# On Jang’s equation and the positive mass theorem for asymptotically
hyperbolic initial data sets with dimensions above three and below eight

David Lundberg  Uppsala University
Department of Mathematics
Box 480, 751 06 UPPSALA<EMAIL_ADDRESS>
###### Abstract.

We solve the Jang equation with respect to asymptotically hyperbolic
“hyperboloidal” initial data in dimensions $n=4,5,6,7$. This gives a non-
spinor proof of the positive mass theorem in the asymptotically hyperbolic
setting in these dimensions. Our work extends an earlier result of [Sak21]
obtained in dimension $n=3$.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Initial data sets and their masses
    2. 2.2 Jang’s equation
  3. 3 Barrier construction
    1. 3.1 Heuristic analysis
    2. 3.2 Barrier construction
  4. 4 The regularized Jang equation as a Dirichlet problem
  5. 5 A geometric solution to Jang’s equation
    1. 5.1 Limit and regularity
    2. 5.2 Topology of the Jang graph
  6. 6 Asymptotic flatness of the Jang graph
    1. 6.1 Estimates for the second fundamental form near infinity
    2. 6.2 Setup and Fermi coordinates
    3. 6.3 Existence of the height function and a priori estimates
    4. 6.4 The Jang equation in terms of the height function
  7. 7 The conformal structure of the Jang graph
    1. 7.1 Conformal change
    2. 7.2 Deformation to Asymptotically Schwarzschildean metric
    3. 7.3 Undarning of the conical singularitites
  8. 8 The positive mass theorem
    1. 8.1 Positivity; $E\geq|P|$
    2. 8.2 Rigidity; $E=0$
  9. A Computations for Wang’s asymptotics
  10. B Geometry of a smooth approximate Jang graph
  11. C The ADM energy of the Jang graph
  12. D Some properties of Fermi coordinates
  13. E Geometric Measure Theory

##  1\. Introduction

General Relativity has had a beautiful and fruitful history of interplay with
Matematics and, in particular, both Geometry and Partial Differential
Equations. A very important example of this is the classical Positive Mass (or
Energy) Theorem. In physical terms, the theorem asserts that the mass of an
isolated gravitational system with non-negative energy density is non-
negative. At first glance, this may seem to be purely physical statement, but
it is very geometrical and has far reaching mathematical consequences not only
within General Relativity.
Central to the Positive Energy Theorem is the notions of initial data sets
$(M^{n},g,k)$. Roughly speaking, this is a Riemannian manifold $(M^{n},g)$ and
a symmetric $(0,2)$-tensor $k$ that is thought of a ”constant time slice” in
some spacetime with $k$ as its second fundamental form. An initial data set
$(M^{n},g,k)$ is said to be asymptotically Euclidean if the metric tends to
the Euclidean metric in a chart at infinity, that is $g\rightarrow\delta$ at a
certain rate. Such initial data sets are used in General Relativity to model
isolated systems. Defined by Arnowitt, Deser and Misner [ADM59] in 1959, the
ADM energy is a flux-integral computed at infinity of $(M^{n},g,k)$, which
turns out to be a coordinate invariant. The question whether this quantity is
non-negative under physically reasonable assumptions is known as the positive
energy conjecture. Results proving this conjecture were obtained by Schoen and
Yau in [SY79b], [SY81b] and Witten [Wit81]. The result of [SY79b] was obtained
for dimension $n=3$ and $k=0$ (the so-called Riemannian or time-symmetric
case, where the dominant energy condition implies $R_{g}\geq 0$) using minimal
surface techniques and the result of [SY81b] was proven for general $k$ (the
so-called spacetime case) by reduction to the $k=0$ case using the Jang
equation. The result of [Wit81] holds in all dimensions but requires the
assumption that $(M^{n},g)$ be spin, which imposes additional restrictions on
the topology in dimensions above $3$. The results of [SY79b] and [SY81b] have
been extended to dimensions $3\leq n\leq 7$ in [SY79a] and [Eic13],
respectively.
Since the original work of Schoen and Yau in [SY79b] and [SY81b] the
positivity of mass for asymptotically Euclidean manifolds and asyptotically
Euclidean initial data sets have remained an active area of research. Very
recently, many new methods have been introduced to the field, for example the
level set methods (see for instance [AMO21], [BKKS22] and [HKK22]) and
$\mu$-bubbles of Gromov (see [LUY21]). Furthermore, the analogue of the
minimal surface tecnhique of [SY79b] has been developed for initial data sets
in [EHLS16]. We would also like to highlight the recent proofs of optimal
rigidity results characterizing the case when the mass is zero, see [HL20] and
[HL23].
Another important class of initial data is the so-called asymptotically
hyperbolic initial data which is characterized by $g\rightarrow b$, where $b$
is the hyperbolic metric. Two main models that are used to define
asymptotically hyperbolic initial data are upper unit hyperboloid of Minkowski
space, which is an umbilic hypersurface (that is $k=g$), and the
$\\{t=0\\}$-slice of the anti-de Sitter spacetime, which is a totally geodesic
hypersurface (that is $k=0$). Mass for asymptotically hyperbolic manifolds was
first defined by Wang in [Wan01] and a positive mass theorem was proven under
the assumption that $M^{n}$ be spin. The result in [Wan01] was subsequently
extended by Chrusciel and Herzlich in [CH03] to the case of more general
asymptotics. These Riemannian results can be interpreted as positive mass
theorems for asymptotically anti-de Sitter initial data sets with $k=0$
alternatively asymptotically hyperboloidal initial data sets with $k=g$, where
in both cases the dominant energy condition is equivalent to
$R_{g}\geq-n(n-1)$. Positivity of mass for asymptotically hyperbolic manifolds
$(M^{n},g)$ was also proven in [ACG08] in dimensions $3\leq n\leq 7$ under
aditional assumptions on the geometry at infinity. These assumptions have
recently been removed in [CGNP18].
As already mentioned, in the spacetime hyperbolic setting we can have either
$k\rightarrow 0$ or $k\rightarrow g$. Results have been obtained under the
spin assumption in both cases. See, for instance, [Zha99], [CJL04], [CM06],
[Mae06], [CMT06], [Zha04], [XZ08], [WX15].
In this work, we establish a positive mass theorem for asymptotically
hyperboloidal initial data. For this, we use a technique introduced in [SY81b]
known as _Jang equation reduction_. In this procedure one considers the
Riemannian product $(M^{n}\times\mathbb{R},g+dt^{2})$ and solves a certain
prescribed mean curvature equation $\mathcal{J}(f)=0$. One then performs some
additional deformations on the graph of the solution
$f:M^{n}\rightarrow\mathbb{R}$, after which the minimal surface proof used in
[SY79b] can be applied, to conclude that the energy is non-negative. We prove
the following result, which extends that of [Sak21] obtained in dimension
$n=3$.

######  Theorem 1.1.

Let $(M^{n},g,k)$ be initial data of type $(\ell,\alpha,\tau,\tau_{0})$, where
$4\leq n\leq 7$, $\ell\geq 6$, $0\leq\alpha<1$, $\frac{n}{2}<\tau<n$ and
$\tau_{0}>0$. If the dominant energy condition $\mu\geq|J|_{g}$ holds, then
the mass vector is future pointing causal, $E\geq|\vec{P}|$.
If, in addition, $(M^{n},g,k)$ has Wang’s asymptotics and $E=0$ then
$(M^{n},g,k)$ can be isometrically embedded into Minkowski space
$\mathcal{M}^{n+1}$ as a spacelike hypersurface with second fundamental form
$k$.

Again, the cornerstone of the proof of Theorem 1.1 is the method of Jang
equation reduction originating from [SY81b]. However, the setting of [SY81b]
is asymptotically Euclidean and the dimension is $n=3$, whereas the current
work deals with the asymptotically hyperbolic setting and dimensions $4\leq
n\leq 7$. This requires, in many cases, more recent approaches to the Jang
equation reduction. In particular, our construction of the geometric solution
relies on geometric measure theory methods developed by Eichmair in [Eic13],
and we also mostly follow [Eic13] when dealing with the blow ups and blow
downs of Jang’s equation. Regarding the construction of barriers for Jang’s
equation, we use the ”asymptotic ODE” method of [Sak21], developed
specifically to deal with the more complicated asymptotics encountered in the
asymptotically hyperbolic setting. We also use the ”graph over barrier”
approach from [Sak21] to show that the Jang graph has asymptotically Euclidean
asymptotics.
We would like to point out other applications of Jang’s equation besides the
proof of positive mass theorem. As discussed in Section 5 the blow ups of the
geometric solution occur on the so-called marginally outer (inner) trapped
surfaces (MOTS or MITS). This feature has been used to prove existence of
MOTS/MITS in certain initial data sets, see [Eic09] and [AEM11]. Furthermore,
[BKS21] have applied the Jang’s equation to obtain results on stability of the
spacetime positive mass theorem in the spherically symmetric setting. There is
also a plethora of reduction arguments for various geometric inequalities,
starting from [BK11].
The paper is organized as follows. In Section 2 we clarify the notations and
discuss preliminaries required for this work. In Section 3 we construct
barriers for Jang’s equation assuming Wang’s asymptotics. In Section 4 we
solve the regularized Jang equation $\mathcal{J}(f_{\tau})=\tau f_{\tau}$ on
coordinate balls of radius $R$. In Section 5 we use techniques from geometric
measure theory to obtain a geometric solutions of Jang’s equation, that is to
say a limit hypersurface for $\tau\rightarrow 0$ and $R\rightarrow\infty$ and
discuss its possible blow up sets. In Section 6 we prove that the obtained
geometric solution is asymptotically Euclidean. In Section 7 we conformally
change the metric to achieve zero scalar curvature and we also improve the
asymptotics near infinity. Furthermore, we discuss how to deal with conical
singularities that arise when compactifying the cylindrical ends by a
conformal change. Finally, the positive mass theorem is proven in Section 8.
Some supplementary results are collected in the appendices.

### Acknowledgments

This project was suggested and supervised by Anna Sakovich, who contributed
many important suggestions and constructive comments over its span. The
project was partly supported by the Swedish Research Council’s grant dnr.
2016-04511.
The current version of this article is part of the author’s thesis at Uppsala
University.

##  2\. Preliminaries

###  2.1. Initial data sets and their masses

In this work we adopt the following definition of initial data sets:

######  Definition 2.1.

Let $n\geq 3$ and $(M^{n},g)$ be an orientable, $n$-dimensional
$C^{2}$-regular Riemannian manifold without boundary. Let $k\in
C^{1}(\text{Sym}^{2}(T^{\ast}M^{n}))$ be a symmetric $(0,2)$-tensor. Then the
triple $(M^{n},g,k)$ is called an _initial data set_. The equations

$\begin{split}R_{g}-|k|^{2}_{g}+\operatorname{trace}^{g}(k)^{2}&=2\mu,\\\
\text{div}^{g}\big{(}k-\operatorname{trace}^{g}(k)\cdot
g\big{)}&=J,\end{split}$ (2.1)

are called the _constraint equations_ , where $\mu$ is the _local mass
density_ and $J$ is the _local current density_. The condition

$\mu\geq|J|_{g}$ (2.2)

is called the _dominant energy condition_. The so-called _Riemannian_ (or
_time-symmetric_) setting is characterized by $k\equiv 0$.

We recall the ”hyperboloidal” model for $n$-dimensional hyperbolic space
$\mathbb{H}^{n}=(\mathbb{R}^{n},b)$ where $\mathbb{H}^{n}$ arises as the graph

$\\{(t,r,\theta)\>|\>t=\sqrt{1+r^{2}}\\}$ (2.3)

in Minkowski space
$\mathcal{M}^{n+1}=(\mathbb{R}\times\mathbb{R}^{n},\eta=-dt^{2}+dr^{2}+r^{2}\Omega)$,
where $r$ and $\theta$ are spherical coordinates on $\mathbb{R}^{n}$ and
$\Omega$ is the Euclidean metric $\delta$ induced on $\mathbb{S}^{n-1}$. In
this case, the induced metric $g$ on the graph and the second fundamental form
$k$ of the graph satisfy $g=k=b$, where

$b=\frac{dr^{2}}{1+r^{2}}+r^{2}\Omega.$ (2.4)

In this work we use the same definition of asymptotically hyperbolic
”hyperboloidal” initial data sets as in [DS21]. The reader is referred to
[DS21] for the definition of the weighted Hölder spaces
$C^{\ell,\alpha}_{\tau}$ used below.

######  Definition 2.2.

Let $(M^{n},g,k)$ be initial an data set. We say that $(M^{n},g,k)$ is
_asymptotically hyperbolic of type_ $(\ell,\alpha,\tau,\tau_{0})$ for
$\ell\geq 2$, $0\leq\alpha<1$, $\tau>\frac{n}{2}$ and $\tau_{0}>0$, if $g\in
C^{\ell,\alpha}_{\tau}(M^{n})$, $k\in C^{\ell-1,\alpha}_{\tau}(M^{n})$ and
there is a compact set $K\subset M^{n}$ and a diffeomorphism
$\Psi:M^{n}\setminus K\rightarrow\mathbb{R}^{n}\setminus\overline{B}_{1}(0)$
such that

  1. (1)

$e=\Psi_{\ast}(g)-b\in
C^{\ell,\alpha}_{\tau}(\mathbb{H}^{n}\setminus\bar{B}_{1}(0);\text{Sym}^{2}(T^{\ast}\mathbb{H}^{n}))$,

  2. (2)

$\eta=\Psi_{\ast}(k-g)\in
C^{\ell-1,\alpha}_{\tau}(\mathbb{H}^{n}\setminus\bar{B}_{1}(0);\text{Sym}^{2}(T^{\ast}\mathbb{H}^{n}))$,

  3. (3)

$\Psi_{\ast}(\mu),\Psi_{\ast}(J)\in
C^{\ell-2,\alpha}_{\tau_{0}+n}(\mathbb{H}^{n}\setminus\bar{B}_{1}(0))$.

In this work it will be sufficient to work with simpler asymptotics similar to
those used in [Wan01].

######  Definition 2.3.

Let $(M^{n},g,k)$ be an asymptotically hyperbolic initial data of type
$(\ell,\alpha,\tau=n,\tau_{0})$ as in Definition 2.2. Then $(M^{n},g,k)$ is
said to have _Wang’s asymptotics_ if

  1. (1)

$\Psi_{\ast}(g)-b=\textbf{m}r^{-(n-2)}+\mathcal{O}^{\ell,\alpha}(r^{-(n-1)})$,

  2. (2)

$\big{(}\Psi_{\ast}(k)-b\big{)}|_{T\mathbb{S}^{n-1}\times
T\mathbb{S}^{n-1}}=\textbf{p}r^{-(n-2)}+\mathcal{O}^{\ell-1,\alpha}(r^{-(n-1)})$,

where $\textbf{m},\textbf{p}\in
C^{\ell,\alpha}(\mathbb{S}^{n-1};\text{Sym}^{2}(T^{\ast}\mathbb{S}^{n-1}))$
are symmetric $(0,2)$-tensors on $\mathbb{S}^{n-1}$ and $\Omega$ the standard
Euclidean metric on $\mathbb{S}^{n-1}$. The expressions
$\mathcal{O}^{\ell,\alpha}(r^{-\tau})$ are symmetric tensors in
$C^{\ell,\alpha}(\mathbb{S}^{n-1};\text{Sym}^{2}(T^{\ast}\mathbb{S}^{n-1}))$
with norms in $C^{\ell,\alpha}_{\tau}(\mathbb{H}^{n})$.

Throughout this work we will suppress the dependence on the chart and write,
for instance, $\Psi^{\ast}(g)=g$ as long as there is no risk for confusion.
We now discuss the notion of mass for asymptotically hyperbolic initial datas.
Let

$\mathcal{N}=\\{V\in C^{\infty}(\mathbb{H}^{n})\>|\>\text{Hess}^{b}V=Vb\\}.$
(2.5)

Then

$\mathcal{N}=\text{span}_{\mathbb{R}}\\{V_{0}=\sqrt{1+r^{2}},V_{1}=\hat{x}^{1}r,\ldots,V_{n}=\hat{x}^{n}r\\},$
(2.6)

where $\hat{x}^{i}=\frac{x^{i}}{r}$ is the $i^{th}$ coordinate of
$\mathbb{R}^{n}$ restricted to the unit sphere. These functions may be
interpreted as the coordinate functions of Minkowski space restricted to the
upper unit hyperboloid, as defined in (2.3).

######  Definition 2.4.

Let $(M^{n},g,k)$ be an asymptotically hyperbolic initial data set as in
Definition 2.2 with respect to a chart $\Psi$ at infinity. The map
$\mathcal{M}_{\Psi}:\mathcal{N}\rightarrow\mathbb{R}$ defined as the integral
at infinity

$\begin{split}\mathcal{M}_{\Psi}(V)&=\lim_{R\rightarrow\infty}\int_{\\{r=R\\}}\bigg{(}V\big{(}\text{div}^{b}(e)-d\operatorname{trace}^{b}(e)\big{)}\\\
&\qquad+\operatorname{trace}^{b}(e)dV-(e+2\eta)(\nabla^{b}V,\cdot)\bigg{)}(\vec{n}_{r}^{b})d\mu^{b}\end{split}$
(2.7)

is called the _mass functional_. Here
$\vec{n}^{b}_{r}=\sqrt{1+r^{2}}\partial_{r}$ is the outward pointing unit
normal with respect to the hyperbolic metric. The vector $(E,\vec{P})$, where

$E=\frac{\mathcal{M}_{\Psi}(V_{0})}{2(n-1)\omega_{n-1}}\qquad\text{and}\qquad
P^{i}=\frac{\mathcal{M}_{\Psi}(V_{i})}{2(n-1)\omega_{n-1}},$ (2.8)

is called the _mass vector_. Its Minkowskian length
$m=\sqrt{-|(E,\vec{P})|^{2}_{\eta}}=\sqrt{E^{2}-|\vec{P}|^{2}}$ is called the
_mass_.

The mass vector $(E,\vec{P})$ is clearly a coordinate dependent. Moreover, for
an isometry $A$ of hyperbolic space $\mathbb{H}^{n}$ it can be shown that if
$\Psi$ is a chart at infinity as in Definition 2.2 the composition
$A\circ\Psi$ is also such a chart. It follows immediately from the definition
that the mass functional transforms equivariantly under such a composition,
that is $\mathcal{M}_{A\circ\Psi}(V)=\mathcal{M}_{\Psi}(V\circ A)$. In
particular, this shows that the mass is a coordinate invariant. For further
details on this, we refer to [CN01], [CH03] and [Mic11].
If the chart $\Psi$ in Definition 2.2 is such that the mass vector takes the
form $(E,\vec{0})$ we use terminology coined in [CCS16] and say that $\Psi$ is
_balanced_. If the mass vector is causal it is possible to find such a chart.
The following Theorem 2.5 is a density theorem, proven in [DS21], important
for our work.

######  Theorem 2.5.

Let $(M^{n},g,k)$ be initial data as in Definition 2.2 of type
$(\ell,\alpha,\tau,\tau_{0})$, where $\ell\geq 3$, $0<\alpha<1$,
$\frac{n}{2}<\tau<n$ and $0<\tau_{0}$, and the dominant energy condition
$\mu\geq|J|_{g}$ holds. Then, for any $\epsilon>0$ there exists an initial
data set $(M^{n},\hat{g},\hat{k})$ of type $(\ell-1,\alpha,n,\hat{\tau}_{0})$,
where $\hat{\tau}_{0}>0$, with Wang’s asymptotics (possibly with respect to a
different chart $\hat{\Psi}$) such that the strict dominant energy condition
holds:

$\hat{\mu}>|\hat{J}|_{g},$ (2.9)

and

$|E-\hat{E}|<\epsilon.$ (2.10)

For future reference we include the following definition.

######  Definition 2.6.

Let $(M^{n},g)$ be an $n$-dimensional Riemannian manifold. We say that
$(M^{n},g)$ is _asymptotically flat_ if there is a compact set $K\subset
M^{n}$ and a diffeomorphism $\Psi:M^{n}\setminus
K\rightarrow\mathbb{R}^{n}\setminus\overline{B}_{1}(0)$ such that in the
Cartesian coordinates induced by $\Psi$, we have

$|g_{ij}-\delta_{ij}|+r|g_{ij,k}|+r^{2}|g_{ij,k\ell}|=\mathcal{O}(r^{-(n-2)}),\qquad\text{as}\qquad
r\rightarrow\infty,$ (2.11)

which in the coordinate free form reads
$|g-\delta|_{\delta}=\mathcal{O}_{2}(r^{-(n-2)})$. If the scalar curvature
$R_{g}$ is integrable we define the ADM energy:

$E_{ADM}=\lim_{R\rightarrow\infty}\frac{1}{2(n-1)\omega_{n-1}}\int_{\\{r=R\\}}\big{(}\text{div}^{\delta}(g)-d(\operatorname{trace}^{\delta}g)\big{)}(\vec{n}^{\delta}_{r})d\mu^{\delta}.$
(2.12)

If, furthermore, $m\in\mathbb{R}$ and $g$ has the asymptotics

$g=\bigg{(}1+\frac{m}{2r^{n-2}}\bigg{)}^{\frac{4}{n-2}}\delta+\mathcal{O}_{2}(r^{-(n-1)}),\qquad\text{as}\qquad
r\rightarrow\infty,$ (2.13)

we say that $(M^{n},g)$ is _asymptotically Schwarzschildean_. In this case we
have $E_{ADM}=m$.

Note that the asymptotics used in Definition 2.6 are not the most general
ones, but they will be sufficient for this work.

###  2.2. Jang’s equation

Let $(M^{n},g,k)$ be an initial data. For local coordinates
$(x^{1},\ldots,x^{n})$ we let the metric be $g=g_{ij}dx^{i}\otimes dx^{j}$ and
$k=k_{ij}dx^{i}\otimes dx^{j}$. We use the Einstein summation throughout, so
that $g^{i\ell}g_{j\ell}=\delta^{i}_{j}$. Further, let $f_{,i}=\partial_{i}f$
denote the $i^{th}$ coordinate derivative.
$f^{,i}=(\nabla^{g}f)^{i}=g^{ij}f_{,j}$ is the $i^{th}$ component of the
gradient of $f$. The covariant Hessian of $f$ is given by
$\text{Hess}_{ij}^{g}(f)=f_{,ij}-\Gamma^{k}_{ij}f_{,k}$, where $\Gamma$ are
the Christoffel symbols associated to $g$.
For $f\in C^{2}_{loc}(U)$, where $U\subset M^{n}$, we consider the equation

$\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}\bigg{(}k_{ij}-\frac{\text{Hess}_{ij}^{g}(f)}{\sqrt{1+|df|^{2}_{g}}}\bigg{)}=0,$
(2.14)

known as _Jang’s equation_ introduced in [Jan78]. Throughout this text we will
refer to this equation as

$\mathcal{J}(f)=0.$ (2.15)

Jang’s equation may be geometrically interpreted as follows. We consider the
Riemannian product $(M^{n}\times\mathbb{R},g+dt^{2})$ and the graph
$(\hat{M}^{n},\hat{g})$ of a function111As we will see in Section 5, we will
not in general obtain a global graph due to blowups/blow-downs.
$f:M^{n}\rightarrow\mathbb{R}$. Then the induced metric $\hat{g}=g+df\otimes
df$ on the graph has components $\hat{g}_{ij}=g_{ij}+f_{,i}f_{,j}$, and its
inverse is

$\hat{g}^{ij}=\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}.$
(2.16)

Furthermore, the (downward pointing) unit normal is

$\vec{n}=\frac{-\partial_{t}+\nabla^{g}f}{\sqrt{1+|df|_{g}^{2}}}$ (2.17)

and the second fundamental form, given by
$\hat{A}(X,Y)=g(\nabla_{X}Y,\vec{n})$, has components

$\hat{A}_{ij}=\frac{\text{Hess}^{g}_{ij}(f)}{\sqrt{1+|df|_{g}^{2}}},$ (2.18)

Consequently, the mean curvature
$H_{\hat{M}^{n}}=\operatorname{trace}^{\hat{g}}\hat{A}$ of
$(\hat{M}^{n},\hat{g})$ is

$H_{\hat{M}^{n}}=\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}\frac{\text{Hess}_{ij}^{g}(f)}{\sqrt{1+|df|^{2}_{g}}},$
(2.19)

which also equals the divergence of the downward pointing unit normal.
Extending $k$ trivially to a symmetric $(0,2)$-tensor on
$M^{n}\times\mathbb{R}$ by $k(\cdot,\partial_{t})=0$, we obtain

$\operatorname{trace}_{\hat{g}}(k)=\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}k_{ij}.$
(2.20)

Hence, Jang’s equation $\mathcal{J}(f)=0$ may be viewed as a prescribed mean
curvature equation

$H_{\hat{M}^{n}}=\operatorname{trace}^{\hat{g}}(k).$ (2.21)

It is a quasilinear second order PDE and it follows from the positivity of the
metric $\hat{g}$ that it is elliptic.
The reader is referred to [AEM11] for an extensive summary on Jang’s equation
and its applications.

##  3\. Barrier construction

In this section we construct barriers for Jang’s equation (2.14) in the case
when the initial data has Wang’s asymptotics as in Definition 2.3. For this we
perform in Section 3.1 a heuristic analysis of Jang’s equation in order to
better understand the asymptotic behaviour of the solutions. Subsequently, in
Section 3.2 we use these results to obtain barriers with the desired
asymptotics.
Throughout this work we divide the coordinate indices $i,j$ of $M^{n}$ into
radial and tangential, and use greek letters for the latter.

###  3.1. Heuristic analysis

We start with the following elementary Example 3.1.

######  Example 3.1.

We consider the standard $n$-dimensional hyperbolic space
$(\mathbb{R}^{n},b,k=b)$ and show that $f(r)=\sqrt{1+r^{2}}$ is a solution to
the Jang equation. It is not difficult to see222Compare to the calculations
done in Section 3.2 below. that

$\frac{\text{Hess}_{ij}^{b}(f)}{\sqrt{1+|df|_{b}^{2}}}=b_{ij}$ (3.1)

for all $i,j$. Thus $f(r)=\sqrt{1+r^{2}}$ solves Jang’s equation.

For $n\geq 4$, $0<\epsilon<1$ and $\alpha,\Psi$ and $q$ smooth functions we
make the following ansatz:

$f(r,\theta)=\sqrt{1+r^{2}}+\psi(\theta)+\frac{\alpha(\theta)}{r^{n-3}}+q(r,\theta),$
(3.2)

where $q(r,\theta)=\mathcal{O}(r^{-(n-2-\epsilon)})$ with derivatives that
decay one order faster per derivative in the $r$-direction; that is
$q_{,\mu}(r,\theta),q_{,\mu\nu}(r,\theta)=\mathcal{O}(r^{-(n-2-\epsilon)})$,
$q_{,r}(r,\theta),q_{,r\mu}(r,\theta)=\mathcal{O}(r^{-(n-1-\epsilon)})$ and
$q_{,rr}(r,\theta)=\mathcal{O}(r^{-(n-\epsilon)})$ and higher order
derivatives decay as indicated. Below in Lemma 3.2 we see the implications of
the requirement $\mathcal{J}(f)=\mathcal{O}(r^{-(n+1-\epsilon)})$.

######  Lemma 3.2.

If the function

$f(r,\theta)=\sqrt{1+r^{2}}+\frac{\alpha(\theta)}{r^{n-3}}+\psi(\theta)+q(r,\theta),$
(3.3)

satsfies $\mathcal{J}(f)=\mathcal{O}(r^{-(n+1-\epsilon)})$, then
$\psi(\theta)$ is a constant and $\alpha(\theta)$ is the (unique) solution of

$\Delta^{\Omega}\alpha-(n-3)\alpha=\bigg{(}\frac{n-2}{2}\bigg{)}\operatorname{trace}^{\Omega}(\textbf{m})+\operatorname{trace}^{\Omega}(\textbf{p}),$
(3.4)

where $\Omega$ is the standard round metric on the sphere $\mathbb{S}^{n-1}$.

###### Proof.

We omit the details of the computation for brevity333The calculations are very
similar in nature to the ones performed in Section 3.2. and merely show the
result obtained when inserting $f(r,\theta)$ as in (3.2) into
$\mathcal{J}(f)$:

$\begin{split}\mathcal{J}(f)&=\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)\alpha-\bigg{(}\frac{n-2}{2}\bigg{)}\operatorname{trace}^{\Omega}(\textbf{m})-\operatorname{trace}^{\Omega}(\textbf{p})\bigg{)}r^{-n}\\\
&\qquad+\frac{\Delta^{\Omega}(\psi)}{r^{2}}\frac{1}{\sqrt{1+r^{2}+\frac{|d\psi|_{\Omega}^{2}}{r^{2}}}}\\\
&\qquad-\frac{1}{(1+r^{2}+\frac{|d\psi|_{\Omega}^{2}}{r^{2}})^{3/2}}g^{\mu\lambda}g^{\nu\rho}\psi_{,\rho}\psi_{,\lambda}\text{Hess}_{\mu\nu}^{\Omega}(\psi)\\\
&\qquad+2\frac{\sqrt{1+r^{2}}}{(1+r^{2}+\frac{|d\psi|_{\Omega}^{2}}{r^{2}})^{3/2}}\frac{|d\psi|^{2}_{\Omega}}{r^{2}}\\\
&\qquad+\frac{1}{1+r^{2}+\frac{|d\psi|_{\Omega}^{2}}{r^{2}}}\bigg{(}\frac{\sqrt{1+r^{2}}}{\sqrt{1+r^{2}+\frac{|d\psi|_{\Omega}^{2}}{r^{2}}}}-1\bigg{)}\\\
&\qquad+(n-1)\bigg{(}\sqrt{\frac{1+r^{2}}{1+r^{2}+\frac{|d\psi|^{2}_{\Omega}}{r^{2}}}}-1\bigg{)}+\mathcal{O}(r^{-(n+1-\epsilon)})\\\
\end{split}$ (3.5)

Requiring that the $\mathcal{O}(r^{-3})$-term vanishes implies
$\Delta^{\Omega}\psi=0$. It is well-known that the harmonic functions on the
sphere $(\mathbb{S}^{n-1},\Omega)$ are precisely the constants. Requiring that
the $\mathcal{O}(r^{-n})$-term vanishes implies that $\alpha$ must solve
(3.4). Multiplying the left hand side of (3.4) by $\alpha$, integrating over
$(\mathbb{S}^{n-1},\Omega)$ and integrating by parts, we see that the
homogeneous problem has only the trivial solution, and existence of a unique
solution $\alpha$ of (3.4) follows from Fredholm alternative (see e.g.
[Bes08], Appendix I).

∎

Properties of the associated graph of $f$ in $M^{n}\times\mathbb{R}$ are
stated in Appendix B.

######  Remark 3.3.

We recall for comparison that the corresponding result in [Sak21] (Proposition
2.6) is

$f(r,\theta,\varphi)=\sqrt{1+r^{2}}+\alpha(\theta,\varphi)\ln(r)+\psi(\theta,\varphi)+q(r,\theta,\varphi),$
(3.6)

where $\alpha$ is the constant

$\alpha=\frac{1}{8\pi}\int_{\mathbb{S}^{2}}\big{(}\operatorname{trace}^{\Omega}(\textbf{m})+2\operatorname{trace}^{\Omega}(\textbf{p})\big{)}d\mu^{\Omega}$
(3.7)

and

$\Delta^{\Omega}(\psi)=\frac{1}{2}\operatorname{trace}^{\Omega}(\textbf{m})+\operatorname{trace}^{\Omega}(\textbf{p})-\alpha.$
(3.8)

###  3.2. Barrier construction

In this subsection we construct the barriers for Jang’s equation (see
Definition 3.4 below), assuming that the initial data has Wang’s asymptotics
as in Definition 2.3. The significance of the barriers is that they ”squeeze”
the solution to Jang’s equation near infinity, providing the asymptotic
control. In the asymptotically hyperbolic setting the construction of barriers
is much more involved compared to the explicit functions used in the
asymptotically Euclidean setting of [SY81b] and [Eic13].

######  Definition 3.4.

Let $(M^{n},g,k)$ be given initial data. A function $f_{+}\in
C^{2}_{\text{loc}}(M^{n}_{r_{0}})$ (respectively $f_{-}\in
C^{2}_{\text{loc}}(M^{n}_{r_{0}})$), where $M^{n}_{r_{0}}=\\{r\geq
r_{0}\\}\subset M^{n}$, is said to be a _upper barrier_ (respectively _lower
barrier_) if it satisfies

$f_{+,r}(r_{0})=+\infty\qquad(\text{respectively}\>f_{-,r}(r_{0})=-\infty)$
(3.9)

and is _supersolution_ (respectively _subsolutions_), that is

$\mathcal{J}(f_{+})<0\qquad(\text{respectively}\>\mathcal{J}(f_{-})>0)$ (3.10)

for $r>r_{0}$.

We refer the reader to [Sak21] for the construction of barriers when $n=3$ and
perform a related construction in dimensions $n\geq 4$. For this, we will
transform the Jang equation to an asymptotic ODE in the radial variable and
construct upper and lower barriers $f_{+}$ and $f_{-}$ via a change of
variables considered in [MOM04] in the spherically symmetric setting. As in
the previous section, $\theta$ denotes a coordinate system on
$\mathbb{S}^{n-1}$. Based on the results in Lemma 3.2 we choose to make the
ansatz 444This is similar to [Sak21], where the anzats is
$f(r,\varphi,\theta)=\varphi(r)+\psi(\varphi,\theta)$.

$f(r,\theta)=\frac{\alpha(\theta)}{r^{n-3}}+\varphi(r),$ (3.11)

for the barriers, where $\varphi_{,r}(r)\rightarrow 1$ as
$r\rightarrow\infty$. As in [MOM04], we define

$\begin{split}k(r)&=\frac{\sqrt{1+r^{2}}\varphi_{,r}}{\sqrt{1+(1+r^{2})\varphi_{,r}^{2}}}.\end{split}$
(3.12)

For reasons that will become clear below we define

$\begin{split}\Pi&=\frac{1+(1+r^{2})\varphi_{,r}^{2}}{1+|df|_{g}^{2}}\\\
\end{split}$ (3.13)

and note that a straightforward calculation shows

$\begin{split}\Pi&=\bigg{(}1-2\sqrt{1+r^{2}}\alpha(n-3)r^{-(n-2)}k\sqrt{1-k^{2}}\\\
&\qquad+(1+r^{2})(n-3)^{2}\alpha^{2}r^{-2(n-2)}(1-k^{2})+r^{-2(n-3)}|d\alpha|_{g}^{2}(1-k^{2})\bigg{)}^{-1}.\end{split}$
(3.14)

We now rewrite Jang’s equation asymmptotically in terms of $k$. Lemmas 3.5\-
3.8 below contain some preliminary computations.

######  Lemma 3.5.

With the ansatz in (3.11), the trace term in Jang’s equation is

$\begin{split}\operatorname{trace}_{\hat{g}}(k)&=\Pi(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})\\\
&\qquad+(n-1)+\frac{\operatorname{trace}^{\Omega}(\textbf{p})-\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.15)

where the implicit constant in the $\mathcal{O}$-term does not depend on
$\varphi$.

###### Proof.

The trace term is explicitly

$\operatorname{trace}_{\hat{g}}(k)=\hat{g}^{rr}k_{rr}+2\hat{g}^{r\mu}k_{r\mu}+\hat{g}^{\mu\nu}k_{\mu\nu}$
(3.16)

and we expand the terms. For the radial term, using $g_{rr}=b_{rr}$, it is not
difficult to see that the radial metric component is

$\begin{split}\hat{g}^{rr}&=(1+r^{2})\bigg{(}1-\frac{(1+r^{2})f_{,r}^{2}}{1+|df|_{g}^{2}}\bigg{)}\\\
&=(1+r^{2})\bigg{(}\frac{1+r^{-2(n-3)}|d\alpha|_{g}^{2}}{1+|df|_{g}^{2}}\bigg{)}.\end{split}$
(3.17)

With the definition of $\Pi$ it is not difficult to see that

$\begin{split}\bigg{(}\frac{1+r^{-2(n-3)}|d\alpha|_{g}^{2}}{1+|df|_{g}^{2}}\bigg{)}&=\Pi(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})\end{split}$
(3.18)

and so, since $k_{rr}=\frac{1}{1+r^{2}}+\mathcal{O}(r^{-(n+1)})$ from
Definition 2.3, we find

$\begin{split}\hat{g}^{rr}k_{rr}&=\Pi(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.19)

where the implicit constant in the $\mathcal{O}$-term does not depend on
$\varphi$.
As for the mixed term $\hat{g}^{r\mu}k_{r\mu}$ we observe that since both
$g^{r\mu}=0$, $f^{,r}=\mathcal{O}(r^{2})$, $f^{,\mu}=\mathcal{O}(r^{-(n-1)})$
and $k_{r\mu}=\mathcal{O}(r^{-n})$ we immedately obtain
$\hat{g}^{r\mu}k_{r\mu}=\mathcal{O}(r^{-(n+1)})$.
We compute the asymptotics of the tangential term
$\hat{g}^{\mu\nu}k_{\mu\nu}$:

$\hat{g}^{\mu\nu}k_{\mu\nu}=\bigg{(}g^{\mu\nu}-\frac{f^{,\mu}f^{,\nu}}{1+|df|_{g}^{2}}\bigg{)}\bigg{(}b_{\mu\nu}+\frac{\textbf{p}_{\mu\nu}}{r^{n-2}}+\mathcal{O}(r^{-(n-1)})\bigg{)}.\\\
$ (3.20)

Since $f^{,\mu}=\mathcal{O}(r^{-(n-1)})$ it follows that

$\hat{g}^{\mu\nu}=b^{\mu\nu}-\frac{\textbf{m}^{\mu\nu}}{r^{n-2}}+\mathcal{O}(r^{-(n+3)}),$
(3.21)

where indices on m are raised with $g$. From this and the expression

$k_{\mu\nu}=b_{\mu\nu}+\frac{\textbf{p}_{\mu\nu}}{r^{n-2}}+\mathcal{O}(r^{-(n-1)})$
(3.22)

it is immediate that

$\begin{split}\hat{g}^{\mu\nu}k_{\mu\nu}&=(n-1)+\frac{\operatorname{trace}^{\Omega}(\textbf{p})-\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}+\mathcal{O}(r^{-(n+1)})\end{split}$
(3.23)

and so the assertion follows.

∎

######  Lemma 3.6.

With the ansatz in (3.11), the radial Hessian term in Jang’s equation is

$\begin{split}\hat{g}^{rr}\frac{\text{Hess}_{rr}^{g}(f)}{\sqrt{1+|df|_{g}^{2}}}&=\sqrt{1+r^{2}}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})\Pi^{3/2}\\\
&\times\bigg{(}k^{\prime}+(1-k^{2})^{3/2}\sqrt{1+r^{2}}(n-3)^{2}\alpha
r^{-(n-1)}+\mathcal{O}(r^{-(n+1)})\bigg{)},\end{split}$ (3.24)

where the implicit constant in the $\mathcal{O}$-term does not depend on
$\varphi$.

###### Proof.

From the definition of $k$ it follows that

$k^{\prime}(r)=\frac{\sqrt{1+r^{2}}}{(1+(1+r^{2})\varphi_{,r}^{2})^{3/2}}\bigg{(}\varphi_{,rr}+\frac{r}{1+r^{2}}\varphi_{,r}\bigg{)}.$
(3.25)

From the proof of Lemma 3.5 the radial metric component $\hat{g}^{rr}$ is

$\begin{split}\hat{g}^{rr}&=(1+r^{2})\bigg{(}\frac{1+r^{-2(n-3)}|d\alpha|_{g}^{2}}{1+|df|_{g}^{2}}\bigg{)}.\end{split}$
(3.26)

Using expressions for the Christoffel symbols of $g$ obtained in Lemma A.1 we
find

$\begin{split}\text{Hess}_{rr}^{g}(f)&=\varphi_{,rr}+\frac{r}{1+r^{2}}\varphi_{,r}+\text{Hess}_{rr}^{g}\bigg{(}\frac{\alpha}{r^{n-3}}\bigg{)}\end{split}$
(3.27)

where, in turn,

$\begin{split}\text{Hess}_{rr}^{g}\bigg{(}\frac{\alpha}{r^{n-3}}\bigg{)}&=\bigg{(}\frac{\alpha}{r^{n-3}}\bigg{)}_{,rr}+\frac{r}{1+r^{2}}\bigg{(}\frac{\alpha}{r^{n-3}}\bigg{)}_{,r}\\\
&=(n-3)^{2}\frac{\alpha}{r^{n-1}}+\mathcal{O}(r^{-(n+1)}).\end{split}$ (3.28)

Hence, with $\hat{g}^{rr}$ from the proof of Lemma 3.5,

$\begin{split}\hat{g}^{rr}\frac{\text{Hess}_{rr}^{g}(f)}{\sqrt{1+|df|_{g}^{2}}}&=(1+r^{2})\frac{1+r^{-2(n-3)}|d\alpha|_{g}^{2}}{(1+|df|_{g}^{2})^{3/2}}\text{Hess}_{rr}^{g}(f)\\\
&=\frac{1+r^{2}}{(1+(1+r^{2})\varphi_{,r}^{2})^{3/2}}\Pi^{3/2}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})\text{Hess}_{rr}^{g}(f)\\\
&=\frac{1+r^{2}}{(1+(1+r^{2})\varphi_{,r}^{2})^{3/2}}\Pi^{3/2}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})\bigg{(}\varphi_{,rr}+\frac{r}{1+r^{2}}\varphi_{,r}\bigg{)}\\\
&\qquad+(1+r^{2})(1-k^{2})^{3/2}\Pi^{3/2}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})\\\
&\qquad\times\bigg{(}(n-3)^{2}\frac{\alpha}{r^{n-1}}+\mathcal{O}(r^{-(n+1)})\bigg{)}\\\
&=\sqrt{1+r^{2}}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})\Pi^{3/2}\\\
&\qquad\times\bigg{(}k^{\prime}+(1-k^{2})^{3/2}\sqrt{1+r^{2}}(n-3)^{2}\frac{\alpha}{r^{n-1}}+\mathcal{O}(r^{-(n+1)})\bigg{)},\end{split}$
(3.29)

as asserted.

∎

Similar to the proof of Lemmas 3.5 and 3.6 calculations yield Lemmas 3.7 and
3.8 below.

######  Lemma 3.7.

With $f$ as in (3.11), the mixed Hessian term in Jang’s equation is

$\hat{g}^{\mu r}\frac{\text{Hess}_{\mu
r}^{g}(f)}{\sqrt{1+|df|_{g}^{2}}}=\mathcal{O}(r^{-(n+1)}),$ (3.30)

where the implicit constant in the $\mathcal{O}$-term does not depend on
$\varphi$.

######  Lemma 3.8.

With $f$ as in (3.11), the tangential Hessian term in Jang’s equation is

$\begin{split}\hat{g}^{\mu\nu}\frac{\text{Hess}_{\mu\nu}^{g}(f)}{\sqrt{1+|df|_{g}^{2}}}&=\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)(n-1)(1+r^{2})\alpha\bigg{)}\sqrt{\Pi}\sqrt{1-k^{2}}r^{-(n-1)}\\\
&\qquad+\bigg{(}\frac{\sqrt{1+r^{2}}}{r}(n-1)-\frac{\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}\frac{n}{2}\bigg{)}\sqrt{\Pi}k+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.31)

where the implicit constant in the $\mathcal{O}$-term does not depend on
$\varphi$.

Combining Lemmas 3.5 \- 3.8 we obtain the following result:

######  Lemma 3.9.

With $f$ as in (3.11), we have

$\begin{split}\frac{\mathcal{J}(f)}{\Pi^{3/2}}&=\sqrt{1+r^{2}}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})k^{\prime}\\\
&\qquad+\sqrt{1+r^{2}}\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}k-\frac{r}{\sqrt{1+r^{2}}}\bigg{)}-(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})\\\
&\qquad+(n-3)\frac{\alpha}{r^{n-2}}\sqrt{1-k^{2}}\bigg{(}(1-k^{2})\frac{1+r^{2}}{r}(n-3)+\frac{1}{r}\\\
&\qquad\qquad-(n-1)\frac{1+r^{2}}{r}-2(n-1)\frac{1+r^{2}}{r}k^{2}\\\
&\qquad\qquad+\sqrt{1+r^{2}}k(1-k^{2})+3(n-1)\sqrt{1+r^{2}}k\bigg{)}\\\
&\qquad+\bigg{(}\bigg{(}\frac{n-2}{2}\bigg{)}\operatorname{trace}^{\Omega}(\textbf{m})+\operatorname{trace}^{\Omega}(\textbf{p})\bigg{)}\bigg{(}\frac{\sqrt{1-k^{2}}}{r^{n-1}}-\frac{1}{r^{n}}\bigg{)}\\\
&\qquad+\frac{\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}\frac{n}{2}(1-k)+\Lambda+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.32)

where

$\begin{split}\Lambda&=\frac{\sqrt{1-k^{2}}}{r^{n-1}}\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)(n-1)\alpha\bigg{)}\bigg{(}1-\frac{1}{\Pi}\bigg{)}\\\
&\qquad-\frac{|d\alpha|_{\Omega}^{2}}{r^{2(n-2)}}\frac{(1-k^{2})}{2}\bigg{(}-2(n-1)k+(1-k^{2})+3(n-1)\bigg{)}\\\
&\qquad+\frac{\alpha^{2}}{r^{2(n-2)}}(1+r^{2})(1-k^{2})(n-3)^{2}\bigg{(}-2(n-1)k\frac{r^{2}}{1+r^{2}}+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\\\
&\qquad\qquad-(1-k^{2})\frac{1}{2}\bigg{(}1+3k^{2}\bigg{)}-(n-1)\frac{3}{2}(1+k^{2})\bigg{)}\\\
&\qquad+\frac{\alpha^{3}}{r^{3(n-2)}}(1+r^{2})^{3/2}(1-k^{2})^{3/2}(n-3)^{3}\\\
&\qquad\qquad\times\bigg{(}(n-1)\frac{\sqrt{1+r^{2}}}{r}-\frac{k}{2}(3+5k^{2})-\frac{k}{2}(3+k^{2})\bigg{)}\\\
&\qquad-\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}(1-k^{2})^{2}(1+r^{2})^{2}\\\
&\qquad\qquad\times\bigg{(}(1-k^{2})\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}+(n-1)\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}\bigg{)}\\\
&\qquad+\mathcal{O}(r^{-(n+1)}).\end{split}$ (3.33)

###### Proof.

We divide the terms from Lemmas 3.5, 3.6, 3.7 and 3.8 by $\Pi^{3/2}$ and
expand. Since $\Pi^{-1}=1+\mathcal{O}(r^{-(n-3)})$, the contribution of the
mixed Hessian term is of order $\mathcal{O}(r^{-(n+1)})$. To estimate the
tangential Hessian term and trace term, which contain powers
$\Pi^{-1},\Pi^{-1/2},\Pi^{-3/2}$, we rewrite these in terms of functions
$\gamma_{1},\ldots,\gamma_{4}$ in a manner explained below. For the tangential
Hessian term, we recall (3.14) and define two functions $\gamma_{1}$ and
$\gamma_{2}$ as follows:

$\gamma_{1}=(1+r^{2})(n-3)^{2}\alpha^{2}r^{-2(n-2)}(1-k^{2})+r^{-2(n-3)}|d\alpha|_{g}^{2}(1-k^{2})$
(3.34)

and it follows that $\gamma_{1}=\mathcal{O}(r^{-2(n-3)})$. We furthermore
rewrite

$\begin{split}\frac{1}{\Pi}&=\bigg{(}1-2\sqrt{1+r^{2}}\alpha(n-3)r^{-(n-2)}k\sqrt{1-k^{2}}+\gamma_{1}\bigg{)}\\\
&=1-\gamma_{2}.\end{split}$ (3.35)

and $\gamma_{2}=\mathcal{O}(r^{-(n-3)})$. The first term in the right hand
side of (3.31) is

$\begin{split}\bigg{(}\Delta^{\Omega}(\alpha)&-(n-3)(n-1)(1+r^{2})\alpha\bigg{)}\frac{\sqrt{1-k^{2}}}{\Pi}r^{-(n-1)}=\mathcal{O}(r^{-(n-3)}).\\\
\end{split}$ (3.36)

Furthermore, we have that the second and third terms of the right hand side of
(3.31) are

$\begin{split}\frac{\sqrt{1+r^{2}}}{r}(n-1)\frac{k}{\Pi}&=\frac{\sqrt{1+r^{2}}}{r}(n-1)k-2(n-1)(n-3)(1+r^{2})\frac{\alpha}{r^{n-1}}k^{2}\sqrt{1-k^{2}}\\\
&\qquad+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\gamma_{1}\end{split}$ (3.37)

and

$\frac{\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}\frac{n}{2}\frac{k}{\Pi}=\frac{\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}\frac{n}{2}k+\mathcal{O}(r^{-(n+1)}).$
(3.38)

Combining these estimates we find that the tangential Hessian term in (3.31)
divided by $\Pi^{3/2}$ is

$\begin{split}\hat{g}^{\mu\nu}\frac{\text{Hess}^{g}_{\mu\nu}(f)}{\Pi^{3/2}\sqrt{1+|df|^{2}_{g}}}&=\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)(n-1)(1+r^{2})\alpha\bigg{)}\frac{\sqrt{1-k^{2}}}{r^{n-1}}\\\
&\qquad+\bigg{(}\frac{\sqrt{1+r^{2}}}{r}(n-1)-\frac{\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}\frac{n}{2}\bigg{)}k\\\
&\qquad+\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)(n-1)(1+r^{2})\alpha\bigg{)}\frac{\sqrt{1-k^{2}}}{r^{n-1}}\gamma_{2}\\\
&\qquad-2(n-1)(n-3)(1+r^{2})\alpha k^{2}\frac{\sqrt{1-k^{2}}}{r^{n-1}}\\\
&\qquad+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\gamma_{1}+\mathcal{O}(r^{-(n+1)}).\\\
\end{split}$ (3.39)

We similarly expand the trace term divided by $\Pi^{-3/2}$

$\begin{split}\frac{\operatorname{trace}_{\hat{g}}(k)}{\Pi^{3/2}}&=\frac{(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})}{\sqrt{\Pi}}\\\
&\qquad+\frac{(n-1)}{\Pi^{3/2}}+\frac{\operatorname{trace}^{\Omega}(\textbf{p})-\operatorname{trace}^{\Omega}(\textbf{m})}{\Pi^{3/2}r^{n}}+\mathcal{O}(r^{-(n+1)})\end{split}$
(3.40)

by rewriting $\Pi^{-1/2}$ and $\Pi^{-3/2}$ in terms of functions $\gamma_{3}$
and $\gamma_{4}$:

$\begin{split}\frac{1}{\sqrt{\Pi}}&=\bigg{(}1-\frac{\alpha}{r^{n-2}}(n-3)\sqrt{1+r^{2}}k\sqrt{1-k^{2}}+\gamma_{3}\bigg{)}\\\
\frac{1}{\Pi^{3/2}}&=\bigg{(}1-3\frac{\alpha}{r^{n-2}}\sqrt{1+r^{2}}(n-3)k\sqrt{1-k^{2}}+\gamma_{4}\bigg{)},\end{split}$
(3.41)

where $\gamma_{3}=\mathcal{O}(r^{-2(n-3)})$ and
$\gamma_{4}=\mathcal{O}(r^{-2(n-3)})$. With this, we rewrite the terms in the
trace term:

$\begin{split}\frac{1-k^{2}}{\sqrt{\Pi}}&=(1-k^{2})-\frac{\alpha}{r^{n-2}}(n-3)\sqrt{1+r^{2}}k(1-k^{2})^{3/2}+(1-k^{2})\gamma_{3}\end{split}$
(3.42)

and similarly

$\begin{split}\frac{n-1}{\Pi^{3/2}}&=(n-1)-3\frac{\alpha}{r^{n-2}}(n-3)(n-1)\sqrt{1+r^{2}}k\sqrt{1-k^{2}}+(n-1)\gamma_{4}.\end{split}$
(3.43)

Adding terms and simplifying we compute that the contribution of the trace
term to $\frac{\mathcal{J}(f)}{\Pi^{3/2}}$ is

$\begin{split}\frac{\operatorname{trace}_{\hat{g}}(k)}{\Pi^{3/2}}&=(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})-\frac{\alpha}{r^{n-2}}(n-3)\sqrt{1+r^{2}}k(1-k^{2})^{3/2}\\\
&\qquad+(1-k^{2})\gamma_{3}+(n-1)-3\frac{\alpha}{r^{n-2}}(n-3)(n-1)\sqrt{1+r^{2}}k\sqrt{1-k^{2}}\\\
&\qquad+(n-1)\gamma_{4}+\frac{\operatorname{trace}^{\Omega}(\textbf{p})-\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}+\mathcal{O}(r^{-(n+1)}).\end{split}$
(3.44)

We define $\Lambda$:

$\begin{split}\Lambda&=\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)(n-1)(1+r^{2})\alpha\bigg{)}\sqrt{1-k^{2}}\gamma_{2}r^{-(n-1)}\\\
&\qquad+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\gamma_{1}-(1-k^{2})\gamma_{3}-(n-1)\gamma_{4}.\\\
\end{split}$ (3.45)

and collect terms:

$\begin{split}\frac{\mathcal{J}(f)}{\Pi^{3/2}}&=\sqrt{1+r^{2}}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})k^{\prime}\\\
&\qquad+\sqrt{1+r^{2}}\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}k-\frac{r}{\sqrt{1+r^{2}}}\bigg{)}-(1+r^{-2(n-3)}|d\alpha|_{g}^{2})(1-k^{2})\\\
&\qquad+(n-3)\frac{\alpha}{r^{n-2}}\sqrt{1-k^{2}}\bigg{(}(1-k^{2})\frac{1+r^{2}}{r}(n-3)+\frac{1}{r}\\\
&\qquad-(n-1)\frac{1+r^{2}}{r}-2(n-1)\frac{1+r^{2}}{r}k^{2}\\\
&\qquad+\sqrt{1+r^{2}}k(1-k^{2})+3(n-1)\sqrt{1+r^{2}}k\bigg{)}\\\
&\qquad+\bigg{(}\bigg{(}\frac{n-2}{2}\bigg{)}\operatorname{trace}^{\Omega}(\textbf{m})+\operatorname{trace}^{\Omega}(\textbf{p})\bigg{)}\bigg{(}\frac{\sqrt{1-k^{2}}}{r^{n-1}}-\frac{1}{r^{n}}\bigg{)}\\\
&\qquad+\frac{\operatorname{trace}^{\Omega}(\textbf{m})}{r^{n}}\frac{n}{2}\big{(}1-k\big{)}+\Lambda+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.46)

where we used (3.4) to expand the $\Delta^{\Omega}(\alpha)$-term stemming from
the tangential Hessian. Obviously we at least have
$\Lambda=\mathcal{O}(r^{-1})$, but it will be useful to know the first terms
explicitly. We have from (3.14) and the definition of $\gamma_{2}$:

$\begin{split}\gamma_{2}&=2\sqrt{1+r^{2}}\frac{\alpha}{r^{n-2}}(n-3)k\sqrt{1-k^{2}}-(1+r^{2})(n-3)^{2}\frac{\alpha^{2}}{r^{2(n-2)}}(1-k^{2})\\\
&\qquad-\frac{|d\alpha|_{g}^{2}}{r^{2(n-3)}}(1-k^{2})\end{split}$ (3.47)

and we compute

$\begin{split}\gamma_{2}^{2}&=4(1+r^{2})\frac{\alpha^{2}}{r^{2(n-2)}}(n-3)^{2}k^{2}(1-k^{2})-4(1+r^{2})^{3/2}(n-3)^{3}\frac{\alpha^{3}}{r^{3(n-2)}}k(1-k^{2})^{3/2}\\\
&\qquad-(1+r^{2})^{2}(n-3)^{4}\frac{\alpha^{4}}{r^{4(n-2)}}(1-k^{2})^{2}+\mathcal{O}(r^{-(n+1)}).\end{split}$
(3.48)

Similarly, we have

$\begin{split}\gamma_{2}^{3}&=8(1+r^{2})^{3/2}\frac{\alpha^{3}}{r^{3(n-2)}}(n-3)^{3}k^{3}(1-k^{2})^{3/2}\\\
&\qquad-12(1+r^{2})^{2}(n-3)^{4}\frac{\alpha^{4}}{r^{4(n-2)}}k^{2}(1-k^{2})^{2}+\mathcal{O}(r^{-(n+1)})\\\
\end{split}$ (3.49)

and

$\gamma_{2}^{4}=16(1+r^{2})^{2}\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}k^{4}(1-k^{2})^{2}+\mathcal{O}(r^{-(n+1)}).$
(3.50)

Hence, we may expand:

$\begin{split}\frac{1}{\sqrt{\Pi}}&=(1-\gamma_{2})^{1/2}\\\
&=1-\frac{\gamma_{2}}{2}+\frac{3}{8}\gamma_{2}^{2}-\frac{5}{16}\gamma_{2}^{3}+\frac{35}{128}\gamma_{2}^{4}+\mathcal{O}(r^{-(n+1)})\\\
&=1-\sqrt{1+r^{2}}\frac{\alpha}{r^{n-2}}(n-3)k\sqrt{1-k^{2}}+\frac{1}{2}\frac{|d\alpha|_{g}^{2}}{r^{2(n-3)}}(1-k^{2})\\\
&\qquad+\frac{\alpha^{2}}{r^{2(n-2)}}(1+r^{2})(1-k^{2})(n-3)^{2}\frac{1}{2}\bigg{(}1+3k^{2}\bigg{)}\\\
&\qquad+\frac{\alpha^{3}}{r^{3(n-2)}}(1+r^{2})^{3/2}(1-k^{2})^{3/2}(n-3)^{3}\frac{1}{2}\bigg{(}-3k-5k^{3}\bigg{)}\\\
&\qquad+\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}(1-k^{2})^{2}(1+r^{2})^{2}\frac{1}{8}\bigg{(}-3+30k^{2}+35k^{4}\bigg{)}\\\
&\qquad+\mathcal{O}(r^{-(n+1)}),\\\ \end{split}$ (3.51)

so that we can read off

$\begin{split}\gamma_{3}&=\frac{1}{2}\frac{|d\alpha|_{g}^{2}}{r^{2(n-3)}}(1-k^{2})\\\
&\qquad+\frac{\alpha^{2}}{r^{2(n-2)}}(1+r^{2})(1-k^{2})(n-3)^{2}\frac{1}{2}\bigg{(}1+3k^{2}\bigg{)}\\\
&\qquad+\frac{\alpha^{3}}{r^{3(n-2)}}(1+r^{2})^{3/2}(1-k^{2})^{3/2}(n-3)^{3}\frac{1}{2}\bigg{(}-3k-5k^{3}\bigg{)}\\\
&\qquad+\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}(1-k^{2})^{2}(1+r^{2})^{2}\frac{1}{8}\bigg{(}-3+30k^{2}+35k^{4}\bigg{)}\\\
&\qquad+\mathcal{O}(r^{-(n+1)})\end{split}$ (3.52)

Similarly, we have

$\begin{split}\frac{1}{\Pi^{3/2}}&=(1-\gamma_{2})^{3/2}\\\
&=1-\frac{3}{2}\gamma_{2}+\frac{3}{8}\gamma_{2}^{2}-\frac{1}{16}\gamma_{2}^{3}+\frac{3}{128}\gamma_{2}^{4}+\mathcal{O}(r^{-(n+1)})\\\
&=1-3\sqrt{1+r^{2}}\frac{\alpha}{r^{n-2}}(n-3)k\sqrt{1-k^{2}}+\frac{3}{2}\frac{|d\alpha|_{g}^{2}}{r^{2(n-3)}}(1-k^{2})\\\
&\qquad+\frac{\alpha^{2}}{r^{2(n-2)}}(1+r^{2})(1-k^{2})(n-3)^{2}\frac{3}{2}\bigg{(}1+k^{2}\bigg{)}\\\
&\qquad+\frac{\alpha^{3}}{r^{3(n-2)}}(1+r^{2})^{3/2}(1-k^{2})^{3/2}(n-3)^{3}\frac{1}{2}\bigg{(}-3k-k^{3}\bigg{)}\\\
&\qquad+\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}(1-k^{2})^{2}(1+r^{2})^{2}\frac{3}{4}\bigg{(}-\frac{1}{2}+k^{2}+\frac{1}{2}k^{4}\bigg{)}\\\
&\qquad+\mathcal{O}(r^{-(n+1)}),\\\ \end{split}$ (3.53)

so that we can read off

$\begin{split}\gamma_{4}&=\frac{3}{2}\frac{|d\alpha|_{g}^{2}}{r^{2(n-3)}}(1-k^{2})\\\
&\qquad+\frac{\alpha^{2}}{r^{2(n-2)}}(1+r^{2})(1-k^{2})(n-3)^{2}\frac{3}{2}\bigg{(}1+k^{2}\bigg{)}\\\
&\qquad+\frac{\alpha^{3}}{r^{3(n-2)}}(1+r^{2})^{3/2}(1-k^{2})^{3/2}(n-3)^{3}\frac{1}{2}\bigg{(}-3k-k^{3}\bigg{)}\\\
&\qquad+\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}(1-k^{2})^{2}(1+r^{2})^{2}\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}+\mathcal{O}(r^{-(n+1)}).\\\
\end{split}$ (3.54)

With this at hand we can make the asymptotics of $\Lambda$ in (3.45) more
precise. We find that

$\begin{split}\bigg{(}\Delta^{\Omega}(\alpha)&-(n-3)(n-1)(1+r^{2})\alpha\bigg{)}\frac{\sqrt{1-k^{2}}}{r^{n-1}}\gamma_{2}\\\
&=-(n-3)(n-1)\alpha\frac{\sqrt{1-k^{2}}}{r^{n-3}}\gamma_{2}+\frac{\sqrt{1-k^{2}}}{r^{n-1}}\lambda\\\
&=(n-3)^{2}(n-1)(1-k^{2})\bigg{(}-2\frac{\alpha^{2}}{r^{2(n-3)}}k+\sqrt{1-k^{2}}(1+r^{2})\frac{\alpha^{3}}{r^{3(n-2)-1}}\bigg{)}\\\
&\qquad+\frac{\sqrt{1-k^{2}}}{r^{n-1}}\lambda+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.55)

where

$\lambda=\bigg{(}\Delta^{\Omega}(\alpha)-(n-3)(n-1)\alpha\bigg{)}\gamma_{2}.$
(3.56)

Clearly, $\lambda=\mathcal{O}(r^{-(n-3)})$. The second term in (3.45) is

$\begin{split}\frac{\sqrt{1+r^{2}}}{r}(n-1)k\gamma_{1}&=\frac{\sqrt{1+r^{2}}}{r}(1+r^{2})(n-3)^{2}(n-1)k\frac{\alpha^{2}}{r^{2(n-2)}}(1-k^{2})\\\
&\qquad+(n-1)k\frac{|d\alpha|_{\Omega}^{2}}{r^{2(n-2)}}(1-k^{2})+\mathcal{O}(r^{-(n+1)}),\end{split}$
(3.57)

where we used (3.34). In summary, we obtain

$\begin{split}\Lambda&=\frac{\sqrt{1-k^{2}}}{r^{n-1}}\lambda-\frac{|d\alpha|_{\Omega}^{2}}{r^{2(n-2)}}\frac{(1-k^{2})}{2}\bigg{(}-2(n-1)k+(1-k^{2})+3(n-1)\bigg{)}\\\
&\qquad+\frac{\alpha^{2}}{r^{2(n-2)}}(1+r^{2})(1-k^{2})(n-3)^{2}\bigg{(}-2(n-1)k\frac{r^{2}}{1+r^{2}}+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\\\
&\qquad\qquad-(1-k^{2})\frac{1}{2}\bigg{(}1+3k^{2}\bigg{)}-(n-1)\frac{3}{2}(1+k^{2})\bigg{)}\\\
&\qquad+\frac{\alpha^{3}}{r^{3(n-2)}}(1+r^{2})^{3/2}(1-k^{2})^{3/2}(n-3)^{3}\\\
&\qquad\qquad\times\bigg{(}(n-1)\frac{\sqrt{1+r^{2}}}{r}-\frac{k}{2}(3+5k^{2})-\frac{k}{2}(3+k^{2})\bigg{)}\\\
&\qquad-\frac{\alpha^{4}}{r^{4(n-2)}}(n-3)^{4}(1-k^{2})^{2}(1+r^{2})^{2}\\\
&\qquad\qquad\times\bigg{(}(1-k^{2})\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}+(n-1)\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}\bigg{)}\\\
&\qquad+\mathcal{O}(r^{-(n+1)}),\end{split}$ (3.58)

which completes our assertion.

∎

With this result at hand we can estimate the left hand side of Jang’s equation
from above and from below, thereby obtaining the eququations for sub- and
supersolutions.

######  Lemma 3.10.

There exists positive constants $C_{1},C_{2},C_{3},C_{4}$ such that

$\begin{split}\frac{\mathcal{J}(f)}{\sqrt{1+r^{2}}}&\frac{1}{\Pi^{3/2}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})}\leq\mathcal{J}_{+}(k)\\\
&=k^{\prime}+\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}k-\frac{r}{\sqrt{1+r^{2}}}\bigg{)}-\frac{1-k^{2}}{\sqrt{1+r^{2}}}\\\
&\qquad+\frac{\sqrt{1-k^{2}}}{\sqrt{1+r^{2}}}\frac{C_{1}}{r^{n-2}}\bigg{|}(1-k^{2})\frac{1+r^{2}}{r}(n-3)+\frac{1}{r}\\\
&\qquad\qquad-(n-1)\frac{1+r^{2}}{r}-2(n-1)\frac{1+r^{2}}{r}k^{2}\\\
&\qquad\qquad+\sqrt{1+r^{2}}k(1-k^{2})+3(n-1)\sqrt{1+r^{2}}k\bigg{|}\\\
&\qquad+\frac{C_{1}^{2}}{r^{2(n-2)}}\sqrt{1+r^{2}}(1-k^{2})\bigg{|}-2(n-1)k\frac{r^{2}}{1+r^{2}}+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\\\
&\qquad\qquad-(1-k^{2})\frac{1}{2}\bigg{(}1+3k^{2}\bigg{)}-(n-1)\frac{3}{2}(1+k^{2})\bigg{|}\\\
&\qquad+\frac{C_{1}^{3}}{r^{3(n-2)}}(1+r^{2})(1-k^{2})^{3/2}\\\
&\qquad\qquad\times\bigg{|}(n-1)\frac{\sqrt{1+r^{2}}}{r}-\frac{k}{2}(3+5k^{2})-\frac{k}{2}(3+k^{2})\bigg{|}\\\
&\qquad+\frac{C_{1}^{4}}{r^{4(n-2)}}(1-k^{2})^{2}(1+r^{2})^{3/2}\\\
&\qquad\qquad\times\bigg{|}\bigg{(}(1-k^{2})\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}+(n-1)\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}\bigg{)}\bigg{|}\\\
&\qquad+C_{2}\bigg{|}\frac{\sqrt{1-k^{2}}}{r^{n}}-\frac{1}{r^{n+1}}\bigg{|}\\\
&\qquad+\frac{C_{3}}{r^{n+1}}|1-k|+\frac{C_{4}}{r^{n+2}}\end{split}$ (3.59)

and

$\begin{split}\frac{\mathcal{J}(f)}{\sqrt{1+r^{2}}}&\frac{1}{\Pi^{3/2}(1+r^{-2(n-3)}|d\alpha|_{g}^{2})}\geq\mathcal{J}_{-}(k)\\\
&=k^{\prime}+\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}k-\frac{r}{\sqrt{1+r^{2}}}\bigg{)}-\frac{1-k^{2}}{\sqrt{1+r^{2}}}\\\
&\qquad-\frac{\sqrt{1-k^{2}}}{\sqrt{1+r^{2}}}\frac{C_{1}}{r^{n-2}}\bigg{|}(1-k^{2})\frac{1+r^{2}}{r}(n-3)+\frac{1}{r}\\\
&\qquad\qquad-(n-1)\frac{1+r^{2}}{r}-2(n-1)\frac{1+r^{2}}{r}k^{2}\\\
&\qquad\qquad+\sqrt{1+r^{2}}k(1-k^{2})+3(n-1)\sqrt{1+r^{2}}k\bigg{|}\\\
&\qquad-\frac{C_{1}^{2}}{r^{2(n-2)}}\sqrt{1+r^{2}}(1-k^{2})\bigg{|}-2(n-1)k\frac{r^{2}}{1+r^{2}}+\frac{\sqrt{1+r^{2}}}{r}(n-1)k\\\
&\qquad\qquad-(1-k^{2})\frac{1}{2}\bigg{(}1+3k^{2}\bigg{)}-(n-1)\frac{3}{2}(1+k^{2})\bigg{|}\\\
&\qquad-\frac{C_{1}^{3}}{r^{3(n-2)}}(1+r^{2})(1-k^{2})^{3/2}\\\
&\qquad\qquad\times\bigg{|}(n-1)\frac{\sqrt{1+r^{2}}}{r}-\frac{k}{2}(3+5k^{2})-\frac{k}{2}(3+k^{2})\bigg{|}\\\
&\qquad-\frac{C_{1}^{4}}{r^{4(n-2)}}(1-k^{2})^{2}(1+r^{2})^{3/2}\\\
&\qquad\qquad\times\bigg{|}\bigg{(}(1-k^{2})\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}+(n-1)\frac{3}{8}\bigg{(}-1+2k^{2}+k^{4}\bigg{)}\bigg{)}\bigg{|}\\\
&\qquad-C_{2}\bigg{|}\frac{\sqrt{1-k^{2}}}{r^{n}}-\frac{1}{r^{n+1}}\bigg{|}\\\
&\qquad-\frac{C_{3}}{r^{n+1}}|1-k|-\frac{C_{4}}{r^{n+2}}\end{split}$ (3.60)

###### Proof.

The assertion follows directly from Lemma 3.9.

∎

The following Proposition will be useful.

######  Proposition 3.11.

Suppose $y\in C^{1}_{loc}([r_{0},\infty))$ solves the a first order ODE

$y^{\prime}+F(r,y)=0,\qquad y(r_{0})=y_{0},$ (3.61)

on $[r_{0},\infty)$ and suppose furthermore that there exist sub- and
supersolutions $y_{-}$ and $y_{+}$ on $[r_{0},\infty)$ with $y_{-}(r_{0})\leq
y_{0}$ and $y_{0}\leq y_{+}(r_{0})$. Then

$y_{-}\leq y\leq y_{+}$ (3.62)

on $[r_{0},\infty)$.

###### Proof.

We follow the proof of [Sak21, Lemma 3.3]. We have

$y^{\prime}(r)+F(r,y(r))<y_{+}^{\prime}(r)+F(r,y_{+}(r))$ (3.63)

and to see that $y_{-}(r)\leq y_{+}(r)$ for $r\geq r_{0}$ we assume that for
some $r_{1}>r_{0}$ we have $y_{-}(r_{1})>y_{+}(r_{1})$. Let
$r_{2}=\inf\\{r>r_{0}\>|\>y_{-}(r)>y_{+}(r)\\}$. Then $r_{0}\leq r_{2}<r_{1}$
and $y(r_{2})=y_{+}(r_{2})$. On the one hand, it follows from (3.63) that
$y^{\prime}(r_{2})<y_{+}^{\prime}(r_{2})$ so that
$y(r_{2}+\epsilon)<y_{+}(r_{2}+\epsilon)$ for sufficiently small $\epsilon>0$.
On the other hand, this contradicts the definition of $r_{2}$.
Arguing similarly, we obtain $y_{-}\leq y$.

∎

We can now prove the existence of sub- and supersolutions.

######  Lemma 3.12.

Let $C_{1},C_{2},C_{3},C_{4}$ be as in Lemma 3.10. For $r_{0}$ large enough,
there exists a solution $k_{+}:[r_{0},\infty)\rightarrow\mathbb{R}$ to
$\mathcal{J}_{+}(k_{+})=0$ such that $k_{+}(r_{0})=-1$ and $|k_{+}|<1$ for
$r>r_{0}$. Similarly, there exists a solution
$k_{-}:[r_{0},\infty)\rightarrow\mathbb{R}$ to $\mathcal{J}_{-}(k_{-})=0$ such
that $k_{-}(r_{0})=+1$ and $|k_{-}|<1$ for $r>r_{0}$.

###### Proof.

We start with $k_{+}$. We observe that $k_{+}^{+}=+1$ and $k_{+}^{-}=-1$ yield
by Equations (3.60) and (3.59) $\mathcal{J}_{+}(k_{+}^{+})>0$ and
$\mathcal{J}_{+}(k_{+}^{-})<0$ for large enough $r_{0}$. Furthermore
$\mathcal{J}_{+}(k_{+})=0$ is on the form $k^{\prime}(r)+F(r,k)=0$, where
$F(r,k)$ is continuous in both variables for large enough $r_{0}$, and so by
Peano’s Existence Theorem [Har02, Theorem 2.1] we have a solution. By
Proposition 3.11 we have $|k_{+}|\leq 1$. To verify that $|k_{+}|<1$ we
observe that if there is a point $r_{1}>r_{0}$ such that $k(r_{1})=+1$ we
would have $k^{\prime}(r_{1})<0$. By the continuity of $k$ there must be
$k(r_{1}-\epsilon)>1$ for any sufficiently small $\epsilon>0$, but this
contradicts $|k|<1$. Similarly, there can be no points $r_{1}$ with
$k_{+}(r_{1})=-1$. This together with [Har02, Corollary 3.1], shows that
$k_{+}$ is defined for all $r\geq r_{0}$.
Existence of $k_{-}$ with asserted properties is proved in a similar way.

∎

In Lemma 3.13 we establish the asymptotics of $k_{+}$ and $k_{-}$.

######  Lemma 3.13.

For $\epsilon>0$ small enough, there exists $r_{0}>0$ such that $k_{\pm}$ from
Lemma 3.12 satisfy

$k_{\pm}(r)=\frac{r}{\sqrt{1+r^{2}}}+\mathcal{O}(r^{-(n+1-\epsilon)}).$ (3.64)

###### Proof.

We will do finite induction to show the asymptotics. We start with the
equation $\mathcal{J}_{+}(k_{+})=0$ and define

$h(r)=1-2\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon},$ (3.65)

for $\epsilon>0$ small. Clearly $h(r_{0})=-1$ and further we have
$h^{\prime}=2(2-\epsilon)\frac{r_{0}^{2-\epsilon}}{r^{3-\epsilon}}$ and

$\begin{split}\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}h-\frac{r}{\sqrt{1+r^{2}}}\bigg{)}&=\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}-2\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}+\frac{1}{\sqrt{1+r^{2}}(r+\sqrt{1+r^{2}})}\bigg{)}\\\
&\leq-(n-1)\frac{r_{0}^{2-\epsilon}}{r^{3-\epsilon}}+\bigg{(}\frac{n-1}{2}\bigg{)}r^{-3},\end{split}$
(3.66)

as well as

$1-h^{2}=4\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}\bigg{(}1-\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}\bigg{)}.$
(3.67)

In particular, $\sqrt{1-h^{2}}\leq 2(\frac{r_{0}}{r})^{1-\epsilon/2}$. The
$C_{1}$-term of (3.59) may be estimated as follows:

$\begin{split}\frac{C_{1}}{r^{n-2}}&\frac{\sqrt{1-h^{2}}}{\sqrt{1+r^{2}}}\bigg{|}(n-3)(1-h^{2})\frac{1+r^{2}}{r}+\frac{1}{r}-(n-1)\frac{1+r^{2}}{r}-2(n-1)\frac{1+r^{2}}{r}h^{2}\\\
&\qquad+\sqrt{1+r^{2}}h(1-h^{2})+3(n-1)\sqrt{1+r^{2}}h\bigg{|}\\\
&\leq\frac{C_{1}}{r^{n-1}}2\bigg{(}\frac{r_{0}}{r}\bigg{)}^{1-\epsilon/2}\bigg{|}\frac{1}{r}-3(n-1)\frac{1+r^{2}}{r}+3(n-1)\sqrt{1+r^{2}}\bigg{|}\\\
&\qquad+\frac{C_{1}}{r^{n-2}}2\bigg{(}\frac{r_{0}}{r}\bigg{)}^{1-\epsilon/2}(1-h^{2})\bigg{|}(3n-5)\frac{\sqrt{1+r^{2}}}{r}+h\bigg{|}\\\
&\qquad+\frac{C_{1}}{r^{n-1}}12\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-3\epsilon/2}\\\
&\leq\frac{C_{8}}{r^{n-2}}.\end{split}$ (3.68)

The $C_{1}^{2}$-, $C_{1}^{3}$\- and $C_{1}^{4}$-terms in (3.59) are all easily
estimated to decay as $\mathcal{O}(r^{-(n-2)})$. The $C_{2}$-, $C_{3}$\- and
$C_{4}$-terms are similarly estimated to $\mathcal{O}(r^{-{n+2}})$. Inserting
these estimates into $\mathcal{J}_{+}(h)$ yields

$\begin{split}\mathcal{J}_{+}(h)&\leq
2(2-\epsilon)\frac{r_{0}^{2-\epsilon}}{r^{3-\epsilon}}+\bigg{(}\frac{n-1}{2}\bigg{)}\frac{1}{r^{3}}-2(n-1)\frac{r_{0}^{2-\epsilon}}{r^{3-\epsilon}}+\frac{C_{6}}{r^{5}}\\\
&\qquad-4\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}\bigg{(}1-\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}\bigg{)}\frac{1}{\sqrt{1+r^{2}}}+\frac{C_{7}}{r^{n-2}}\\\
&=\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}\frac{2}{r}\bigg{(}\epsilon+2\bigg{(}\frac{r_{0}}{r}\bigg{)}^{2-\epsilon}-(n-1)\bigg{)}+\mathcal{O}(r^{-3})+\mathcal{O}(r^{-(n-2)}).\end{split}$
(3.69)

We see that the sum in the paranthesis of the first term is negative for small
$\epsilon$ and so we have a subsolution for large enough $r_{0}$. We have
already seen that $k_{+}^{+}=+1$ is a supersolution. By Proposition 3.11 we
have $h\leq k_{+}\leq 1$.
Now we write $k_{+}=1+g$, where $g=\mathcal{O}_{1}(r^{-(2-\epsilon)})$ from
the previous step, where again $\epsilon>0$ is small. Clearly
$k_{+}^{\prime}=g^{\prime}$ and straightforward calculations yield

$\begin{split}\bigg{(}\frac{n-1}{r}\bigg{)}\bigg{(}k_{+}-\frac{r}{\sqrt{1+r^{2}}}\bigg{)}&=\bigg{(}\frac{n-1}{2}\bigg{)}\frac{1}{r^{3}}+\bigg{(}\frac{n-1}{r}\bigg{)}g+\mathcal{O}(r^{-5}),\\\
-\frac{1-k_{+}^{2}}{\sqrt{1+r^{2}}}&=\frac{2}{r}g+\mathcal{O}(r^{-(5-2\epsilon)}).\end{split}$
(3.70)

We estimate the $C_{1}$-term of (3.59):

$\begin{split}\frac{\sqrt{1-k_{+}^{2}}}{\sqrt{1+r^{2}}}\frac{C_{1}}{r^{n-2}}&\bigg{|}(1-k_{+}^{2})\frac{1+r^{2}}{r}(n-3)+\frac{1}{r}-(n-1)\frac{1+r^{2}}{r}\\\
&\qquad\qquad-2(n-1)\frac{1+r^{2}}{r}k_{+}^{2}\\\
&\qquad\qquad+\sqrt{1+r^{2}}k_{+}(1-k_{+}^{2})+3(n-1)\sqrt{1+r^{2}}k_{+}\bigg{|}\\\
&=\mathcal{O}(r^{-(n+1-3\epsilon/2)})\end{split}$ (3.71)

Similarly, we estimate the $C_{1}^{2}$-, $C_{1}^{3}$-, $C_{1}^{4}$-terms to be
$\mathcal{O}(r^{-(n+2)})$, the $C_{2}$-term to be
$\mathcal{O}(r^{-(n+1-\epsilon/2)})$ and the $C_{3}$-term to be
$\mathcal{O}(r^{-(n+2)})$. Inserting into $\mathcal{J}_{+}(1+g)=0$ yields

$\begin{split}0&=g^{\prime}+\bigg{(}\frac{n-1}{2}\bigg{)}\frac{1}{r^{3}}+\bigg{(}\frac{n-1}{r}\bigg{)}g+\mathcal{O}(r^{-5})+2\frac{g}{r}+\mathcal{O}(r^{-(5-2\epsilon)})\\\
&=g^{\prime}+\bigg{(}\frac{n+1}{r}\bigg{)}g+\bigg{(}\frac{n-1}{2}\bigg{)}\frac{1}{r^{3}}+\mathcal{O}(r^{-(5-2\epsilon)}).\end{split}$
(3.72)

for any $n\geq 4$. We multiply with $r^{n+1}$ and integrate from $r_{0}$ to
$r$:

$\begin{split}0&=\int_{r_{0}}^{r}\bigg{(}(s^{n+1}g)^{\prime}+\bigg{(}\frac{n-1}{2}\bigg{)}s^{n-2}+\mathcal{O}(r^{n-4+2\epsilon})\bigg{)}ds\\\
&=r^{n+1}g(r)-r_{0}^{n+1}g(r_{0})+\frac{r^{n-1}}{2}-\frac{r^{n-1}_{0}}{2}+\mathcal{O}(r^{n-3+2\epsilon})\end{split}$
(3.73)

so that in turn we find $g(r)$:

$g(r)=-\frac{1}{2r^{2}}+\mathcal{O}(r^{-2(2-\epsilon)}),$ (3.74)

for any $n\geq 4$.
We repeat the argument inductively, but now we assume that

$k_{+}=\frac{r}{\sqrt{1+r^{2}}}+g,$ (3.75)

where $g=\mathcal{O}(r^{-(2m-\epsilon)})$, for integer $m\geq 2$. $2m=n+1$ is
our assertion, so we assume that $2m\leq n$. Straightforward calculations
yield

$k_{+}^{\prime}=\frac{1}{(1+r^{2})^{3/2}}+g^{\prime},\\\ $ (3.76)

and

$-\frac{1-k_{+}^{2}}{\sqrt{1+r^{2}}}=-\frac{1}{(1+r^{2})^{3/2}}+\frac{2}{r}g+\mathcal{O}(r^{-(2m+3-\epsilon)}).$
(3.77)

To estimate the $C_{1}$-term in (3.59) we first observe that
$\sqrt{1-k_{+}^{2}}=\mathcal{O}(r^{-1})$. With this at hand we get

$\begin{split}\frac{\sqrt{1-k^{2}_{+}}}{\sqrt{1+r^{2}}}&\frac{C_{1}}{r^{n-2}}\bigg{|}(1-k_{+}^{2})\frac{1+r^{2}}{r}(n-3)+\frac{1}{r}-(n-1)\frac{1+r^{2}}{r}-2(n-1)\frac{1+r^{2}}{r}k_{+}^{2}\\\
&\qquad+\sqrt{1+r^{2}}k(1-k_{+}^{2})+3(n-1)\sqrt{1+r^{2}}k_{+}\bigg{|}\\\
&=\mathcal{O}(r^{-n})\times\mathcal{O}(r^{-(2m-1-|\epsilon|)})\\\
&=\mathcal{O}(r^{-(n+2)}).\end{split}$ (3.78)

We estimate the $C_{1}^{2}$-, $C_{1}^{3}$-, $C_{1}^{4}$-terms to be
$\mathcal{O}(r^{-(n+2)})$. We estimate the $C_{2}$-term to be
$\mathcal{O}(r^{-(n+1-\epsilon/2)})$ and the $C_{3}$-term to be
$\mathcal{O}(r^{-(n+2)})$. Clearly both the $C_{2}$\- and $C_{3}$-terms are
$\mathcal{O}(r^{-(n+2)})$. Inserting into $\mathcal{J}_{+}(k_{+})=0$ as above
yields

$0=g^{\prime}+\frac{g}{r}(n+1)+\mathcal{O}(r^{-(2m+3-\epsilon)})+\mathcal{O}(r^{-(n+2)}).$
(3.79)

First, let us consider the case when $2m+3\leq n+2$ so that the first
$\mathcal{O}$-term dominates. We multiply by $r^{n+1}$ and integrate from
$r_{0}$ to $r$ as above to find:

$0=r^{n+1}g(r)-r_{0}^{n+1}g(r_{0})+\mathcal{O}(r^{-(2m+2-(n+1)-\epsilon)})$
(3.80)

from which it follows that

$g(r)=\mathcal{O}(r^{-2(m+1-\epsilon)}),$ (3.81)

or equivalently that

$k_{+}=\frac{r}{\sqrt{1+r^{2}}}+\mathcal{O}(r^{-2(m+1-\epsilon)}).$ (3.82)

Finally, in the case of $2m+3>n+2$, we recall that $2m\leq n$, which implies
$2m=n$. We multiply by $s^{n+1}$ and integrate from $r_{0}$ to $r$:

$0=r^{n+1}g(r)-r_{0}^{n+1}g(r_{0})+\mathcal{O}(1)$ (3.83)

so that

$g(r)=\mathcal{O}(r^{-(n+1)}),$ (3.84)

which is even stronger than asserted.
It is not difficult to see that the same procedure works for $k_{-}$. The only
difference is that the $C_{i}^{k}$-terms are negative in this case, but the
evaluations corresponding to Equations (3.69), (3.72) and (3.79) for the
asymptotics in (3.60) remain the same and so the argument still works.

∎

We are now ready to construct the barriers as in Definition 3.4.

######  Proposition 3.14.

For $r_{0}$ sufficiently large, there exists
$f_{+},f_{-}:\mathbb{S}^{n-1}\times[r_{0},\infty)\rightarrow\mathbb{R}$ such
that

  1. (1)

$f_{+}$ (respectively $f_{-}$) is an upper (respectively lower) barrier in the
sense of Definition 3.4;

  2. (2)

the asymptotics of $f_{+}$ and $f_{-}$ is

$f_{\pm}=\sqrt{1+r^{2}}+\frac{\alpha}{r^{n-3}}+\mathcal{O}(r^{-(n-2-\epsilon)}),$
(3.85)

where

$\Delta^{\Omega}(\alpha)-(n-3)\alpha=\bigg{(}\frac{n-2}{2}\bigg{)}\operatorname{trace}^{\Omega}(\textbf{m})+\operatorname{trace}^{\Omega}(\textbf{p});$
(3.86)

  3. (3)

$f_{-}\leq f_{+}$.

###### Proof.

We let $\epsilon>0$ be given and take $r_{0}$, $k_{\pm}$ as in Lemmas 3.9 and
3.13. Clearly $k^{\prime}_{\pm}(r_{0})\not=0$ and hence, by continuity,
$k_{\pm}^{\prime}\not=0$ on $[r_{0},r_{0}+\delta]$ for some $\delta>0$. Hence
$1\pm k_{\pm}(r)\geq C(r-r_{0})$, for some $C>0$, when
$r\in[r_{0},r_{0}+\delta]$. It follows that

$\varphi_{\pm}^{\prime}(r)=\frac{k_{\pm}(r)}{\sqrt{(1-k_{\pm}(r)^{2})(1+r^{2})}}$
(3.87)

defines (modulo constants) the continuous functions $\varphi_{\pm}(r)$ on
$[r_{0},\infty)$, both of which are $C^{2}((r_{0},\infty))$. From the
asymptotics in Lemma 3.9 it follows that

$\varphi^{\prime}_{\pm}(r)=\frac{r}{\sqrt{1+r^{2}}}+\mathcal{O}(r^{-(n-1-\epsilon)})$
(3.88)

and since the Jang equation is invariant under vertical translations we assume

$\varphi_{\pm}(r)=\sqrt{1+r^{2}}+\mathcal{O}(r^{-(n-2-\epsilon)}).$ (3.89)

We define

$f_{\pm}=\varphi_{\pm}+\frac{\alpha}{r^{n-3}}.$ (3.90)

By Lemmas 3.10 and 3.12 $f_{\pm}$ will satisfy

$\mathcal{J}(f_{+})<0,\qquad\mathcal{J}(f_{-})>0.$ (3.91)

Furthermore, we have

$f_{+,r}(r_{0})=-\infty\qquad\text{and}\qquad f_{-,r}(r_{0})=+\infty,$ (3.92)

since $k_{+}(r_{0})=-1$ and $k_{-}(r_{0})=+1$.
It remains only to show that $f_{-}\leq f_{+}$. We use a version of the
Bernstein trick as in [SY81b], Proposition 3. Clearly, the difference
$f_{+}-f_{-}$ does not depend on $\theta$ and there must exist a constant
$L_{0}\geq 0$ such that $f_{+}-f_{-}>-L_{0}$ when $r\geq r_{0}$. We let
$L_{0}$ be the infimum of such constants and show that $L_{0}=0$. Then

$(f_{+}-f_{-})(r)\geq-L_{0}$ (3.93)

for all $r\in[r_{0},\infty)$ and either the equality is attained at some fixed
$r_{1}\in[r_{0},\infty)$ or

$\lim_{r\rightarrow\infty}(f_{+}-f_{-})(r)=-L_{0}.$ (3.94)

In the latter case, we must obviously have $L_{0}=0$. In the former case, we
first suppose that $r_{1}=r_{0}$. But then $(f_{+}-f_{-})(r_{0})\geq 0$ which
directly contradicts the properties of the barriers. Now assume that
$r_{1}>r_{0}$. We let $p\in M^{n}$ be a point with $r(\Psi(p))=r_{1}$ out in
the chart at infinity. Since $(f_{+}-f_{-})_{,r}(r_{0})=0$ and $f_{+}-f_{-}$
is radially symmetric we must have $f_{+,i}=f_{-,i}$ at $p$ and hence also

$g^{ij}-\frac{f_{+}^{,i}f_{+}^{,j}}{1+|df_{+}|_{g}^{2}}=g^{ij}-\frac{f_{-}^{,i}f_{-}^{,j}}{1+|df_{-}|_{g}^{2}}=\hat{g}_{\pm}^{ij}$
(3.95)

at $p$. Since $\hat{g}_{\pm}^{ij}$ is the inverse matrix of $\hat{g}_{ij}$,
which is positive definite, it is itself positive definite. Clearly, the same
arguments also implies that $|df_{+}|_{g}^{2}=|df_{-}|_{g}^{2}$. Furthermore,
since $p$ is a local minimum, it must follow that the matrix
$(f_{+}-f_{-})_{,ij}(p)$ is non-negative definite. Finally, since $f_{+}$ (and
$f_{-}$) is a supersolution (respectively subsolution) we must have

$\begin{split}0&>\mathcal{J}(f_{+})-\mathcal{J}(f_{-})\\\
&=\bigg{(}g^{ij}-\frac{f_{+}^{,i}f_{+}^{,j}}{1+|df_{+}|_{g}^{2}}\bigg{)}\bigg{(}\frac{\text{Hess}_{ij}^{g}(f_{+})}{\sqrt{1+|df_{+}|_{g}^{2}}}-k_{ij}\bigg{)}\\\
&\qquad-\bigg{(}g^{ij}-\frac{f_{-}^{,i}f_{-}^{,j}}{1+|df_{-}|_{g}^{2}}\bigg{)}\bigg{(}\frac{\text{Hess}_{ij}^{g}(f_{-})}{\sqrt{1+|df_{-}|_{g}^{2}}}-k_{ij}\bigg{)}\\\
&=\hat{g}^{ij}_{\pm}\frac{\text{Hess}_{ij}^{g}(f_{+}-f_{-})}{\sqrt{1+|df_{\pm}|_{g}^{2}}}\\\
&=\hat{g}^{ij}_{\pm}\frac{(f_{+}-f_{-})_{,ij}}{\sqrt{1+|df_{\pm}|_{g}^{2}}},\end{split}$
(3.96)

at $p$. We see that on the one hand the last term must be negative but on the
other hand it is the trace of a positive definite matrix $\hat{g}_{\pm}^{ij}$
with a non-negative definite matrix $(f_{+}-f_{-})_{,ij}(p)$ and we have a
contradiction to $f_{+}-f_{-}$ having a local minimum. We have shown that
$L_{0}=0$ and hence $f_{+}\geq f_{-}$.

∎

##  4\. The regularized Jang equation as a Dirichlet problem

In this section we perform the first step in solving Jang’s equation (2.14)
$\mathcal{J}(f)=0$ which is to solve the regularized equation
$\mathcal{J}(f)=\tau f$, where $\tau>0$ is small, on bounded sets. This
circumvents the lack of zeroth order derivatives in $\mathcal{J}(f)$ and
yields a priori $\tau$-dependent supremum estimates $\sup_{M^{n}}|f|$.
Consequently, we can solve the regularized equation on compact sets
$\overline{\Omega}\subset M^{n}$, provided that $\partial\Omega$ satisfies a
certain geometric condition. The procedure is well-known, (see for instance
[AEM11]), but we include it for completeness.

######  Definition 4.1.

Let $\Omega\subset M^{n}$ be a bounded subset with boundary $\partial\Omega$
and let $H_{\partial\Omega}$ be the mean curvature of $\partial\Omega$
computed as the divergence of the outward pointing unit normal. If

$H_{\partial\Omega}>|\operatorname{trace}_{\partial\Omega}(k)|,$ (4.1)

then $\Omega$ is said to fulfill the _trapping condition_.

Let $\tau>0$ be small. We want to solve the Dirichlet problem

$\begin{split}\mathcal{J}(f_{\tau})&=\tau f_{\tau}\qquad\text{on}\>\Omega,\\\
f_{\tau}&=\varphi\>\>\qquad\>\text{on}\>\partial\Omega,\end{split}$ (4.2)

for $\Omega$ as in Definition 4.1. For the remainder of this section, we will
supress the index $\tau$ on $f_{\tau}$ and refer to the solution of (4.2) as
$f$ for brevity.
The solution to Problem 4.2 is obtained by the continuity method, where we
define the parametrized Jang operator
$\mathcal{J}_{s}(f)=H(f)-s\operatorname{trace}(k)(f)$, with $s\in[0,1]$, and
consider the following parametrized problem:

$\begin{split}\mathcal{J}_{s}(f_{s})&=\tau f_{s}\qquad\text{in}\>\Omega,\\\
f_{s}&=s\varphi\qquad\>\text{on}\>\partial\Omega,\end{split}$ (4.3)

where $\varphi\in C^{2,\alpha}(\partial\Omega)$. Here we take $\alpha$ fixed;
we will at the end of the proof of Lemma 4.2 find a $0<\beta\leq 1$ and
throughout this section we fix $0<\alpha<\beta$. Let $\mathcal{S}\subset[0,1]$
denote the subset of parameters $s$ such that Equations (4.3) has a solution
in $C^{2,\alpha}(\bar{\Omega})$. In Lemmas 4.2 and 4.3 below, we will show
that $\mathcal{S}$ is both relatively open and closed.

######  Lemma 4.2.

Let $\Omega$ satisfy the trapping condition in (4.1). Then $\mathcal{S}$ is
closed.

###### Proof.

We start by establishing a uniform $C^{1}(\Omega)$-bound in $s$, which will be
subsquently upgraded to a $C^{2,\alpha}(\Omega)$-bound via standard theory for
elliptic equations.
First, we apply a maximum principle argument to show the uniform estimate
$\tau|f_{s}|\leq C$ on $\bar{\Omega}$, where $C$ is a constant depending only
on the initial data $(M^{n},g,k)$ and not on $s$. If $f_{s}$ achieves its
maximum at some interior point $p$, then $df_{s}=0$ so that $\hat{g}=g$ and
$|df_{s}|^{2}_{g}=0$ at $p$. Further, the Hessian at $p$ reduces to
$\text{Hess}_{ij}^{g}(f_{s})=(f_{s})_{,ij}$ and is non-positive definite
there. Hence, at a point $p$ we have

$\begin{split}\tau f_{s}&=\mathcal{J}_{s}(f_{s})\\\
&=g^{ij}\text{Hess}_{ij}^{g}(f_{s})-s\operatorname{trace}^{\hat{g}_{s}}(k)\\\
&\leq-s\operatorname{trace}^{g}(k)\\\ &\leq|\operatorname{trace}^{g}(k)|\\\
\end{split}$ (4.4)

as $g^{ij}$ is positive definite. Similarly, if $f_{s}$ has a minimum at $p$,
we get $\tau f_{s}\geq-|\operatorname{trace}^{g}(k)|$. Conclusively, we have
shown that $\tau|f_{s}|\leq|\operatorname{trace}^{g}(k)|$ at $p$ and hence it
follows that

$\tau|f_{s}|\leq\max\bigg{(}\sup_{\bar{\Omega}}|\operatorname{trace}^{g}(k)|,\>\sup_{\partial\Omega}\tau|\varphi|\bigg{)},$
(4.5)

which only depends on the initial data $(M^{n},g,k)$ and the boundary data
$\varphi$.
We now establish a bound, uniform in $s$, for the gradient $df_{s}$. We start
with an interior estimate $|df_{s}|_{g}$ in $\Omega$. Suppose
$|df_{s}|_{g}^{2}$ achieves its maximum at some point $p\in\Omega$. We take
the covariant derivative of both sides of $\mathcal{J}_{s}(f_{s})=\tau f_{s}$
and contract with $\nabla^{g}f_{s}$ to recover an expression for
$|df_{s}|^{2}_{g}$. Straightforward calculations show that

$\nabla_{k}|df_{s}|_{g}^{2}=2f^{,i}\text{Hess}_{ik}^{g}(f_{s})$ (4.6)

and

$\begin{split}(\nabla_{k}\hat{g}_{s})^{ij}&=-\frac{f_{s}^{,j}g^{im}\text{Hess}_{mk}^{g}(f_{s})}{1+|df_{s}|_{g}^{2}}-\frac{f_{s}^{,i}g^{jm}\text{Hess}_{mk}^{g}(f_{s})}{1+|df_{s}|_{g}^{2}}\\\
&\qquad+2\frac{f_{s}^{,i}f_{s}^{,j}}{(1+|df_{s}|_{g}^{2})^{2}}f^{,\ell}_{s}\text{Hess}_{\ell
k}^{g}(f_{s}).\\\ \end{split}$ (4.7)

For our calculations below, it will be convenient to recast the Hessian term
in divergence form. We straightforwardly get

$\begin{split}\bigg{(}\nabla_{m}\bigg{(}\frac{\nabla^{g}f_{s}}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}\bigg{)}^{k}&=\bigg{(}g^{k\ell}-\frac{f_{s}^{,k}f_{s}^{,\ell}}{1+|df_{s}|_{g}^{2}}\bigg{)}\frac{\text{Hess}_{\ell
m}^{g}(f_{s})}{\sqrt{1+|df_{s}|_{g}^{2}}}.\end{split}$ (4.8)

Contracting over $k$ and $m$ and recalling (2.19) yields the familiar
divergence form for the mean curvature:

$\begin{split}\text{div}^{g}\bigg{(}\frac{\nabla^{g}f_{s}}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}&=H_{\hat{M}^{n}_{s}},\end{split}$
(4.9)

where $H_{\hat{M}^{n}_{s}}$ denotes the mean curvature of $\hat{M}^{n}_{s}$,
the graph of $f_{s}$ over $\Omega$. Differentiating we find

$\begin{split}H_{,k}^{\hat{M}^{n}_{s}}&=\bigg{(}\nabla_{k}\nabla\bigg{(}\frac{\nabla^{g}f_{s}}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}\bigg{)}^{i}_{i}.\\\
\end{split}$ (4.10)

Using the definition of the Riemann tensor as the commutation of second order
derivatives it follows that

$\begin{split}H_{,k}^{\hat{M}^{n}_{s}}&=\bigg{(}\nabla_{i}\nabla\bigg{(}\frac{\nabla^{g}f_{s}}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}\bigg{)}^{i}_{k}-\text{Ric}^{g}_{k\ell}\frac{f^{,\ell}_{s}}{\sqrt{1+|df_{s}|^{2}_{g}}}\\\
&=\nabla_{i}\bigg{(}\bigg{(}g^{mi}-\frac{f_{s}^{,i}f_{s}^{,m}}{1+|df_{s}|^{2}_{g}}\bigg{)}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}-\text{Ric}^{g}_{k\ell}\frac{f_{s}^{,\ell}}{\sqrt{1+|df_{s}|^{2}_{g}}}.\end{split}$
(4.11)

Differentiating the trace term we obtain

$\operatorname{trace}_{\hat{g}}(k)_{,k}=(\nabla_{k}\hat{g})^{ij}k_{ij}+\hat{g}^{ij}(\nabla_{k}k)_{ij},$
(4.12)

where

$\begin{split}(\nabla_{k}\hat{g})^{ij}k_{ij}&=-2\bigg{(}g^{ij}-\frac{f_{s}^{,i}f_{s}^{,j}}{1+|df_{s}|_{g}^{2}}\bigg{)}\frac{\text{Hess}_{jk}^{g}(f_{s})f_{s}^{,\ell}k_{i\ell}}{1+|df_{s}|_{g}^{2}}.\end{split}$
(4.13)

In summary, differentiating the regularized Jang equation
$\mathcal{J}_{s}(f_{s})=\tau f_{s}$ yields

$\begin{split}\tau
f^{s}_{,k}&=\nabla_{i}\bigg{(}\bigg{(}g^{mi}-\frac{f_{s}^{,i}f_{s}^{,m}}{1+|df_{s}|^{2}_{g}}\bigg{)}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}-\text{Ric}^{g}_{k\ell}\frac{f_{s}^{,\ell}}{\sqrt{1+|df_{s}|^{2}_{g}}}\\\
&\qquad-s\bigg{(}-2\bigg{(}g^{ij}-\frac{f_{s}^{,i}f_{s}^{,j}}{1+|df_{s}|_{g}^{2}}\bigg{)}\frac{\text{Hess}_{jk}^{g}(f_{s})f_{s}^{,\ell}k_{i\ell}}{1+|df_{s}|_{g}^{2}}+\bigg{(}g^{ij}-\frac{f_{s}^{,i}f_{s}^{,j}}{1+|df_{s}|_{g}^{2}}\bigg{)}(\nabla_{k}k)_{ij}\bigg{)}.\end{split}$
(4.14)

We multiply this equation by $f_{s}^{,k}$ and sum over $k$. To estimate the
first term in the right hand side of the resulting equation, we observe that

$\begin{split}\nabla_{i}\bigg{(}\hat{g}^{mi}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}f^{,k}\bigg{)}&=\nabla_{i}\bigg{(}\hat{g}^{mi}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}f_{s}^{,k}+\hat{g}^{mi}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}g^{k\ell}\text{Hess}_{\ell
i}^{g}(f_{s})\\\
&=\nabla_{i}\bigg{(}\hat{g}^{mi}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}f_{s}^{,k}+\frac{\operatorname{trace}(\hat{g}^{-1}\text{Hess}^{g}(f)\>g^{-1}\text{Hess}^{g}(f))}{\sqrt{1+|df_{s}|^{2}_{g}}}\\\
&\geq\nabla_{i}\bigg{(}\hat{g}^{mi}\frac{\text{Hess}_{mk}^{g}(f_{s})}{\sqrt{1+|df_{s}|^{2}_{g}}}\bigg{)}f_{s}^{,k},\end{split}$
(4.15)

From the second term in the right hand side of the resulting equation we
obtain

$\begin{split}\text{Ric}^{g}_{k\ell}\frac{f_{s}^{,\ell}}{\sqrt{1+|df_{s}|^{2}_{g}}}f_{s}^{,k}&\leq\frac{|\text{Ric}^{g}|_{g}|df_{s}\otimes
df_{s}|_{g}}{\sqrt{1+|df_{s}|^{2}_{g}}}\\\ &\leq C|df_{s}|_{g},\end{split}$
(4.16)

where the constant $C$ depends only on the initial data $(M^{n},g,k)$. As for
the third term, we note that

$\begin{split}2\hat{g}^{ij}\frac{\text{Hess}_{jk}^{g}(f_{s})f^{,\ell}k_{i\ell}}{1+|df_{s}|_{g}^{2}}f^{,k}&=\hat{g}^{ij}\frac{(|df_{s}|_{g}^{2})_{,j}g^{m\ell}f^{s}_{,m}k_{i\ell}}{1+|df_{s}|_{g}^{2}}\\\
&\leq B^{k}(|df_{s}|_{g}^{2})_{,k},\end{split}$ (4.17)

where $B^{k}$ is bounded. Finally, an estimation of
$p_{k}=\hat{g}^{ij}(\nabla_{k}k)_{ij}$ yields

$\begin{split}p_{k}f_{s}^{,k}&=\langle p,df_{s}\rangle_{g}\\\
&\leq|p|_{g}|df_{s}|_{g}\\\ \end{split}$ (4.18)

Defining $u=|df_{s}|^{2}_{g}$ and adding all terms we arrive at the inequality

$\tau u\leq\nabla_{i}(A^{ij}u_{,j})+B^{k}u_{,k}+C\sqrt{u},$ (4.19)

where $A^{ij}$ is positive definite, $B^{k}$ is bounded, $C$ is a constant and
for all $A^{ij}$, $B^{k}$ and $C$ depend only on the initial data
$(M^{n},g,k)$.
At an interior maximum point $p\in\Omega$ of $u$ we must have $u_{,k}=0$.
Thus, by (4.19), we obtain $\tau|df_{s}|_{g}\leq C$ at $p$, where $C$ only
depends on the initial data $(M^{n},g,k)$.
We now proceed to obtain the boundary gradient estimate. Since $\varphi\in
C^{2,\alpha}(\partial\Omega)$ we trivially have a bound on the gradient in the
tangential direction. To estimate the gradient in the normal direction we
employ the barrier method, where suitable barrier functions $w^{-}$ and
$w^{+}$ are used to control the normal derivative. Explicitly we require that
$\mathcal{J}_{s}(w^{+})<\tau w^{+}$, $\mathcal{J}_{s}(w^{-})>\tau w^{-}$ near
$\partial\Omega$ and $w^{\pm}=s\varphi$ on $\partial\Omega$. From the
comparison principle of [GT01], Chapter 10 (but see also Appendix B of
[Sak21]) it follows that in this case barriers satisfy $w^{-}\leq f_{s}\leq
w^{+}$, which gives

$\frac{w^{-}(p)-w^{-}(p_{0})}{d_{g}(p,p_{0})}\leq\frac{f_{s}(p)-f_{s}(p_{0})}{d_{g}(p,p_{0})}\leq\frac{w^{+}(p)-w^{+}(p_{0})}{d_{g}(p,p_{0})},$
(4.20)

where $p_{0}\in\partial\Omega$ and $p\in\Omega$. It follows that

$\frac{\partial w^{-}}{\partial\vec{n}}(p_{0})\leq\frac{\partial
f}{\partial\vec{n}}(p_{0})\leq\frac{\partial w^{+}}{\partial\vec{n}}(p_{0}),$
(4.21)

where $\vec{n}$ is the inward pointing unit normal to $\partial\Omega$, and so
the full boundary gradient estimate would follow.
In order to construct the barriers we invoke _Fermi coordinates_ 555We will
use Fermi coordinates again in Section 6, where a more detailed description is
found. (or _normal geodesic coordinates_). Namely, we let
$\rho=\text{dist}(\cdot,\partial\Omega)$ and denote by $N_{\rho}$ the
hypersurfaces (or the leaves) of constant $\rho$, for $\rho$ sufficiently
small. Using coordinates $x^{\mu}$, where $\mu=1,\ldots,n-1$ on
$\partial\Omega$, we have a coordinate system $(\rho,x^{1},\ldots,x^{n-1})$
defined in a neighbourhood $\\{\rho<\rho_{0}\\}$ of $\partial\Omega$ where we
can write $g=g_{\rho}+d\rho^{2}$, where $g_{\rho}$ is the induced metric on
$N_{\rho}$. By (4.1) and the continuity of $\varphi$, there exists a small
number $\rho_{0}$, such that in the neighbourhood $\\{\rho<\rho_{0}\\}$ we
have $H_{N^{\rho}}-|\operatorname{trace}^{N_{\rho}}(k)|>\tau|\varphi|$.
We note that in this coordinate system we have
$(A^{\rho})_{\mu\nu}=\Gamma^{\rho}_{\mu\nu}$, where $A^{\rho}$ denotes the
second fundamental form of $N_{\rho}$. In particular, we have
$H_{\partial\Omega}=g^{\mu\nu}A^{0}_{\mu\nu}$.
We let $\varphi$ be extended trivially along the $\rho$-coordinate and claim
that $w^{\pm}=s\varphi\pm\rho B$ are barriers, where $B$ is a sufficiently
large positive constant. We define $Q(f)=\mathcal{J}_{s}(f)-\tau f$ and need
to show that $\pm\mathcal{J}_{s}(w^{\pm})\mp\tau w^{\pm}<0$ holds for large
enough $B$. We have
$1+|dw^{\pm}|^{2}_{g}=1+B^{2}+s^{2}|\varphi|^{2}_{g^{\rho}}=\mathcal{O}(B^{2})$.
For the mean curvature term we have

$\begin{split}\bigg{(}g^{ij}-\frac{w^{\pm,i}w^{\pm,j}}{1+|dw^{\pm}|_{g}^{2}}\bigg{)}\bigg{(}\frac{w^{\pm}_{,ij}}{\sqrt{1+|dw^{\pm}|_{g}^{2}}}\bigg{)}&=\mathcal{O}(B^{-1}),\end{split}$
(4.22)

since all derivatives except for tangential vanish. We also have

$\begin{split}-\bigg{(}g^{ij}-\frac{w^{\pm,i}w^{\pm,j}}{1+|dw^{\pm}|_{g}^{2}}\bigg{)}\frac{\Gamma^{k}_{ij}w^{\pm}_{,k}}{\sqrt{1+|dw^{\pm}|_{g}^{2}}}&=-g^{\mu\nu}\frac{\Gamma^{\rho}_{\mu\nu}(\pm
B)}{B+\mathcal{O}(B^{-1})}+\mathcal{O}(B^{-1})\\\
&=-H_{N^{\rho}}+\mathcal{O}(B^{-1}).\end{split}$ (4.23)

A straightforward calculation estimates the trace term:

$\begin{split}\bigg{(}g^{ij}-\frac{w^{\pm,i}w^{\pm,j}}{1+|dw^{\pm}|_{g}^{2}}\bigg{)}k_{ij}&=\operatorname{trace}_{N^{\rho}}(k)+\mathcal{O}(B^{-1}).\end{split}$
(4.24)

Taken together, this yields

$\begin{split}\mathcal{J}_{s}(w^{+})-\tau
w^{+}&=-H_{N^{\rho}}-s\operatorname{trace}^{N_{\rho}}(k)-\tau
s\varphi-\tau\rho B+\mathcal{O}(B^{-1})\\\ &\leq-
H_{N^{\rho}}+|\operatorname{trace}^{N_{\rho}}(k)|+\tau|\varphi|+\mathcal{O}(B^{-1})\\\
&<0\\\ \end{split}$ (4.25)

for $B$ sufficiently large. A similar estimate shows
$\mathcal{J}_{s}(w^{-})-\tau w^{-}>0$ so that $w^{\pm}$ are barriers. The
barriers have normal derivatives $\partial_{\vec{n}}w^{\pm}=\pm B$ and since
$B$ only depends on the initial data it follows that we have a uniform in $s$
$C^{1}$-estimate of $f_{s}$, which we denote by $||f_{s}||_{C^{1}(\Omega)}\leq
K_{\tau}$.
It is clear that there is a uniform lower bound of the eigenvalues of
$\hat{g}_{s}^{ij}:$ $\lambda_{K_{\tau}}\leq\lambda(x,f_{s},df_{s})$ and
rewriting Jang’s equation as
$Qf_{s}=a^{ij}(x,f_{s},df_{s})f_{s,ij}+b(x,f_{s},df_{s})=0$ where

$\begin{split}a^{ij}(x,f_{s},df_{s})&=\bigg{(}g^{ij}-\frac{f_{s}^{,i}f_{s}^{,j}}{1+|df_{s}|^{2}_{g}}\bigg{)},\\\
b(x,f_{s},df_{s})&=\bigg{(}g^{ij}-\frac{f_{s}^{,i}f_{s}^{,j}}{1+|df_{s}|^{2}_{g}}\bigg{)}\bigg{(}-\Gamma^{k}_{ij}f_{s,k}-k_{ij}\bigg{)},\end{split}$
(4.26)

it is also straightforward to see that there exists a constant
$\mu_{K_{\tau}}$ (uniform in $s$) such that

$\begin{split}|a^{ij}(x,z,\vec{p})|+|a^{ij}(x,z,\vec{p})_{,p^{k}}|&+|a^{ij}(s,z,\vec{p})_{,z}|\\\
&\qquad+|a^{ij}(x,z,\vec{p})_{,x^{k}}|+|b(x,z,\vec{p})|\leq\mu_{K_{\tau}}.\end{split}$
(4.27)

From the global Hölder estimate of Ladyzhenskaya and Ural’tseva (cf. Chapter
13 in [GT01]) we then get a uniform bound of the Hölder coefficient
$[df_{s}]\leq C(n,\Omega,K,\mu_{K_{\tau}}/\lambda_{K_{\tau}})$ over $\Omega$
and in turn a global bound in $C^{1,\beta}(\overline{\Omega})$, for some
$0<\beta<1$. Applying global Schauder estimates (cf. [GT01] Chapter 6) we get
a uniform over $s$ estimate in $C^{2,\beta}(\overline{\Omega})$.
We now let $\\{s_{n}\\}\subset\mathcal{S}$ be any sequence converging to
$s\in[0,1]$. It is well-known that the Arzela-Ascoli theorem implies
compactness of the embedding
$C^{2,\beta}(\bar{\Omega})\xrightarrow{}C^{2,\alpha}(\bar{\Omega})$ for
$0<\alpha<\beta$. In turn, the uniform
$C^{2,\beta}(\overline{\Omega})$-estimate hence gives subconvergence in
$C^{2,\alpha}(\bar{\Omega})$ to some $f_{s}\in C^{2,\alpha}(\bar{\Omega})$.
The smoothness of the convergence implies that $\mathcal{J}_{s}(f_{s})=\tau
f_{s}$ so that $s\in\mathcal{S}$. Hence $\mathcal{S}$ is closed.

∎

With Lemma 4.3 we show that $\mathcal{S}$ is open.

######  Lemma 4.3.

Let $\mathcal{S}\subset[0,1]$ be the set of $s$ such that (4.3) has a solution
$f_{s}\in C^{2,\alpha}(\overline{\Omega})$. Then $\mathcal{S}$ is open.

###### Proof.

We aim to show that $\mathcal{S}$ is open with the Implicit Function Theorem.
We consider the operator
$T:C^{2,\alpha}(\overline{\Omega})\times\mathbb{R}\rightarrow
C^{0,\alpha}(\overline{\Omega})\times
C^{0,\alpha}(\partial\Omega)\times\mathbb{R}$ given

$T(f,s)=(H(f)-s\operatorname{trace}(k)(f)-\tau
f,f|_{\partial\Omega}-s\varphi,s),$ (4.28)

Suppose that $f_{0}$ is a solution of Equations (4.3) for some $s_{0}$, that
is to say $T(f_{0},s_{0})=(0,0,s_{0})$. The linearization of $T$ at
$(f_{0},s_{0})$ is

$L_{(f_{0},s_{0})}(h,t)=\bigg{(}A^{ij}\text{Hess}_{ij}^{g}(h)+B^{k}h_{,k}-\tau
h-s_{0}\operatorname{trace}_{g}(k)(f_{0}),h|_{\partial\Omega}-s_{0}\varphi,t\bigg{)},$
(4.29)

where

$\begin{split}A^{ij}&=\frac{1}{\sqrt{1+|df_{0}|_{g}^{2}}}\bigg{(}g^{ij}-\frac{f_{0}^{i}f_{0}^{j}}{1+|df_{0}|_{g}^{2}}\bigg{)}\\\
B^{k}&=(\text{div}^{g}A)^{k}+2s_{0}\frac{1}{\sqrt{1+|df_{0}|_{g}^{2}}}A^{ik}f_{0}^{j}k_{ij}.\end{split}$
(4.30)

In order to apply the Implicit Function Theorem we need to show that
$L_{(f_{0},s_{0})}$ is an isomorphism. But if $F\in C^{0,\alpha}(\Omega)$ and
$G\in C^{2,\alpha}(\partial\Omega)$, then it is known from the theory of
linear elliptic partial differential equations (cf. Chapter 6 of [GT01]) that
the problem

$\begin{split}A^{ij}\text{Hess}_{ij}^{g}(h)+B^{k}h_{,k}-\tau
h&=s_{0}\operatorname{trace}_{g}(k)(f_{0})+F,\qquad\text{in}\qquad\Omega,\\\
h&=s_{0}\varphi+G,\qquad\quad\qquad\qquad\text{on}\qquad\partial\Omega\end{split}$
(4.31)

has a unique solution $h\in C^{2,\alpha}(\overline{\Omega})$. From the
Implicit Function Theorem there is an $\epsilon_{0}>0$ and a $C^{1}$-map
$s\rightarrow f_{s}$ defined on $|s-s_{0}|<\epsilon_{0}$, so that $f_{s}$
solves Equations (4.3). Hence $\mathcal{S}$ is open. ∎

We may now prove the main result of this section.

######  Proposition 4.4.

Let $\varphi\in C^{2,\alpha}(\partial\Omega)$ and suppose $\Omega$ satisfies
the trapping condition in (4.1). Then, for $\tau>0$ so small so as to ensure
that
$H_{\partial\Omega}-|\operatorname{trace}_{\partial\Omega}(k)|>\tau\varphi$,
the Dirichlet problem

$\begin{split}\mathcal{J}(f_{\tau})&=\tau f_{\tau}\qquad\text{on}\>\Omega,\\\
f_{\tau}&=\varphi\qquad\>\>\>\>\text{on}\>\partial\Omega\end{split}$ (4.32)

has a solution in $C^{2,\alpha}(\overline{\Omega})$. Moreover, if
$f_{-}\leq\varphi\leq f_{+}$ on $\partial\Omega$ then $f_{-}\leq f_{\tau}\leq
f_{+}$ on $\\{r_{0}\leq r\\}\cap\bar{\Omega}$, where $f_{\pm}$ are the
barriers obtained in Proposition 3.14.

###### Proof.

The proof is immediate from Lemmas 4.2 and 4.3, as for $s=0$ the trivial
function solves Equations (4.3) so that $\mathcal{S}$ is non-empty and hence
$\mathcal{S}=[0,1]$. In particular, a solution exists for $s=1$, which solves
the Dirichlet Problem in (4.2).
To show the assertion about $f_{-}\leq f_{\tau}\leq f_{+}$ we first note that
if $\tau>0$ is small enough, then

$\begin{split}\mathcal{J}(f_{+})-\tau f_{+}&>0\\\ \mathcal{J}(f_{-})-\tau
f_{\tau}&<0.\end{split}$ (4.33)

A similar argument as in the proof of Proposition 3.14 now applies to show
that $f_{-}\leq f_{\tau}\leq f_{+}$ on $\\{r_{0}\leq r\\}\cap\bar{\Omega}$.

∎

We abuse notation slightly and write $S_{R}=\Psi^{-1}(S_{R})\subset M^{n}$,
where $S_{R}\subset\mathbb{R}^{n}$ is the standard coordinate sphere and
$\Psi$ is the diffeomorphism of the initial data. With the following Lemma we
show that the coordinate spheres $S_{R}$ satisfy the trapping condition of
Definition 4.1.

######  Lemma 4.5.

Let $S_{R}$ be a coordinate sphere with $R>r_{0}$. Then $S_{R}$ satisfies the
trapping condition of Definition 4.1 for sufficiently large $R$.

###### Proof.

Straightforward compuations show that both

$H_{S_{R}}=(n-1)+\bigg{(}\frac{n-1}{2}\bigg{)}R^{-2}+\mathcal{O}(R^{-4})$
(4.34)

and

$\operatorname{trace}_{S_{R}}{k}=(n-1)+\mathcal{O}(R^{-n})$ (4.35)

which proves the assertion.

∎

##  5\. A geometric solution to Jang’s equation

In this section we obtain a geometric solution to Jang’s equation (2.14) by
subconverging the graphs obtained in Section 4. More specifically, this is
done using Geometric Measure Theory as summarized in Appendix E. The arguments
follow Section 2 of [Eic13] very closely, but we include them for
completeness.

###  5.1. Limit and regularity

We let $R$ be large so that the trapping condition in Definition 4.1 is
satisfied as per Lemma 4.5. For small enough $\tau>0$ and
$\varphi=\frac{1}{2}(f_{+}+f_{-})$, where $f_{\pm}$ are the barriers obtained
in Proposition 3.14, we get a solution $f_{\tau}$ satisfying
$\mathcal{J}(f_{\tau})=\tau f_{\tau}$ on $\bar{B}_{R}=\\{r\leq R\\}\subset
M^{n}$ by Proposition 4.4. For any sequences $\\{R_{k}\\}_{k=1}^{\infty}$ and
$\\{\tau_{k}\\}_{k=1}^{\infty}$ such that $R_{k}\rightarrow\infty$ and
$\tau_{k}\rightarrow 0$ as $k\rightarrow\infty$ we let
$\bar{B}_{k}=\bar{B}_{R_{k}}$ and denote by $\\{f_{k}\\}_{k=1}^{\infty}$ the
solutions obtained from Proposition 4.4. Further, denote the graphs of
$\\{f_{k}\\}_{k=1}^{\infty}$ over $\bar{B}_{k}$ by
$\\{\hat{M}^{n}_{k}\\}_{k=1}^{\infty}$. For our choice of boundary data
$\varphi$, Proposition 3.14 implies that we have $\varphi_{k}\leq 2R_{k}$ near
infinity, and so it is possible to chose $\\{R_{k}\\}_{k=1}^{\infty}$ and
$\\{\tau_{k}\\}_{k=1}^{\infty}$ such that $\tau_{k}R_{k}\leq A$, uniformly
over $k$. In turn, the estimate of $\tau_{k}\sup_{\bar{B}_{k}}|f_{k}|$ in the
proof of Lemma 4.2 then implies a uniform over $k$ estimate
$\tau_{k}\sup_{\bar{B}_{k}}|f_{k}|\leq C$, where $C$ depends only on the
initial data and the uniform constant $A$. Finally, by the Cauchy-Schwarz
inequality and the estimate $|\hat{g}|_{g}\leq\sqrt{n}$, we have

$\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}k_{ij}\leq\sqrt{n}|k|_{g}.$
(5.1)

Thus, it follows that the graphs $\\{\hat{M}^{n}_{k}\\}$ have uniformly
bounded mean curvature by some $\lambda$ depending on the initial data and
$A$.
We recall the following Harnack Principle, which appears as Lemma 2.3 in
[Eic09]:

######  Proposition 5.1.

(”Harnack principle”) Let $f_{k}:\Omega\rightarrow\mathbb{R}$ be
$C^{3}$-functions with open and connected domain $\Omega$, such that for some
$\beta>0$,

$\Delta^{k}(\gamma_{k}^{-1})\leq\beta\gamma_{k}^{-1}+\langle
X,d\big{(}\gamma_{k}^{-1}\big{)}\rangle_{k},$ (5.2)

where $\gamma_{k}=\sqrt{1+|df_{k}|^{2}}$, $X$ is a locally bounded vector
field and $\Delta_{k}$ is the Laplace-Beltrami operator of the graphs
$G_{k}=\text{graph}(f_{k})$. Suppose the graphs $G_{k}$ converge in $C^{3}$ to
a submanifold $G\subset\Omega\times\mathbb{R}$. Then, on each component of
$G$, $\gamma$ is either everywhere positive or everywhere vanishing.

The following Proposition is similar to Proposition 4 in [SY81b] and
Proposition 7 in [Eic13].

######  Proposition 5.2.

Let $(M^{n},g,k)$ be asymptotically hyperbolic initial data with Wang’s
asymptotics of type $(\ell,\alpha,\tau=n,\tau_{0}>0)$ as in Definition 2.3,
where $4\leq n\leq 7$. There exists an embedded
$C^{3,\alpha}_{loc}$-hypersurface
$(\hat{M}^{n},\hat{g})\subset(M^{n}\times\mathbb{R},g+dt^{2})$, with the
following properties:

  1. (1)

$\hat{M}^{n}$ is the boundary of an open set $\Omega$. We have
$H_{\hat{g}}-\operatorname{trace}_{\hat{g}}(k)=0$, where the mean curvature
$H_{\hat{g}}$ is computed as the tangential divergence the downward pointing
unit normal. Moreover, $\hat{M}^{n}=\partial\Omega$ is a $\lambda$-minimizing
boundary.

  2. (2)

$\hat{M}^{n}$ has finitely many connected components. Each component of
$\hat{M}^{n}$ is either cylindrical of the form $C_{\ell}\times\mathbb{R}$,
where $C_{\ell}$ is a closed and properly embedded $C^{3,\alpha}$-hypersurface
in $M^{n}$, or the graph of a function $f$, which solves the Jang equation
$\mathcal{J}(f)=0$, on an open subset $U_{f}\subset M^{n}$.

  3. (3)

The boundary $\partial U_{f}$ is a closed properly embedded
$C^{3,\alpha}$-hypersurface in $M^{n}$. More specifically, $\partial U_{f}$ is
the disjoint union of components $C^{+}_{\ell}$ and $C_{\ell}^{-}$, where
$f(p)\rightarrow\pm\infty$ uniformly as $p\rightarrow C_{\ell}^{\pm}$ from
$U_{f}$. These hypersurfaces $C_{\ell}^{\pm}$ satisfy
$H_{C_{\ell}}\mp\operatorname{trace}_{C_{\ell}}(k)=0$, where the mean
curvature is computed as the tangential divergence of the outward from $U_{f}$
pointing unit normal. There exists a $T\geq 1$ such that each component of
$\hat{M}^{n}\cap\\{|t|\geq T\\}$ is a graphs over $C_{\ell}\times[T,\infty)$.
Finally, the graphs $\text{graph}(f-A)$ converge in $C^{3,\alpha}_{loc}$ to
$C_{\ell}^{\pm}\times\mathbb{R}$ as $A\rightarrow\pm\infty$.

  4. (4)

$\hat{M}^{n}$ contains a graphical component with domain

$\\{p\in M^{n}\>|\>r>r_{0}\\}\subset U_{f},$ (5.3)

where $r_{0}$ is as in Proposition 3.14. Furthermore, $f$ has the
asymptotics666At this stage, we do not show the asymptotic flatness of the
graph. This is done in Section 6. as in (3.85) on this set.

###### Proof.

We use the results from the Geometric Measure Theory summmarized in Appendix E
to show the existence of the limit and its regularity. We let
$\\{R_{k}\\}_{k=1}^{\infty}$ and $\\{\tau_{k}\\}_{k=1}^{\infty}$ be as
explained previously in this subsection and denote by
$\\{f_{k}\\}_{k=1}^{\infty}$ be the functions obtained from Proposition 4.4.
As explained, the graphs $\\{\hat{M}^{n}_{k}\\}_{k=1}^{\infty}$ of
$\\{f_{k}\\}_{k=1}^{\infty}$ over $\\{\bar{B}_{k}\\}_{k=1}^{\infty}$ have
uniformly bounded mean curvature by some $\lambda$ depending only on the
initial data $(M^{n},g,k)$.
By the Nash-embedding Theorem there exists an isometric embedding
$F:M^{n}\times\mathbb{R}\rightarrow\mathbb{R}^{n+\ell}$ for some $\ell>0$. We
denote $F(M^{n}\times\mathbb{R})$ by $N^{n+1}$ to agree with the notation of
Appendix E. The graphs $\\{\hat{M}^{n}_{k}\\}_{k=1}^{\infty}$ are viewed as
currents $\\{T_{k}\\}_{k=1}^{\infty}$ which have multiplicity one and are
boundaries $T_{k}=\partial[[E_{k}]]$ that are $\lambda$-minimizing. Hence
$T_{k}\in\mathcal{F}_{\lambda}$ and by Theorem E.4 there is a subsequence
(denoted by the same index for notational convenience) such that
$\\{T_{k}\\}_{k=1}^{\infty}$ converges as currents to some
$T\in\mathcal{F}_{\lambda}$. By Theorem E.9 the limit graph $\hat{M}^{n}$ is
regular in the sense of Definition E.8 for $4\leq n\leq 6$ and we refer the
reader to Remark 4.1 in [Eic09] for the explanation why it is also regular for
$n=7$. From Lemma E.10 it follows that the convergence is in
$C^{1,\alpha}_{loc}$.
The limit satisfies Jang’s equation distributionally, as for each
$\hat{M}^{n}_{k}$ the mean curvature term in divergence form integrates to

$\int_{M^{n}}\text{div}_{g}\bigg{(}\frac{\nabla^{g}(f_{k})}{\sqrt{1+|df_{k}|^{2}_{g}}}\bigg{)}\varphi
d\mu_{g}=-\int_{M^{n}}\bigg{\langle}\frac{\nabla^{g}(f_{k})}{\sqrt{1+|df_{k}|^{2}_{g}}},d\varphi\bigg{\rangle}d\mu_{g},$
(5.4)

for $\varphi\in C^{\infty}_{c}(M^{n})$, and it follows from the
$C^{1}_{loc}$-convergence of $\\{f_{k}\\}$ that $\\{f_{k}\\}$ convergens to
$f$ distributionally. Hence, $f$ satisfies the (non-regularized) Jang equation
weakly. Standard elliptic regularity theory gives regularity up to order
$C^{3,\alpha}_{loc}$.
We now use the Schauder estimates to get $C^{3,\alpha}_{loc}$-convergence. We
have that $f_{k}\rightarrow f$ in $C^{1,\alpha}_{loc}$ and for $u\in
C^{2,\alpha}(M^{n})$ we define

$\begin{split}L(u)&=\hat{g}^{ij}u_{,ij}-\hat{g}^{ij}\Gamma^{\ell}_{ij}u_{,\ell},\qquad\text{where}\\\
\hat{g}^{ij}&=\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}\qquad\text{and}\\\
L_{k}(u)&=\hat{g}^{ij}_{k}u_{,ij}-\hat{g}^{ij}_{k}\Gamma^{\ell}_{ij}u_{,\ell}-\tau_{k}u,\qquad\text{where}\\\
\hat{g}^{ij}_{k}&=\bigg{(}g^{ij}-\frac{f^{,i}_{k}f^{,j}_{k}}{1+|df_{k}|^{2}_{g}}\bigg{)},\end{split}$
(5.5)

where $f$ solves the Jang equation $\mathcal{J}(f)=0$ and $f_{k}$ solves the
regularized Jang equation $\mathcal{J}(f_{k})=\tau_{k}f_{k}$ on $\bar{B}_{k}$.
The $C^{1,\alpha}_{loc}$-convergence of $f_{k}$ implies uniform bounds of the
coefficients of the differential operator $L_{k}$. We note that $f-f_{k}$
satisfies $L(f-f_{k})=F_{k}$, where

$F_{k}=\operatorname{trace}_{\hat{g}_{k}}(k)-\operatorname{trace}_{\hat{g}}(k)+L(f_{k})-L_{k}(f_{k}).$
(5.6)

If we show that $||F_{k}||_{C^{0,\alpha}_{loc}}\rightarrow 0$, as
$k\rightarrow\infty$, then $f_{k}\rightarrow f$ in $C^{2,\alpha}_{loc}$ will
follow by the interior Schauder estimate. The $C^{1,\alpha}_{loc}$-convergence
established above yields $C^{0,\alpha}_{loc}$-converence to zero of the trace
terms. Furthermore, from the equations that $f_{k}$ satisfy we have a uniform
$C^{2,\alpha}_{loc}$-bound on $f_{k}$ and from this it follows that
$L(f_{k})-L_{k}(f_{k})\rightarrow 0$ in $C^{0,\alpha}_{loc}$. Iterating this
argument we obtain the convergence in $C^{3,\alpha}_{loc}$.
It remains to show that the limit contains at least one graphical component.
We write $\gamma_{k}=\sqrt{1+|df_{k}|_{g}^{2}}$ and recall that the Jacobi
equation [SY81b, Equation (2.18)] holds for $\gamma_{k}$ on
$(\hat{M}^{n}_{k},\hat{g}_{k})$:

$\Delta^{\hat{g}_{k}}\big{(}\gamma^{-1}_{k}\big{)}+\big{(}|\hat{A}_{k}|^{2}_{\hat{g}}+\text{Ric}^{M^{n}\times\mathbb{R}}(\vec{n}_{k},\vec{n}_{k})+\vec{n}_{k}(H_{\hat{M}^{n}_{k}})\big{)}\gamma_{k}^{-1}=0,$
(5.7)

where $\hat{A}_{k}$ is the second fundamental form of
$(\hat{M}^{n},\hat{g}_{k})$, we think of $H_{\hat{M}^{n}_{k}}$ as a function
trivially extended from the graph $(\hat{M}^{n}_{k},\hat{g}_{k})$ to all of
$M^{n}\times\mathbb{R}$ and $\vec{n}_{k}$ is the downward pointing unit normal
of $\hat{M}^{n}_{k}$ in $M^{n}\times\mathbb{R}$. We have

$\vec{n}_{k}(H_{\hat{M}^{n}_{k}})=\vec{n}_{k}(\operatorname{trace}^{\hat{g}_{k}}k)+\frac{\tau_{k}|df_{k}|_{g}^{2}}{\sqrt{1+|df_{k}|_{g}^{2}}},$
(5.8)

and we estimate the first term on the right hand side following the proof of
[EM16, Lemma A.1]. Firstly, we have

$\begin{split}\vec{n}_{k}(\operatorname{trace}^{\hat{g}_{k}}k)&=\vec{n}_{k}^{\ell}(\langle\hat{g}^{k},k\rangle_{g})_{,\ell}\\\
&=\vec{n}_{k}^{\ell}\langle\nabla_{\ell}\hat{g}^{k},k\rangle_{g}+\vec{n}_{k}^{\ell}\langle\hat{g}^{k},\nabla_{\ell}k\rangle_{g}\\\
&=\langle\nabla\hat{g}^{k},\vec{n}^{k}\otimes
k\rangle_{g}+\langle\theta^{k}\otimes\hat{g}^{k},\nabla
k\rangle_{g}\end{split}$ (5.9)

where the first line holds since $\operatorname{trace}^{\hat{g}_{k}}k$ has no
$t$-dependence, $\theta^{k}=df_{k}/\sqrt{1+|df_{k}|^{2}_{g}}$ is the $1$-form
$g$-dual to the part of $\vec{n}_{k}$ tangential to $M^{n}$ and $\hat{g}^{k}$
is the $(0,2)$-tensor obtained by lowering the indices of $\hat{g}^{-1}$ with
$g$, so that $\hat{g}^{k}=g-\frac{df_{k}\otimes df_{k}}{1+|df_{k}|_{g}^{2}}$.
We estimate the second term in (5.9) as follows:

$\begin{split}\langle\theta^{k}\otimes\hat{g}^{k},\nabla
k\rangle_{g}&\leq|\theta^{k}\otimes\hat{g}^{k}|_{g}|\nabla k|_{g}\\\
&=|\theta^{k}|_{g}|\hat{g}^{k}|_{g}|\nabla k|_{g}\\\ &\leq\sqrt{n}|\nabla
k|_{g},\end{split}$ (5.10)

where the tensor $\theta^{k}\otimes\hat{g}^{k}$ has components
$(\theta^{k}\otimes\hat{g}^{k})_{\ell ij}=\theta^{k}_{\ell}\hat{g}^{k}_{ij}$.
To estimate first term in (5.9), we first note (recalling that
$\hat{g}^{ij}_{k}=g^{ij}-\vec{n}^{i}_{k}\vec{n}^{j}_{k}$) that

$(\nabla_{\ell}\hat{g})_{ij}=-(\nabla_{\ell}\theta^{k})_{i}\theta^{k}_{j}-(\nabla_{\ell}\theta^{k})_{j}\theta^{k}_{i},$
(5.11)

It follows that

$\begin{split}\langle\nabla\hat{g}^{k},\theta^{k}\otimes
k\rangle_{g}&=-2\langle\nabla\theta^{k}\otimes\theta^{k},\theta^{k}\otimes
k\rangle_{g}\\\
&=-2\vec{n}_{k}^{i}\vec{n}_{k}^{d}\hat{g}_{k}^{mb}\frac{\text{Hess}^{g}_{mi}f_{k}}{\sqrt{1+|df_{k}|_{g}^{2}}}k_{bd}\\\
&=2k_{bd}\vec{n}_{k}^{d}\hat{g}_{k}^{mb}(\gamma_{k}^{-1})_{,m}\\\ &=\langle
X,d\big{(}\gamma_{k}^{-1}\big{)}\rangle_{\hat{g}_{k}},\end{split}$ (5.12)

where we used (4.6),(4.9) and defined $X_{\ell}=-2k_{\ell i}\vec{n}_{k}^{i}$.
We note that

$\begin{split}|X|_{\hat{g}}&=4\hat{g}_{k}^{ab}\vec{n}^{i}_{k}\vec{n}^{j}_{k}k_{ai}k_{bj}\\\
&=4|X|_{g}^{2}-4k(\vec{n}_{k},\vec{n}_{k})k(\vec{n}_{k},\vec{n}_{k})\\\ &\leq
4|X|_{g}^{2}\\\ &=16g^{ab}k_{ai}\vec{n}_{k}^{i}k_{bj}\vec{n}_{k}^{j}\\\
&=16\langle k\otimes k,g\otimes\theta_{k}\otimes\theta_{k}\rangle_{g}\\\ &\leq
16|k\otimes k|_{g}|g\otimes\theta_{k}\otimes\theta_{k}|_{g}\\\
&=16|k|_{g}^{2}\sqrt{n}|\theta_{k}|_{g}^{2}\\\ &\leq C|k|_{g}^{2},\end{split}$
(5.13)

so that $|X|_{\hat{g}}$ is bounded. Straightforward calculations show that
$\text{Ric}^{M^{n}\times\mathbb{R}}_{tt}=0$,
$\text{Ric}^{M^{n}\times\mathbb{R}}_{ti}=0$ and
$\text{Ric}^{M^{n}\times\mathbb{R}}_{ij}=\text{Ric}_{ij}^{M^{n}}$ and so from
Lemma A.1 we obtain
$|\text{Ric}^{M^{n}\times\mathbb{R}}|_{ds^{2}}=|\text{Ric}^{M^{n}}|_{g}\leq
C(M^{n},g,k)$ (where we write $ds^{2}=g+dt^{2}$). Hence, the Cauchy-Schwarz
inequality gives the estimate
$\text{Ric}^{M^{n}\times\mathbb{R}}(\vec{n}_{k},\vec{n}_{k})\leq
C(M^{n},g,k)$. Consequently, there exists some $\beta\geq 0$ such that

$\Delta^{\hat{g}_{k}}\big{(}\gamma_{k}^{-1}\big{)}\leq\beta\gamma_{k}^{-1}+\langle
X,d\big{(}\gamma_{k}^{-1}\big{)}\rangle_{\hat{g}_{k}}.$ (5.14)

From Proposition 5.1 it now follows that the limit $\hat{M}^{n}$ consists of
graphical components and cylindrical components. In particular there exists an
outermost graphical component over some open subset $U_{f}\subset M^{n}$. This
shows (2.).
To prove assertion $(3)$, we note that depending on whether $f$ blows up or
down near some $C_{k}$, the upward pointing unit normal blows outward or
inward, respectively. Hence blow-up hypersurfaces $C_{k}^{\pm}$ satisfy
$H_{C_{k}^{\pm}}\mp\operatorname{trace}_{C_{k}^{\pm}}(k)=0$, where the mean
curvature is computed as the tangential divergence of the unit normal pointing
out of $M^{n}$. To show that $\hat{M}^{n}$ is graphical over
$C_{\ell}^{\pm}\times[T,\infty)$ we note that from Allard’s theorem, we obtain
a uniform radius $R>0$ such that $\hat{M}^{n}$ is graphical over the geodesic
ball in the tangenplane $T_{p}\hat{M}^{n}$ at each point $p$. This rules out
the possibility of the downward pointing unit normal $\vec{n}$ being parallel
with the cylinder sufficiently close to $\partial U_{f}$, so that
$\hat{M}^{n}$ is graphical over $C_{\ell}\times\mathbb{R}$ there.
Assertion $(4)$ follows from Proposition 3.14.

∎

###  5.2. Topology of the Jang graph

The discussion in this subsection follows that of [Eic13] rather closely. The
Jang graph $(\hat{M}^{n},\hat{g})$ has the asymptotics calculated in
Proposition 3.14 and we denote by $\hat{N}^{n}$ the end of $\hat{M}^{n}$ over
the asymptotically hyperbolic end $N^{n}$. Furthermore we have $\ell$
cylindrical ends $\hat{C}_{1},\ldots,\hat{C}_{\ell}$. We focus only on the
graphical component. The boundary of $U_{f}$ is the disjoint union of closed
hypersurfaces $\partial U_{f}=C_{1}\cup\ldots\cup C_{\ell}$, where each
$\hat{C}_{i}$ is asymptotic to $C_{i}\times\mathbb{R}$. We denote by
$\sigma_{i}=g|_{C_{i}}$ the induced metric on $C_{i}$.
We begin by discussing some properties of the closed manifolds $C_{i}$. In
[SY81b] it was shown that when the strict dominant energy condition holds in a
neighbourhood of each $C_{i}$ they are topologically spheres by the Gauss-
Bonnet Theorem. In Lemma 5.3 we show that the analogue of the result holds in
dimensions $n\geq 4$.

######  Lemma 5.3.

Suppose that the strict dominant energy condition $\mu>|J|_{g}$ holds locally
around a component $C_{i}\subset\partial U_{f}$. Then the spectrum of the
conformal Laplacian of each $(C_{i},\sigma_{i})$

$L=-\Delta^{C_{i}}+c_{n-1}R_{C_{i}},\qquad c_{n}=\frac{n-2}{4(n-1)},$ (5.15)

is positive. In particular, $C_{i}$ has positive Yamabe type.

###### Proof.

We recall the _Schoen-Yau identity_ :

$R_{\hat{g}}=2(\mu-J(\omega))+|\hat{A}-k|_{\hat{g}}^{2}+2|q|_{\hat{g}}^{2}-2\text{div}^{\hat{g}}(q)$
(5.16)

where $\hat{A}$ is the second fundamental form of the Jang graph
$(\hat{M}^{n},\hat{g})$,

$\omega=\frac{\nabla^{g}f}{\sqrt{1+|df|_{g}^{2}}},\qquad\text{and}\qquad
q_{i}=\frac{f^{,j}}{\sqrt{1+|df|_{g}^{2}}}(\hat{A}_{ij}-k_{ij}).$ (5.17)

The estimate $J(\omega)\leq|J|_{g}$ follows from the Cauchy-Schwarz
inequality, which together with the inequality of arithmetic and geometric
means also implies

$-\langle
d(\varphi^{2}),q\rangle_{\hat{g}}\leq|d\varphi|_{\hat{g}}^{2}+\varphi^{2}|q|_{\hat{g}}^{2}.$
(5.18)

For $\varphi\in C_{c}^{1}(\hat{M}^{n})$, we multiply (5.16) by $\varphi^{2}$
and integrate by parts over $\hat{M}^{n}$:

$\begin{split}\int_{\hat{M}^{n}}\bigg{(}2(\mu-J(\omega))+|\hat{A}-k|_{\hat{g}}^{2}\bigg{)}\varphi^{2}d\mu^{\hat{g}}&=\int_{\hat{M}^{n}}\bigg{(}R_{\hat{g}}-2|q|_{\hat{g}}^{2}+2\text{div}^{\hat{g}}(q)\bigg{)}\varphi^{2}d\mu^{\hat{g}}\\\
&=\int_{\hat{M}^{n}}\bigg{(}R_{\hat{g}}\varphi^{2}-2|q|_{\hat{g}}^{2}\varphi^{2}-2\langle
d(\varphi^{2}),q\rangle_{\hat{g}}\bigg{)}d\mu^{\hat{g}}\\\
&\leq\int_{\hat{M}^{n}}\bigg{(}R_{\hat{g}}\varphi^{2}+2|d\varphi|_{\hat{g}}^{2}\bigg{)}d\mu^{\hat{g}},\end{split}$
(5.19)

where we used (5.18) in the last line and that $\hat{M}^{n}$ has no boundary
for the partial integration. Hence, the last integral is positive as a
consequence of the strict dominant energy condition $2(\mu-J(\omega))\geq
2(\mu-|J|_{g})\geq\delta>0$. Note also that for a compactly supported function
$\varphi\in C_{c}^{1}(C_{i}\times\mathbb{R})$ we get from the converge in
$C^{3,\alpha}_{loc}$ of $\text{graph}(f-A)$ as $A\rightarrow\pm\infty$ to
$C_{\ell}^{\pm}\times\mathbb{R}$ in Proposition 5.2 that

$\begin{split}\delta\int_{C_{i}\times\mathbb{R}}\varphi^{2}d\mu^{\sigma_{i}+dt^{2}}&\leq\int_{C_{i}\times\mathbb{R}}\bigg{(}R_{C_{i}}\varphi^{2}+2|d\varphi|_{\sigma_{i}}^{2}\bigg{)}d\mu^{\sigma_{i}+dt^{2}}\\\
&\leq\int_{C_{i}\times\mathbb{R}}\bigg{(}R_{C_{i}}\varphi^{2}+c_{n-1}^{-1}|d\varphi|_{\sigma_{i}+dt^{2}}^{2}\bigg{)}d\mu^{\sigma_{i}+dt^{2}},\end{split}$
(5.20)

where we used $2c_{n-1}\leq 1$ in the last line.
We now let $\varphi(p,t)=\xi(p)\chi(t)$, for $p\in C_{i}$ and with $\chi$ a
smooth cutoff function such that $\chi(t)=1$ for $|t|\leq T$, $\chi(t)=0$ for
$|t|\geq T+2$ and $|\chi_{,t}|\leq 2$. We have

$\begin{split}0&<\delta\int_{C_{i}\times\mathbb{R}}c_{n-1}\varphi^{2}d\mu^{\sigma_{i}+dt^{2}}\\\
&\leq\int_{C_{i}\times\mathbb{R}}c_{n-1}R_{C_{i}}\varphi^{2}d\mu^{\sigma_{i}+dt^{2}}+\int_{C_{i}\times\mathbb{R}}|d\varphi|^{2}_{\sigma_{i}+dt^{2}}d\mu^{\sigma_{i}+dt^{2}}\\\
&=\bigg{(}\int_{C_{i}}c_{n-1}R_{C_{i}}\xi^{2}d\mu^{\sigma_{i}}\bigg{)}\bigg{(}\int_{\mathbb{R}}\chi^{2}dt\bigg{)}+\int_{C_{i}\times\mathbb{R}}\bigg{(}\chi^{2}|d\xi|_{\sigma_{i}}^{2}+\xi^{2}\chi_{,t}^{2}\bigg{)}d\mu^{\sigma_{i}+dt^{2}}\\\
&\leq\bigg{(}\int_{C_{i}}R_{C_{i}}\xi^{2}d\mu^{\sigma_{i}}\bigg{)}\bigg{(}\int_{\mathbb{R}}\chi^{2}dt\bigg{)}+\bigg{(}\int_{C_{i}}|d\xi|_{\sigma_{i}}^{2}d\mu^{\sigma_{i}}\bigg{)}\bigg{(}\int_{\mathbb{R}}\chi^{2}dt\bigg{)}\\\
&\qquad+\bigg{(}\int_{C_{i}}\xi^{2}d\mu^{\sigma_{i}}\bigg{)}\bigg{(}\int_{\mathbb{R}}\chi_{,t}^{2}dt\bigg{)}.\end{split}$
(5.21)

We divide both sides by the integral $\int^{\infty}_{-\infty}\chi^{2}dt$ and
after letting $T\rightarrow\infty$ we get

$0<\int_{C_{i}}c_{n-1}R_{C_{i}}\xi^{2}d\mu^{\sigma_{i}}+\bigg{(}\int_{C_{i}}|d\xi|_{\sigma_{i}}^{2}d\mu^{\sigma_{i}}\bigg{)}.$
(5.22)

By choosing the function $\xi$ to vanish on all but components of $\partial
U_{f}$ except $C_{i}$, we get that the integral of the scalar curvature
$R_{C_{i}}$ is positive. It is standard that the first eigenfunction of the
conformal Laplacian $-\Delta^{C_{i}}+c_{n-1}R_{C_{i}}$ is positive and so it
follows that the conformal Laplacian has positive spectrum on each $C_{i}$. In
particular, its first eigenfunction $\varphi_{i}>0$ can be used to conformally
change the metric on $C_{i}$ to $\varphi_{i}^{\frac{4}{n-2}}\sigma_{i}$, which
will have positive scalar curvature. It is standard theory that the existence
of a positive scalar curvature metric is equivalent to the Yamabe type being
positive (we refer the reader to [Sch89] for a comprehensive discussion on
this).

∎

Next we deform the cylindrical ends $\hat{C}_{1},\ldots,\hat{C}_{\ell}$ to
_exact cylinders_. Clearly, there exists a $T\geq 1$ such that
$\hat{M}^{n}\cap\\{|t|\geq T\\}$ can be perturbed to agree _exactly_ with the
disjoint union $C_{1}\times\mathbb{R}\cup\ldots\cup C_{\ell}\times\mathbb{R}$.
We denote the induced, complete metric by $\tilde{g}$. In the case when
$U_{f}=M^{n}$, we let $\tilde{g}=\hat{g}$. In both cases $\tilde{g}=\hat{g}$
on $\hat{N}^{n}$. We refer to the exact cylindrical ends as
$\tilde{C}_{1},\ldots,\tilde{C}_{\ell}$ and the deformed graph as
$\tilde{M}^{n}$. There is a compact set $\tilde{K}\subset\tilde{M}^{n}$ such
that
$\tilde{M}^{n}=\tilde{K}\cup\hat{N}^{n}\cup\tilde{C}_{1}\cup\ldots\cup\tilde{C}_{\ell}$.
We note that $(\tilde{M}^{n},\tilde{g})$ will not satisfy the mean curvature
equations but will satisfy the necessary inequalities derived in Lemma 5.4
below. Since $\hat{C}_{i}$ and $C_{i}\times(T,\infty)$ are diffeomorphic we
may equivalently pull up the metric $\tilde{g}$ to $\hat{C}_{i}$ and consider
the manifold $(\hat{M}^{n},\tilde{g})$ (with slight abuse of notation). There
is a compact set $\hat{K}$ so that
$\hat{M}^{n}=\hat{K}\cup\hat{N}^{n}\cup\hat{C}_{1}\cup\ldots\hat{C}_{\ell}$.
Clearly, $\tilde{g}$ and $\hat{g}$ are uniformly equivalent777We recall that
two metrics $g_{1}$ and $g_{2}$ are uniformly equivalent if there exists a
constant $C\geq 1$ such that $C^{-1}g_{1}\leq g_{2}\leq Cg_{1}$ as quadratic
forms.. We will alternate between the conventions after convenience.
For each exact cylindrical end $(\tilde{C}_{i},\tilde{g})$, we define a
function $\Psi_{i}=\exp(-\sqrt{\lambda_{i}}t_{i})\varphi_{i}$, where
$\lambda_{i}>0$ is the principal eigenvalue and $\varphi_{i}$ is the
respective principal (positive) eigenfunction, of the operator
$-\Delta^{C_{i}}+c_{n}R_{C_{i}}$ in Lemma 5.3. We arrange so that the
coordinate $t_{i}$ carries sign so that $\Psi_{i}\rightarrow 0$ in the
cylindrical infinities and the cylinders are exact for $t_{i}\in(T,\infty)$.
We define a positive function $\Psi>0$ via

$\Psi(p)=\begin{cases}1,\qquad&\text{if}\qquad p\in\hat{K}\cup\hat{N}^{n},\\\
\Psi_{i},&\text{if}\qquad p\in\tilde{C}_{i},\>t_{i}\geq 1.\end{cases}$ (5.23)

In Lemma 5.4 we establish some of the properties of
$\tilde{g}_{\Psi}=\Psi^{\frac{4}{n-2}}\tilde{g}$.

######  Lemma 5.4.

The scalar curvature of $\tilde{g}_{\Psi}$ vanishes on the exact cylinders
$\tilde{C}_{i}$, that is to say $R_{\tilde{g}_{\Psi}}|_{\tilde{C}_{i}}=0$.
Moreover, $(\tilde{C}_{i},\tilde{g}_{\Psi})$ is isometric to

$\bigg{(}C_{i}\times\bigg{(}0,\frac{n-2}{2\sqrt{\lambda_{i}}}\bigg{)},\varphi_{i}^{\frac{4}{n-2}}\bigg{(}\frac{4\lambda_{i}s_{i}^{2}}{(n-2)^{2}}\sigma_{i}+ds_{i}^{2}\bigg{)}\bigg{)}$
(5.24)

and is uniformly equivalent to the metric cone

$\bigg{(}C_{i}\times\bigg{(}0,\frac{n-2}{2\sqrt{\lambda_{i}}}\bigg{)},s_{i}^{2}\sigma_{i}+ds_{i}^{2}\bigg{)}.$
(5.25)

Furthermore, let $\varphi\in C^{1}(\hat{M}^{n})$ be such that
$\text{supp}(\varphi)\cap(\tilde{C}_{1}\cup\ldots\cup\tilde{C}_{\ell})$ is
compact. Then

$\frac{1}{2}\int_{\hat{M}^{n}}|d\varphi|_{\tilde{g}}^{2}d\mu^{\tilde{g}}+c_{n}\int_{\hat{N}^{n}}|\hat{A}-k|_{\hat{g}}^{2}d\mu^{\hat{g}}\leq\int_{\hat{M}^{n}}\bigg{(}|d\varphi|_{\tilde{g}}^{2}+c_{n}R_{\tilde{g}}\varphi^{2}\bigg{)}d\mu^{\tilde{g}}.$
(5.26)

Finally,

$\int_{\hat{M}^{n}}\frac{1}{2}\Psi^{-2}|d(\Psi\varphi)|^{2}_{\tilde{g}_{\Psi}}d\mu^{\tilde{g}_{\Psi}}+c_{n}\int_{\hat{N}^{n}}|\hat{A}-k|^{2}_{\hat{g}}\varphi^{2}d\mu^{\hat{g}}\leq\int_{\hat{M}^{n}}\bigg{(}|d\varphi|^{2}_{\tilde{g}\Psi}+c_{n}R_{\tilde{g}_{\Psi}}\varphi^{2}\bigg{)}d\mu^{\tilde{g}_{\Psi}},$
(5.27)

###### Proof.

We recall the well-known formula for the conformal transformation of scalar
curvature:

$c_{n}R_{\tilde{g}_{\Psi}}=\Psi^{-\frac{n+2}{n-2}}\big{(}-\Delta^{\tilde{g}}\Psi+c_{n}\Psi
R_{\tilde{g}}\big{)}.$ (5.28)

Using the equation that $\lambda_{i}$ and $\varphi_{i}$ satisfy it is
straightforward to check that $\Delta^{\tilde{g}}\Psi=\Psi c_{n}R_{\tilde{g}}$
on $C_{i}\times\mathbb{R}$. Hence the scalar curvature
$R_{\tilde{g}_{\Psi}}=0$ in
$(C_{i}\times\mathbb{R},\Psi^{\frac{4}{n-2}}(\sigma_{i}+dt^{2}))$. It is
furthermore easily seen that $R_{C_{i}\times\mathbb{R}}=R_{C_{i}}$ with the
product metric and so the scalar curvature $R_{\tilde{g}_{\Psi}}=0$ in
$(C_{i},\Psi^{\frac{4}{n-2}}\sigma_{i})$.
For the assertion about the isometry, we consider the map
$\Phi:C_{i}\times\mathbb{R}\rightarrow C_{i}\times\mathbb{R}$ explicitly
defined by

$\Phi(\theta,t)=\bigg{(}\theta,s_{i}=\frac{n-2}{2\sqrt{\lambda_{i}}}\exp\bigg{(}-\frac{2\sqrt{\lambda_{i}}t}{n-2}\bigg{)}\bigg{)}.$
(5.29)

It is straightforward to verify that $\Phi$ is an isometry from
$(\tilde{C}_{i},\Psi^{\frac{4}{n-2}}\tilde{g})$ to the manifold in (5.24),
which also implies the uniform equivalence to the metric cone.
We note that far enough into the cylindrical ends we have from the Schoen-Yau
identity (5.16) and the strict dominant energy condition that

$R_{\tilde{g}}-|\hat{A}-k|_{\tilde{g}}^{2}-2|q|_{\tilde{g}}^{2}+2\text{div}^{\tilde{g}}(q)\geq(\mu-|J|_{g}).$
(5.30)

The same treatment of this equation as in (5.19) in the proof of Lemma 5.3
yields

$\int_{\hat{M}^{n}}c_{n}\bigg{(}(\mu-|J|^{2}_{\tilde{g}})+|\hat{A}-k|^{2}_{\tilde{g}}\bigg{)}\varphi^{2}d\mu^{\tilde{g}}\leq\int_{\hat{M}^{n}}c_{n}\bigg{(}R_{\tilde{g}}\varphi^{2}+2|d\varphi|^{2}_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}}.$
(5.31)

By the dominant energy condition and the inclusion
$\hat{N}^{n}\subset\hat{M}^{n}$ it follows that

$\int_{\hat{N}^{n}}c_{n}|\hat{A}-k|^{2}_{\tilde{g}}\varphi^{2}d\mu^{\tilde{g}}\leq\int_{\hat{M}^{n}}c_{n}\bigg{(}R_{\tilde{g}}\varphi^{2}+2|d\varphi|^{2}_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}}.$
(5.32)

Adding
$\int_{\hat{M}^{n}}\frac{1}{2}|d\varphi|^{2}_{\tilde{g}}d\mu^{\tilde{g}}$ to
both sides we get

$\begin{split}\int_{\hat{M}^{n}}\frac{1}{2}|d\varphi|^{2}_{\tilde{g}}d\mu^{\tilde{g}}+\int_{\hat{N}^{n}}c_{n}|\hat{A}-k|^{2}_{\tilde{g}}\varphi^{2}d\mu_{\tilde{g}}&\leq\int_{\hat{M}^{n}}\bigg{(}c_{n}R_{\tilde{g}}\varphi^{2}+\bigg{(}\frac{1}{2}+2c_{n}\bigg{)}|d\varphi|^{2}_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}}\\\
&\leq\int_{\hat{M}^{n}}\bigg{(}c_{n}R_{\tilde{g}}\varphi^{2}+|d\varphi|^{2}_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}}\end{split}$
(5.33)

since $\frac{1}{2}+2c_{n}\leq 1$.
To show (5.27), we replace $\varphi$ by $\Psi\varphi$ in (5.33). Clearly, we
have $d\mu^{\tilde{g}_{\Psi}}=\Psi^{\frac{2n}{n-2}}d\mu^{\tilde{g}}$. Hence,
it follows straightforwardly that the first term in the left hand side of
(5.33) is

$\int_{\hat{M}^{n}}\frac{1}{2}|d(\Psi\varphi)|^{2}_{\tilde{g}}d\mu^{\tilde{g}}=\int_{\hat{M}^{n}}\frac{1}{2}\Psi^{-2}|d(\Psi\varphi)|^{2}_{\tilde{g}\Psi}d\mu^{\tilde{g}_{\Psi}}.$
(5.34)

The second term remains the same as $\Psi=1$ on $\hat{N}^{n}$. Similarly, we
obtain

$\begin{split}\int_{\hat{M}^{n}}|d(\Psi\varphi)|_{\tilde{g}}^{2}d\mu^{\tilde{g}}&=\int_{\hat{M}^{n}}\bigg{(}\Psi^{2}|d\varphi|^{2}_{\tilde{g}}+2\Psi\varphi\langle
d\varphi,d\Psi\rangle_{\tilde{g}}+\varphi^{2}|d\Psi|^{2}_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}}\\\
&=\int_{\hat{M}^{n}}|d\varphi|^{2}_{\tilde{g}_{\Psi}}d\mu_{\tilde{g}_{\Psi}}+\int_{\hat{M}}\bigg{(}2\Psi\varphi\langle
d\varphi,d\Psi\rangle^{\tilde{g}}+\varphi^{2}|d\Psi|^{2}_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}},\\\
\end{split}$ (5.35)

and recalling the equation $\Psi$ satisfies we get

$\begin{split}\int_{\hat{M}^{n}}c_{n}R_{\tilde{g}}(\Psi\varphi)^{2}d\mu^{\tilde{g}}&=\int_{\hat{M}^{n}}\bigg{(}\Delta^{\tilde{g}}\Psi+c_{n}R_{\tilde{g}_{\Psi}}\Psi^{\frac{n+2}{n-2}}\bigg{)}\Psi\varphi^{2}d\mu^{\tilde{g}}\\\
&=\int_{\hat{M}^{n}}\Psi\varphi^{2}\Delta^{\tilde{g}}\Psi
d\mu^{\tilde{g}}+\int_{\hat{M}^{n}}c_{n}R_{\tilde{g}_{\Psi}}\varphi^{2}d\mu^{\tilde{g}_{\Psi}}\\\
\end{split}$ (5.36)

Finally, since $\Psi=1$ on $\tilde{K}^{n}\cup\hat{N}^{n}$ and
$f\Delta^{\tilde{g}}h=\text{div}_{\tilde{g}}(hdf)+\langle
dh,df\rangle_{\tilde{g}}$, we have

$\begin{split}\int_{\hat{M}^{n}}\Psi\varphi^{2}\Delta^{\tilde{g}}\Psi
d\mu^{\tilde{g}}&=\int_{\hat{M}^{n}}\bigg{(}\text{div}_{\tilde{g}}(\Psi\varphi^{2}d\Psi)-\langle
d(\Psi\varphi^{2}),d\Psi\rangle_{\tilde{g}}\bigg{)}d\mu^{\tilde{g}}\\\
&=\int_{\tilde{C}_{1}\cup\ldots\cup\tilde{C}_{\ell}}\text{div}^{\tilde{g}}(\Psi\varphi^{2}d\Psi)d\mu_{\tilde{g}}-\int_{\hat{M}^{n}}\bigg{(}\varphi^{2}|d\Psi|^{2}_{\tilde{g}}+2\Psi\varphi\langle
d\varphi,d\Psi\rangle_{\tilde{g}}\bigg{)}d\mu_{\tilde{g}}\\\ \end{split}$
(5.37)

The first term vanishes since $\varphi$ is compactly supported in
$\tilde{C}_{1}\cup\ldots\cup\tilde{C}_{\ell}$. This shows (5.27).

∎

At this point it is convenient to introduce a new ”distance” function $s\in
C^{3,\alpha}_{loc}(\tilde{M}^{n})$ on $\tilde{M}^{n}$. We let

$s(p)=\begin{cases}r(p),\qquad&\text{when}\qquad p\in\hat{N}^{n},\\\
\frac{n-2}{2\sqrt{\lambda_{i}}}\exp(-\frac{2\sqrt{\lambda_{i}}t_{i}}{n-2}),&\text{when}\qquad
p\in C_{i}.\end{cases}$ (5.38)

When we have no blow-ups or blow-downs, we simply let $s(p)=r(p)$ globally. In
the case of blow-ups or blow-downs, this distance function will measure the
distances to the infinities at the exact cylinders.
By adding a point at the infinities we obtain a complete metric
$(\tilde{M}^{n},\tilde{g}_{\Psi})$, where $s$ tends to zero at the points.

######  Remark 5.5.

At this stage it is convenient to define the following function:

$\chi_{\epsilon}(p)=\begin{cases}0\qquad&0\leq s(p)\leq\epsilon,\\\
-1+s(p)/\epsilon&\epsilon\leq s(p)\leq 2\epsilon,\\\ 1&2\epsilon\leq
s(p)\end{cases}$ (5.39)

for $\epsilon>0$. It is not difficult to see that
$|d\chi_{\epsilon}|_{\tilde{g}_{\Psi}}=\mathcal{O}(\epsilon^{-1})$ so that

$\int_{\hat{M}^{n}}|d\chi_{\epsilon}|^{2}_{\tilde{g}_{\Psi}}d\mu^{\tilde{g}_{\Psi}}=\mathcal{O}(\epsilon^{n-2}).$
(5.40)

It follows that the conical singularities have vanishing
$\tilde{g}_{\Psi}$-harmonic capacity.

##  6\. Asymptotic flatness of the Jang graph

In this section we show that the induced metric on the Jang graph is
asymptotically flat, that is
$|\hat{g}-\delta|_{\delta}=\mathcal{O}_{2}(r^{-(n-2)})$ (see Definition 2.6).
To achieve this, we will show that

$f=\sqrt{1+r^{2}}+\frac{\alpha}{r^{n-3}}+\mathcal{O}^{3}(r^{-(n-2-\epsilon)}).$
(6.1)

As explained in [Sak21, Section 6] we cannot directly apply the rescaling
technique as in [SY81b] for this purpose. Instead, we follow the argument in
[Sak21]. We will write $f$ as a height function $h$ over an asymptotically
Euclidean manifold (roughly the graph of the lower barrier constructed in
Proposition 3.14) sufficiently near the hyperbolic infinity of $(M^{n},g)$.
The rescaling technique can then be applied and in turn this will be
translated into the desired fall-off properties of $f$.
As opposed to [Sak21], we will firstly estimate the norm of the second
fundamental form $\hat{A}$ of the graph of $f$ near the hyperbolic infinity
$N^{n}$ of $M^{n}$.

###  6.1. Estimates for the second fundamental form near infinity

In this section we want to establish an asymptotic estimate on the second
fundamental form $\hat{A}$ of the Jang graph obtained in Proposition 5.2. We
state for convenience the expected asymptotic behaviour of $\hat{A}$ from
Lemma B.4:

$|\hat{A}|^{2}_{\hat{g}}=\frac{1}{(1+r^{2})^{2}}+(n-1)+\mathcal{O}(r^{-(n+1-\epsilon)}).$
(6.2)

Having shown that the graph $\hat{M}^{n}$ is asymptotically Euclidean as in
Definition 2.6 the asymptotic decay of $\hat{A}$ in (6.2) will be immediate.
However at this stage we have only the asymptotic control of $\hat{M}^{n}$
obtained with the barriers.
We will first derive an interior gradient estimate following [Eic09] in the
asymptotically flat setting. Then we prove an estimate for $\hat{A}$ in
Proposition 6.2.

######  Lemma 6.1.

Let $(\hat{M}^{n},\hat{g})$ be the Jang graph obtained in Proposition 5.2. Fix
$p_{0}\in U_{f}$ and $\rho\in(0,\text{inj}_{p_{0}}(M^{n},g))$ such that
$B_{\rho}(p_{0})\subset U_{f}$, where $B_{\rho}(p_{0})$ is the geodesic ball
in $(M^{n},g)$ centered at $p_{0}$ with radius $\rho$. Suppose there exists a
real number $T>0$ such that either $f(p)\leq f(p_{0})+T$ or $f(p_{0})-T\leq
f(p)$ for every $p\in B_{\rho}(p_{0})$. Then $|df|_{g}(p_{0})\leq C$, where
$C$ depends only on $T$, $\rho$ and the restrictions of $g$ and $k$ to
$B_{\rho}(p_{0})$.

###### Proof.

The following argument was used in the proofs of [Eic09, Lemma 2.1] and [EM16,
Lemma 2.1 in Appendix A] (see also [Spr07, Theorem 1.1]).
We will only consider the case when $f(p)\leq f(p_{0})+T$ as the case when
$f(p)\geq f(p_{0})-T$ is similar. For a point $p\in B_{\rho}(p_{0})$, let

$\varphi(p)=f(p)-f(p_{0})+\rho-\frac{T+\rho}{\rho^{2}}\text{dist}^{2}(p_{0},p),$
(6.3)

where $\text{dist}(p_{0},p)$ is the geodesic distance between $p_{0}$ and $p$
in $(M^{n},g)$. Further, let

$\Omega=\\{p\in B_{\rho}(p_{0})\>|\>\varphi(p)>0\\}.$ (6.4)

Then $\varphi=0$ on $\partial\Omega$, $p_{0}\in\Omega$, $\varphi\leq T+\rho$
on $B_{\rho}(p_{0})$ and $\varphi\leq 0$ on $\partial B_{\rho}(p_{0})$ so that
$\Omega\subset B_{\rho}(p_{0})$. We let $\gamma=\sqrt{1+|df|_{g}^{2}}$, extend
$\gamma$ trivially to $U_{f}\times\mathbb{R}$ to be constant in the
$t$-variable and view $\gamma$ as a function on the graph $\hat{M}^{n}$. We
note that

$\begin{split}0&=\Delta^{\hat{g}}1\\\
&=\gamma^{-1}\Delta^{\hat{g}}\gamma+2\langle
d\gamma,d(\gamma^{-1})\rangle_{\hat{g}}+\gamma\Delta^{\hat{g}}(\gamma^{-1})\\\
&=\gamma^{-1}\Delta^{\hat{g}}\gamma-\frac{2}{\gamma^{2}}|d\gamma|_{\hat{g}}+\gamma\Delta^{\hat{g}}(\gamma^{-1}).\end{split}$
(6.5)

Let $K\geq 1$ be a real number to be specified later and consider the cutoff-
function $\eta=e^{K\varphi}-1$. Clearly $\eta\geq 0$ in $\Omega$ and $\eta=0$
on $\partial\Omega$. In what follows, we view $\eta$ as a function on the
graph $\hat{M}^{n}$. At a point in $\Omega$ where $\eta\gamma$ attains a
maximum, we have both $0=d(\eta\gamma)=\eta d\gamma+\gamma d\eta$ and
$d\eta=\eta\gamma d(\gamma^{-1})$. Combining these equations with (6.5) we
obtain

$\begin{split}0&\geq\gamma^{-1}\Delta^{\hat{g}}(\eta\gamma)\\\
&=\frac{\eta}{\gamma}\Delta^{\hat{g}}\gamma+\frac{2}{\gamma}\langle
d\gamma,d\eta\rangle_{\hat{g}}+\Delta^{\hat{g}}\eta\\\
&=\frac{2\eta}{\gamma^{2}}|d\gamma|_{\hat{g}}-\gamma\eta\Delta^{\hat{g}}\gamma^{-1}+\frac{2}{\gamma}\langle
d\gamma,d\eta\rangle_{\hat{g}}+\Delta^{\hat{g}}\eta\\\
&=-\gamma\eta\Delta^{\hat{g}}(\gamma^{-1})+\Delta^{\hat{g}}\eta\\\
&\geq-\beta\eta-\gamma\eta\langle
X,d(\gamma^{-1})\rangle_{\hat{g}}+\Delta^{\hat{g}}\eta\\\ &=-\beta\eta-\langle
X,d\eta\rangle_{\hat{g}}+\Delta^{\hat{g}}\eta\end{split}$ (6.6)

where the vector field $X$ and the constant $\beta$ are defined before (5.13)
and in (5.14), respectively, in the proof of Proposition 5.2. We have

$\begin{split}\langle
X,d\eta\rangle_{\hat{g}}&\leq|X|_{\hat{g}}|d\eta|_{\hat{g}}\\\
&=|X|_{\hat{g}}e^{K\varphi}K|d\varphi|_{\hat{g}}\\\
&\leq\sqrt[4]{n}Ke^{K\varphi}|X|_{g}|d\varphi|_{\hat{g}}\\\ &\leq
DKe^{K\varphi}|d\varphi|_{\hat{g}},\end{split}$ (6.7)

where the second inequality follows in a similar fashion as (5.1) and the last
inequality is shown in (5.13). A computation shows that

$\Delta^{\hat{g}}\eta=K^{2}e^{K\varphi}|d\varphi|^{2}_{\hat{g}}+Ke^{K\varphi}\Delta^{\hat{g}}\varphi.$
(6.8)

and so (6.6) implies

$\begin{split}0&\geq-\beta\eta-\langle
X,d\eta\rangle_{\hat{g}}+\Delta^{\hat{g}}\eta\\\
&\geq\beta+\big{(}K\Delta^{\hat{g}}\varphi+K^{2}|d\varphi|^{2}_{\hat{g}}-KD|d\varphi|_{\hat{g}}-\beta\big{)}e^{K\varphi}.\\\
\end{split}$ (6.9)

We compute

$\begin{split}|d\varphi|^{2}_{\hat{g}}&=\bigg{(}g^{ij}-\frac{f^{,i}f^{,j}}{1+|df|^{2}_{g}}\bigg{)}\varphi_{,i}\varphi_{,j}\\\
&=|d\varphi|^{2}_{g}-\frac{\langle
df,d\varphi\rangle_{g}^{2}}{1+|df|^{2}_{g}}\\\
&=\bigg{|}df-\bigg{(}\frac{T+\rho}{\rho^{2}}\bigg{)}d\big{(}\text{dist}^{2}(p_{0},\cdot)\big{)}\bigg{|}_{g}^{2}\\\
&\qquad-\frac{\big{\langle}df,df-\big{(}\frac{T+\rho}{\rho^{2}}\big{)}d\big{(}\text{dist}^{2}(p_{0},\cdot)\big{)}\big{\rangle}_{g}^{2}}{1+|df|^{2}_{g}}\\\
&=\frac{|df|_{g}^{2}-1+\big{(}1-\big{(}\frac{T+\rho}{\rho^{2}}\big{)}\langle
df,d\big{(}\text{dist}^{2}(p_{0},\cdot)\big{)}\rangle_{g}\big{)}^{2}}{1+|df|_{g}^{2}}\\\
&\qquad+\bigg{(}\frac{T+\rho}{\rho^{2}}\bigg{)}^{2}\big{|}d\big{(}\text{dist}^{2}(p_{0},\cdot)\big{)}\big{|}_{g}^{2}\end{split}$
(6.10)

and note that
$1\leq\liminf_{|df|_{g}\rightarrow\infty}|d\varphi|^{2}_{\hat{g}}$. To compute
$\Delta^{\hat{g}}\varphi$, we note that for any function $u$ defined on
$M^{n}$ and viewed as a function on $\hat{M}^{n}$, we have

$\text{Hess}^{\hat{g}}_{ij}u=\text{Hess}^{g}_{ij}u-f^{,k}u_{,k}\frac{\text{Hess}^{g}_{ij}f}{1+|df|^{2}_{g}}$
(6.11)

and so, from (2.18), we obtain

$\Delta^{\hat{g}}u=\hat{g}^{ij}\text{Hess}^{g}_{ij}u-du(\nabla^{g}f)\frac{H_{\hat{M}^{n}}}{\sqrt{1+|df|^{2}_{g}}}.$
(6.12)

In particular, when $u=f$, this yields

$\Delta^{\hat{g}}f=\frac{H_{\hat{M}^{n}}}{\sqrt{1+|df|_{g}^{2}}},$ (6.13)

so that $f$ is $\hat{g}$-harmonic precisely when the graph
$(\hat{M}^{n},\hat{g})$ is minimal. Furthermore, we note that

$\begin{split}\Delta^{\hat{g}}\big{(}\text{dist}^{2}(p_{0},\cdot)\big{)}&=2\text{dist}(p_{0},\cdot)\Delta^{\hat{g}}\big{(}\text{dist}(p_{0},\cdot)\big{)}+2|d\big{(}\text{dist}(p_{0},\cdot)\big{)}|_{\hat{g}}^{2}\\\
&\leq
2\rho\Delta^{\hat{g}}\big{(}\text{dist}(p_{0},\cdot)\big{)}+2\sqrt{n}|d\big{(}\text{dist}(p_{0},\cdot)\big{)}|_{g}^{2}.\end{split}$
(6.14)

Inserting $\text{dist}(p_{0},\cdot)$ into (6.12) we obtain

$\begin{split}\Delta^{\hat{g}}\text{dist}(p_{0},\cdot)&=\hat{g}^{ij}\text{Hess}^{g}_{ij}\text{dist}(p_{0},\cdot)-d\big{(}\text{dist}(p_{0},\cdot)\big{)}(\nabla^{g}f)\frac{H_{\hat{M}^{n}}}{\sqrt{1+|df|^{2}_{g}}}\\\
&\leq\sqrt{n}|\text{Hess}^{g}\text{dist}(p_{0},\cdot)|_{g}+|d\big{(}\text{dist}(p_{0},\cdot)\big{)}|_{g}|df|_{g}\frac{H_{\hat{M}^{n}}}{\sqrt{1+|df|^{2}_{g}}}\\\
&\leq\sqrt{n}C+|H_{\hat{M}^{n}}|\\\ &\leq\sqrt{n}(C+|k|_{g}),\end{split}$
(6.15)

where we used the Hessian comparison theorem to estimate
$|\text{Hess}^{g}\big{(}\text{dist}(p_{0},\cdot)\big{)}|_{g}\leq C$ and
$|d\big{(}\text{dist}(p_{0},\cdot)\big{)}|_{g}^{2}=1$ in the second inequality
and (5.1) was used in the last inequality. From the estimates (6.13) - (6.15),
we conclude that $|\Delta^{\hat{g}}\varphi|\leq\tilde{C}$, where $\tilde{C}$
is a constant depending only on the initial data and $\rho$. Inserting this
estimate into (6.9), we obtain

$0\geq K|d\varphi|^{2}_{\hat{g}}-D|d\varphi|_{\hat{g}}-(\beta+\tilde{C}).$
(6.16)

Choosing $K$ large enough so that $|d\varphi|_{\hat{g}}<1$ and recalling
(6.10), we conclude that $|df|_{g}$ is bounded in terms of $T$, $\rho$ and the
restriction of $g$ and $k$ to $B_{\rho}(p_{0})$ at a point $q$ in $\Omega$
where $\eta\gamma$ attains a local maximum. Denote this constant by $A$, so
that $|df|_{g}(q)\leq A$. Then the estimate
$\eta\gamma(p_{0})\leq\eta\gamma(q)$ together with the bound $\varphi\leq
T+\rho$ implies
$(e^{K\rho}-1)\sqrt{1+|df|_{g}^{2}(p_{0})}\leq(e^{K(\rho+T)}-1)\sqrt{1+A^{2}}$,
which proves our claim.

∎

######  Proposition 6.2.

Let $(\hat{M}^{n},\hat{g})$ the Jang graph obtained in Proposition 5.2 and
$\hat{A}$ its second fundamental form. Then $|\hat{A}|_{\hat{g}}\leq
C(M^{n},g,k)$ sufficiently far out in $\hat{N}^{n}$.

###### Proof.

We get the estimate of $|\hat{A}|_{\hat{g}}$ as follows; the estimate in Lemma
6.1 in $B_{\rho}(p_{0})$ improves to a $C^{1,\alpha}$-bound as in Section 4,
except that since $B_{\rho}(p_{0})$ is open we restrict to the closed ball
$\overline{B}_{\rho/2}(p)$ which is a compact manifold with boundary (and the
coefficient estimates necessary follow from the $C^{1}$-estimate). The
arguments in Lemma 4.2 yields a $C^{2,\alpha}$-bound on $f$:
$||f||_{C^{2,\alpha}(\bar{B}_{\rho/2})}\leq C(M^{n},g,k)$. In particular, we
obtain an estimate on $|\text{Hess}^{g}(f)|_{g}$ and so the bound on
$|\hat{A}|_{g}$ follows by considering the coordinate expression for $\hat{A}$
in terms of $f$ as given explicitly in (2.14).
It remains to verify the bound on the intrinsic norm $|\hat{A}|_{\hat{g}}$. It
is easily seen that

$\begin{split}|\hat{A}|^{2}_{\hat{g}}&=\langle\hat{g}\otimes\hat{g},\hat{A}\otimes\hat{A}\rangle\\\
&\leq|\hat{g}\otimes\hat{g}|_{g}|\hat{A}\otimes\hat{A}|_{g}\\\
&=|\hat{g}|^{2}_{g}|\hat{A}|^{2}_{g},\end{split}$ (6.17)

so that the bound follows since $|\hat{g}|^{2}_{g}\leq\sqrt{n}$.
Finally, from the fall-off properties of the barriers and the properties of
the initial data the estimate is uniform sufficiently far out in $N^{n}$.

∎

###  6.2. Setup and Fermi coordinates

We now describe how to rewrite the Jang graph in terms of a height function
$h$ over an asymptotically Euclidean manifold. Firstly, from Section 3 we know
that

$f=\sqrt{1+r^{2}}+\frac{\alpha}{r^{n-3}}+\mathcal{O}(r^{-(n-2-\epsilon)}),$
(6.18)

and that there are functions $f_{\pm}$ such that $f_{-}\leq f\leq f_{+}$ with
the same asymptotics defined on $M^{n}_{r_{0}}=\\{r\geq r_{0}\\}$. It is clear
from the properties of $f_{\pm}$ that there exists functions with the same
asymptotics but with derivatives of better fall-off properties, defined on
$M^{n}_{r_{1}}$, for $r_{1}>r_{0}$. We let these functions be denoted by
$f_{\pm}$ (as we will not use the barriers of Proposition 3.14 again). The
functions are such that $f_{-}\leq f\leq f_{+}$ and

$f_{\pm}=\sqrt{1+r^{2}}+\frac{\alpha}{r^{n-3}}+\mathcal{O}^{3}(r^{-(n-2-\epsilon)})$
(6.19)

and we denote them by _upper-_ and _lower barriers_. We denote the graphs of
the barriers $f_{\pm}$ by $\hat{M}^{n}_{\pm}$.
Next, we invoke _Fermi coordinates_ (or _normal geodesic coordinates_) adapted
to $\hat{M}^{n}_{-}$. Let $\Psi=(x^{1},\ldots,x^{n})$ be an asymptotically
Euclidean Cartesian coordinate system on $\hat{M}^{n}_{-}$. From Proposition
6.2 we know that $\hat{M}^{n}$ has uniformly bounded geometry sufficiently far
out in $\hat{N}^{n}$. By [Eld13, Chapter 2] $\hat{N}^{n}$ has an open
neighbourhood in $M^{n}\times\mathbb{R}$, denoted by
$N_{\gamma}(\hat{M}^{n}_{-})$, and a diffeomorphism
$y:N_{\gamma}(\hat{M}^{n}_{-})\simeq\hat{M}^{n}_{-}\times(-\gamma,\gamma)$ for
some positive radius $\gamma>0$. We take the coordinates $y$ on
$N_{\gamma}(\hat{M}^{n})$ to be $y(\cdot,0)=\Psi(\cdot)$ and $\frac{\partial
y}{\partial\rho}=\vec{n}^{\rho}$, where $\vec{n}^{\rho}$ is the upward
pointing unit normal to the $\rho$-level set
$\hat{M}^{n}_{\rho}=y(\cdot,\rho)$. In these coordinates the metric $g+dt^{2}$
takes the form $d\rho^{2}+\hat{g}_{\rho}$, where $\hat{g}_{\rho}$ is the
induced metric on $\hat{M}^{n}_{\rho}$. The induced metric $\hat{g}_{0}$ on
$\hat{M}^{n}_{0}=\hat{M}^{n}_{-}$ will be denoted by $\hat{g}_{-}$. We will
denote by $\langle\cdot,\cdot\rangle$ the metric $ds^{2}=g+dt^{2}$ on
$M^{n}\times\mathbb{R}$. The second fundamental form of $\hat{M}^{n}_{\rho}$
is denoted by $\hat{A}_{\rho}$, where the convention is
$(\hat{A}_{\rho})_{ij}=\langle\nabla_{\partial_{i}}\partial_{j},\vec{n}^{\rho}\rangle$.
We will show that we can write the Jang solution constructed in Section 5 as
the graph of $h$:

$\hat{M}^{n}\cap\\{r\geq
r_{2}\\}=\\{(p,h(p))\in\hat{M}^{n}_{-}\times(-\gamma,\gamma)\>|\>p\in\hat{M}^{n}_{-}\\}\cap\\{r\geq
r_{2}\\},$ (6.20)

where $h:\hat{M}^{n}_{-}\rightarrow[0,\gamma)$ and the relevance of $r_{2}\geq
r_{1}$ will be explained in Corollary 6.5.
The following proposition, which is proven in Appendix D using the bound on
$\hat{A}$ obtained in Proposition 6.2, will be useful.

######  Proposition 6.3.

There exists constants $\rho_{0}>0$ and $C\geq 1$ such that
$|\hat{A}_{\rho}|_{\hat{g}_{\rho}}<C$ and $C^{-1}\delta\leq\hat{g}^{\rho}\leq
C\delta$ for any $0\leq\rho\leq\rho_{0}$. Furthermore, all partial derivatives
of $(\hat{g}_{\rho})_{ij}$ and $(\hat{A}_{\rho})^{i}_{j}$ up to order $3$ in
the Fermi coordinates are bounded.

###  6.3. Existence of the height function and a priori estimates

In this section we obtain the height function
$h:\hat{M}^{n}_{-}\rightarrow[0,\gamma)$ as described above. Moreover we
derive a priori estimates of $h$.
As a first step, we obtain the following ”tilt-excess” estimate or the normal.

######  Lemma 6.4.

Let $\vec{n}$ and $\vec{n}_{-}$ be the upward pointing unit normals of