_{\epsilon,b}\,ds\bigg{|}-\int_{r}^{r_{*}}\frac{1}{s+\epsilon}\,ds+4\pi\int_{r_{*}}^{b}(s+\epsilon)(\partial_{r}\,\overline{h}_{\epsilon,b})^{2}ds,$

it then follows that

$(r+\epsilon)\big{(}|\overline{h}_{\epsilon,b}|^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\big{)}(u,r)\leq
2\,|\overline{h}_{\epsilon,b}(u,r)|^{2}e^{-\sqrt{\pi}\,|\overline{h}_{\epsilon,b}(u,r)|}\leq\frac{8}{e^{2}\pi}.$
(B.7)

Therefore, from (4.13) and (B.7) we find

$\displaystyle
c^{2}\int_{0}^{b}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,dr$
$\displaystyle=c^{2}\int_{0}^{1}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,dr+c^{2}\int_{1}^{b}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,dr,$
(B.8) $\displaystyle\leq
8\max\Big{(}\frac{c^{2}}{e^{2}\pi},\,\frac{m_{\epsilon,b}(u,b)}{\pi}\Big{)}+\frac{m_{\epsilon,b}(u,b)}{2\pi}.$

We now consider the last term of (B.2). Setting

$J_{\epsilon}:=-4\pi\,\int_{0}^{b}\partial_{r}\,\overline{h}_{\epsilon,b}\bigg{(}\int_{0}^{r}\overline{g}_{\epsilon,b}\partial_{r}\,\overline{h}_{\epsilon,b}\,ds+c^{2}\int_{0}^{r}(s+\epsilon)\,\overline{h}_{\epsilon,b}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,ds\bigg{)}\,dr,$

we have

$\displaystyle J_{\epsilon}$
$\displaystyle=-4\pi\,\overline{h}_{\epsilon,b}(b)\int_{0}^{b}\overline{g}_{\epsilon,b}\frac{h_{\epsilon,b}-\overline{h}_{\epsilon,b}}{s+\epsilon}\,ds+4\pi
c^{2}\overline{h}_{\epsilon,b}(b)\int_{0}^{b}(s+\epsilon)\,\overline{h}_{\epsilon,b}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,ds$
$\displaystyle\quad+2\pi\,\int_{0}^{b}\partial_{r}(\overline{h}_{\epsilon,b}^{2})\overline{g}_{\epsilon,b}dr+4\pi
c^{2}\int_{0}^{b}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}dr$
$\displaystyle\leq-2\pi\,\bigg{(}\overline{h}_{\epsilon,b}^{2}(b)\overline{g}_{\epsilon,b}(b)-2\overline{h}_{\epsilon,b}(b)\int_{0}^{b}\overline{h}_{\epsilon,b}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{s+\epsilon}ds\bigg{)}+4\pi
c^{2}|\overline{h}_{\epsilon,b}(b)|\sqrt{b}\bigg{(}\int_{0}^{b}(s+\epsilon)^{2}\overline{h}_{\epsilon,b}^{2}e^{2(\nu_{\epsilon,b}+\lambda_{\epsilon,b})}\,ds\bigg{)}^{1/2}$
$\displaystyle\quad-2\pi\,\int_{0}^{b}\overline{h}_{\epsilon,b}^{2}\partial_{r}\overline{g}_{\epsilon,b}dr+4\pi
c^{2}\int_{0}^{b}(r+\epsilon)\,\overline{h}_{\epsilon,b}^{2}e^{\nu_{\epsilon,b}+\lambda_{\epsilon,b}}\,dr,$

hence

$\displaystyle J_{\epsilon}$ $\displaystyle\leq
2m_{\epsilon,b}(u,b)+32\max\big{\\{}c^{2}e^{-2},m_{\epsilon,b}(u,b)\big{\\}}+33\pi
c^{2}\bigg{(}\max\big{\\{}4\pi
A^{4}(u,0),\,e^{-2\sqrt{2}\pi^{1/4}}\big{\\}}\bigg{)}$ (B.9)
$\displaystyle\quad+4\pi\bigg{(}\frac{A^{2}(u,0)}{b}\int_{0}^{b}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{(r+\epsilon)^{2}}\,dr+\frac{|c|A(u,0)}{\sqrt{b}}\sqrt{\frac{m_{\epsilon,b}(u,b)}{2\pi}}\bigg{)},$

where we the first inequality follows from Hölder’s inequality and the second
inequality follows from Lemma 4.3, (4.18), (4.13) and (B.8).

Now, taking (B.3), (B.8) and (B.9) into (B.2), we obtain

$\displaystyle\partial_{u}\bigg{(}\int_{0}^{b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\,dr\bigg{)}\leq\frac{1}{4\pi}+\frac{17}{\pi}\max\Big{(}c^{2}e^{-2},\,m_{\epsilon,b}(u,b)\Big{)}+\frac{33c^{2}}{4}\bigg{(}\max\big{\\{}4\pi
A^{4}(u,0),\,e^{-2\sqrt{2}\pi^{1/4}}\big{\\}}\bigg{)}$
$\displaystyle+\frac{1}{4\pi}\bigg{(}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{2}+2\pi\,\overline{g}_{\epsilon,b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\bigg{)}(u,b)+\bigg{(}\frac{A^{2}(u,0)}{b}\int_{0}^{b}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{(r+\epsilon)^{2}}\,dr+\frac{|c|A(u,0)}{\sqrt{b}}\sqrt{\frac{m_{\epsilon,b}(u,b)}{2\pi}}\bigg{)},$

hence

$\displaystyle\int_{0}^{b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\,dr$
$\displaystyle\leq\int_{0}^{b}(h_{0}-\overline{h}_{0})^{2}\,dr+\frac{u}{4\pi}+\frac{17}{\pi}\int_{0}^{u}\bigg{(}\max\big{\\{}c^{2}e^{-2},\,m_{\epsilon,b}(u_{1},b)\big{\\}}\bigg{)}\,du_{1}$
$\displaystyle\quad+\frac{33c^{2}}{4}\int_{0}^{u}\bigg{(}\max\big{\\{}4\pi
A(u_{1},0)^{4},\,e^{-2\sqrt{2}\pi^{1/4}}\big{\\}}\bigg{)}\,du_{1}+\frac{1}{4\pi}\int_{0}^{u}\bigg{(}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{2}+2\pi\,\overline{g}_{\epsilon,b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\bigg{)}(u_{1},b)\,du_{1}$
$\displaystyle\quad+\int_{0}^{u}\bigg{(}\frac{A^{2}(u_{1},0)}{b}\int_{0}^{b}\frac{g_{\epsilon,b}-\overline{g}_{\epsilon,b}}{(r+\epsilon)^{2}}\,dr+\frac{A(u_{1},0)}{\sqrt{b}}\sqrt{\frac{m_{\epsilon,b}(u_{1},b)}{\pi}}\bigg{)}\,du_{1}.$

Therefore, by Lemma 4.3 we have

$\displaystyle\int_{0}^{+\infty}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\,dr$
$\displaystyle\leq\int_{0}^{b}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\,dr+\int_{b}^{b+2}(h_{\epsilon,b}-\overline{h}_{\epsilon,b})^{2}\,dr+(b+2+\epsilon)^{2}|\overline{h}_{\epsilon,b}|^{2}(u,b+2+\epsilon)\int_{b+2}^{+\infty}\frac{1}{(r+\epsilon)^{2}}\,dr$
(B.10)
$\displaystyle\leq\int_{0}^{b}(h_{0}-\overline{h}_{0})^{2}\,dr+\frac{u}{4\pi}+\frac{17}{\pi}\int_{0}^{u}\bigg{(}\max\big{\\{}c^{2}e^{-2},\,m_{\epsilon,b}(u_{1},b)\big{\\}}\bigg{)}\,du_{1}$
$\displaystyle\quad+\frac{33c^{2}}{4}\int_{0}^{u}\bigg{(}\max\Big{(}4\pi
A(u_{1},0)^{4},\,e^{-2\sqrt{2}\pi^{1/4}}\Big{)}\bigg{)}\,du_{1}+B_{2}(u,b,c_{1}).$

This completes the proof of Proposition 4.5.

##  Appendix C Proof of technical lemmas

###### Proof of Lemma 6.1.

By the Hölder inequality we have
$|\overline{h}_{\epsilon}|\leq\frac{1}{r}\int_{0}^{r}|h_{\epsilon}(u,s)|\,ds\leq\frac{1}{\sqrt{r}}\left(\int_{0}^{r}h^{2}_{\epsilon}(u,s)\,ds\right)^{1/2},$
hence

$\sup_{r\geq\delta}|\overline{h}_{\epsilon}(u,r)|\leq\frac{1}{\sqrt{\delta}}\left(\int_{0}^{+\infty}h^{2}_{\epsilon}(u,s)\,ds\right)^{1/2},$

and the result then follows from the inequality (3) in (2.7). ∎

###### Proof of Lemma 6.2.

From the proof of Theorem 2.1 it follows that

$\lim_{r\to+\infty}\big{(}\sup_{u\in[0,u_{0}]}|h_{\epsilon}|\big{)}=\lim_{r\to+\infty}\big{(}\sup_{u\in[0,u_{0}]}|\partial_{r}h_{\epsilon}|\big{)}=0,$

so it only remains to check that

$\lim_{r\to+\infty}\big{(}\sup_{u\in[0,u_{0}]}|(r+\epsilon)\,\overline{h}_{\epsilon}(u,r)|\big{)}=0.$
(C.1)

We argue by contradiction. Suppose that there exists a sequence
$(u_{i},r_{i})$ in $[0,u_{0}]\times[0,+\infty)$ converging to
$(u_{\infty},+\infty)$ such that

$\alpha:=\lim\big{(}|(r_{i}+\epsilon)\,\overline{h}_{\epsilon}|(u_{i},r_{i})\big{)}>0.$

For any $\zeta>0$, let $u_{\zeta}\in[0,u_{0}]$ be such that
$|u_{\zeta}-u_{\infty}|\leq\zeta$. Observe that the inequalities (3) and (5)
in (2.7) tell us that for, each $u$, $(r+\epsilon)h_{\epsilon}(u_{\zeta},r)\in
W^{1,2}$, hence by the Sobolev embedding theorem

$\int_{0}^{+\infty}h_{\epsilon}(u_{\zeta},r)\,dr=0.$ (C.2)

Next, similarly to (4.6), we have

$(r_{i}+\epsilon)|\overline{h}_{\epsilon}|(u_{i},r_{i})\leq\Big{\|}\int_{0}^{r_{i}}h_{\epsilon}(u_{\zeta},r)\,dr\Big{\|}_{C^{0}([r_{i},+\infty))}+\sqrt{\frac{\zeta
M_{\epsilon}(0)}{2\pi}},$

which, in combination with (C.2), gives us

$\alpha=\lim\big{(}|(r_{i}+\epsilon)\,\overline{h}_{\epsilon}|(u_{i},r_{i})\big{)}\leq\sqrt{\frac{\zeta
M_{\epsilon}(0)}{2\pi}},$

which is a contradiction if $\zeta\leq\frac{\alpha^{2}\pi}{M_{\epsilon}(0)}$.
∎

###### Proof of Lemma 6.3.

In view of the inequality (2) in (2.6), (5.19), (5.20) and (5.24), we obtain
from (5.3) on $\mathcal{O}_{\delta}$ that

$\big{|}D_{\epsilon}(\partial_{r}h_{\epsilon})\big{|}\leq K_{2}(u,\delta),$
(C.3)

where

$\displaystyle K_{2}(u,\delta)$
$\displaystyle:=\frac{2M_{\epsilon}(0)}{\delta}B_{8}(u,\delta)+\frac{c^{2}}{2}B_{5}(u,\delta)+\left(B_{5}+\frac{\mathcal{A}_{\epsilon}}{\delta}\right)\left(\frac{2\pi
c^{2}\mathcal{A}_{\epsilon}}{\delta}\left(B_{5}+\frac{\mathcal{A}_{\epsilon}}{\delta}\right)+\frac{B_{7}}{2\delta}+\frac{3M_{\epsilon}(0)}{\delta^{3}}\right)(u,\delta).$

Thanks to (C.3), we only need to prove that $\\{\partial_{r}h_{\epsilon}\\}$
is equicontinuous in $\mathcal{O}_{\delta}$ with respect to $r$. In fact,
uniform continuity with respect to $u$ would then follow by the following
arguments. Given $u_{1},r_{1}$ and $\Delta u_{1}>0$, let $r_{1}+\Delta
r_{1,\epsilon}$ be the value of $r$ at which the characteristic
$\chi_{\epsilon}(u;u_{1},r_{1})$ intersects the line $u=u_{1}-\Delta u_{1}$.
Observe that $\Delta r_{1,\epsilon}\leq 1/2\Delta u_{1}$ and then

$\displaystyle\big{|}\partial_{r}h_{\epsilon}(u_{1},r_{1})-\partial_{r}h_{\epsilon}(u_{1}-\Delta
u_{1},r_{1})\big{|}$ $\displaystyle\leq\int_{u_{1}-\Delta
u_{1}}^{u_{1}}|D_{\epsilon}(\partial_{r}h_{\epsilon})|du+\big{|}\partial_{r}h_{\epsilon}(u_{1}-\Delta
u_{1},r_{1}+\Delta r_{1,\epsilon})-\partial_{r}h_{\epsilon}(u_{1}-\Delta
u_{1},r_{1})\big{|}$ $\displaystyle\leq K_{2}(u_{0},\delta)\Delta
u_{1}+\eta\Delta r_{1,\epsilon}\leq K_{2}(u_{0},\delta)\Delta
u_{1}+\frac{1}{2}\eta\Delta u_{1},$

where $\eta$ denotes the common modulus of continuity of the family
$\\{\partial_{r}h_{\epsilon}\\}$ with respect to $r$.

Let now $\delta\leq r_{1}\leq r_{1}^{\prime}$. Given $u_{1}\leq u_{0}$, we see
that the characteristics $\chi_{\epsilon}(u;u_{1},r_{1})$ and
$\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})$ are contained in
$\mathcal{O}_{\delta}$ for $0\leq u\leq u_{1}$. Define

$\psi_{\epsilon}(u):=\partial_{r}h_{\epsilon}(u,\chi_{\epsilon}(u;u_{1},r_{1}))-\partial_{r}h_{\epsilon}(u,\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})).$

It then follows from (5.23) that

$\displaystyle\partial_{u}\psi_{\epsilon}$
$\displaystyle=\big{(}\partial_{r}\overline{g}_{\epsilon}\big{)}_{(u,\chi_{\epsilon}(u;u_{1},r_{1}))}\psi_{\epsilon}(u)+\Big{(}\big{(}\partial_{r}\overline{g}_{\epsilon}\big{)}_{(u,\chi_{\epsilon}(u;u_{1},r_{1}))}-\big{(}\partial_{r}\overline{g}_{\epsilon}\big{)}_{(u,\chi_{\epsilon}(u;u_{1},r_{1}^{\prime}))}\Big{)}\big{(}\partial_{r}h_{\epsilon}\big{)}_{(u,\chi_{\epsilon}(u;u_{1},r_{1}^{\prime}))}$
(C.4)
$\displaystyle\quad+f_{\epsilon}(u,\chi_{\epsilon}(u;u_{1},r_{1}))-f_{\epsilon}(u,\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})),$

where

$f_{\epsilon}:=\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{2(r+\epsilon)}\partial_{r}g_{\epsilon}-\frac{3}{2(r+\epsilon)^{2}}(g_{\epsilon}-\overline{g}_{\epsilon})(h_{\epsilon}-\overline{h}_{\epsilon})-\frac{c^{2}}{2}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\big{(}h_{\epsilon}+4\pi\overline{h}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}\big{)}.$

By straightforward calculations, we find

$\displaystyle\partial_{rr}\overline{g}_{\epsilon}$
$\displaystyle=\frac{\partial_{r}g_{\epsilon}}{r+\epsilon}-2\frac{g_{\epsilon}-\overline{g}_{\epsilon}}{(r+\epsilon)^{2}},$
$\displaystyle\partial_{rr}g_{\epsilon}$
$\displaystyle=4\pi\,\partial_{r}g_{\epsilon}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{r+\epsilon}-12\pi\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{(r+\epsilon)^{2}}g_{\epsilon}+8\pi\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{r+\epsilon}g_{\epsilon}\partial_{r}h_{\epsilon}$
$\displaystyle\quad-8\pi
c^{2}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\big{(}h_{\epsilon}^{2}+(r+\epsilon)\,\overline{h}_{\epsilon}\partial_{r}h_{\epsilon}+4\overline{h}_{\epsilon}h_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}\big{)},$
$\displaystyle\partial_{r}f_{\epsilon}$
$\displaystyle=-\frac{3}{2(r+\epsilon)^{2}}(g_{\epsilon}-\overline{g}_{\epsilon})\partial_{r}h_{\epsilon}+\Big{(}\frac{5}{2(r+\epsilon)^{2}}(h_{\epsilon}-\overline{h}_{\epsilon})+\frac{\partial_{r}h_{\epsilon}}{2(r+\epsilon)}\Big{)}\partial_{r}g_{\epsilon}$
$\displaystyle\quad+\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{2(r+\epsilon)}\partial_{rr}g_{\epsilon}+\frac{6}{(r+\epsilon)^{3}}(g_{\epsilon}-\overline{g}_{\epsilon})(h_{\epsilon}-\overline{h}_{\epsilon})$
$\displaystyle\quad-\frac{c^{2}}{2}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\Big{(}\partial_{r}h_{\epsilon}+4\pi\,\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{3}}{r+\epsilon}\Big{)}-2\pi
c^{2}e^{\nu_{\epsilon}+\lambda_{\epsilon}}\Big{(}h_{\epsilon}+4\pi\,\overline{h}_{\epsilon}(h_{\epsilon}-\overline{h}_{\epsilon})^{2}\Big{)}\frac{(h_{\epsilon}-\overline{h}_{\epsilon})^{2}}{r+\epsilon}.$

Therefore, by Lemma 5.2 we have
$|\partial_{rr}\overline{g}_{\epsilon}|,|\partial_{r}f_{\epsilon}|\leq
K_{3}(u_{0},\delta)$ for some constant $K_{3}(u_{0},\delta)>0$ independent of
$\epsilon$, hence

$\displaystyle\Big{|}\big{(}\partial_{r}\overline{g}_{\epsilon}\big{)}_{(u,\chi_{\epsilon}(u;u_{1},r_{1}))}-\big{(}\partial_{r}\overline{g}_{\epsilon}\big{)}_{(u,\chi_{\epsilon}(u;u_{1},r_{1}^{\prime}))}\Big{|}$
$\displaystyle\leq
K_{3}(u_{0},\delta)\big{(}\chi_{\epsilon}(u;u_{1},r_{1})-\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})\big{)},$
(C.5)
$\displaystyle\Big{|}f_{\epsilon}(u,\chi_{\epsilon}(u;u_{1},r_{1}))-f_{\epsilon}(u,\chi_{\epsilon}(u;u_{1},r_{1}^{\prime}))\Big{|}$
$\displaystyle\leq
K_{3}(u_{0},\delta)\big{(}\chi_{\epsilon}(u;u_{1},r_{1})-\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})\big{)}.$

Combining this result with (5.19), in view of (C.4) we obtain

$|\partial_{u}\psi_{\epsilon}|\leq\frac{K_{4}(u_{0},\delta)}{\delta}|\psi_{\epsilon}|+K_{5}(u_{0},\delta)\big{(}\chi_{\epsilon}(u;u_{1},r_{1})-\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})\big{)},$
(C.6)

where

$\displaystyle K_{4}(u_{0},\delta):$ $\displaystyle=\max\Big{(}1,2(1+2\pi
c^{2}\mathcal{A}^{2}_{\epsilon}(u_{0},\delta))\Big{)},\qquad
K_{5}(u_{0},\delta):=K_{3}(u_{0},\delta)\big{(}1+B_{5}(u_{0},\delta)\big{)}.$

Therefore, integrating (C.6) yields

$|\psi_{\epsilon}(u_{1})|\leq
e^{\frac{K_{4}(u_{0},\delta)u_{1}}{\delta}}\Big{(}\psi_{\epsilon}(0)+K_{5}(u_{0},\delta)\int_{0}^{u_{1}}\big{(}\chi_{\epsilon}(u;u_{1},r_{1})-\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})\big{)}du\Big{)}.$
(C.7)

Now, since both characteristics $\chi_{\epsilon}(u;u_{1},r_{1})$ and
$\chi_{\epsilon}(u;u_{1},r_{1}^{\prime})$ are contained in
$\mathcal{O}_{\delta}$, the following inequality holds (thanks to (5.19) and
the mean value theorem):

$\displaystyle\chi_{\epsilon}(u;u_{1},r_{1})-\chi_{\epsilon}(u;u_{1},r^{\prime}_{1})$
$\displaystyle=(r_{1}-r_{1}^{\prime})e^{\frac{1}{2}\int_{u}^{u_{1}}\big{(}\frac{g_{\epsilon}-\overline{g}_{\epsilon}}{r+\epsilon}\big{)}_{(u^{\prime},\chi_{\epsilon}(u^{\prime};u_{1},s))}\,du^{\prime}}\leq(r_{1}-r^{\prime}_{1})e^{\frac{(u_{1}-u)M_{\epsilon}(0)}{\delta^{2}}}$

for some $s\in[r_{1}^{\prime},r_{1}]$. Hence, from (C.7) we have

$|\psi_{\epsilon}(u_{1})|\leq
e^{\frac{K_{4}(u_{0},\delta)u_{0}}{\delta}}\Big{(}\eta_{0}+K_{5}(u_{0},\delta)u_{0}e^{\frac{u_{0}M_{\epsilon}(0)}{\delta^{2}}}\Big{)}(r_{1}-r_{1}^{\prime}),$

where $\eta_{0}$ is a modulus of continuity of $\partial_{r}h_{0}$. ∎

###### Proof of Lemma 6.4.

We have
$|\partial_{r}\,\overline{h}_{\epsilon}|=\bigg{|}\frac{h_{\epsilon}-\overline{h}_{\epsilon}}{r+\epsilon}\bigg{|}\leq
K_{6}(u_{0},\delta),$ for all $(u,r)\in\mathcal{O}_{\delta}$, in which
$K_{6}(u_{0},\delta):=\frac{B_{5}\delta+A(u_{0},\delta)}{\delta^{2}}$.
Consequently, $\\{\overline{h}_{\epsilon}\\}$ is equicontinuous with respect
to $r$ in $\mathcal{O}_{\delta}$. We now show equi-continuity with respect to
$u$. In fact, given $u_{1},r_{1}$ and $\Delta u_{1}>0$, let $r_{1}+\Delta
r_{1,\epsilon}$ be the value of $r$ at which the characteristic
$\chi_{\epsilon}(u;u_{1},r_{1})$ intersects the line $u=u_{1}-\Delta u_{1}$.
Observe that $\Delta r_{1,\epsilon}\leq 1/2\Delta u_{1}$. Then, we have

$\displaystyle\big{|}\overline{h}_{\epsilon}(u_{1},r_{1})-\overline{h}_{\epsilon}(u_{1}-\Delta
u_{1},r_{1})\big{|}$ $\displaystyle\leq\Big{|}\int_{u_{1}-\Delta
u_{1}}^{u_{1}}\Big{(}D_{\epsilon}\overline{h}_{\epsilon}\Big{)}_{\chi_{\epsilon,u_{1}}}\,du\Big{|}+\big{|}\overline{h}_{\epsilon}(u_{1}-\Delta
u_{1},r_{1}+\Delta r_{1,\epsilon})-\overline{h}_{\epsilon}(u_{1}-\Delta
u_{1},r_{1})\big{|}$
$\displaystyle\leq\frac{1}{\delta}\Big{|}\int_{u_{1}-\Delta
u_{1}}^{u_{1}}\Big{(}(r+\epsilon)D_{\epsilon}\overline{h}_{\epsilon}\Big{)}_{\chi_{\epsilon,u_{1}}}\,du\Big{|}+K_{6}(u_{0},\delta)\Delta
r_{1,\epsilon}$ $\displaystyle\leq\frac{\sqrt{\Delta
u_{1}}}{\delta}\Big{(}\int_{u_{1}-\Delta
u_{1}}^{u_{1}}\Big{(}(r+\epsilon)^{2}(D_{\epsilon}\overline{h}_{\epsilon})^{2}\Big{)}_{\chi_{\epsilon,u_{1}}}\Big{)}^{1/2}+\frac{K_{6}(u_{0},\delta)}{2}\Delta
u_{1}\leq\frac{1}{2\delta}\sqrt{\frac{M_{\epsilon}(0)\Delta
u_{1}}{\pi}}+\frac{K_{6}(u_{0},\delta)}{2}\Delta u_{1},$

which implies the equi-continuity with respect to $u$ as claimed. ∎