# Comparison geometry for substatic manifolds and a weighted Isoperimetric
Inequality

Stefano Borghini  S. Borghini, Università degli Studi di Trento, via Sommarive
14, 38123 Povo (TN), Italy<EMAIL_ADDRESS>and  Mattia Fogagnolo
M. Fogagnolo, Università degli Studi di Padova , via Trieste 63, 35121 Padova
(PD), Italy<EMAIL_ADDRESS>
###### Abstract.

Substatic Riemannian manifolds with minimal boundary arise naturally in
General Relativity as spatial slices of static spacetimes satisfying the Null
Energy Condition. Moreover, they constitute a vast generalization of
nonnegative Ricci curvature. In this paper we will prove various geometric
results in this class, culminating in a sharp, weighted Isoperimetric
inequality that quantifies the area minimizing property of the boundary. Its
formulation and proof will build on a comparison theory partially stemming
from a newly discovered conformal connection with $\mathrm{CD}(0,1)$ metrics.

###### Key words and phrases:

Keywords: substatic manifolds, comparison geometry, isoperimetric inequality.

MSC (2020): 49Q10, 53C21, 53E10,

##  1\. Introduction

In this paper we are interested in the study of triples $(M,g,f)$, where
$(M,g)$ is a Riemannian manifold of dimension $n\geq 3$ with (possibly empty)
compact boundary $\partial M$ and $f:M\to\mathbb{R}$ is a smooth function that
is positive in the interior of $M$ and zero on $\partial M$, satisfying the
following inequality

$f{\rm Ric}-\nabla^{2}f+(\Delta f)g\geq 0,$ (1.1)

where ${\rm Ric}$ is the Ricci tensor of the metric $g$, $\nabla^{2}$ is the
Hessian and $\Delta={\rm tr}\nabla^{2}$ is the Laplace–Beltrami operator with
respect to the Levi-Civita connection $\nabla$ of $g$. We will refer to such
triples $(M,g,f)$ as substatic triples or simply substatic manifolds. We say
that a substatic manifold has horizon boundary if $\partial M$ is either empty
or it is a minimal hypersurface and $\lvert\nabla f\rvert\neq 0$ on $\partial
M$.

Condition (1.1) arises naturally in the study of static spacetimes satisfying
the Null Energy Condition, as already observed in [WWZ17]. More precisely, a
Lorentzian manifold $(L,\mathfrak{g})$ of the form

$L=\mathbb{R}\times M\,,\qquad\mathfrak{g}\,=\,-f^{2}dt\otimes dt+g,$

happens to be a solution to the Einstein Field Equation

${\rm
Ric}_{\mathfrak{g}}\,+\,\left(\Lambda-\frac{1}{2}{\mathrm{R}}_{\mathfrak{g}}\right)\mathfrak{g}\,=\,T\,,$

subject to $T(X,X)\geq 0$ for any vector field $X$ satisfying
$\mathfrak{g}(X,X)=0$, exactly when $f$ and $g$ satisfy (1.1). A minimal
boundary represents, in this framework, the event horizon of a black hole. For
the reader’s sake, we included the computations in Section A.1. The class of
substatic manifolds obviously includes the very large and thoroughly studied
class of manifolds with nonnegative Ricci curvature, where $f$ is just
constant, and consequently the minimal boundary is empty. However, even
considering explicit model warped products only, a whole new zoo of examples
arises.

As an example, we recall that the following family of triples $(M,g,f)$ is in
fact a family of substatic triples:

$M=I\times\Sigma\,,\qquad g=\frac{dr\otimes dr}{f^{2}}+r^{2}g_{\Sigma},\qquad
f=\sqrt{1-\frac{2\Lambda}{n(n-1)}r^{2}-\frac{2m}{r^{n-2}}+\frac{q^{2}}{r^{2n-4}}},$
(1.2)

where $(\Sigma,g_{\Sigma})$ is a closed $(n-1)$-dimensional Riemannian
manifold satisfying ${\rm Ric}_{g_{\Sigma}}\geq(n-2)g_{\Sigma}$,
$\Lambda,q\in\mathbb{R}$, $m\geq 0$ and $I\subseteq[0,+\infty)$ is the maximal
interval such that the quantity in square root in (1.2) is nonnegative for all
$r\in I$. According to the sign of $\Lambda$, the case $m=q=0$ corresponds to
the space forms. If instead $m>0$, $q=0$, one obtains the families of the
Schwarzschild, Schwarzschild–de Sitter and Schwarzschild–Anti de Sitter black
holes, again with respect to $\Lambda$ being vanishing, positive or negative.
If $m>0$ and $q\neq 0$, one gets the Reissner–Nordström versions of these last
spaces. From a physical point of view, $\Lambda$ is the cosmological constant,
$m$ is the mass and $q$ is the charge of the black hole.

We will always tacitly assume that $(M,g)$ is complete as a metric space. This
holds true for the models (1.2), provided the absolute value of the charge $q$
is not too big. For instance, for $\Lambda=0$, the solution has a singularity
at $r=0$ when $\lvert q\rvert>m$.

The main achievement of the present work is the following sharp Isoperimetric
Inequality, taking place in a relevant subclass of substatic triples. It is
saturated by warped product metrics only, such as the ones in (1.2).

######  Theorem A (Substatic $f$-isoperimetric inequality).

Let $(M,g,f)$ be a substatic triple of dimension $n\leq 7$, with horizon
boundary and one uniform $f$-complete end. Assume there exists an exhaustion
of nonmimimal outward minimizing hypersurfaces homologous to the boundary.
Then, for any bounded domain $\Omega_{\Sigma}$ with smooth boundary
$\partial\Omega_{\Sigma}=\partial M\sqcup\Sigma$ it holds

$\lvert\Sigma\rvert^{\frac{n}{n-1}}-\lvert\partial M\rvert^{\frac{n}{n-1}}\geq
n\left[\mathrm{AVR}(M,g,f)\lvert\mathbb{S}^{n-1}\rvert\right]^{\frac{1}{n-1}}\lvert\Omega_{\Sigma}\rvert_{f}.$
(1.3)

Moreover, in the case $\mathrm{AVR}(M,g,f)>0$,

  * •

if $\partial M\neq\mathchar 31\relax$, the equality holds in (1.3) if and only
if $\partial M$ is connected and $(M,g)$ is isometric to

$\left([\overline{s},+\infty)\times\partial M,\,\frac{ds\otimes
ds}{f(s)^{2}}+\frac{s^{2}}{\overline{s}^{2}}\,g_{\partial M}\right),$ (1.4)

where $g_{\partial M}$ is the metric induced by $g$ on $\partial M$ and
$\Sigma$ is a level set of $s$. In particular, $f=f(s)$ is a function of $s$
alone.

  * •

If $\partial M=\mathchar 31\relax$, the equality holds in (1.3) if and only if
$(M,g)$ is isometric to

$\left([0,+\infty)\times\mathbb{S}^{n-1},\,\frac{ds\otimes
ds}{f(s)^{2}}+\frac{s^{2}}{f(x)^{2}}\,g_{\mathbb{S}^{n-1}}\right),$ (1.5)

where $g_{\mathbb{S}^{n-1}}$ is the round metric on the $(n-1)$-dimensional
sphere $\mathbb{S}^{n-1}$ and $x\in\Omega_{\Sigma}$. In this case, $\Sigma$ is
a level set of $s$ homothetic to the round sphere. The function $f=f(s)$
depends on $s$ alone also in this case.

The asymptotic assumptions entering in the above statement will be better
understood in the next Subsection, in connection with the comparison results
presented below. Concerning the quantities appearing in (1.3), we have denoted
by $\lvert\Omega_{\Sigma}\rvert_{f}$ the weighted volume
$\int_{\Omega_{\Sigma}}f\,d\mu$, whereas $\mathrm{AVR}(M,g,f)$ is a suitable
substatic generalization of the classical Asymptotic Volume Ratio for
nonnegative Ricci curvature, see 1.12. When it is nonzero, inequality 1.3 in
particular yields a quantitative information about the minimal boundary being
in fact area minimizing, in terms of a suitable weighted volume. Observe that
a priori the boundary is not even assumed to be area minimizing at all. From a
more analytical point of view, formula (1.3) constitutes a nonstandard
weighted isoperimetric inequality, as the perimeter is actually unweighted.
The geometric intuition behind it will be given by the end of the following
Subsection. One can interpret the very thoroughly recently studied
Isoperimetric Inequality in nonnegative Ricci curvature [AFM20, Bre21, Ant+22,
BK22, Joh21, CM22, CM22a, Poz23] as a special case of 1.3, obtained when the
boundary is empty and $f$ is constant. The rigidity statement accordingly
generalizes the one of the nonnegative Ricci curvature case.

We point out that Theorem A is particularly meaningful and perfectly sharp
already in the above recalled Reissner–Nordström and Schwarzschild metrics,
consisting in 1.2 for $\Lambda=0$, $m>0$, $|q|<m$, and more generally in
_asymptotically flat_ substatic manifolds. With asymptotically flat we mean
that the manifold converges to the Euclidean space (in a very weak sense) and
that $f$ goes to $1$ at infinity, see Definition 4.8. From the definition, it
follows that an asymptotically flat end is automatically uniform and
$f$-complete, it possesses a natural exhaustion of coordinate spheres and it
is possible to compute ${\rm AVR}(M,g,f)=1$. Thus, the above statement
simplifies significantly.

######  Corollary 1.1.

Let $(M,g,f)$ be a substatic triple of dimension $n\leq 7$, with horizon
boundary and one asymptotically flat end. Then, for any bounded domain
$\Omega_{\Sigma}$ with smooth boundary $\partial\Omega_{\Sigma}=\partial
M\sqcup\Sigma$ it holds

$\lvert\Sigma\rvert^{\frac{n}{n-1}}-\lvert\partial M\rvert^{\frac{n}{n-1}}\geq
n\lvert\mathbb{S}^{n-1}\rvert^{\frac{1}{n-1}}\lvert\Omega_{\Sigma}\rvert_{f}.$
(1.6)

The same rigidity statement as in Theorem A applies in case of equality.

To our knowledge, even in the model cases, inequality 1.6 was never observed
before, and does not seem to be inferable from the characterization of
classical isoperimetric sets resulting from the work of Brendle [Bre13], or
the earlier [Bra97, BM02] about the Schwarzschild case.

###  1.1. Substatic comparison geometry

Our analysis begins with the aim of working out a satisfactory substatic
comparison theory, inspired by the classical nonnegative Ricci case. While, in
such case, the model to be compared with is $\mathbb{R}^{n}$, or more
generally a cone, in the substatic generalization the model should be
constituted by the large family of substatic warped products in fact appearing
in the rigidity statement of Theorem A.

To pursue our goal, an initial step consists in comparing the mean curvature
of geodesic spheres with that of the models. Interestingly, in order to obtain
a manageable Riccati equation ruling such comparison, we are led to work in
the metric $\tilde{g}=g/f^{2}$. This is no accident: the metric $\tilde{g}$
happens to fulfil the ${\rm CD}(0,1)$ condition, consisting in a metric
subject to

${\rm Ric}_{\tilde{g}}+\widetilde{\nabla}^{2}\psi+\frac{1}{n-1}d\psi\otimes
d\psi\geq 0$ (1.7)

for some smooth function $\psi$. To our knowledge, such explicit conformal
relation was not pointed out in literature yet. However, a remarkable link is
described by Li–Xia [LX17]: they come up with a family of connections with
Ricci curvatures interpolating between the tensor in the left-hand side of 1.7
and the tensor in the left-hand side of 1.1. We discuss the $\mathrm{CD}(0,1)$
conformal change and Li-Xia connections in more details in Section A.3. We
also point out that the conformal metric $\tilde{g}$ has a natural physical
interpretation in the context of static spacetimes, where it is referred to as
optical metric. We give some more details on this point at the end of Section
A.1.

We will denote by $\rho$ the $\tilde{g}$-distance from a point $p\in M$, or
the signed $\tilde{g}$-distance from a smooth strictly mean-convex
hypersurface $\Sigma$ homologous to the boundary. We give some more details on
this second case, which is slightly less classical but will be crucial for the
Willmore-type inequality (1.13) discussed below and in turn for the proof of
Theorem A. With homologous to the boundary we mean that there exists a compact
domain $\Omega$ with boundary $\partial\Omega=\partial M\sqcup\Sigma$, and by
strictly mean-convex we understand that $\Sigma$ has pointwise positive
$g$-mean curvature ${\rm H}$ with respect to the normal pointing towards
infinity. We always choose the signed distance $\rho$ to be positive in the
noncompact region $M\setminus\Omega$, that is,

$\rho(x)\,=\,\begin{cases}{\rm d}_{\tilde{g}}(x,\Sigma)&\hbox{if
$x\not\in\Omega$},\\\ -{\rm d}_{\tilde{g}}(x,\Sigma)&\hbox{if
$x\in\Omega$}.\end{cases}$ (1.8)

Both in the case of the distance from a point and in the case of the signed
distance from a hypersurface, through an analysis of the evolution of the mean
curvature of the level sets of $\rho$, we come up (see Theorem 2.5 and
Proposition 2.7) with the following inequality

$0\,<\,\frac{{\rm H}}{f}\,=\,\Delta\rho+\frac{1}{f}\langle\nabla
f\,|\,\nabla\rho\rangle\,\leq\,\frac{n-1}{\eta}\,,$ (1.9)

where ${\rm H}$ denotes the $g$-mean curvature of a level set of the
$\tilde{g}$-distance $\rho$, and $\eta$ denotes an useful auxiliary function
that will be called reparametrized distance. It is defined by the first order
PDE 2.10, when the distance from a point is concerned, and in 2.25 when $\rho$
is the $\tilde{g}$-distance from a hypersurface. The function $\eta$
represents the distance along the radial geodesics computed with respect to
the metric $\overline{g}=f^{2}g=f^{4}\tilde{g}$. This third conformal metric
will not play a prominent role in the paper, but we will take some advantage
from this along the proof of Theorem A. More details on this point and further
comments on $\eta$ (in particular its relation with the weighted connection
introduced by Li–Xia) may be found in Remark 2.4.

We remark that 1.9 could be derived also from [Wyl17, Theorem 3.2], rewriting
it in the substatic setting thanks to the conformal relation with
$\mathrm{CD}(0,1)$-metrics. Nevertheless, we have preferred to include a full
proof of it, in order to emphasize the role of $\eta$ and to show the
substatic point of view.

A main consequence we draw out of the Laplacian Comparison Theorem above is a
Bishop–Gromov Monotonicity Theorem. We state here a version substantially
gathering Theorem 2.9 and Theorem 2.11 below.

######  Theorem B (Substatic Bishop–Gromov).

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\rho$ be the $\tilde{g}$-distance function from a point or the signed
$\tilde{g}$-distance function from a strictly mean-convex hypersurface
$\Sigma$ homologous to $\partial M$ and disjoint from it. Let $\eta$ be the
corresponding reparametrized distance, defined by 2.10 or by 2.25, and let
${\rm Cut}^{\tilde{g}}$ be the cut locus of the point/hypersurface. Then, for
any $k>0$, the functions

$A(t)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\int_{\\{\rho=t\\}\setminus{\rm
Cut}^{\tilde{g}}}\frac{1}{\eta^{n-1}}d\sigma\,,\qquad
V(t)\,=\,\frac{1}{|\mathbb{B}^{n}|t^{k}}\int_{\\{0\leq\rho\leq
t\\}}\frac{\rho^{k-1}}{f\eta^{n-1}}d\mu\,,$ (1.10)

are well defined and monotonically nonincreasing. Furthermore:

  * •

if $A(t_{1})=A(t_{2})$ for $0<t_{1}<t_{2}$, then the set $\\{t_{1}\leq\rho\leq
t_{2}\\}$ is isometric to $[t_{1},t_{2}]\times\Sigma$, for some
$(n-1)$-dimensional manifold $(\Sigma,g_{0})$, with metric

$g=f^{2}\,d\rho\otimes d\rho+\eta^{2}g_{0}\,;$

  * •

if $V(t_{1})=V(t_{2})$ for $0<t_{1}<t_{2}$, then the set $\\{0\leq\rho\leq
t_{2}\\}$ is isometric to $[0,t_{2}]\times\Sigma$, for some
$(n-1)$-dimensional manifold $(\Sigma,g_{0})$, with metric

$g=f^{2}\,d\rho\otimes d\rho+\eta^{2}g_{0}\,;$

in the case where $\rho$ is the distance from a point $x$, then $f$ and $\eta$
are functions of $\rho$ only and $g_{0}=f(x)^{-2}g_{\mathbb{S}^{n-1}}$.

A first $\mathrm{CD}(0,1)$-version of Bishop–Gromov monotonicity has been
obtained by Wylie–Yeroshkin [WY16]. Various further generalizations then
arose; dropping any attempt to be complete, we mention [KL22, LMO22, KS23],
and leave the interested reader to the references therein and to Ohta’s
monograph [Oht21]. However, even appealing to the conformal change $g/f^{2}$,
formally relating the $\mathrm{CD}(0,1)$ to the substatic condition, it does
not seem straightforward to deduce a Bishop–Gromov statement in the form
above, that will be ruling the Willmore-type inequality 1.13 below and in turn
1.3. To the authors’ knowledge, the rigidity statements contained in Theorem B
have not been considered in literature yet.

In the case where $f$ is constant equal to $1$, then both $\rho$ and $\eta$
coincide with the $g$-distance, hence we recover the standard Bishop–Gromov
monotonicity for nonnegative Ricci tensor. Remarkably, in contrast with the
standard Bishop–Gromov, in our setting we do not have to require the boundary
$\partial M$ to be empty. This is because, since we have $f=0$ at $\partial
M$, the boundary becomes an end with respect to the conformal metric
$\tilde{g}=g/f^{2}$, see Lemma 3.4. As a consequence, we do not have issues
when the geodesic spheres $\\{\rho=t\\}$ intersect the boundary, simply
because the boundary is at infinite $\tilde{g}$-distance so the intersection
is always empty. On the other hand, in general $\tilde{g}$-geodesics may have
finite length when going towards the ends of $M$. To avoid this, we added the
assumption of $\tilde{g}$-geodesic completeness in the statement. The main
type of ends considered in this paper (the $f$-complete ends introduced just
below) will be $\tilde{g}$-geodesically complete by definition.

Proceeding in analogy with the nonnegative Ricci curvature case, we are
interested in defining a suitable Asymptotic Volume Ratio, motivated by the
Bishop–Gromov monotonicity above. In order to get a satisfactory notion, we
first aim at understanding basic properties of the ends of substatic
manifolds, at least under asymptotic assumptions on the potential of $f$. A
fundamental tool for this kind of study in the classical theory is
Cheeger–Gromoll Splitting Theorem [CG72].

Wylie [Wyl17] in fact exploited the Laplacian Comparison Theorem to prove a
splitting theorem in the ${\rm CD}(0,1)$ setting, that will in turn provide
surprising pieces of information in the conformal substatic setting. For our
main geometric goals, the kind of end that we will be mostly interested in is
that of $f$-complete ends. We say that an end is $f$-complete if for any
$g$-unit speed curve $\gamma:[0,+\infty)\to M$ going to infinity along the end
it holds

$\lim_{t\to+\infty}\rho(\gamma(t))\,=\,+\infty\,,\qquad\int_{0}^{+\infty}f(\gamma(t))dt\,=\,+\infty\,.$
(1.11)

The first condition is essentially asking that the end remains an end even
with respect to the conformal metric $\tilde{g}=g/f^{2}$. In other words, the
first condition is equivalent to the requirement that the end is geodesically
complete with respect to the metric $\tilde{g}$. The second condition instead
is a way to ensure that the reparametrized distance $\eta$ diverges to
$+\infty$ and is connected to an analogous definition given in [Wyl17] in the
${\rm CD}(0,1)$ framework. Further discussion and comments on this definition
can be found after Definition 3.1. It is easy to check that an end is
$f$-complete whenever there exists a constant $c$ such that $cr^{-k}<f<cr^{k}$
for $0<k<1$ at sufficiently large distances, where $r$ is the $g$-distance
from a point (see Proposition 3.2).

We provide here the full statement of the substatic Splitting Theorem.

######  Theorem C (Substatic Splitting Theorem).

Let $(M,g,f)$ be a substatic triple with ends that are all $f$-complete. If
there is more than one end, then $(M,g)$ is isometric to

$(\mathbb{R}\times\Sigma,\,f^{2}\,ds\otimes ds+g_{\Sigma})\,,$

for some $(n-1)$-dimensional Riemannian manifold $(\Sigma,g_{\Sigma})$. In
particular, if $\partial M$ is nonempty, then $(M,g,f)$ has only one end.

Finally, we point out that similar arguments can be performed also for a
different kind of ends, known as conformally compact ends, see Theorems 3.7
and 3.8. In particular we will show in Theorem 3.7 that conformally compact
substatic manifolds necessarily have connected conformal infinity,
generalizing a known result in the literature of _static vacuum solutions_
[CS01]. Static vacuum solutions are in fact substatic triples such that 1.1 is
satisfied with equality on the whole space.

Focusing now for simplicity on $f$-complete substatic triples that have only
one end, one would then be led to define the Asymptotic Volume Ratio as

${\rm
AVR}(M,g,f)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\lim_{t\to+\infty}\int_{\\{\rho=t\\}}\frac{1}{\eta^{n-1}}d\sigma\,=\,\frac{1}{|\mathbb{B}^{n}|}\lim_{t\to+\infty}\frac{1}{t^{n}}\int_{\\{\rho\leq
t\\}}\frac{\rho^{n-1}}{f\eta^{n-1}}d\mu\,.$ (1.12)

with $\rho$ denoting the $g/f^{2}$-distance from a mean-convex hypersurface
homologous to $\partial M$ or from a point, if the boundary is empty. The fact
that both limits above give the same result is easy to establish. However, we
have to make sure that such quantity is independent of the initial
hypersurface. We accomplish this task under the assumption of _uniformity of
the end_ , meaning that the quotient $\eta_{x}/\eta_{y}$ of the reparametrized
distances with respect to two different points $x,y$ converges uniformly to
$1$ at infinity. Again, such condition is immediately checked to be fulfilled
in the asymptotically flat regime. More generally, it can in fact be inferred
under a natural decay condition on the gradient of $f$ only, see Proposition
4.3.

Exploiting the global features of our Bishop–Gromov Theorem we obtain the
following Willmore-like inequality for mean-convex hypersurfaces homologous to
$\partial M$.

######  Theorem D (Substatic Willmore inequality).

Let $(M,g,f)$ be a substatic triple with one uniform $f$-complete end. Let
$\Sigma$ be a hypersurface homologous to the boundary. Suppose that the mean
curvature ${\rm H}$ of $\Sigma$ with respect to the normal pointing towards
infinity satisfies ${\rm H}>0$ pointwise. Then

$\int_{\Sigma}\left[\frac{{\rm H}}{(n-1)f}\right]^{n-1}d\sigma\,\geq\,{\rm
AVR}(M,g,f)\,|\mathbb{S}^{n-1}|\,.$ (1.13)

Furthermore, if the equality holds, then the noncompact connected component
$U$ of $M\setminus\Sigma$ is isometric to $[0,+\infty)\times\Sigma$ with
metric

$g\,=\,f^{2}d\rho\otimes d\rho+\eta^{2}g_{0}\,,$

where $g_{0}$ is a metric on $\Sigma$.

Notice that if there are multiple ends, the above inequality is trivial, since
in this case the Asymptotic Volume Ratio of each end is zero as pointed out in
Lemma 4.7.

In the classical nonnegative Ricci curvature setting, the above result was
obtained in [AFM20]. However, the more elementary proof we propose displays
more resemblances with the alternative argument of Wang [Wan23].

The validity of the above Willmore-type inequality very naturally suggests the
isoperimetric inequality 1.3. Indeed, assume that smooth area minimizers exist
among hypersurfaces $\Sigma_{V}$ enclosing a given weighted volume
$\int_{\Omega}fd\mu=V$ with $\partial M$, for any given value $V$. We will
call such $\Sigma_{V}$ $f$-isoperimetric. Then, through standard variation
formulas, there must exist a Lagrange multiplier $\lambda\in\mathbb{R}$ such
that

$\int_{\Sigma}({\rm H}-\lambda f)\varphi d\sigma=0$ (1.14)

for any $\varphi\in\mathscr{C}^{\infty}_{c}(\Sigma)$, implying that the mean
curvature of $\Sigma$ satisfies ${\rm H}/f=\lambda$. Letting
$I_{f}(V)=\lvert\Sigma_{V}\rvert$, one has that $(I_{f})^{\prime}(V)$ is
proportional to $\lambda$. If this multiplier happens to positive, it is
sharply estimated in terms of $I_{f}(V)$ by 1.13. The resulting differential
inequality leads to

$\lvert\Sigma_{V}\rvert^{\frac{n}{n-1}}-\lvert\Sigma_{V_{0}}\rvert^{\frac{n}{n-1}}\geq
n(\mathrm{AVR}(M,g,f)\lvert\mathbb{S}^{n-1}\rvert)^{\frac{1}{n-1}}(V-V_{0})$
(1.15)

for any $V_{0}<V$. Now, if $\partial M$ happens to be, in addition to minimal,
also area-minimizing, 1.15 directly implies 1.3 for $\Sigma_{V}$ in the limit
as $V_{0}\to 0^{+}$. But $\Sigma_{V}$ being the best competitor, it holds for
any $\Sigma$ as in the claimed statement.

The assumptions of $f$-completeness and uniformity at infinity are obviously
added in order to count on the validity of Theorem D. The additional
requirement of existence of a nonminimal outward minimizing exhaustion (see
Section 5) is ultimately added in order to overcome the problem of the
possible nonexistence of $f$-isoperimetric $\Sigma_{V}$’s. We follow the
general strategy devised by Kleiner [Kle92], and reinterpreted in the
nonnegative Ricci curvature setting in [FM22], that consists in considering
$f$-isoperimetric sets constrained in mean-convex boxes, in our case given by
the exhaustion. However, some new geometric difficulties arise, mostly given
by the new portion of boundary $\partial M$. They will be overcome by
exploiting the fact that such boundary is in turn _a priori_ outermost area-
minimizing, a new piece of information that we obtain through an argument
involving the Mean Curvature Flow of the outward minimizing exhaustion (see
Proposition 5.1), and by discovering that the constrained $f$-isoperimetric
sets crucially never touch $\partial M$, see Theorem 5.2. We will in the end
have all the tools at hand to run the argument sketched above, under the usual
dimensional threshold ensuring the constrained $f$-isoperimetric sets to be
regular enough. The strong rigidity statement contained in Theorem A will stem
from the fact that in case of equality all of the $\Sigma_{V}$ must satisfy
the equality in 1.13. The rigidity statement of Theorem D will be thus
complemented with the additional information given by ${\rm H}=\lambda f$,
forcing the metric to split as 1.4.

###  1.2. Further directions

The results presented in this paper raise a number of natural questions,
especially out of our main geometric inequalities, Theorem A and Theorem D.
The Willmore-type inequality in nonnegative Ricci curvature [AFM20, Theorem
1.1] has been first obtained with a completely different technique, involving
the evolution of the initial hypersurface along the level sets of a harmonic
potential function. Understanding a version of such a route in the substatic
context may have various interests. First of all, it may allow to remove the
mean-convexity assumption on $\Sigma$ we have in Theorem D in favour of the
absolute value of the mean curvature in 1.13. Secondly, and more
interestingly, it would suggest the viability of a suitable version of the
analysis through $p$-harmonic functions performed in nonnegative curvature in
[BFM22], likely leading to a new substatic Minkowski inequality, potentially
stronger than our Willmore-type. Moreover, studying the behaviour of such
substatic $p$-harmonic functions may have implications in the existence of the
weak Inverse Mean Curvature Flow [HI01] in the substatic regime, furnishing a
vast extension of the important existence results in nonnegative Ricci
curvature [MRS22]. Recalling the outward minimizing properties of the evolving
hypersurfaces [HI01, Minimizing Hull Property 1.4], the existence of the weak
IMCF would imply the a priori existence of the outward minimizing exhaustion
requested in Theorem A. In the special case of asymptotically flat static
vacuum solutions, the weak IMCF has already been introduced and employed to
prove Minkowski-type inequalities in [Wei18, McC18, HW23]. Such inequalities
are lower bound on the integral of $f{\rm H}$, and, as such, do not seem
related to 1.13.

It would also be rather interesting to explore other approaches for the proof
of the $f$-Isoperimetric Inequality 1.3 as well, possibly allowing to remove
the dimensional threshold. Antonelli–Pasqualetto–Pozzetta–Semola [Ant+22,
Theorem 1.1] provided a natural and very strong proof in the nonnegative Ricci
curvature case taking advantage of a generalized compactness result [Nar14,
AFP22, ANP22] for isoperimetric minimizing sequences in the nonsmooth
$\mathrm{RCD}$ setting. This immediately invites to study the nonsmooth
counterpart of the substatic condition. A possible key for this may lie in the
recent optimal transport equivalent definition of $\mathrm{CD}(0,1)$ given in
[Sak21].

Another, completely different approach one may undertake consists in Brendle’s
[Bre21], building on the ABP method applied to a torsion problem with Neumann
condition. A substatic version of such approach promises to deal with the PDE
considered in [LX17, FP22] in relation with the Heintze-Karcher inequality. We
also point out that both these alternative approaches should have consequences
in going beyond the dimensional threshold we imposed.

From the comparison geometry point of view, the validity of the Splitting
Theorem and of the Bishop–Gromov monotonicity strongly suggests that other
classical results, such as the Cheng eigenvalue estimate and Cheng–Yau
gradient estimate, should have analogues in the substatic setting. A promising
advance in this direction has been obtained in the $\mathrm{CD}(0,1)$ setting
[Fuj22].

It may also be interesting to study compact substatic triples. Important
models for this class of manifolds are given by static solutions with positive
cosmological constant, most notably the Schwarzschild–de Sitter and
Reissner–Nordström–de Sitter spacetimes, corresponding to (1.2) with $\Lambda$
positive and $(\Sigma,g_{\Sigma})$ a round sphere. Another natural direction
is to investigate what can be said for the more general problem of studying
triples $(M,g,f)$ satisfying

$f{\rm Ric}-\nabla^{2}f+(\Delta f)g\geq-\mu g\,,\quad\mu\in\mathbb{R}\,.$
(1.16)

The case $\mu=0$ corresponds to the substatic condition. The case $\mu\neq 0$
is also of interest: triples that saturate (1.16) for $\mu\neq 0$ are called
$V$-static and are connected with the critical point equation and the Besse
conjecture, see [FY19, He21] and references therein for more details on these
topics. We mention that inequality (1.16) has been considered in [Zen22],
where an almost-Schur inequality has been proved and exploited to generalize
results in [Che14, LX19].

###  1.3. Structure of the paper

In Section 2 we compute the evolution of the mean curvature of geodesic
spheres, leading to the aforementioned Laplacian Comparison Theorem, formula
(1.9) (see Theorem 2.5). Building on it, we prove Theorem B, first for the
functional $A$ (Theorem 2.9) and then for the functional $V$ (Theorem 2.11).
Section 3 is dedicated to the proof of the Splitting Theorem. We first analyze
the most important case of $f$-complete ends and prove Theorem C (Subsection
3.2), then we discuss analogous results for conformally compact ends as well,
see Theorems 3.7 and 3.8. In Section 4 we introduce the notions of uniform
ends (Definition 4.1) and Asymptotic Volume Ratio (Definition 4.4) and prove
the Willmore Inequality (see Theorem 4.10). Finally, in Section 5 we prove
Theorem A. We include an Appendix encompassing the physical motivation for the
substatic condition, the conformal relation with the $\mathrm{CD}(0,1)$
curvature-dimension condition and some additional comments.

### Acknowledgements

The work was initiated during the authors’ participation at the conference
_Special Riemannian Metrics and Curvature Functionals_ held at _Centro De
Giorgi_ in Pisa in 2022. A substantial part of the work has been carried out
during the authors’ attendance to the _Thematic Program on Nonsmooth
Riemannian and Lorentzian Geometry_ that took place at the Fields Institute in
Toronto. They warmly thank the staff, the organizers and the colleagues for
the wonderful atmosphere and the excellent working conditions set up there.

During the preparation of the work, M. F. was supported by the European Union
– NextGenerationEU and by the University of Padova under the 2021 STARS
Grants@Unipd programme “QuASAR”.

The authors are members of Gruppo Nazionale per l’Analisi Matematica, la
Probabilità e le loro Applicazioni (GNAMPA), which is part of the Istituto
Nazionale di Alta Matematica (INdAM), and are partially funded by the GNAMPA
project “Problemi al bordo e applicazioni geometriche”.

The authors are grateful to Lucas Ambrozio, Gioacchino Antonelli, Luca
Benatti, Camillo Brena, Philippe Castillion, Nicola Gigli, Marc Herzlich,
Lorenzo Mazzieri, Marco Pozzetta and Eric Woolgar for their interest in their
work and for various useful conversations.

##  2\. Riccati comparison and Bishop-Gromov Theorem

###  2.1. Evolution of the mean curvature

Let $(M,g,f)$ be a substatic solution. Since $f$ does not vanish inside $M$ by
definition, the metric $\tilde{g}=g/f^{2}$ is well defined in
$M\setminus\partial M$. Let $\rho$ be the distance function from a point $p\in
M\setminus\partial M$ with respect to the metric $\tilde{g}$ and consider
Riemannian polar coordinates $(\rho,\theta^{1},\dots,\theta^{n-1})$ in a
neighborhood $U$ of $p$. In this subsection we focus our computation in $U$
and we assume that $U$ does not contain points in the cut locus of $p$. This
guarantees that $\rho$ is smooth in $U$, that $|d\rho|_{\tilde{g}}=1$ at all
points of $U$ and that the metric $\tilde{g}$ has the following form:

$\tilde{g}\,=\,d\rho\otimes
d\rho+\tilde{g}_{ij}(\rho,\theta^{1},\dots,\theta^{n-1})d\theta^{i}\otimes
d\theta^{j}\,.$ (2.1)

We denote by $\widetilde{\nabla}$ the Levi-Civita connection with respect to
$\tilde{g}$. It is well known that the Hessian of a distance function
satisfies the inequality

$\big{|}\widetilde{\nabla}^{2}\rho\big{|}^{2}_{\tilde{g}}\geq\frac{(\Delta_{\tilde{g}}\rho)^{2}}{n-1}\,.$
(2.2)

We now write down this inequality in terms of the original metric. To this
end, we first compute

$\displaystyle|\nabla\rho|^{2}$ $\displaystyle=\,\frac{1}{f^{2}}\,,$ (2.3)
$\displaystyle\widetilde{\nabla}^{2}\rho$
$\displaystyle=\,\nabla^{2}\rho\,+\,\frac{1}{f}\left(d\rho\otimes df+df\otimes
d\rho-\langle\nabla\rho\,|\,\nabla f\rangle g\right)\,,$
$\displaystyle\Delta_{\tilde{g}}\rho$
$\displaystyle=\,f^{2}\Delta\rho-(n-2)f\langle\nabla\rho\,|\,\nabla
f\rangle\,.$

Using these identities, with some computations we can rewrite (2.2) in terms
of the original metric as

$|\nabla^{2}\rho|^{2}\,\geq\,\frac{(\Delta\rho)^{2}}{n-1}-\frac{n-2}{n-1}\frac{1}{f^{2}}\langle\nabla\rho\,|\,\nabla
f\rangle^{2}+\frac{2}{n-1}\frac{1}{f}\Delta\rho\langle\nabla\rho\,|\,\nabla
f\rangle+\frac{2}{f^{4}}|\nabla f|^{2}\,.$ (2.4)

From this formula onwards we just focus on the original metric $g$. From
(2.1), with respect to the coordinates $(\rho,\theta^{1},\dots,\theta^{n-1})$,
we have

$g\,=\,f^{2}\,d\rho\otimes
d\rho+g_{ij}(\rho,\theta^{1},\dots,\theta^{n-1})d\theta^{i}\otimes
d\theta^{j}\,.$

We are interested in the evolution of the mean curvature ${\rm H}$ of the
level sets of $\rho$ with respect to the metric $g$. We have

${\rm
H}\,=\,\frac{\Delta\rho}{|\nabla\rho|}\,-\,\frac{\nabla^{2}\rho(\nabla\rho,\nabla\rho)}{|\nabla\rho|^{3}}\,=\,f\Delta\rho\,-\,\frac{1}{2}f^{3}\langle\nabla
f\,|\,\nabla\rho\rangle\,=\,f\Delta\rho\,+\,\langle\nabla
f\,|\,\nabla\rho\rangle\,.$

On the other hand, using the fact that $|\nabla\rho|=1/f$ and the Bochner
formula we compute

$\frac{6}{f^{4}}|\nabla f|^{2}\,-\,\frac{2}{f^{3}}\Delta
f\,=\,\Delta|\nabla\rho|^{2}\,=\,2\,|\nabla^{2}\rho|^{2}\,+\,2\,{\rm
Ric}(\nabla\rho,\nabla\rho)\,+\,2\,\langle\nabla\Delta\rho\,|\,\nabla\rho\rangle\,.$

Combining this with (1.1), some computations lead to

$\langle\nabla\Delta\rho\,|\,\nabla\rho\rangle\,\leq\,\Delta\rho\langle\nabla\rho\,|\,\nabla
f\rangle-\frac{1}{f}\nabla^{2}f(\nabla\rho,\nabla\rho)\,-\,|\nabla^{2}\rho|^{2}$

We can use this information to find the evolution of ${\rm H}$:

$\displaystyle\langle\nabla{\rm H}\,|\,\nabla\rho\rangle\,$
$\displaystyle\leq\,-f|\nabla^{2}\rho|^{2}\,+\,\frac{2}{f^{3}}|\nabla
f|^{2}\,+\,\Delta\rho\langle\nabla\rho\,|\,\nabla f\rangle$ (2.5)
$\displaystyle\leq\,-f\frac{(\Delta\rho)^{2}}{n-1}\,+\,\frac{n-3}{n-1}\Delta\rho\langle\nabla\rho\,|\,\nabla
f\rangle\,+\,\frac{n-2}{n-1}\frac{1}{f}\left(\langle\nabla\rho\,|\,\nabla
f\rangle\right)^{2}$ (2.6) $\displaystyle=\,-\frac{1}{n-1}\frac{1}{f}{\rm
H}^{2}\,+\,\frac{1}{f}{\rm H}\langle\nabla\rho\,|\,\nabla f\rangle\,,$ (2.7)

where in the second inequality we have used estimate (2.4). This formula can
be rewritten as

$\bigg{\langle}\nabla\left(\frac{{\rm
H}}{f}\right)\,\bigg{|}\,\nabla\rho\bigg{\rangle}\,\leq\,-\frac{1}{n-1}\left(\frac{{\rm
H}}{f}\right)^{2}\,.$

In other words, we have found the following formula for the evolution of the
mean curvature of the level sets of $\rho$.

######  Lemma 2.1.

In the notations above, at any point of $U$ the evolution of the mean
curvature ${\rm H}$ along $\rho$ satisfies

$\frac{\partial}{\partial\rho}\left(\frac{{\rm
H}}{f}\right)\,\leq\,-\frac{1}{n-1}\,{\rm H}^{2}\,.$ (2.8)

###  2.2. Bounds on the mean curvature and Laplacian comparison

Let $p\in M\setminus\partial M$ and $\rho$ be the distance function from $p$
with respect to the metric $\tilde{g}=g/f^{2}$. We assume that
$M\setminus\partial M$ is geodesically complete with respect to the metric
$\tilde{g}$. We denote by ${\rm Cut}^{\tilde{g}}_{p}(M)$ the cut locus of $p$,
again with respect to $\tilde{g}$. The function $\rho$ is smooth in
$U=M\setminus({\rm Cut}^{\tilde{g}}_{p}(M)\cup\\{p\\})$, where ${\rm
Cut}^{\tilde{g}}_{p}(M)$ is the cut locus of $p$ with respect to the metric
$\tilde{g}$.

For every $\theta\in\mathbb{S}^{n-1}\subset T_{p}M$, we denote by
$\sigma_{\theta}$ the geodesic starting from $p$ in the direction $\theta$ and
by $\tau(\theta)\in(0,+\infty]$ the smallest positive value such that
$\sigma_{\theta}(\tau(\theta))\in{\rm Cut}^{\tilde{g}}_{p}(M)$. We recall from
[Car92, Proposition 2.9, Chapter 13] that
$\tau:\mathbb{S}^{n-1}\to(0,+\infty]$ is a continuous function. Notice that
there is a diffeomorphism between $U$ and the set

$\\{(\rho,\theta)\in(0,+\infty)\times\mathbb{S}^{n-1}\,:\,\rho<\tau(\theta)\\}\,,$

hence we can use $\rho,\theta$ as coordinates in $U$.

We can now exploit Lemma 2.1 to find a bound for ${\rm H}$ in $U$. To this
end, given a positive function $\eta\in\mathscr{C}^{\infty}$ we use (2.8) to
compute

$\frac{\partial}{\partial\rho}\left(\frac{f}{{\rm
H}}-\frac{\eta}{n-1}\right)\,=\,-\frac{f^{2}}{{\rm
H}^{2}}\frac{\partial}{\partial\rho}\left(\frac{{\rm
H}}{f}\right)-\frac{1}{n-1}\frac{\partial\eta}{\partial\rho}\,\geq\,\frac{1}{n-1}\left(f^{2}-\frac{\partial\eta}{\partial\rho}\right)\,.$
(2.9)

We then choose $\eta$ so that the the right hand side vanishes pointwise.
Since $f$ is smooth, the equation

$\frac{\partial\eta}{\partial\rho}\,=\,f^{2}$

can be solved and yields a unique solution once we fix its value on a level
set of $\rho$.

######  Proposition 2.2.

There exists a unique function $\eta\in\mathscr{C}^{\infty}(U)$ satisfying

$\begin{dcases}\frac{\partial\eta}{\partial\rho}\,=\,f^{2}\,,\\\ \lim_{\rho\to
0^{+}}\eta\,=\,0\,.\end{dcases}$ (2.10)

######  Remark 2.3.

It should be remarked that $\eta$ is not even necessarily continuous outside
$U$. In fact, if there are two minimizing geodesics from $p$ to $q\in{\rm
Cut}^{\tilde{g}}_{p}(M)$, the function $\eta$ may behave differently on the
two geodesics, hence the limit of $\eta$ as we approach $q$ along the two
different geodesics would give different results.

###### Proof.

For every $\varepsilon>0$, consider the function $\eta_{\varepsilon}$ defined
by

$\begin{dcases}\frac{\partial\eta_{\varepsilon}}{\partial\rho}\,=\,f^{2}&\hbox{
in }\\{\rho>\varepsilon\\}\cap U\,,\\\ \eta_{\varepsilon}\,=\,0&\hbox{on
}\\{\rho=\varepsilon\\}\,.\end{dcases}$ (2.11)

Since $f$ is smooth, it is well known (see for instance [Eva10, Section
3.2.4]) that (2.11) can be solved and yields a unique $\mathscr{C}^{2}$
solution. Furthermore, by differentiating the first equation we find that the
first derivatives $\partial_{\alpha}\eta$ also solve a first order PDE, namely
$\partial_{\rho}\partial_{\alpha}\eta=\partial_{\alpha}f^{2}$. Since
$\partial_{\alpha}f^{2}$ is smooth, it follows that the derivatives
$\partial_{\alpha}\eta$ are also $\mathscr{C}^{2}$. Proceeding this way, we
deduce that $\eta_{\varepsilon}$ is smooth.

It is now sufficient to pass to the limit as $\varepsilon\to 0$ using
Ascoli–Arzelà. Since the functions $\eta_{\varepsilon}$ (and their derivatives
as well) are uniformly continuous and uniformly bounded on any compact domain
inside $U$, it follows that $\eta_{\varepsilon}$ converge to a smooth function
$\eta$, defined on the whole $U$.

Concerning uniqueness, if there were two different solutions
$\eta_{1},\eta_{2}$ of (2.10), then the difference $\eta_{1}-\eta_{2}$ would
have derivative equal to zero along the direction $\partial/\partial\rho$.
Since the limit as $\rho\to 0$ is zero, one immediately obtains
$\eta_{1}-\eta_{2}=0$. ∎

######  Remark 2.4 (The reparametrized distance $\eta$).

The function $\eta$ is also called reparametrized distance. The reason for
this terminology is that the radial $\tilde{g}$-geodesics from our point $p$,
reparametrized with respect to $\eta$, are geodesics for the weighted
connection

${\rm D}_{X}Y=\nabla_{X}Y+\frac{1}{f}g(X,Y)\nabla f\,.$

The significance of such connection is that the Ricci tensor associated to
${\rm D}$ is nonnegative if and only if the substatic condition is satisfied.
More details on this can be found in Section A.3 and in [LX17] (see also
[KWY19, WY16] for further discussions in the conformally related ${\rm
CD(0,1)}$ setting). Alternatively, as mentioned in the Introduction, $\eta$
can also be seen to represent the distance along radial $\tilde{g}$-geodesics
with respect to the metric $\overline{g}=f^{2}g$. In fact, if $\sigma:[0,S]\to
M$ is a radial $\tilde{g}$-geodesic with $\sigma(0)=p$ and
$\dot{\sigma}=\partial/\partial\rho$, we have
$|\dot{\sigma}(s)|_{\tilde{g}}=1$ and so

${\rm
d}_{\overline{g}}(\gamma(S),p)\,=\,\int_{0}^{S}|\dot{\sigma}(s)|_{\overline{g}}ds\,=\,\int_{0}^{S}f^{2}(\sigma(s))ds\,=\,\eta(S)\,.$

We are now in a position to prove the following crucial bound on the mean
curvature of the level sets of $f$. This bound can be naturally interpreted as
the Laplacian Comparison for the conformal distance function $\rho$. This
result corresponds to [Wyl17, Theorem 3.2] in the ${\rm CD}(0,1)$ framework.

######  Theorem 2.5 (Laplacian Comparison).

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\rho$ be the distance function to a point $p\in M\setminus\partial M$ with
respect to the metric $\tilde{g}=g/f^{2}$ and $\eta$ be the solution to
(2.10). Then the mean curvature ${\rm H}$ of the level sets of $\rho$ with
respect to the metric $g$ satisfies

$0\,<\,\frac{{\rm H}}{f}\,=\,\Delta\rho+\frac{1}{f}\langle\nabla
f\,|\,\nabla\rho\rangle\,\leq\,\frac{n-1}{\eta}$ (2.12)

in the classical sense in the open dense set $U=M\setminus({\rm
Cut}^{\tilde{g}}_{p}(M)\cup\\{p\\})$.

###### Proof.

Let us first prove the thesis working inside the open set $U$. From the
definition of $\eta$ and (2.9), we immediately deduce

$\frac{\partial}{\partial\rho}\left(\frac{f}{{\rm
H}}-\frac{\eta}{n-1}\right)\,\geq\,0\,.$

In other words, the function $f/{\rm H}-\eta/(n-1)$ is nondecreasing.

We then estimate its value near the point $p$. It is well known that, for
small $\rho$’s, the mean curvature ${\rm H}_{\tilde{g}}$ of the geodesic balls
grows as the geodesic balls in Euclidean space, namely ${\rm
H}_{\tilde{g}}=(n-1)/\rho+o(1/\rho)$ when $\rho$ is sufficiently small. Here
we have denoted by ${\rm H}_{\tilde{g}}$ the mean curvature with respect to
the metric $\tilde{g}$. This is related to the mean curvature with respect to
$g$ by ${\rm H}={\rm H}_{\tilde{g}}/f+(n-1)\langle\nabla\rho\,|\,\nabla
f\rangle$. Since $\langle\nabla\rho\,|\,\nabla f\rangle$ is bounded in a
neighborhood of $p$, we obtain

${\rm H}=\frac{n-1}{f(p)\rho}+o(1/\rho)\,,\quad\hbox{as }\rho\to 0\,.$

In particular, $f/{\rm H}\to 0$ as $\rho\to 0$. Since $\eta\to 0$ as well by
definition, we have obtained that $f/{\rm H}-\eta/(n-1)\to 0$ when $\rho\to
0$. From the monotonicity of $f/{\rm H}-\eta/(n-1)$ we then deduce
$(n-1)f/{\rm H}\geq\eta$ on the whole $U$. Since $\eta$ is positive on the
whole manifold by construction, the conclusion follows. ∎

A first important observation is that there is a more effective version of the
above inequality. We now argue that the Laplacian comparison obtained in
Theorem 2.5 and Proposition 2.7 gives us a vector with nonnegative divergence,
which is

$X\,=\,\frac{f}{\eta^{n-1}}\nabla\rho\,,$ (2.13)

defined on the open dense set $U=M\setminus({\rm
Cut}^{\tilde{g}}_{p}(M)\cup\\{p\\})$. In fact:

$\displaystyle{\rm div}\,X\,$
$\displaystyle=\,\frac{f}{\eta^{n-1}}\left[\Delta\rho\,+\,\frac{1}{f}\langle\nabla
f\,|\,\nabla\rho\rangle\right]\,-\,(n-1)\frac{f}{\eta^{n}}\langle\nabla\eta\,|\,\nabla\rho\rangle$
(2.14)
$\displaystyle\leq\,(n-1)\frac{f}{\eta^{n}}\,-\,(n-1)\frac{f}{\eta^{n}}\frac{1}{f^{2}}\frac{\partial\eta}{\partial\rho}\,=\,0\,.$
(2.15)

The inequality ${\rm div}\,X\leq 0$ holds then in the classical sense in the
whole $U$. It is crucial that this inequality actually holds in the
distributional sense in the whole manifold.

######  Theorem 2.6.

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\rho$ be the distance function to a point $p$ with respect to the metric
$\tilde{g}=g/f^{2}$ and $\eta$ be the solution to (2.10). Then the vector $X$
defined in (2.13) has nonpositive divergence in the weak sense in the whole
$M\setminus\\{p\\}$. Namely, for every nonnegative test function
$\chi\in\mathscr{C}_{c}^{\infty}(M)$ with $\chi\equiv 0$ in a neighborhood of
$p$, it holds

$\int_{M}\langle X\,|\,\nabla\chi\rangle d\sigma\,\geq\,0\,.$

###### Proof.

Let $\sqrt{g}$ be the volume element of the metric $g$ with respect to the
coordinates $(\rho,\theta)$. We first observe that at all points of
$U=M\setminus({\rm Cut}^{\tilde{g}}_{p}(M)\cup\\{p\\})$ it holds

$\displaystyle\frac{\partial}{\partial\rho}\left(\frac{\sqrt{g}}{f\eta^{n-1}}\right)\,$
$\displaystyle=\,\frac{\partial}{\partial\rho}\left(\frac{f^{n-1}\sqrt{\tilde{g}}}{\eta^{n-1}}\right)$
(2.16) $\displaystyle=\,\left[\frac{f^{n-1}{\rm
H}_{\tilde{g}}}{\eta^{n-1}}\,-\,(n-1)\frac{f^{n+1}}{\eta^{n}}\,+\,(n-1)\frac{f^{n-2}}{\eta^{n-1}}\frac{\partial
f}{\partial\rho}\right]\sqrt{\tilde{g}}$ (2.17)
$\displaystyle=\,\frac{f^{n}}{\eta^{n-1}}\left[{\rm
H}-(n-1)\frac{1}{f^{2}}\frac{\partial
f}{\partial\rho}\,-\,(n-1)\frac{f}{\eta}\,+\,(n-1)\frac{1}{f^{2}}\frac{\partial
f}{\partial\rho}\right]\sqrt{\tilde{g}}$ (2.18) $\displaystyle\leq\,0\,,$
(2.19)

where the last inequality follows from the Laplacian comparison. Recalling
that in polar coordinates $(\rho,\theta)$ the set $U$ is diffeomorphic to the
set of pairs $(\rho,\theta)$ with $\rho<\tau(\theta)$ for a suitable
continuous function $\tau:\mathbb{S}^{n-1}\to(0,+\infty]$, we can then compute

$\displaystyle\int_{M}\langle X\,|\,\nabla\chi\rangle d\mu\,$
$\displaystyle=\,\int_{\mathbb{S}^{n-1}}\int_{0}^{\tau(\theta)}\frac{f}{\eta^{n-1}}\langle\nabla\rho\,|\,\nabla\chi\rangle\sqrt{g}d\rho
d\theta$ (2.20)
$\displaystyle=\,\int_{\mathbb{S}^{n-1}}\int_{0}^{\tau(\theta)}\frac{\partial\chi}{\partial\rho}\frac{\sqrt{g}}{f\eta^{n-1}}d\rho
d\theta$ (2.21)
$\displaystyle=\,-\int_{\mathbb{S}^{n-1}}\\!\int_{0}^{\tau(\theta)}\\!\\!\chi\,\frac{\partial}{\partial\rho}\left(\frac{\sqrt{g}}{f\eta^{n-1}}\right)\\!d\rho
d\theta+\\!\int_{\\{\theta\in\mathbb{S}^{n-1}:\,\tau(\theta)<\infty\\}}\left[\chi\frac{\sqrt{g}}{f\eta^{n-1}}\right]\\!(\tau(\theta),\theta)d\theta$
(2.22)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\lim_{\varepsilon\to
0^{+}}\int_{\mathbb{S}^{n-1}}\left[\chi\frac{\sqrt{g}}{f\eta^{n-1}}\right]\\!(\varepsilon,\theta)d\theta\,.$
(2.23)

In the last identity, the first integral is nonnegative thanks to (2.19)
whereas the second integral is nonnegative by construction. Concerning the
third and final integral, notice that $\eta$ behaves near $p$ as
$f^{2}(p)\rho$ by definition, hence the integral is easily seen to converge to
$\chi(p)|\mathbb{S}^{n-1}|/f^{2n-1}(p)$ as $\varepsilon\to 0$. Since we are
assuming that $\chi$ vanishes in a neighborhood of $p$, the third integral
goes to zero as $\varepsilon\to 0$. The conclusion follows. ∎

###  2.3. Evolution of the mean curvature of hypersurfaces

Let $\Sigma$ be a compact smooth hypersurface. We are interested to the case
where $\Sigma$ is homologous to the boundary: namely, there exists a compact
domain $\Omega$ such that $\partial\Omega=\partial M\sqcup\Sigma$. Let $\rho$
be the $\tilde{g}$-distance from $\Sigma$ in $M\setminus\Omega$. If $\Sigma$
is smooth, then so is $\rho$ in a collar of $\Sigma$. Under the usual
assumption of $\tilde{g}$-geodesic completeness of $M\setminus\partial M$, it
is known from [MM03, Proposition 4.6] that the open set
$U=(M\setminus\Omega)\setminus{\rm Cut}^{\tilde{g}}_{\Sigma}(M)$ of the points
where $\rho$ is smooth is dense in $M\setminus\Omega$. In particular, there is
a function $\tau:\Sigma\to(0,+\infty]$ such that the gradient flow of
$\partial/\partial\rho$ gives a diffeomorphism between $U$ and

$\\{(\rho,x)\,:\,x\in\Sigma\,,\ \rho\in(0,\tau(x))\\}\,.$ (2.24)

It is convenient to estimate the evolution of the geometry of $\Sigma$ with
respect to these coordinates. If ${\rm H}>0$ at the point $x\in\Sigma$, the
evolution of the mean curvature ${\rm H}(0,x)$ is quite similar to the one
described for geodesic spheres in Subsection 2.1 and 2.2. One can then define
the function $\eta\in\mathscr{C}^{\infty}(U)$ as the solution to

$\begin{dcases}\frac{\partial\eta}{\partial\rho}(\rho,x)\,=\,f^{2}(\rho,x)\,,\\\
\eta(0,x)\,=\,(n-1)\frac{f(0,x)}{{\rm H}(0,x)}\,.\end{dcases}$ (2.25)

As in Proposition 2.2, one shows that $\eta$ is well defined and smooth in
$U$. We can then replicate the proof of Theorem 2.5 to find ${\rm
H}/f(\rho,x)\leq(n-1)/\eta(\rho,x)$ for any $0\leq\rho<\tau(x)$. In this case,
the proof is actually even easier, since $f/{\rm H}-\eta/(n-1)$ vanishes at
$x$ by construction, without the need of proving it.

Notice that from (2.8) we can get interesting information on the evolution of
the mean curvature also in the case where ${\rm H}$ is nonpositive at a point
$x\in\Sigma$. If ${\rm H}<0$ at $x$, we deduce from (2.8) that ${\rm H}$ must
remain negative for all times, whereas if ${\rm H}=0$ then ${\rm H}$ remains
nonpositive. Furthermore, even when ${\rm H}$ is nonpositive one can still
define $\eta$ by (2.25) and find $\eta/(n-1)\leq f/{\rm H}<0$ at the points
$(\rho,x)$. From this fact we can obtain even more information under the
assumption that the end is $f$-complete. We recall that an end is $f$-complete
if (1.11) is satisfied, see also Definition 3.1 below for a more extensive
discussion about this notion. The main feature of $f$-complete ends is that
the end is complete with respect to the metric $\tilde{g}$ (so that $\rho$
goes to $+\infty$) and $\eta$ grows to $+\infty$ along the end. Since
$\eta(x)$ is negative when the mean curvature at $x$ is negative, in
particular $\eta$ must reach the value zero, at which point the bound
$\eta/(n-1)\leq f/{\rm H}<0$ fails. This implies that the line
$\rho\mapsto(\rho,x)$ must reach the cut locus before $\eta$ hits zero.
Finally, if ${\rm H}=0$ then from (2.8) we would get that ${\rm H}$ remains
nonpositive. If at some point ${\rm H}$ becomes negative, then the previous
argument applies and the line $\rho\mapsto(\rho,x)$ must reach the cut locus.
Summarizing, we have obtained the following:

######  Proposition 2.7.

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\rho$ be the distance from an hypersurface $\Sigma$ homologous to the
boundary with respect to the metric $\tilde{g}=g/f^{2}$, and let $\eta$ be the
solution to (2.25). Finally, let $x\in\Sigma$ and consider the evolution ${\rm
H}(\rho,x)$ of the mean curvature in the direction of the end.

  * $(i)$

If ${\rm H}(0,x)>0$ then for any $0<\rho<\tau(x)$ it holds

$0\,<\,\frac{{\rm H}}{f}(\rho,x)\,\leq\,\frac{n-1}{\eta(\rho,x)}\,.$ (2.26)

  * $(ii)$

If ${\rm H}(0,x)<0$, then for every $0<\rho<\tau(x)$ it holds

$\frac{{\rm H}}{f}(\rho,x)\,\leq\,\frac{n-1}{\eta(\rho,x)}\,<\,0\,.$ (2.27)

Furthermore, if the ends of $M$ are $f$-complete then $\tau(x)<+\infty$.

  * $(iii)$

If ${\rm H}(0,x)=0$, then ${\rm H}(\rho,x)\leq 0$ for every $0<\rho<\tau(x)$.
Furthermore, if the ends of $M$ are $f$-complete and $\tau(x)=+\infty$, then
${\rm H}(\rho,x)=0$ for all $\rho\geq 0$.

In the following we will focus on the case where the hypersurface $\Sigma$ is
homologous to $\partial M$ and strictly mean-convex, meaning that $\Sigma$ has
positive mean curvature ${\rm H}$ with respect to the normal pointing outside
$\Omega$. In this case, Proposition 2.7-$(i)$ tells us that the bound ${\rm
H}/f\leq(n-1)/\eta$ is in place on the whole $U=M\setminus(\Omega\cup{\rm
Cut}^{\tilde{g}}_{\Sigma})$. Furthermore, the vector field (2.13), that we
recall here for convenience,

$X\,=\,\frac{f}{\eta^{n-1}}\nabla\rho\,,$ (2.28)

is also well defined on $U$. We can now proceed exactly as in the proof of
Theorem 2.6 to show that the vector field $X$ has nonnegative divergence in
the weak sense.

######  Theorem 2.8.

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\Sigma$ be a strictly mean-convex hypersurface homologous to $\partial M$ and
disjoint from it. Suppose that the mean curvature ${\rm H}$ of $\Sigma$ with
respect to the normal pointing towards infinity satisfies ${\rm H}>0$
pointwise. Let $\rho$ be the $\tilde{g}$-distance function from $\Sigma$ and
$\eta$ be the solution to (2.25). Then the vector $X$ defined in (2.28) has
nonpositive divergence in the weak sense in the whole $M\setminus\Omega$.
Namely, for every nonnegative test function
$\chi\in\mathscr{C}_{c}^{\infty}(M\setminus\Omega)$ it holds

$\int_{M\setminus\Omega}\langle X\,|\,\nabla\chi\rangle d\sigma\,\geq\,0\,.$

###  2.4. Growth of weighted areas and volumes

In this subsection, we exploit the monotonicity of the mean curvature of the
level sets to deduce a Bishop–Gromov-type theorem for the behaviour of areas
and volumes of geodesic spheres. We first study the monotonicity of the
following functional

$A(t)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\int_{\\{\rho=t\\}\setminus{\rm
Cut}^{\tilde{g}}}\frac{1}{\eta^{n-1}}d\sigma\,,$ (2.29)

where $\rho$ is the distance function from a point or the signed distance from
a strictly mean-convex hypersurface homologous to the boundary, with respect
to the metric $\tilde{g}$, whereas ${\rm Cut}^{\tilde{g}}$ is the cut locus of
the point/hypersurface with respect to $\tilde{g}$. It is important that we
remove the cut locus from the domain of the integral, as we have observed in
Remark 2.3 that the function $\eta$ is not well defined on it. When $\rho$ is
the distance from a point, the function $A$ can be written in polar
coordinates as

$A(t)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\int_{\\{\theta\in\mathbb{S}^{n-1}\,:\,\tau(\theta)>t\\}}\frac{\sqrt{g}(t,\theta)}{\eta^{n-1}(t,\theta)}d\theta\,,$
(2.30)

where we recall that $\tau(\theta)$ is the minimum value of $\rho$ such that
the point with coordinate $(\rho,\theta)$ belongs to the cut locus. An
analogous definition can of course be given for the distance from an
hypersurface using coordinates (2.24). Notice that $\tau$ is a continuous
function, hence the domain of the integral in (2.30) is measurable, meaning
that the function $A$ is well defined for all $t\in(0,+\infty)$. The domain of
the integral shrinks as $t$ increases, whereas the integrand is positive and
continuous, hence it is easily seen that for all values $a\in(0,+\infty)$ it
holds

$\liminf_{t\to a^{-}}A(t)\,\geq\,A(a)\,\geq\,\limsup_{t\to a^{+}}A(t)\,.$
(2.31)

Furthermore, notice that if the cut locus intersects $\\{\rho=a\\}$ in a set
with positive $\mathscr{H}^{n-1}$-measure, then the first inequality is
strict, that is $\liminf_{t\to a^{-}}A(t)\,\geq\,A(a)$. We are finally ready
to state the first main result of this subsection.

######  Theorem 2.9.

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\rho$ be the $\tilde{g}$-distance function from a point or the signed
$\tilde{g}$-distance function from a strictly mean-convex hypersurface
$\Sigma$ homologous to $\partial M$ and disjoint from it. Let $\eta$ be the
corresponding reparametrized distance, defined by 2.10 or by 2.25, and let
${\rm Cut}^{\tilde{g}}$ be the cut locus of the point/hypersurface. Then the
function

$A(t)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\int_{\\{\rho=t\\}\setminus{\rm
Cut}^{\tilde{g}}}\frac{1}{\eta^{n-1}}d\sigma$

is monotonically nonincreasing.

Furthermore, if $A(t_{1})=A(t_{2})$ for any $0<t_{1}<t_{2}$, then the set
$U=\\{t_{1}\leq\rho\leq t_{2}\\}$ is isometric to $[t_{1},t_{2}]\times\Sigma$
with metric

$g\,=\,f^{2}d\rho\otimes d\rho+\eta^{2}g_{0}\,,$

where $g_{0}$ is a metric on the level set $\Sigma$. In $U$ the functions $f$
and $\eta$ satisfy

$\frac{1}{\eta}\frac{\partial\eta}{\partial\theta^{i}}\,=\,\psi\eta-\varphi\frac{1}{\eta^{n-1}}\,,\qquad\frac{1}{f}\frac{\partial
f}{\partial\theta^{i}}\,=\,\psi\eta+\frac{n-2}{2}\varphi\frac{1}{\eta^{n-1}}\,,$
(2.32)

where $\varphi,\psi$ are independent of $\rho$.

######  Remark 2.10.

If we set $f\equiv 1$ then $\rho$ is the distance function with respect to $g$
and $\eta=\rho$, hence $A(t)=|\\{\rho=t\\}\setminus{\rm
Cut}^{\tilde{g}}|/(|\mathbb{S}^{n-1}|t^{n-1})$ and the above monotonicity
becomes completely analogous to the standard Bishop–Gromov monotonicity of the
areas of geodesic spheres when ${\rm Ric}\geq 0$, which concerns the function
$|\partial\\{\rho\leq t\\}|/(|\mathbb{S}^{n-1}|t^{n-1})$. Clearly, the two
functions coincide almost everywhere, except at the values $t$ such that
$\\{\rho=t\\}\cap{\rm Cut}^{\tilde{g}}$ has nonzero
$\mathscr{H}^{n-1}$-measure. Notice that the number of values for which this
may happen is necessarily countable. A way to see this is to observe that, as
mentioned below formula (2.31), every value such that $\\{\rho=t\\}\cap{\rm
Cut}^{\tilde{g}}$ has nonzero $\mathscr{H}^{n-1}$-measure must correspond to a
jump of $A$, and these are at most countable since $A$ has bounded variation
(this is shown in the proof below).

###### Proof.

If the cut locus were empty, the proof of the monotonicity of $A$ would follow
easily by integrating the inequality ${\rm div}\,X\leq 0$ between two level
sets of $\rho$, where $X$ is the vector field defined in (2.13), and then
applying the Divergence Theorem. In the general case, in order to take into
account the lack of smoothness of $\rho$ at the cut locus, we will need a more
refined analysis, that we now discuss.

Since $\rho$ is a $\tilde{g}$-distance function, in particular it is locally
Lipschitz, hence its gradient is well defined almost everywhere. Furthermore,
as highlighted in [BFM22, Proposition 2.1], $\rho$ being Lipschitz also
implies that the coarea formula can be applied to the level sets of $\rho$. In
particular, for any $0<a<b<+\infty$ we have

$\int_{a}^{b}A(t)dt\,=\,\int_{a}^{b}\left(\int_{\\{\rho=t\\}\setminus{\rm
Cut}^{\tilde{g}}}\frac{1}{\eta^{n-1}}d\sigma\right)dt\,=\,\int_{\\{a<\rho<b\\}}\frac{1}{f\eta^{n-1}}d\mu\,<\,+\infty\,.$

In the last integral we did not have to specify that we are not integrating on
${\rm Cut}^{\tilde{g}}$, since ${\rm Cut}^{\tilde{g}}$ is negligible when
integrating on a volume. The above tells us that $A$ is locally integrable.
Consider now a test function $\chi\in\mathscr{C}^{\infty}_{c}((0,+\infty))$
and let $X$ be the vector field defined by (2.13). We then compute

$\displaystyle\int_{M}\langle\nabla(\chi\circ\rho)\,|\,\,X\rangle\,d\mu$
$\displaystyle=\,\int_{M}\chi^{\prime}(\rho)\langle X\,|\,\nabla\rho\rangle
d\mu$ (2.33)
$\displaystyle=\,\int_{M}\chi^{\prime}(\rho)\frac{f}{\eta^{n-1}}|\nabla\rho|^{2}d\mu$
(2.34)
$\displaystyle=\,\int_{0}^{+\infty}\int_{\\{\rho=t\\}}\chi^{\prime}(t)\frac{f}{\eta^{n-1}}|\nabla\rho|\,d\sigma\,dt$
(2.35)
$\displaystyle=\,|\mathbb{S}^{n-1}|\int_{0}^{+\infty}\chi^{\prime}(t)A(t)\,dt\,.$
(2.36)

On the other hand, Theorem 2.6 (when $\rho$ is the distance from a point) and
Theorem 2.8 tells us that the first integral in the above chain of identities
is nonnegative whenever the test function $\chi$ is nonnegative. More
precisely, from (2.23), since $\chi(0)=0$, we have

$\int_{M}\langle\nabla(\chi\circ\rho)\,|\,X\rangle
d\mu\,=-\int_{\mathbb{S}^{n-1}}\\!\int_{0}^{\tau(\theta)}\\!\\!\\!\chi(\rho)\,\frac{\partial}{\partial\rho}\\!\left(\frac{\sqrt{g}}{f\eta^{n-1}}\right)d\rho
d\theta\\\
\,+\int_{\\{\theta\in\mathbb{S}^{n-1}:\,\tau(\theta)<\infty\\}}\\!\left[\chi\frac{\sqrt{g}}{f\eta^{n-1}}\right]\\!\\!(\tau(\theta),\theta)d\theta\,\geq\,0\,,$
(2.37)

where as usual $\tau(\theta)$ is the minimum value of $\rho$ such that the
point with coordinate $(\rho,\theta)$ belongs to the cut locus. Combining
(2.37) with the above chain of identities we have obtained
$\int_{0}^{+\infty}\chi^{\prime}(t)A(t)dt\geq 0$ for any nonnegative test
function. If we knew $A$ to be weakly differentiable, this would force its
weak derivative to be nonpositive thus proving that $A$ is nonincreasing.
However, we have no information on the regularity of the function $A$ at the
moment. In the following we will show that $A$ has bounded variation, which
will be enough to infer the desired monotonicity.

If we fix $0<a<b<+\infty$, for any $\chi\in\mathscr{C}^{\infty}([a,b])$ with
$\|\chi\|_{\infty}=1$ we have from (2.37) the following bound

$\displaystyle\int_{M}\langle\nabla(\chi\circ\rho)\,|\,X\rangle d\mu\,$
$\displaystyle\leq-\int_{\\{\theta\in\mathbb{S}^{n-1}:\,\tau(\theta)>a\\}}\\!\int_{a}^{\min\\{\tau(\theta),b\\}}\\!\\!\frac{\partial}{\partial\rho}\\!\left(\frac{\sqrt{g}}{f\eta^{n-1}}\right)d\rho
d\theta$ (2.38)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,+\int_{\\{\theta\in\mathbb{S}^{n-1}:\,a\leq\tau(\theta)\leq
b\\}}\\!\left[\frac{\sqrt{g}}{f\eta^{n-1}}\right]\\!\\!(\tau(\theta),\theta)\,d\theta$
(2.39)
$\displaystyle=\,\int_{\\{\theta\in\mathbb{S}^{n-1}:\,\tau(\theta)>a\\}}\left[\frac{\sqrt{g}}{f\eta^{n-1}}\right]\\!\\!(a,\theta)\,d\theta\,-\int_{\\{\theta\in\mathbb{S}^{n-1}:\,\tau(\theta)>b\\}}\left[\frac{\sqrt{g}}{f\eta^{n-1}}\right]\\!\\!(b,\theta)\,d\theta\,.$
(2.40)

In other words, the quantity
$\int_{M}\langle\nabla(\chi\circ\rho)\,|\,X\rangle d\mu$, and thus also
$\int_{0}^{+\infty}\chi^{\prime}(t)A(t)\,dt$, is bounded from above by a
constant that depends on $a$, $b$, $f$ and $g$, but not on $\chi$. It follows
that $A$ has bounded variation in $[a,b]$. As a consequence, the signed finite
Radon measure $\mu$ on $(a,b)$ defined by $\mu((c,d))=\lim_{t\to
d^{-}}A(t)-\lim_{t\to c^{+}}A(t)$ for any $a<c<d<b$, is such that

$\int_{0}^{+\infty}\chi^{\prime}(t)A(t)\,dt\,=\,-\int_{0}^{+\infty}\chi(t)\mu(dt)\,.$

Since we have already shown that
$\int_{0}^{+\infty}\chi^{\prime}(t)A(t)\,dt\geq 0$ for any nonnegative test
function $\chi$, it follows that the measure $\mu$ is nonpositive. From the
definition of $\mu$ and (2.31), we deduce then that the function $A$ is
monotonically nonincreasing in $(a,b)$. Since this should hold for any
$0<a<b<+\infty$, it must necessarily hold on the whole $(0,+\infty)$. This
proves that $A$ is monotonically nonincreasing.

It remains to prove the rigidity statement. If $A(t_{1})=A(t_{2})$, then
thanks to the discussion above it follows $A(t)=A(t_{1})$ for all
$t_{1}<t<t_{2}$. As a consequence, for any test function $\chi$ supported in
$[t_{1},t_{2}]$, we get that the last line in the computation (2.36) vanishes,
that is, $\int_{M}\langle\nabla(\chi\circ\rho)\,|\,\,X\rangle\,d\mu=0$. On the
other hand, this integral can also be computed as in (2.37). From the fact
that $\chi\geq 0$ and from (2.19), we know that the two terms on the right
hand side of (2.37) are both nonnegative, hence they must both vanish for all
$\chi\in\mathscr{C}^{\infty}_{c}([0,+\infty))$. This implies that
$\tau(\theta)$ never belongs to $(t_{1},t_{2})$, meaning that the cut locus
does not intersect $\\{t_{1}<\rho<t_{2}\\}$ and that equality is achieved in
(2.19). In other words, the following holds

$\frac{{\rm H}}{f}\,=\,\Delta\rho+\frac{1}{f}\langle\nabla
f\,|\,\nabla\rho\rangle\,=\,\frac{n-1}{\eta}\qquad\hbox{ in
}\\{t_{1}\leq\rho\leq t_{2}\\}\,.$ (2.41)

This identity in turn triggers the equality in the estimates made in
Subsection 2.1, namely

$\big{|}\widetilde{\nabla}^{2}\rho\big{|}^{2}_{\tilde{g}}\,=\,\frac{(\Delta_{\tilde{g}}\rho)^{2}}{n-1}\,,\qquad{\rm
Ric}\big{(}\widetilde{\nabla}\rho,\widetilde{\nabla}\rho\big{)}\,=\,(n-1)\left[\frac{1}{f}\widetilde{\nabla}^{2}f\big{(}\widetilde{\nabla}\rho,\widetilde{\nabla}\rho\big{)}-\frac{2}{f^{2}}\big{\langle}\widetilde{\nabla}f\,|\,\widetilde{\nabla}\rho\big{\rangle}^{2}\right]\,,$
(2.42)

where we recall that $\widetilde{\nabla}$ is the Levi-Civita connection with
respect to $\tilde{g}$. Notice that, since
$|\widetilde{\nabla}\rho|_{\tilde{g}}=1$, for any vector $X$ it holds

$\widetilde{\nabla}^{2}\rho\big{(}\widetilde{\nabla}\rho,X\big{)}\,=\,\big{\langle}\widetilde{\nabla}|\widetilde{\nabla}\rho|_{\tilde{g}}^{2}\,|\,X\big{\rangle}_{\tilde{g}}\,=\,0\,.$

It follows immediately from this and the first equation in (2.42) that, in the
coordinates in which $\tilde{g}$ has the form (2.1), for any $i,j=1,\dots,n-1$
it holds

$\widetilde{\nabla}^{2}_{ij}\rho\,=\,\frac{\Delta_{\tilde{g}}\rho}{n-1}\tilde{g}_{ij}\,=\,\left(\frac{f^{2}}{\eta}-\frac{1}{f}\frac{\partial
f}{\partial\rho}\right)\tilde{g}_{ij}\,.$

where the latter identity makes use of (2.3). On the other hand, from the
definition of Hessian we have
$\widetilde{\nabla}^{2}_{ij}\rho=-\Gamma_{ij}^{\rho}=\partial_{\rho}\tilde{g}_{ij}/2$,
hence

$\frac{\partial\tilde{g}_{ij}}{\partial\rho}\,=\,2\left(\frac{f^{2}}{\eta}-\frac{1}{f}\frac{\partial
f}{\partial\rho}\right)\tilde{g}_{ij}\,=\,2\frac{\partial}{\partial\rho}\left(\log\eta-\log
f\right)\tilde{g}_{ij}\,.$

This identity can be solved explicitly, yielding

$\tilde{g}_{ij}\,=\,\frac{\eta^{2}}{f^{2}}(g_{0})_{ij}\,,$

where $(g_{0})_{ij}$ does not depend on $\rho$. Comparing with (2.1) and
recalling $g=f^{2}\tilde{g}$, we have obtained

$g\,=\,f^{2}d\rho\otimes d\rho+\eta^{2}g_{0}\,.$ (2.43)

The functions $f$ and $\eta$ may be functions of both the radial coordinate
$\rho$ and of $\\{\theta^{1},\dots,\theta^{n-1}\\}$. Any metric $g$ having the
form (2.43) satisfies the substatic condition with equality in the radial
direction, that is:

$f{\mathrm{R}}_{\rho\rho}\,-\,\nabla^{2}_{\rho\rho}f\,+\,(\Delta
f)g_{\rho\rho}\,=\,0\,.$

From this identity and the substatic condition we find out that, for any
vector $X=\partial/\partial\rho+\lambda\partial/\partial_{i}$, it holds

$\displaystyle 0\,$ $\displaystyle\leq\,\left[f{\rm
Ric}\,-\,\nabla^{2}f-(\Delta f)g\right](X,X)$ (2.44)
$\displaystyle=\,\lambda^{2}\,\left[f{\mathrm{R}}_{ii}\,-\,\nabla^{2}_{ii}f\,+\,(\Delta
f)g_{ii}\right]\,+\,2\lambda\,\left[f{\mathrm{R}}_{i\rho}\,-\,\nabla^{2}_{i\rho}f\,+\,(\Delta
f)g_{i\rho}\right]\,.$ (2.45)

Since this inequality holds pointwise for any $\lambda\in\mathbb{R}$, it
follows that

$f{\mathrm{R}}_{i\rho}\,-\,\nabla^{2}_{i\rho}f\,+\,(\Delta
f)g_{i\rho}\,=\,0\,.$ (2.46)

Recalling the expression (2.43) for the metric, a direct computation gives us
that (2.46) is equivalent to

$(n-2)\partial^{2}_{i\rho}G\,+\,\partial^{2}_{i\rho}F-(n-1)\partial_{\rho}G\partial_{i}F\,=\,0\,,$
(2.47)

where $F=\log f$, $G=\log\eta$. On the other hand, since
$\partial_{\rho}\eta=f^{2}$, we have $\partial_{\rho}G=e^{2F-G}$, from which
we compute

$\partial_{i\rho}^{2}G\,-\,2\partial_{i}F\partial_{\rho}G\,+\,\partial_{i}G\partial_{\rho}G\,=\,0\,.$
(2.48)

We will now combine (2.47) and (2.48) in two different ways.

On the one hand, if we subtract $(n-1)$ times equation (2.48) from equation
(2.47), we obtain

$\partial^{2}_{i\rho}F\,+\,(n-1)\partial_{i}F\partial_{\rho}G\,=\,\partial^{2}_{i\rho}G\,+\,(n-1)\partial_{\rho}G\partial_{i}G\,,$

which can be rewritten as

$\partial_{\rho}\left(e^{(n-1)G}\partial_{i}F\right)\,=\,\partial_{\rho}\left(e^{(n-1)G}\partial_{i}G\right)\,.$

In other words, we have found

$e^{(n-1)G}(\partial_{i}F-\partial_{i}G)=\varphi(\theta)\,,\ \ \forall
i=1,\dots,n-1\,.$ (2.49)

On the other hand, subtracting $(n/2-1)$ times equation (2.48) from (2.47), we
obtain

$\partial^{2}_{i\rho}F\,-\,\partial_{i}F\partial_{\rho}G\,=\,-\frac{n-2}{2}\partial^{2}_{i\rho}G\,+\,\frac{n-2}{2}\partial_{\rho}G\partial_{i}G\,,$

which can be rewritten as

$\partial_{\rho}\left(e^{-G}\partial_{i}F\right)\,=\,-\frac{n-2}{2}\partial_{\rho}\left(e^{-G}\partial_{i}G\right)\,,$

that gives

$e^{-G}\left(\partial_{i}F-\frac{n-2}{2}\partial_{i}G\right)=\psi(\theta)\,,\
\ \forall i=1,\dots,n-1\,.$ (2.50)

Writing (2.49) and (2.50) in terms of $\eta$ and $f$, with straightforward
computations we obtain formulas (2.32). ∎

An immediate consequence of the above result is that we can find a bound for
the area functional $A(t)$ by taking its limit as $t\to 0$. In the case where
$\rho$ is the $\tilde{g}$-distance from a hypersurface, then $A(0)$ is in fact
well defined and we have $A(t)\leq A(0)$. This will be exploited later, in
Subsection 4.2 to prove Theorem D.

We focus now briefly on the case where $\rho$ is the $\tilde{g}$-distance from
a point $x$. As $\rho\to 0$ we have $\partial\eta/\partial\rho=f(x)^{2}+o(1)$,
hence $\eta=f(x)^{2}\rho+o(\rho)$. Furthermore,
$d\sigma=(1+o(1))f(x)^{n-1}d\sigma_{\tilde{g}}=(1+o(1))f(x)^{n-1}\rho^{n-1}d\sigma_{\mathbb{S}^{n-1}}$.
It follows that

$\lim_{t\to 0^{+}}A(t)\,=\,\frac{1}{f(x)^{n-1}}\,.$

As a consequence of the monotonicity of $A$ we then deduce that, for every
$t>0$, it holds

$A(t)\,\leq\,\frac{1}{f(x)^{n-1}}\,.$ (2.51)

If the equality holds, then the rigidity statement in Theorem 2.9 applies in
$\\{\rho\leq t\\}$.

Building on Theorem 2.9, we can also show the following volumetric version of
the Bishop–Gromov monotonicity theorem.

######  Theorem 2.11.

Let $(M,g,f)$ be a substatic triple. Suppose that $M\setminus\partial M$ is
geodesically complete with respect to the metric $\tilde{g}=g/f^{2}$. Let
$\rho$ be the $\tilde{g}$-distance function from a point or the signed
$\tilde{g}$-distance function from a strictly mean-convex hypersurface
$\Sigma$ homologous to $\partial M$ and disjoint from it. Let $\eta$ be the
corresponding reparametrized distance, defined by 2.10 or by 2.25. Then, for
any $k>0$, the function

$V(t)\,=\,\frac{1}{|\mathbb{B}^{n}|t^{k}}\int_{\\{0\leq\rho\leq
t\\}}\frac{\rho^{k-1}}{f\eta^{n-1}}d\mu$ (2.52)

is well defined and monotonically nonincreasing.

Furthermore, if $V(t_{1})=V(t_{2})$ for $0<t_{1}<t_{2}$, then the set
$U=\\{0\leq\rho\leq t_{2}\\}$ is isometric to $[0,t_{2}]\times\Sigma$ with
metric

$g\,=\,f^{2}d\rho\otimes d\rho+\eta^{2}g_{0}\,,$

where $g_{0}$ is a metric on the level sets $\Sigma$. In $U$ the functions $f$
and $\eta$ satisfy

$\frac{1}{\eta}\frac{\partial\eta}{\partial\theta^{i}}\,=\,\psi\eta-\varphi\frac{1}{\eta^{n-1}}\,,\qquad\frac{1}{f}\frac{\partial
f}{\partial\theta^{i}}\,=\,\psi\eta+\frac{n-2}{2}\varphi\frac{1}{\eta^{n-1}}\,,$
(2.53)

where $\varphi,\psi$ are independent of $\rho$. If $\rho$ is the distance from
a point $x$, then $g_{0}=f(x)^{-2}g_{\mathbb{S}^{n-1}}$ and $\varphi=\psi=0$
in the whole $U$, that is, both $f$ and $\eta$ are functions of $\rho$ in $U$.

######  Remark 2.12.

Recall that $\eta$ is smooth outside the cut locus. Since the cut locus has
finite $\mathscr{H}^{n-1}$-measure, the functional $V$ in the statement is
well posed. When $k=n$ we have recovered the standard Bishop–Gromov
monotonicity for volumes of geodesic spheres for ${\rm Ric}\geq 0$. Indeed if
one sets $f=1$ in the statement above, so that $\eta=\rho$, one gets
$V(t)=|\\{\rho\leq t\\}|/(|\mathbb{B}^{n}|t^{n})$.

###### Proof.

We start by observing that the coarea formula (together with the fact that
$|\nabla\rho|=1/f$) gives the following relation between the functionals $A$
and $V$:

$V(t)\,=\,\frac{n}{|\mathbb{S}^{n-1}|t^{k}}\int_{0}^{t}\left(\int_{\\{\rho=\tau\\}}\frac{\tau^{k-1}}{\eta^{n-1}}d\sigma\right)d\tau\,=\,\frac{n}{t^{k}}\int_{0}^{t}\tau^{k-1}A(\tau)d\tau\,.$
(2.54)

From Theorem 2.9, we know that for almost every $\tau\leq t$ the area integral
$A(\tau)$ is well defined and that it is nonincreasing in $\tau$. In the case
where $\rho$ is the distance from a point $x$, we have observed in (2.51) that
$A(\tau)$ is bounded by the constant $1/f(x)^{n-1}$. If instead $\rho$ is the
distance from a strictly mean-convex hypersurface, then $A(0)$ is well defined
and we have $A(\tau)\leq A(0)$. In both cases, it holds
$A(\tau)\leq\mathrm{C}$ for some constant $\mathrm{C}$, hence from (2.54),
recalling $k>0$, we compute

$V(t)\,\leq\,\frac{\mathrm{C}}{t^{k}}\int_{0}^{t}\tau^{k-1}d\tau\,=\,\frac{\mathrm{C}}{k}\,.$

As a consequence, $V(t)$ is well defined. Furthermore, the monotonicity of $A$
also implies $A(t)\leq A(\tau).$ Plugging this information in (2.54) gives

$V(t)\,\geq\,\frac{n}{t^{k}}A(t)\int_{0}^{t}\tau^{k-1}d\tau\,=\,\frac{n}{k}A(t)\,.$
(2.55)

With this information at hand, we are ready to compute the derivative of
$V(t)$:

$\displaystyle V^{\prime}(t)\,$ $\displaystyle=\,\lim_{\varepsilon\to
0}[V(t+\varepsilon)-V(t)]/\varepsilon$ (2.56)
$\displaystyle=\,\lim_{\varepsilon\to
0}\frac{1}{\varepsilon|\mathbb{B}^{n}|}\left(\frac{1}{(t+\varepsilon)^{k}}-\frac{1}{t^{k}}\right)\int_{\\{\rho\leq
t+\varepsilon\\}}\frac{\rho^{k-1}}{f\eta^{n-1}}d\mu\,+\,\lim_{\varepsilon\to
0}\frac{1}{t^{k}}\frac{1}{\varepsilon|\mathbb{B}^{n}|}\int_{\\{t\leq\rho\leq
t+\varepsilon\\}}\frac{\rho^{k-1}}{f\eta^{n-1}}d\mu$ (2.57)
$\displaystyle=\,\lim_{\varepsilon\to
0}\frac{t^{k}-(t+\varepsilon)^{k}}{\varepsilon
t^{k}}V(t+\varepsilon)\,+\,\lim_{\varepsilon\to
0}\frac{n}{t^{k}}\frac{1}{\varepsilon}\int_{t}^{t+\varepsilon}\tau^{k-1}A(\tau)d\tau$
(2.58) $\displaystyle=\,-\frac{k}{t}V(t)+\frac{n}{t}A(t)$ (2.59)
$\displaystyle\leq\,0\,.$ (2.60)

We now prove the rigidity statement. If $V(t_{1})=V(t_{2})$ for two values
$0<t_{1}<t_{2}$, then retracing our computations we find out that
$A(\tau)=A(t)$ for all $0<\tau<t_{2}$. From the rigidity statement of Theorem
2.9 we then deduce that in $\\{0\leq\rho\leq t\\}$ the metric writes as

$g\,=\,f^{2}d\rho\otimes d\rho+\eta^{2}g_{0}\,,$

and $f,\eta$ satisfy formulas (2.53).

Finally, we suppose now that $\rho$ is the distance from a point $x$ and we
prove that $f,\eta$ must necessarily depend on $\rho$ only. To this end,
notice that formulas (2.53) must hold up to $\rho=0$ (that is, up to the point
$x$), hence at the limit as $\rho$ goes to zero, the derivative $\partial
f/\partial\theta^{i}$ goes to zero. Since $\eta$ goes to $0$ as $\rho\to 0$,
it follows then from the second formula in (2.53) that $\varphi$ must vanish
identically.

As a consequence, the first formula in (2.53) can be rewritten as

$\frac{1}{\eta^{2}}\frac{\partial\eta}{\partial\theta^{i}}\,=\,\psi\,.$ (2.61)

In particular, taking the derivative with respect to $\rho$ we deduce that
$\partial^{2}(1/\eta)/\partial\rho\partial\theta=0$. In other words,
$1/\eta=\alpha+\beta$, where $\alpha$ is a function of $\rho$ and $\beta$ is a
function of the $\theta^{i}$’s. Substituting this expression for $\eta$ in
(2.61), we deduce that

$\frac{\partial\beta}{\partial\theta^{i}}\,=\,-\psi$

On the other hand, taking the difference of the two formulas in (2.53) we have

$\frac{\partial}{\partial\theta^{i}}\left(\log\eta-\log f\right)\,=\,0\,.$

In other words, the quantity $\eta/f$ must be a function of $\rho$. Recalling
the decomposition $1/\eta=\alpha+\beta$ shown right above, it follows then
that $1/f=\lambda(\alpha+\beta)$, where $\lambda$ is a function of $\rho$.

Notice now that, when $\rho$ goes to zero, the limit of $f$ must go to the
value of $f$ at the point $x$, so that in particular the limit of $1/f$ as
$\rho\to 0$ must not depend on $\theta^{i}$. It follows that $\beta=0$, hence
$\psi=-\partial\beta/\partial\theta^{i}$ vanishes as well. We have proved that
both $\varphi$ and $\psi$ in (2.53) vanish, hence both $f$ and $\eta$ must be
functions of the sole $\rho$ in the whole $\\{\rho\leq t\\}$. Since the metric

$\tilde{g}=d\rho\otimes d\rho+\frac{\eta^{2}}{f^{2}}g_{0}$

is smooth at the point $x$, it follows that $(\eta^{2}/f^{2})g_{0}$ should be
close to $\rho^{2}g_{\mathbb{S}^{n-1}}$ near $x$. From the definition of
$\eta$ it follows $\eta=f(x)^{2}\rho+o(\rho)$ close to $x$, hence
$g_{0}=f(x)^{-2}g_{\mathbb{S}^{n-1}}$ and we conclude the rigidity statement.
∎

##  3\. Wylie’s Splitting Theorem for substatic manifolds

###  3.1. $f$-complete and conformally compact ends

From now on we will study noncompact manifolds with some special behaviour at
infinity, focusing mainly on $f$-complete ends.

######  Definition 3.1.

We say that an end is $f$-complete if for any $g$-unit speed curve
$\gamma:[0,+\infty)\to M$ going to infinity along the end it holds

$\lim_{t\to+\infty}\rho(\gamma(t))\,=\,+\infty\,,\qquad\int_{0}^{+\infty}f(\gamma(t))dt\,=\,+\infty\,,$
(3.1)

where $\rho$ is the distance from a point with respect to $\tilde{g}=g/f^{2}$.

It is clear from the triangle inequality that the definition above does not
depend on the point we are taking the distance $\rho$ from. It also would not
change if we replace the distance from a point with the distance from a
hypersurface.

For all the arguments that follows it would actually be enough to require
(3.1) only along $\tilde{g}$-geodesics. More precisely, it is enough to
require the end to be $\tilde{g}$-complete and satisfying the second condition
in (3.1) along any $\tilde{g}$-geodesic. In fact, the above definition is
analogous to the one given in [Wyl17, Definition 6.2] in the ${\rm CD}(0,1)$
framework: there, a triple $(M,\tilde{g},\psi)$ satisfying the ${\rm CD}(0,1)$
condition is said to be $\psi$-complete if for any $\tilde{g}$-geodesic
$\sigma:[0,+\infty)\to M$ going to infinity along the end it holds

$\int_{0}^{+\infty}e^{-\frac{2\psi(\sigma(t))}{n-1}}dt\,=\,+\infty\,.$

Recalling the relations $\tilde{g}=g/f^{2}$ and $\psi=-(n-1)\log f$ between
the ${\rm CD}(0,1)$ and substatic setting (see Section A.2), it is easily seen
that this integrability condition is equivalent to the second requirement in
(3.1). As already observed in [Wyl17], this integrability condition can be
interpreted as completeness with respect to the metric $f^{2}g$ or,
alternatively, as completeness with respect to the weighted connection
introduced in [WY16] and [LX17] (see Section A.3). For what concerns this
paper however, the only relevance of the second condition in (3.1) is that it
implies that the reparametrized distance $\eta$ defined in Section 2 goes to
infinity along the end. This is easy to show as follows. Let $\rho$ be the
$\tilde{g}$-distance to a point or hypersurface and let $\eta$ be defined by
(2.10) or (2.25) depending on whether we are taking the distance from a point
or hypersurface. Let $\sigma:[0,+\infty)\to M$ be a $\tilde{g}$-geodesic with
$\dot{\sigma}=\widetilde{\nabla}\rho$ and let $\gamma:[0,+\infty)\to M$ be the
reparametrization of $\sigma$ that has $g$-length constant and equal to $1$.
We then have

$\eta(\gamma(t))-\eta(\gamma(0))\,=\,\int_{0}^{t}f^{2}(\gamma(t))|\dot{\gamma}(t)|_{\tilde{g}}dt\,=\,\int_{0}^{t}f(\gamma(t))|\dot{\gamma}(t)|_{g}dt\,=\,\int_{0}^{t}f(\gamma(t))dt\,,$

hence if the second condition in (3.1) holds then $\eta$ goes to $+\infty$.

The family of $f$-complete ends includes a number of interesting examples.
Most notably, asymptotically flat ends are $f$-complete. We say that $(M,g,f)$
is asymptotically flat if there exists a compact set $K$ such that $M\setminus
K$ is diffeomorphic to $\mathbb{R}^{n}$ minus a ball, the metric $g$ converges
to the Euclidean metric and $f$ goes to $1$ at infinity along the end. A
precise definition of asymptotic flatness is given below, see Definition 4.8.
A notable example of asymptotically flat substatic solution is the
Reissner–Nordström solution, corresponding to (1.2) with $\Lambda=0$. In fact,
the family of $f$-complete ends is quite more general: for instance, it is
sufficient to require a suitable behaviour of $f$ at infinity, without any
assumption on the topology and geometry of the end, as clarified by the
following proposition.

######  Proposition 3.2.

Let $(M,g,f)$ be a substatic triple and let $r$ be the $g$-distance from a
point (or more generally from a compact domain). If there exist a compact set
$K\supset\partial M$ and constants $0<c<C$, $0<k<1$ such that

$cr^{-k}<f<Cr^{k}$ (3.2)

at all points in $M\setminus K$, then all ends are $f$-complete.

###### Proof.

Let $\gamma:[0,+\infty)\to M$ be a $g$-unit speed curve going to infinity
along the end and let $\delta$ be the $g$-distance between $\gamma(0)$ and the
point $p$ we are taking the distance $r$ from (if $r$ is the distance from a
compact domain instead, it is sufficient to choose $\delta$ as the maximum
distance between $\gamma(0)$ and the points of the domain; the rest of the
proof is easy to adapt). It is also convenient to assume that $\gamma(0)$ and
$p$ belong to $K$ and that $K$ is geodesically convex with respect to the
metric $\tilde{g}=g/f^{2}$ (this can of course always be achieved by possibly
enlarging $K$).

Since $\gamma$ has unit speed, we have $d(\gamma(0),\gamma(t))<t$, hence by
triangle inequality

$r(\gamma(t))<t+\delta\,.$

If we then denote by $T$ the maximum value of $t$ such that $\gamma(t)\in K$,
estimate (3.2) tells us that for any $t>T$ it holds

$c\,(t+\delta)^{-k}\,<\,f(\gamma(t))\,<\,C\,(t+\delta)^{k}\,.$

In particular, since $0<k<1$ we have

$\int_{T}^{+\infty}f(\gamma(t))dt\,>\,c\,\int_{T}^{+\infty}(t+\delta)^{-k}dt\,>\,c\lim_{t\to+\infty}\frac{(t+\delta)^{1-k}-(T+\delta)^{1-k}}{1-k}\,=\,+\infty\,.$

To conclude, it is sufficient to show that the $\tilde{g}$-distance $\rho$ of
$\gamma(t)$ from a fixed point (we will take $\gamma(0)$ for simplicity) also
goes to $+\infty$ as $t\to+\infty$. For any fixed $t>0$, let $\sigma_{t}$ be
the unit-speed $\tilde{g}$-geodesic from $\gamma(0)$ to $\gamma(t)$. We
reparametrize $\sigma_{t}$ so that it has speed $1$ with respect to the metric
$g$. With a slight abuse of notation, we still denote by $\sigma_{t}$ the
reparametrized curve. We will have $\sigma_{t}:[0,\tau]\to M$, with $\tau\geq
t$. Since we have chosen $K$ to be $\tilde{g}$-geodesically convex, there will
exist a value $T_{t}$ such that $\sigma_{t}(s)\in K$ for all $s\leq T_{t}$ and
$\sigma_{t}(s)\not\in K$ for all $s>T_{t}$. Clearly $\tau-T_{t}>{\rm
d}_{g}(K,\gamma(t))$. Furthermore, since $\sigma_{t}$ restricted to
$[0,T_{t}]$ is $\tilde{g}$-minimizing and both $\sigma_{t}(0)=\gamma(0)$ and
$\sigma_{t}(T_{t})$ belong to $K$, we have

${\rm
diam}_{\tilde{g}}(K)\,>\,\int_{0}^{T_{t}}|\dot{\sigma}_{t}(s)|_{\tilde{g}}ds\,=\,\int_{0}^{T_{t}}\frac{1}{f(\sigma_{t}(s))}ds>\frac{T_{t}}{\max_{K}f}\,.$

We are now ready to estimate the $\tilde{g}$-distance as follows:

$\rho(\gamma(t))=\int_{0}^{\tau}|\dot{\sigma}_{t}(s)|_{\tilde{g}}ds\,=\,\int_{0}^{\tau}\frac{1}{f(\sigma_{t}(s))}|\dot{\sigma}_{t}(s)|_{g}ds\,=\,\int_{0}^{\tau}\frac{1}{f(\sigma_{t}(s))}ds\,>\,\int_{T_{t}}^{\tau}\frac{r(\sigma_{t}(s))^{-k}}{\sqrt{C}}ds\,.$

Since $r(\sigma_{t}(s))<{\rm d}_{g}(\sigma_{t}(s),\gamma(0))+\delta<s+\delta$,
we then have

$\displaystyle\rho(\gamma(t))$
$\displaystyle>\frac{(\tau+\delta)^{1-k}-(T_{t}+\delta)^{1-k}}{\sqrt{C}(1-k)}\,$
(3.3) $\displaystyle>\,\frac{(T_{t}+\delta+{\rm
d}_{g}(\gamma(t),K))^{1-k}-(T_{t}+\delta)^{1-k}}{\sqrt{C}(1-k)}$ (3.4)
$\displaystyle>\,\frac{\left({\rm d}_{g}(\gamma(t),K)\right)^{1-k}-\left({\rm
diam}_{\tilde{g}}(K)\max_{K}f\right)^{1-k}}{\sqrt{C}(1-k)}\,.$ (3.5)

Since $K$ is fixed, the distance ${\rm d}_{g}(\gamma(t),K)$ is going to
$+\infty$, hence $\rho(\gamma(t))\to+\infty$ as $t\to+\infty$ as wished. ∎

Another well studied family of ends is the following:

######  Definition 3.3.

We say that an end of a substatic triple $(M,g,f)$ is conformally compact if a
neighborhood $E$ of it is the interior of a compact manifold $\overline{E}$
with boundary $\partial\overline{E}$ and the metric $\tilde{g}=g/f^{2}$
extends to the boundary of $\overline{E}$ in $\mathscr{C}^{3}$-fashion. We
denote by $\partial E_{\infty}=\partial\overline{E}\setminus M$ the conformal
boundary of the end. Finally, we require $f^{-1}$ to extend to a
$\mathscr{C}^{3}$-function on $\overline{E}$ in such a way that $f^{-1}=0$ and
$d(f^{-1})\neq 0$ on $\partial E_{\infty}$.

It is clear that on conformally compact ends the $\tilde{g}$-distance function
$\rho$ does not grow to infinity along the end. On the other hand, it is
easily seen that $\eta$ goes to infinity along any ray going into a
conformally compact end. An example of this behaviour is given by the
Schwarzschild-Anti de Sitter solution.

The two families of $f$-complete ends and conformally compact ends enclose the
model solutions we are interested in. Furthermore, following [Wyl17], we can
prove a splitting theorem for both these types of ends. The proof, given
below, makes substantial use of the conformal metric $\tilde{g}$. It is then
convenient to remark that our manifold remains complete with respect to this
conformal metric.

######  Lemma 3.4.

Let $(M,g,f)$ be a substatic triple with ends that are either $f$-complete or
conformally compact and let $\partial M_{\infty}$ be the (possibly empty)
conformal infinity (namely, the union of the conformal infinities of the
conformally compact ends). Then the manifold $(M\cup\partial
M_{\infty})\setminus\partial M$ is complete with respect to the metric
$\tilde{g}=g/f^{2}$.

######  Remark 3.5.

We specify that we are referring here to completeness as a metric space, not
to geodesic completeness. Clearly geodesic completeness fails in presence of a
conformal boundary, since $\tilde{g}$-geodesics may end at $\partial
M_{\infty}$.

###### Proof.

The fact that the manifold is complete near the conformal boundary is
immediate from the definition of conformally compact ends. The
$\tilde{g}$-completeness of $f$-complete ends has already been discussed after
Definition 3.1. It remains to prove $\tilde{g}$-completeness near $\partial
M$. To do this, we show that the boundary components become ends with respect
to the conformal metric $\tilde{g}$. Let $\gamma:[0,\ell]\to\mathbb{R}$ be a
$g$-unit speed curve with $\gamma(0)\in\partial M$ and $\gamma(t)$ in the
interior of $M$ for any $t>0$. It is enough to show that its
$\tilde{g}$-length is infinite. From the mean value theorem we know that for
any $t\in(0,\ell]$ there exists $\xi\in(0,t)$ such that

$f(\gamma(t))\,=\,f(\gamma(t))-f(\gamma(0))\,=\,\langle\nabla
f(\gamma(\xi))\,|\,\dot{\gamma}(t)\rangle\,t\,\leq\,|\nabla
f|(\gamma(\xi))\,t\,.$

If $K$ is a compact collar neighborhood of $\partial M$ containing $\gamma$,
we then compute

$\int_{0}^{\ell}|\dot{\gamma}(t)|_{\tilde{g}}dt\,=\,\int_{0}^{\ell}\frac{dt}{f(\gamma(t))}\,\geq\,\int_{0}^{\ell}\frac{1}{\left(\max_{K}|\nabla
f|\right)t}dt=+\infty\,,$

as wished. ∎

###  3.2. Splitting Theorem for $f$-complete ends

In [Wyl17], the author proves a Splitting Theorem in the ${\rm CD}(0,1)$
framework. Here we translate this result in the substatic setting, obtaining
Theorem C. The proof is essentially the one in [Wyl17], but we prefer to show
it for completeness. The strategy is that of exploiting the Laplacian
Comparison given by Theorem 2.5 together with standard techniques for the
Busemann function, to be coupled with a refined splitting argument.

###### Proof of Theorem C.

The first part of the proof follows quite closely the usual proof of the
classical Splitting Theorem for manifolds with nonpositive Ricci curvature, so
we avoid to give all the technical details, that can be found in any standard
Riemannian geometry book (see for instance [Pet16, Theorem 7.3.5]).

We are assuming that there are at least two $f$-complete ends and we know from
Lemma 3.4 that $M\setminus\partial M$ is complete with respect to $\tilde{g}$,
hence we can take points arbitrarily far away in the two ends and connect them
via a $\tilde{g}$-minimizing geodesic. At the limit, we thus produce a
globally minimizing geodesic $\sigma:[-\infty,+\infty]$ going from one end to
the other. For a given $t\in\mathbb{R}$ we then consider the functions

$\beta_{t}(x)={\rm d}_{\tilde{g}}(x,\sigma(t))-t$

and the Busemann functions

$\beta_{+}(x)=\lim_{t\to+\infty}\beta_{t}(x)\,,\qquad\beta_{-}(x)=\lim_{t\to-\infty}\beta_{t}(x)\,.$

Under the assumption of $f$-completeness, we know that the
$\tilde{g}$-distance goes to infinity as we approach the end, hence the limits
are well defined. Theorem 2.5 tells us that
$\Delta\beta_{t}+(1/f)\langle\nabla f\,|\,\nabla\beta_{t}\rangle\leq 1/\eta$
for every $t$. Again using the fact that we are assuming $f$-completeness of
the ends, we have $\eta\to+\infty$ at infinity. Standard arguments then tell
us that $\beta_{\pm}$ satisfy the inequality

$\Delta\beta_{\pm}+\frac{1}{f}\langle\nabla f\,|\,\nabla\beta_{\pm}\rangle\leq
0$

in the barrier sense. In particular, $\beta_{-}+\beta_{+}$ satisfies the same
elliptic inequality. Furthermore $\beta_{-}+\beta_{+}=0$ on $\sigma$ by
construction and a simple application of the triangle inequality shows that
$\beta_{-}+\beta_{+}\geq 0$ on the whole $M$. In follows then by maximum
principle that $\beta_{+}+\beta_{-}=0$. We then conclude that
$\beta=\beta_{+}$ solves

$\Delta\beta+\frac{1}{f}\langle\nabla f\,|\,\nabla\beta\rangle\,=\,0$ (3.6)

in the barrier sense. Standard regularity theory tells us that $\beta$ is in
fact smooth, hence solves (3.6) in the classical sense as well.

We denote by $\widetilde{\nabla}$ the Levi-Civita connection with respect to
the metric $\tilde{g}$. Our next step is to prove that
$\widetilde{\nabla}\beta$ is in fact the splitting direction. We first exploit
(3.6) to compute

$\Delta_{\tilde{g}}\beta\,=\,-(n-1)\frac{1}{f}\big{\langle}\widetilde{\nabla}f\,\big{|}\,\widetilde{\nabla}\beta\big{\rangle}_{\tilde{g}}$

Furthermore, rewriting the substatic condition (1.1) in terms of the new
metric, we find

${\rm
Ric}_{\tilde{g}}\,\geq\,(n-1)\frac{1}{f}\widetilde{\nabla}^{2}f-2(n-1)\frac{1}{f^{2}}df\otimes
df$

The above formulas can be applied in combination with the Bochner formula to
obtain

$\displaystyle\Delta_{\tilde{g}}\big{|}\widetilde{\nabla}\beta\big{|}_{\tilde{g}}^{2}\,$
$\displaystyle=\,\big{|}\widetilde{\nabla}^{2}\beta\big{|}_{\tilde{g}}^{2}+2{\rm
Ric}_{\tilde{g}}\big{(}\widetilde{\nabla}\beta,\widetilde{\nabla}\beta\big{)}+2\big{\langle}\widetilde{\nabla}\Delta_{\tilde{g}}\beta\,\big{|}\,\widetilde{\nabla}\beta\big{\rangle}_{\tilde{g}}$
(3.7)
$\displaystyle\geq\,2\,\left[\big{|}\widetilde{\nabla}^{2}\beta\big{|}_{\tilde{g}}^{2}-\frac{(\Delta_{\tilde{g}}\beta)^{2}}{n-1}\right]-(n-1)\frac{1}{f}\big{\langle}\widetilde{\nabla}\big{|}\widetilde{\nabla}\beta\big{|}_{\tilde{g}}^{2}\,\big{|}\,\widetilde{\nabla}f\big{\rangle}_{\tilde{g}}$
(3.8)

A standard estimate for the Hessian tells us that

$\big{|}\widetilde{\nabla}^{2}\beta\big{|}_{\tilde{g}}^{2}-\frac{(\Delta_{\tilde{g}}\beta)^{2}}{n-1}\,\geq\,-\frac{1}{n-1}\frac{\Delta_{\tilde{g}}\beta}{|\widetilde{\nabla}\beta|_{\tilde{g}}^{2}}\big{\langle}\widetilde{\nabla}\big{|}\widetilde{\nabla}\beta\big{|}_{\tilde{g}}^{2}\,\big{|}\,\widetilde{\nabla}\beta\big{\rangle}_{\tilde{g}}\,.$

Substituting this in the previous inequality, we get

$\Delta_{\tilde{g}}\big{|}\widetilde{\nabla}\beta\big{|}_{\tilde{g}}^{2}\,\geq\,\bigg{\langle}\widetilde{\nabla}\big{|}\widetilde{\nabla}\beta\big{|}_{\tilde{g}}^{2}\,\bigg{|}\,\frac{2}{f}\frac{\langle\widetilde{\nabla}f\,|\,\widetilde{\nabla}\beta\rangle_{\tilde{g}}}{|\widetilde{\nabla}\beta|_{\tilde{g}}^{2}}\widetilde{\nabla}\beta+(n-1)\frac{1}{f}\widetilde{\nabla}f\bigg{\rangle}_{\tilde{g}}$

By Cauchy–Schwarz the vector on the right-hand side is bounded on any compact
set. In particular, $|\widetilde{\nabla}\beta|_{\tilde{g}}^{2}$ satisfies the
maximum principle. To conclude from this that
$|\widetilde{\nabla}\beta|_{\tilde{g}}$ is constant, we first observe via
triangle inequality that $\beta(x)-\beta(y)\leq{\rm d}_{\tilde{g}}(x,y)$ for
any two points $x$, $y$. This immediately implies that
$|\widetilde{\nabla}\beta|_{\tilde{g}}\leq 1$ on $M$. On the other hand, since
$\beta(\sigma(\tau))=\tau$, we conclude that
$|\widetilde{\nabla}\beta|_{\tilde{g}}=1$ on $\sigma$. The strong maximum
principle then implies that $|\widetilde{\nabla}\beta|_{\tilde{g}}=1$ on the
whole manifold. In particular, the previous inequalities must be equalities,
namely

$\displaystyle{\rm
Ric}_{\tilde{g}}\big{(}\widetilde{\nabla}\beta,\widetilde{\nabla}\beta\big{)}\,$
$\displaystyle=\,(n-1)\frac{1}{f}\widetilde{\nabla}^{2}f\big{(}\widetilde{\nabla}\beta,\widetilde{\nabla}\beta\big{)}-2(n-1)\frac{1}{f^{2}}\big{\langle}\widetilde{\nabla}f\,\big{|}\,\widetilde{\nabla}\beta\big{\rangle}_{\tilde{g}}\,,$
(3.9)
$\displaystyle\big{|}\widetilde{\nabla}^{2}\beta\big{|}_{\tilde{g}}^{2}\,$
$\displaystyle=\,\frac{(\Delta_{\tilde{g}}\beta)^{2}}{n-1}\,.$ (3.10)

Furthermore, the fact that $|\widetilde{\nabla}\beta|_{\tilde{g}}=1$ on the
whole manifold grants us that we can use $\beta$ as a coordinate and that the
manifold $M$ is diffeomorphic to $\mathbb{R}\times\Sigma$, for some
$(n-1)$-dimensional manifold $\Sigma$. With respect to coordinates
$\\{\beta,\theta^{1},\dots,\theta^{n-1}\\}$, the conformal metric writes as

$\tilde{g}\,=\,d\beta\otimes d\beta+\tilde{g}_{ij}d\theta^{i}\otimes
d\theta^{j}\,.$

Again, since $|\widetilde{\nabla}\beta|_{\tilde{g}}=1$, for any vector $X$ it
holds

$\widetilde{\nabla}^{2}\beta\big{(}\widetilde{\nabla}\beta,X\big{)}\,=\,\big{\langle}\widetilde{\nabla}\big{|}\widetilde{\nabla}\beta\big{|}_{\tilde{g}}^{2}\,\big{|}\,X\big{\rangle}_{\tilde{g}}\,=\,0\,.$

It follows immediately from this and the identity
$|\widetilde{\nabla}^{2}\beta|_{\tilde{g}}^{2}=(\Delta_{\tilde{g}}\beta)^{2}/(n-1)$
that, in the coordinates in which $\tilde{g}$ has the form (2.1), for any
$i,j=1,\dots,n-1$ it holds

$\widetilde{\nabla}^{2}_{ij}\beta\,=\,\frac{\Delta_{\tilde{g}}\beta}{n-1}\tilde{g}_{ij}\,=\,-\frac{1}{f}\frac{\partial
f}{\partial\beta}\tilde{g}_{ij}\,,$

where the latter identity makes use of (2.3). On the other hand, from the
definition of Hessian we have
$\widetilde{\nabla}^{2}_{ij}\beta=-\Gamma_{ij}^{\beta}=\partial_{\beta}\tilde{g}_{ij}/2$,
hence

$\frac{\partial\tilde{g}_{ij}}{\partial\beta}\,=\,-\frac{2}{f}\frac{\partial
f}{\partial\beta}\tilde{g}_{ij}\,.$

This identity can be solved explicitly, yielding

$\tilde{g}_{ij}\,=\,\frac{1}{f^{2}}(g_{0})_{ij}\,,$

where $(g_{0})_{ij}$ does not depend on $\rho$. Comparing with (2.1) and
recalling $g=f^{2}\tilde{g}$, we have obtained

$g\,=\,f^{2}d\beta\otimes d\beta+g_{0}\,.$

Finally, we remark again that in this proof we did not have to ask for the
boundary of $M$ to be empty since, as observed in Lemma 3.4, with respect to
the conformal metric $\tilde{g}$ the boundary components become ends, hence
they cannot obstruct minimizing geodesics. Therefore, the argument to produce
a line between two $f$-complete ends goes through and the manifold splits. But
then this would imply $\partial M=(-\infty,+\infty)\times\partial\Sigma$,
which contradicts our initial assumption that the boundary is compact. It
follows that the boundary must be empty if there is more than one $f$-complete
end. ∎

######  Remark 3.6.

We point out that it is actually possible to obtain a stronger thesis in
Theorem C above. In fact, proceeding as in the proof of Theorem 2.9, one can
also show that identity (2.46) is in force, and from there deduce that
$f=f_{1}f_{2}$, where $f_{1}$ is a function of $s$ whereas $f_{2}$ does not
depend on $s$. We do not give the details on this computation, which has
already been performed in the conformal ${\rm CD}(0,1)$ framework [Wyl17,
Proposition 2.2]. Recalling the relation between ${\rm CD}(0,1)$ and substatic
discussed in Section A.2, one can easily translate this result in our setting.
One may also write down explicitly the substatic condition in the directions
tangential to the cross section, to obtain some information on the triple
$(\Sigma,g_{0},f_{2})$. Again, in the ${\rm CD}(0,1)$ setting, this has been
done in [Wyl17, Proposition 2.3], where it is shown that the triple
$(\Sigma,g_{0},-(n-1)\log f_{2})$ satisfies the ${\rm CD(0,1)}$ condition (in
fact, it is even ${\rm CD}(0,0)$). It is not immediately clear whether this
fact translates nicely in our setting. These refinements of the thesis of
Theorem C will not be needed in the rest of the paper.

###  3.3. Splitting theorem for conformally compact ends

We now discuss conformally compact ends. For such ends, by definition the
metric extends to the conformal infinity sufficiently smoothly so that the
mean curvature ${\rm H}_{\tilde{g}}$ of the conformal infinity $\partial
E_{\infty}$ is well defined. On the other hand, the mean curvatures ${\rm H}$
and ${\rm H}_{\tilde{g}}$ of a hypersurface with respect to the two different
metrics can be seen to be related by

${\rm H}_{\tilde{g}}\,=\,f{\rm H}-(n-1)\langle\nabla f\,|\,\nu\rangle\,.$

Alternatively, setting $\varphi=1/f$ we can write

${\rm H}\,=\,\varphi{\rm
H}_{\tilde{g}}-(n-1)\big{\langle}\widetilde{\nabla}\varphi\,\big{|}\,\nu_{\tilde{g}}\big{\rangle}_{\tilde{g}}\,.$

By definition of conformal compactness we know that $\varphi$ extend in a
$\mathscr{C}^{3}$ fashion to the conformal boundary by setting $\varphi=0$ on
$\partial E_{\infty}$. In particular $|\widetilde{\nabla}\varphi|_{\tilde{g}}$
is bounded, which implies that the quantity ${\rm H}/f$ can be extended to
zero in a continuous fashion on $\partial E_{\infty}$. Taking now as $\rho$
the $\tilde{g}$-distance from $\partial E_{\infty}$, recalling the Riccati
equation (2.8) we have that

$\frac{\partial}{\partial\rho}\left(\frac{{\rm
H}}{f}\right)\,\leq\,-\frac{1}{n-1}\,{\rm H}^{2}\,,\qquad{\frac{{\rm
H}}{f}}_{|_{\rho=0}}=0\,.$

The assumption of $\mathscr{C}^{3}$-regularity of the conformal boundary made
in Definition 3.3 was needed precisely to make sense of the $\rho$-derivative
of ${\rm H}$. Proceeding exactly as in Subsection 2.3, from this formula on
the evolution of the mean curvature we obtain the Laplacian comparison

$\frac{{\rm H}}{f}\,=\,\Delta\rho+\frac{1}{f}\langle\nabla
f\,|\,\nabla\rho\rangle\,\leq\,0\,.$ (3.11)

This is the main ingredient to prove the Splitting Theorem in the conformally
compact setting:

######  Theorem 3.7.

Let $(M,g,f)$ be a substatic triple with conformally compact ends. Then there
is at most one end.

###### Proof.

Again, a ${\rm CD}(0,1)$-version of this argument can be found in [Wyl17,
Theorem 5.1]. The proof follows closely the one in [Kas83, Theorem B-(1)],
where a splitting theorem for compact manifolds is discussed. Suppose by
contradiction that the conformal infinity has at least two connected
components. Let $S_{-},S_{+}$ be the two components with least distance. Then
there exists a $\tilde{g}$-geodesic $\sigma$ minimizing the distance between
them.

Let $\beta_{-}$ (resp. $\beta_{+}$) be the distance from $S_{-}$ (resp.
$S_{+}$) with respect to $\tilde{g}$. The discussion above grants us that both
$\beta_{-}$ and $\beta_{+}$ satisfy the Laplacian comparison (3.11). In
particular, so does $\beta_{-}+\beta_{+}$. Since by construction
$\beta_{-}+\beta_{+}$ reaches its minimum value ${\rm dist}(S_{-},S_{+})$ on
the geodesic $\sigma$, we then conclude by the strong maximum principle that
$\beta_{-}+\beta_{+}$ is constant and equal to ${\rm dist}(S_{-},S_{+})$ on
the whole manifold. It follows immediately that $\beta=\beta_{+}$ satisfies

$\Delta\beta+\frac{1}{f}\langle\nabla f\,|\,\nabla\beta\rangle\,=\,0$

in the barrier sense. We are now exactly in the same situation reached in the
proof of the Splitting Theorem for $f$-complete ends. We can then proceed
exactly as after formula (3.6) to conclude that $(M,g)$ is isometric to a
twisted product

$\left((a,b)\times\Sigma,\,f^{2}\,ds\otimes ds+g_{\Sigma}\right)\,.$

On the other hand, such a manifold is not conformally compact, as the metric
$\tilde{g}=g/f^{2}$ is degenerate as $s$ approaches $a$ or $b$. We have thus
reached a contradiction, implying that there were not multiple conformally
compact ends in the first place. ∎

This result generalizes [CS01, Theorem I.1], where the same thesis is obtained
for conformally compact vacuum static solutions with negative cosmological
constant. It is interesting to notice that the proof proposed in [CS01] also
makes use of the conformal metric $\tilde{g}=g/f^{2}$, which is exploited to
invoke a spacetime censorship result from [Gal+99, Theorem 2.1].

###  3.4. Splitting Theorem for mixed ends

For completeness, we include here the case where there are ends with different
behaviours. This case was not considered in [Wyl17] but the proof is similar.

######  Theorem 3.8.

Let $(M,g,f)$ be a substatic triple with ends that are either conformally
compact or $f$-complete. If there is at least one $f$-complete end, then there
cannot be any conformally compact end.

###### Proof.

This time the proof follows [Kas83, Theorem C-(2)]. Suppose that there is a
conformally compact end and an $f$-complete end. Then, one constructs a
globally minimizing $\tilde{g}$-geodesic $\sigma$ starting at a connected
component $S$ of the conformal boundary and reaching infinity. Let $\beta_{-}$
be the distance from $S$ and $\beta_{+}$ be the Busemann function relative to
$\sigma$. As in the previous cases, from the Laplacian comparisons for both
$\beta_{-}$ and $\beta_{+}$ and the fact that $\beta_{-}+\beta_{+}$ achieves
its minimum value $0$ on $\sigma$, we deduce that $\beta=\beta_{+}$ satisfies

$\Delta\beta+\frac{1}{f}\langle\nabla f\,|\,\nabla\beta\rangle\,=\,0$

in the barrier sense. We now proceed as in the other cases to show that the
manifold must be a twisted product

$\left((0,+\infty)\times\Sigma,\,f^{2}\,ds\otimes ds+g_{\Sigma}\right)\,.$

Again as in the proof of Theorem 3.7, we observe that the end corresponding to
$s=0$ cannot be conformally compact as the metric $\tilde{g}=g/f^{2}$ becomes
degenerate as $s\to 0$. We have thus reached a contradiction, meaning that it
is impossible to have an $f$-complete end and a conformally compact end at the
same time. ∎

This theorem, together with the other results in this Section (Theorem C and
Theorem 3.7) strongly narrows the acceptable configurations of ends for a
substatic triple. We sum up the topological information we have collected in
the following statement.

######  Corollary 3.9.

Let $(M,g,f)$ be a substatic triple with ends that are either conformally
compact or $f$-complete. If there is more than one end, then there are exactly
two ends, both $f$-complete, and $\partial M=\mathchar 31\relax$.

##  4\. Asymptotic Volume Ratio and Willmore-type inequality

In this section we focus on $f$-complete ends and we introduce the notion of
asymptotic volume ratio (AVR), in analogy with the classical case of
nonpositive Ricci curvature, as the limit of the Bishop–Gromov monotonic
quantity. In order to have a well defined AVR, we will need to focus our
attention to the special case of uniform ends. Building on the notion of AVR
we will finally prove the Willmore-type inequality mentioned in the
introduction.

###  4.1. Uniform $f$-complete ends

Here we introduce and comment the notion of uniformity of an $f$-complete end.
For convenience, instead of working on the whole $M$, we focus our attention
on the end only. In other words, starting from the next definition and for
most of this subsection, instead of working on the whole substatic triple
$(M,g,f)$, we just consider a neighborhood $E$ of our end and we focus our
attention on the restriction $(E,g,f)$, which we refer to as a substatic
$f$-complete end. It is easy to show that the definitions and statements below
do not depend on the choice of the neighborhood $E$ of our end.

######  Definition 4.1.

Let $(E,g,f)$ be a substatic $f$-complete end. We say that $(E,g,f)$ is
_uniform_ , if, for any two compact hypersurfaces $\Sigma_{1}$, $\Sigma_{2}$
contained in the interior of $E$ and every $\delta>0$, there exists a compact
set $K\supset\partial E$ such that for any two unit speed
$\tilde{g}$-geodesics $\sigma_{1}$, $\sigma_{2}$ minimizing the distance
between $\Sigma_{1}$, $\Sigma_{2}$ and a point
$p=\sigma_{1}(t_{1})=\sigma_{2}(t_{2})$ outside $K$, it holds

$\left|\frac{\int_{0}^{t_{1}}f^{2}(\sigma_{1}(t))dt}{\int_{0}^{t_{2}}f^{2}(\sigma_{2}(t))dt}\,-\,1\right|\,\leq\,\delta\,.$
(4.1)

While the definition above is slightly technical, we point out that there are
natural cases in which uniformity is guaranteed. We give here a couple of
easily described families of uniform $f$-complete ends. The following result
for instance guarantees us that an end is uniform as long as $f$ goes to one
at infinity.

######  Proposition 4.2.

Let $(E,g,f)$ be a substatic end. If $f\to 1$ at infinity, then $(E,g,f)$ is
$f$-complete and uniform.

###### Proof.

The fact that the ends are $f$-complete has already been shown in far greater
generality in Proposition 3.2. Let now $\Sigma_{1}$, $\Sigma_{2}$ be two
hypersurfaces. Since $f\to 1$ at infinity, for every $\varepsilon$ the set
$K_{\varepsilon}=\\{|f-1|>\varepsilon\\}$ is compact. In particular
$1-\varepsilon<f<1+\varepsilon$ outside $K_{\varepsilon}$. We consider now a
$\tilde{g}$-geodesically convex compact set $K$ containing $K_{\varepsilon}$
and the two hypersurfaces $\Sigma_{1}$, $\Sigma_{2}$.

Let $\sigma_{1}$, $\sigma_{2}$ be two unit speed $\tilde{g}$-geodesics
minimizing the distance between $\Sigma_{1}$, $\Sigma_{2}$ and a point
$p=\sigma_{1}(t_{1})=\sigma_{2}(t_{2})$ outside $K_{\varepsilon}$. For
$j=1,2$, let $0<T_{j}<t_{j}$ be the largest number such that
$\sigma_{j}(T_{j})\in K_{\varepsilon}$. We then have, for $j=1,2$,

$0\,<\,\int_{0}^{T_{j}}f^{2}(\sigma_{j}(t))dt\,<\,T_{j}\max_{K}f^{2}\,,\qquad(1-\varepsilon)^{2}(t_{j}-T_{j})\,<\,\int_{T_{j}}^{t_{j}}f^{2}(\sigma_{j}(t))dt\,<\,(1+\varepsilon)^{2}(t_{j}-T_{j})\,.$

As a consequence, we estimate:

$\frac{(1-\varepsilon)^{2}(t_{1}-T_{1})}{T_{2}\max_{K}f^{2}+(1-\varepsilon)^{2}(t_{2}-T_{2})}\,<\,\frac{\int_{0}^{t_{1}}f^{2}(\sigma_{1}(t))dt}{\int_{0}^{t_{2}}f^{2}(\sigma_{2}(t))dt}\,<\,\frac{T_{1}\max_{K}f^{2}+(1+\varepsilon)^{2}(t_{1}-T_{1})}{(1-\varepsilon)^{2}(t_{2}-T_{2})}\,.$

Since $f$ is bounded at infinity, the quantity $\max_{K}f^{2}$ is bounded by a
constant independent of $K$. Furthermore, we have $T_{j}<{\rm
diam}_{\tilde{g}}K_{\varepsilon}$, for $j=1,2$, by construction. On the other
hand, if we take the compact set $K$ to be much larger than $K_{\varepsilon}$,
we can make $t_{1}$ and $t_{2}$ arbitrarily large. The uniformity estimate
(4.1) follows then easily. ∎

Another case in which uniformity is guaranteed is under the assumption that
the norm of the gradient of $f$ decays sufficiently fast.

######  Proposition 4.3.

Let $(E,g,f)$ be a substatic $f$-complete end and let $\rho$ be the
$\tilde{g}$-distance from a point, where $\tilde{g}=g/f^{2}$. If for some
$\varepsilon>0$ there exist a compact set $K\supset\partial E$ and a constant
$C>0$ such that

$|\nabla f|<C\rho^{-1-\varepsilon}$

outside $K$, then $(E,g,f)$ is uniform.

###### Proof.

Fix the compact hypersurfaces $\Sigma_{1}$, $\Sigma_{2}$, the point $x$ and
the constant $\varepsilon>0$. Let $K\supset\partial E$ be the compact set such
that $|\nabla f|<C\rho^{-1-\varepsilon}$ outside $K$, where $\rho$ is the
$\tilde{g}$-distance from $x$. Up to enlarging $K$, we can suppose that $x$,
$\Sigma_{1}$ and $\Sigma_{2}$ are inside $K$.

Let $p$ be a point outside $K$. For $i=1,2$, consider the unit speed
$\tilde{g}$-geodesic $\sigma_{i}:[0,t_{i}]\to M$ minimizing the distance
between $\Sigma_{i}$ and $p$ and such that $\sigma_{i}(0)\in\Sigma_{i}$,
$\sigma_{i}(t_{i})=p$. We compare the value of $f$ at a point
$\sigma_{i}(\tau)$ and at the point $p$. Integrating along the geodesic, we
find

$\log f(p)\,=\,\log
f(\sigma_{i}(\tau))+\int_{\tau}^{t_{i}}\big{\langle}\widetilde{\nabla}\log
f\,\big{|}\,\dot{\sigma}_{i}\big{\rangle}_{\tilde{g}}(\sigma_{i}(t))\,dt\,,$

hence

$\left|\log\left(\frac{f(\sigma_{i}(\tau))}{f(p)}\right)\right|\,\leq\,\int_{\tau}^{t_{i}}\frac{1}{f}\big{|}\widetilde{\nabla}f\big{|}_{\tilde{g}}(\sigma_{i}(t))\,dt=\,\int_{\tau}^{t_{i}}|\nabla
f|(\sigma_{i}(t))\,dt\,.$ (4.2)

We now exploit our hypothesis. Assume that the segment
${\sigma_{i}}_{|_{[\tau,t_{i}]}}$ is outside $K$. Then $|\nabla f|\leq
C\rho^{-1-\varepsilon}$ at the points of ${\sigma_{i}}_{|_{[\tau,t_{i}]}}$,
where $\rho$ is the distance from $x$. Notice that for all $\tau\leq t\leq
t_{i}$, $\sigma_{i}(t)$ is at distance $t$ from $\Sigma_{i}$, so that by
triangle inequality, for any $y\in K$ it holds

$t-\max_{y\in\Sigma_{i}}{\rm d}_{\tilde{g}}(x,y)\leq\rho(\sigma_{i}(t))\leq
t+\max_{y\in\Sigma_{i}}{\rm d}_{\tilde{g}}(x,y)\,.$

If we then take

$t\geq 2\max_{y\in\Sigma_{i}}{\rm d}_{\tilde{g}}(x,y)$ (4.3)

we get

$|\nabla
f|(\sigma_{i}(t))\,\leq\,C\rho^{-1-\varepsilon}(\sigma_{i}(t))\,\leq\,C(t-\max_{y\in\Sigma_{i}}{\rm
d}_{\tilde{g}}(x,y))^{-1-\varepsilon}\,\leq\,C\left(\frac{t}{2}\right)^{-1-\varepsilon}=2^{1+\varepsilon}Ct^{-1-\varepsilon}\,.$

If we then suppose that $\tau\geq 2\max_{y\in\Sigma_{i}}{\rm
d}_{\tilde{g}}(x,y)$, we deduce from (4.2) that

$\left|\log\left(\frac{f(\sigma_{i}(\tau))}{f(p)}\right)\right|\,\leq\,2^{1+\varepsilon}C\frac{\tau^{-\varepsilon}-t_{i}^{-\varepsilon}}{\varepsilon}$
(4.4)

that is,

$e^{-\frac{2^{1+\varepsilon}C}{\varepsilon}\tau^{-\varepsilon}}\,\leq\,{\rm
exp}\left[2^{1+\varepsilon}C\frac{t_{i}^{-\varepsilon}-\tau^{-\varepsilon}}{\varepsilon}\right]\,\leq\,\frac{f(\sigma_{i}(\tau))}{f(p)}\,\leq\,{\rm
exp}\left[2^{1+\varepsilon}C\frac{\tau^{-\varepsilon}-t_{i}^{-\varepsilon}}{\varepsilon}\right]\,\leq\,e^{\frac{2^{1+\varepsilon}C}{\varepsilon}\tau^{-\varepsilon}}\,.$
(4.5)

Let now $\kappa>0$ be a large number and consider the compact set

$K_{\kappa}\,=\,B^{\tilde{g}}_{\kappa}(K)\,:=\,\\{y\in M\,:\,{\rm
d}_{\tilde{g}}(K,y)\leq\kappa\\}\,.$

Since $\Sigma_{i}\subset K$, notice that if $\sigma_{i}(t)\not\in K_{\kappa}$
then $t>\kappa$. It follows that, up to taking $\kappa$ large enough,
inequality (4.3) holds and in particular $|\nabla f|(\sigma_{i}(t))\leq
2^{1+\varepsilon}Ct^{-1-\varepsilon}$ for any $t$ such that
$\sigma_{i}(t)\not\in K_{\kappa}$.

Conversely, also notice by triangle inequality that if

$t>\kappa+\max_{y\in\Sigma_{i},z\in\partial K\setminus\partial E}{\rm
d}_{\tilde{g}}(y,z)\,,$

then $\sigma_{i}(t)\not\in K_{\kappa}$. Up to taking $\kappa$ large enough, we
can then also suppose that $\sigma_{i}(t)\not\in K_{\kappa}$ for every
$t>2\kappa$. In particular, for any $\tau>2\kappa$ we can apply estimate
(4.5), obtaining

$e^{-\frac{2C}{\varepsilon}\kappa^{-\varepsilon}}\,\leq\,\frac{f(\sigma_{i}(\tau))}{f(p)}\,\leq\,e^{\frac{2C}{\varepsilon}\kappa^{-\varepsilon}}$
(4.6)

We are finally ready to prove uniformity at infinity with respect to the
compact set $K_{\kappa}$, for $\kappa$ sufficiently large. As already noticed,
$\sigma_{i}(t)\in K_{2\kappa}$ for $t<2\kappa$, hence

$\frac{\int_{0}^{t_{1}}f^{2}(\sigma_{1}(t))dt}{\int_{0}^{t_{2}}f^{2}(\sigma_{2}(t))dt}\,\leq\,\frac{2\kappa\max_{K_{2\kappa}}f^{2}+\int_{2\kappa}^{t_{1}}f^{2}(\sigma_{1}(t))dt}{2\kappa\min_{K_{2\kappa}}f^{2}+\int_{2\kappa}^{t_{2}}f^{2}(\sigma_{2}(t))dt}\,\leq\,\frac{2\kappa\max_{K_{2\kappa}}f^{2}+(t_{1}-2\kappa)f(p)^{2}e^{\frac{2C}{\varepsilon}\kappa^{-\varepsilon}}}{2\kappa\min_{K_{2\kappa}}f^{2}+(t_{2}-2\kappa)f(p)^{2}e^{-\frac{2C}{\varepsilon}\kappa^{-\varepsilon}}}\,.$

Notice that $t_{1}$ and $t_{2}$ are comparable (their difference is bounded
via the triangle inequality by the maximum of the distance between points of
$\Sigma_{1}$ and $\Sigma_{2}$). Therefore, for any $\tilde{\varepsilon}>0$
arbitrarily small, we can find $\tilde{\kappa}$ much larger than $\kappa$ so
that, assuming $p$ is outside $K_{\tilde{\kappa}}$ (in particular
$t_{1},t_{2}$ are also much larger than $\kappa$) it holds

$\frac{\int_{0}^{t_{1}}f^{2}(\sigma_{1}(t))dt}{\int_{0}^{t_{2}}f^{2}(\sigma_{2}(t))dt}\,\leq\,(1+\tilde{\varepsilon})e^{\frac{4C}{\varepsilon}\kappa^{-\varepsilon}}\,.$

Up to choosing $\kappa$ large enough, we can also make the exponential term in
the inequality above as close to $1$ as necessary. Of course, exchanging the
roles of $\sigma_{1}$ and $\sigma_{2}$ we also find the opposite bound. This
proves uniformity. ∎

Equipped with the notion of uniformity of the ends, we are now ready to define
the substatic version of the asymptotic volume ratio.

######  Definition 4.4.

Let $(M,g,f)$ be a substatic solution and let $E$ be a uniform $f$-complete
end. Let $\rho$ be the distance function to a point or a hypersurface with
respect to the metric $\tilde{g}=g/f^{2}$ and $\eta$ be the solution to (2.10)
or (2.25), respectively. The Asymptotic Volume Ratio $\mathrm{AVR}(E,g,f)$ of
$E$ is defined as

${\rm
AVR}(E,g,f)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\lim_{t\to+\infty}\int_{\\{\rho=t\\}\cap
E}\frac{1}{\eta^{n-1}}d\sigma\,.$

If $(M,g,f)$ has a unique end $E$, we refer to $\mathrm{AVR}(E,g,f)$ with
$\mathrm{AVR}(M,g,f)$.

The following basic fact motivates the introduction of the notion of uniform
ends.

######  Proposition 4.5.

The substatic Asymptotic Volume Ratio is well-defined on any uniform
$f$-complete end. In other words, its definition does not depend on the choice
of the point/hypersurface we are taking the distance $\rho$ from.

###### Proof.

Let $\rho$ be the $\tilde{g}$-distance from a point or a hypersurface. We
consider the functional $V(t)$ defined in (2.52), that we recall here for the
reader’s convenience:

$V(t)\,=\,\frac{1}{|\mathbb{B}^{n}|t^{k}}\int_{\\{0\leq\rho\leq
t\\}}\frac{\rho^{k-1}}{f\eta^{n-1}}d\mu\,,$

where $k>0$ is a constant. A simple application of L’Hôpital’s rule tells us
immediately that

$\displaystyle\lim_{t\to+\infty}V(t)\,$
$\displaystyle=\,\frac{1}{|\mathbb{B}^{n}|}\lim_{t\to+\infty}\frac{1}{t^{k}}\int_{\\{\rho\leq
t\\}}\frac{\rho^{k-1}}{f\eta^{n-1}}d\mu$ (4.7)
$\displaystyle=\,\frac{n}{k|\mathbb{S}^{n-1}|}\lim_{t\to+\infty}\int_{\\{\rho=t\\}}\frac{1}{\eta^{n-1}}d\sigma$
(4.8) $\displaystyle=\,\frac{n}{k}\lim_{t\to+\infty}A(t)\,.$ (4.9)

In order to conclude the proof, it is then enough to show that
$\lim_{t\to+\infty}V(t)$ is independent of the choice of the
point/hypersurface. In the rest of the proof, it is convenient to set $k=1$ in
our functional $V$. Let $\eta_{1}$, $\eta_{2}$ be reparametrized distances
with respect to two different points (resp. two different hypersurfaces) and
let $\delta$ be the distance between the two points (resp. the maximum
distance between points of the two hypersurfaces) with respect to the metric
$\tilde{g}=g/f^{2}$. By triangle inequality we have the inclusion
$\\{\rho_{1}\leq t-\delta\\}\subset\\{\rho_{2}\leq t\\}$, therefore

$\displaystyle V_{1}(t)-V_{2}(t)$
$\displaystyle=\frac{1}{|\mathbb{B}^{n}|t}\int_{\\{\rho_{1}\leq
t\\}}\frac{1}{f\eta_{1}^{n-1}}d\mu-\frac{1}{|\mathbb{B}^{n}|t}\int_{\\{\rho_{2}\leq
t\\}}\frac{1}{f\eta_{2}^{n-1}}d\mu$ (4.10)
$\displaystyle\leq\frac{1}{|\mathbb{B}^{n}|t}\int_{\\{\rho_{1}\leq
t\\}}\frac{1}{f\eta_{1}^{n-1}}d\mu-\frac{1}{|\mathbb{B}^{n}|t}\int_{\\{\rho_{1}\leq
t-\delta\\}}\frac{1}{f\eta_{2}^{n-1}}d\mu$ (4.11)
$\displaystyle\leq\frac{1}{|\mathbb{B}^{n}|t}\int_{\\{t-\delta\leq\rho_{1}\leq
t\\}}\frac{1}{f\eta_{1}^{n-1}}d\mu+\frac{1}{|\mathbb{B}^{n}|t}\int_{\\{\rho_{1}\leq
t-\delta\\}}\frac{1}{f\eta_{1}^{n-1}}\left(1-\frac{\eta_{1}^{n-1}}{\eta_{2}^{n-1}}\right)d\mu\,.$
(4.12)

Concerning the first integral, applying again L’Hôpital’s rule we find that
its limit is the same as the limit of $nA(t)-nA(t-\delta)$ at $t\to+\infty$,
where $A$ is the usual area functional with respect to the distance
$\rho_{1}$. Since $A(t)$ has a finite limit at infinity and $\delta$ is fixed,
this limit is zero. From the uniformity of the end and the fact that $V(t)$ is
bounded, we deduce that the second integral also goes to zero. Hence, we have
found that $\lim_{t\to+\infty}[V_{1}(t)-V_{2}(t)]\leq 0$. Switching the roles
of $V_{1}$ and $V_{2}$ we find that the opposite inequality is also in place,
hence the limits of $V_{1}(t)$ and $V_{2}(t)$ are the same, as wished. ∎

######  Remark 4.6.

As noted in Remark 2.4, $\eta$ represents the distance along radial
$\tilde{g}$-geodesics with respect to the metric $\overline{g}=f^{2}g$.
Providing a suitable Bishop-Gromov-type Theorem in terms of the
$\overline{g}$-distance in place of $\eta$ may be useful to cook a notion of
Asymptotic Volume Ratio that does not need the notion of uniformity to be well
defined.

The following is a basic yet fundamental consequence of the Splitting Theorem
C.

######  Lemma 4.7.

Let $(M,g,f)$ be a substatic triple with $f$-complete ends. If there is more
than one uniform $f$-complete end, then all ends have vanishing asymptotic
volume ratio.

###### Proof.

Suppose that there is more than one uniform $f$-complete end. Then the
Splitting Theorem C implies that the manifold splits as a twisted product

$(\mathbb{R}\times\Sigma,\,f^{2}\,ds\otimes ds+g_{\Sigma})\,,$

for some $(n-1)$-dimensional Riemannian manifold $(\Sigma,g_{\Sigma})$. Let
$\rho$ be the $\tilde{g}$-distance from the cross section $\\{s=0\\}$ and
$\eta$ be defined by (2.25), as usual. Notice that the level sets of $\rho$
and $\eta$ are also level sets of $s$, hence in particular the metric induced
on any level set of $\rho$ is $g_{\Sigma}$. It follows then that

${\rm
AVR}(E,g,f)\,=\,\frac{1}{|\mathbb{S}^{n-1}|}\lim_{t\to+\infty}\int_{\\{\rho=t\\}\cap
E}\frac{1}{\eta^{n-1}}d\sigma\,=\,\lim_{t\to+\infty}\frac{|\Sigma|}{|\mathbb{S}^{n-1}|\eta_{|_{\\{\rho=t\\}}}^{n-1}}\,.$

Since the end is $f$-complete, we have $\eta\to+\infty$ at infinity, hence the
above limit vanishes. ∎

In light of the above Lemma, our main geometric inequalities 1.3 and 1.13 will
only involve one end and the global $\mathrm{AVR}(M,g,f)$.

In this framework, we now discuss some cases in which we are able to give more
precise estimates for the Asymptotic Volume Ratio. A first simple estimate, in
the case where the boundary is empty, is obtained from (2.51). Taking the
limit of this formula as $t\to+\infty$, assuming that the boundary is empty
(so that the term $A(t)$ appearing in that formula converges to the asymptotic
volume ratio) we find the following

${\rm AVR}(M,g,f)\,f(p)^{n-1}\,\leq\,1\,.$

This must hold for any point $p\in M$. In particular, it follows that if
$\partial M=\mathchar 31\relax$ and $f$ is not bounded then the Asymptotic
Volume Ratio must vanish.

An important family of substatic manifolds having nonzero AVR is that of
asymptotically flat triples, that we now define precisely.

######  Definition 4.8.

A substatic triple $(M,g,f)$ is said to be asymptotically flat if

  * $(i)$

there exists a compact domain $K\supset\partial M$ and a diffeomorphism
(called chart at infinity) between $M\setminus K$ and $\mathbb{R}^{n}$ minus a
ball.

  * $(ii)$

in the chart at infinity, it holds $|g_{ij}-\delta_{ij}|=o(1)$ and
$|f-1|=o(1)$ as $|x|\to+\infty$.

We remark that the usual definition of asymptotic flatness requires a higher
degree of convergence of the metric $g$ to the Euclidean one. However, the
above definition is sufficient to compute precisely the asymptotic volume
ratio.

######  Proposition 4.9.

Let $(M,g,f)$ be an asymptotically flat substatic triple. Then ${\rm
AVR}(M,g,f)=1$.

###### Proof.

The fact that the end is $f$-complete and uniform follows from Proposition
4.2. Let $K$ be a compact set as in Definition 4.8 and let $S=\\{|x|=R\\}$ be
a large coordinate sphere contained in the chart at infinity. From Proposition
4.5 we know that the asymptotic volume ratio does not depend on the
hypersurface we are taking the distance from. It is then convenient to work
with the $\tilde{g}$-distance $\rho$ from the coordinate sphere $S$. As it
follows from (4.9), we can also compute the AVR via the following limit

${\rm
AVR}(M,g,f)\,=\,\frac{1}{|\mathbb{B}^{n}|}\lim_{t\to+\infty}\frac{1}{t^{n}}\int_{\\{0\leq\rho\leq
t\\}}\frac{\rho^{n-1}}{f\eta^{n-1}}d\mu\,.$ (4.13)

If the radius $R$ of the coordinate sphere $S$ is large, then we can assume
$|f-1|<\varepsilon$ and $|g_{ij}-\delta_{ij}|<\varepsilon$ in $\\{|x|>R\\}$
for some fixed small $\varepsilon$. It is then easily seen that there exists
$\delta=\delta(\varepsilon)$ such that

$\\{R\leq|x|\leq(R+t)(1-\delta)\\}\subset\\{0\leq\rho\leq
t\\}\subset\\{R\leq|x|\leq(R+t)(1+\delta)\\}$

for all $t$. Since $\eta$ grows as $f^{2}$ along $\tilde{g}$-geodesics, we
have $(1-\varepsilon)^{2}\rho<\eta<(1+\varepsilon)^{2}\rho$. It follows that
the integral in (4.13) grows as the Euclidean volume of the annulus
$\\{R<|x|<R+t\\}$, or more explicitly as:

$\left[(R+t)^{n}-R^{n}\right]\,|\mathbb{B}^{n}|\,\cong\,|\mathbb{B}^{n}|t^{n}\,.$

The wished result follows easily. ∎

###  4.2. Willmore inequality

As a consequence of our definition of AVR and the Bishop–Gromov monotonicity
of the area functional $A(t)$ (Theorem 2.9), we obtain the Willmore inequality
for hypersurfaces with nonnegative mean curvature of Theorem D. The following
statement provides more details about the equality case.

######  Theorem 4.10 (Willmore inequality).

Let $(M,g,f)$ be a substatic solution with a uniform $f$-complete end. Let
$\Sigma$ be a hypersurface that is homologous to the boundary. Suppose that
the mean curvature ${\rm H}$ of $\Sigma$ with respect to the normal pointing
towards infinity satisfies ${\rm H}>0$ pointwise. Then

$\int_{\Sigma}\left[\frac{{\rm H}}{(n-1)f}\right]^{n-1}d\sigma\,\geq\,{\rm
AVR}(M,g,f)\,|\mathbb{S}^{n-1}|\,.$ (4.14)

If the equality holds, then the set $U=\\{\rho>0\\}$ is isometric to
$[0,+\infty)\times\Sigma$ with metric

$g\,=\,f^{2}d\rho\otimes d\rho+\eta^{2}g_{0}\,,$

where $g_{0}$ is a metric on the level set $\Sigma$. Furthermore, in $U$ the
functions $f$ and $\eta$ satisfy

$\eta\,=\,(\alpha+\beta)^{\frac{1}{n-1}}\,,\qquad
f^{2}\,=\,\frac{\dot{\alpha}}{(n-1)(\alpha+\beta)^{\frac{n-2}{n-1}}}$

where $\alpha$ is a function of $\rho$ and $\beta$ is a function on $\Sigma$.

###### Proof.

We recall from Theorem 2.9 that $A(t)$ is monotonically nonincreasing. Taking
the limit as $t\to+\infty$ we then get

$\frac{1}{|\mathbb{S}^{n-1}|}\int_{\Sigma}\left[\frac{{\rm
H}}{(n-1)f}\right]^{n-1}d\sigma\,=\,A(0)\,\geq\,\lim_{t\to+\infty}A(t)\,=\,{\rm
AVR}(M,g,f)\,.$

This proves the inequality.

Furthermore, if the equality holds, then from the rigidity statement in
Theorem 2.9 it follows

$\frac{1}{\eta}\frac{\partial\eta}{\partial\theta^{i}}\,=\,\psi\eta-\varphi\frac{1}{\eta^{n-1}}\,,\qquad\frac{1}{f}\frac{\partial
f}{\partial\theta^{i}}\,=\,\psi\eta+\frac{n-2}{2}\varphi\frac{1}{\eta^{n-1}}\,,$

where $\varphi$ and $\psi$ are functions on $\Sigma$. From the first of these
equations, in particular we get

$\frac{\partial}{\partial\theta^{i}}\left(\frac{1}{\eta}\right)\,=\,\psi-\varphi\frac{1}{\eta^{n}}\,.$

We now focus on this identity near infinity: since our end is $f$-complete, we
know that $\eta$ is going to $+\infty$. Furthermore, from the uniformity at
infinity we can also prove that $1/\eta$ goes to zero uniformly at infinity.
To prove that, it is enough to apply the uniformity at infinity property to
the two hypersurfaces $\Sigma$ and $\Sigma_{\delta}=\\{\rho=\delta\\}$. Notice
that then the function $\eta_{\delta}$ associated to $\Sigma_{\delta}$ differs
from $\eta$ just by a constant $k$, that is $\eta_{\delta}=\eta+k$. Then
uniformity at infinity implies precisely that for any $\varepsilon$ there
exists a compact set such that

$\left|\frac{k}{\eta}\right|\,=\,\left|\frac{k+\eta}{\eta}-1\right|\,\leq\,\varepsilon\,.$

Hence, $1/\eta$ goes to $0$ uniformly at infinity, as wished.

Given $\varepsilon>0$, fix $R>0$ big enough so that $1/\eta<\varepsilon$ in
$[R,+\infty)\times V$. If $\psi$ is not everywhere vanishing, then there is an
open set $V\subset\Sigma$ such that $\psi>\delta$ in $V$ (the case
$\psi<-\delta$ is done in the exact same way). Therefore we would get

$\frac{\partial}{\partial\theta^{i}}\left(\frac{1}{\eta}\right)\,>\,\delta-|\varphi|\,\varepsilon^{n}\,\geq\,\delta-\varepsilon^{n}\max_{\Sigma}|\varphi|=\tilde{\delta}$

in $[R,+\infty)\times V$. Up to taking $\varepsilon$ small enough, we can
assume that $\tilde{\delta}>0$. But then, for any two points
$p_{\rho}=(\rho,\theta^{1},\dots,\theta^{i},\dots,\theta^{n})$,
$q_{\rho}=(\rho,\theta^{1},\dots,\theta^{i}+\lambda,\dots,\theta^{n})$
belonging to $[R,+\infty)\times V$, we would deduce

$\frac{1}{\eta}(p_{\rho})\,=\,\frac{1}{\eta}(q_{\rho})+\int_{0}^{\lambda}\frac{\partial}{\partial\theta^{i}}{\left(\frac{1}{\eta}\right)}_{|_{(\rho,\theta^{1},\dots,\theta^{i}+t,\dots,\theta^{n})}}dt\,\geq\,\lambda\tilde{\delta}\,,$

which in turn would imply
$\lim_{\rho\to+\infty}(1/\eta)(p_{\rho})\geq\lambda\tilde{\delta}>0$,
contradicting the fact that $1/\eta\to 0$ at infinity. It follows then that

$\psi=0\,.$

Our constraints on $f$ and $\eta$ then become

$\frac{1}{\eta}\frac{\partial\eta}{\partial\theta^{i}}\,=\,-\varphi\frac{1}{\eta^{n-1}}\,,\qquad\frac{1}{f}\frac{\partial
f}{\partial\theta^{i}}\,=\,\frac{n-2}{2}\varphi\frac{1}{\eta^{n-1}}\,.$ (4.15)

The first equation can be rewritten as

$\frac{\partial}{\partial\theta^{i}}\eta^{n-1}\,=\,-(n-1)\varphi\,.$

Since $\varphi$ does not depend on $\rho$, it follows then that

$\eta^{n-1}\,=\,\alpha+\beta\,,$

where $\alpha$ is a function of $\rho$ and $\beta$ is a function on $\Sigma$.
Taking the derivative with respect to $\rho$ of this formula and using the
fact that $\partial_{\rho}\eta=f^{2}$, we then deduce

$f^{2}\,=\,\frac{\dot{\alpha}}{(n-1)(\alpha+\beta)^{\frac{n-2}{n-1}}}$

It is easy to check that for any $f$ and $\eta$ of this form (that is, for any
choice of $\alpha$ and $\beta$), formulas (4.15) are satisfied. ∎

##  5\. Isoperimetric Inequality for Substatic manifolds

In this Section, we focus our attention on substatic manifolds $(M,g,f)$
admitting an _exhaustion of outward minimising hypersurfaces_ homologous to
$\partial M$. A hypersurface $\Sigma$ homologous to $\partial M$ is outward
minimizing if, denoting by $\Omega$ the compact domain with
$\partial\Omega=\Sigma\sqcup\partial M$, we have

$P(\Omega)\leq P(F)$ (5.1)

for any bounded set $F\supset\Omega$. We say that a sequence
$(S_{j})_{j\in\mathbb{N}}$ of hypersurfaces homologous to $\partial M$ exhaust
$M$ if, given a compact set $K\subset M$, there exists an element $S$ in the
sequence such that $K\subset\Omega$, for $\Omega$ satisfying
$\partial\Omega=S\sqcup\partial M$. Conditions ensuring the existence of such
an exhaustion are discussed in [FM22].

We start from showing that $\partial M$ is a priori area minimizing. In
showing so, we also derive that $\partial M$ is _outermost_ , that is, there
exist no minimal submanifolds homologous to $\partial M$ other than the
boundary itself. These facts are the first main reason why we require the
existence of (nonminimal) outward minimizing sets homologous to $\partial M$.
Since the following auxiliary result does not need any a priori growth at
infinity assumption, we think it may have an independent interest.

######  Proposition 5.1 (The boundary is outermost area-minimizing).

Let $(M,g,f)$ be a substatic triple with horizon boundary. Assume that there
exists an outward minimizing smooth hypersurface $S$ homologous to $\partial
M$. Then, the horizon is outward minimizing, meaning that

$\lvert\Sigma\rvert\geq\lvert\partial M\rvert$ (5.2)

for any hypersurface $\Sigma$ homologous to $\partial M$. Moreover, it is
outermost, that is there exists no other minimal hypersurfaces homologous to
$\partial M$.

###### Proof.

Let $\Omega$ be such that $\partial\Omega=S\sqcup\partial M$. Indeed $S$ being
outward minimizing is mean-convex, and consequently the Maximum Principle
implies it is disjoint from $\partial M$ (see e.g. [Lee19, Corollary 4.2]; it
is a consequence of the strong comparison principle for quasilinear
equations). We flow $\Omega$ by weak Mean Curvature Flow, referring to the
notion considered in [Ilm92]. In particular the analysis carried out by White
[Whi00] applies. Moreover, observe that the mean curvature of $S$ is
necessarily nonnegative, and in particular $\Omega$ is mean-convex in the