# Decay of solutions
to the Klein-Gordon equation
on some expanding cosmological spacetimes

José Natário  CAMGSD, Departamento de Matemática
Instituto Superior Técnico
Universidade de Lisboa
Portugal<EMAIL_ADDRESS>and  Amol Sasane  Department of Mathematics
London School of Economics
Houghton Street
London WC2A 2AE
United Kingdom<EMAIL_ADDRESS>
###### Abstract.

The decay of solutions to the Klein-Gordon equation is studied in two
expanding cosmological spacetimes, namely

  * •

the de Sitter universe in flat Friedmann-Lemaître-Robertson-Walker (FLRW)
form, and

  * •

the cosmological region of the Reissner-Nordström-de Sitter (RNdS) model.

Using energy methods, for initial data with finite higher order energies,
decay rates for the solution are obtained. Also, a previously established
decay rate of the time derivative of the solution to the wave equation, in an
expanding de Sitter universe in flat FLRW form, is improved, proving Rendall’s
conjecture. A similar improvement is also given for the wave equation in the
cosmological region of the RNdS spacetime.

###### Key words and phrases:

Klein-Gordon equation, general relativity, FLRW models, Schwarzschild-de
Sitter, Reissner-Nordström-de Sitter

###### 2010 Mathematics Subject Classification:

Primary 58J45; Secondary 35B40, 35L15, 83C57, 83F05

###### Contents

  1. 1 Introduction
    1. 1.1 Relation of our results to previous work
  2. 2 Decay in the de Sitter universe in flat FLRW form; $m=0$
  3. 3 Decay in the de Sitter universe in flat FLRW form
    1. 3.1 Preliminary energy function and estimates
    2. 3.2 The auxiliary function $\psi$ and its PDE
    3. 3.3 The case $|m|>\frac{n}{2}$
    4. 3.4 The case $|m|<\frac{n}{2}$
    5. 3.5 The case $|m|=\frac{n}{2}$
  4. 4 Decay in the cosmological region of the $\mathrm{RNdS}$ spacetime
    1. 4.1 Preliminary energy function
    2. 4.2 The auxiliary function $\psi$ and its PDE
    3. 4.3 The case $|m|>\frac{n}{2}$
    4. 4.4 The case $|m|\leq\frac{n}{2}$
    5. 4.5 Redshift estimates
  5. 5 Decay in $\mathrm{RNdS}$ when $m=0$, the wave equation
  6. 6 Appendix A: Fourier modes (de Sitter in flat FLRW form)
  7. 7 Appendix B
  8. 8 Appendix C: Sharpness of bound when $|m|=\frac{n}{2}$ in Theorem 3.1

##  1\. Introduction

The aim of this article is to obtain exact decay rates for solutions to the
Klein-Gordon equation in a fixed background of some expanding cosmological
spacetimes. The two spacetimes we will consider are the de Sitter universe in
flat Friedmann-Lemaître-Robertson-Walker (FLRW) form, and the cosmological
region of the Reissner-Nordström-de Sitter (RNdS) model. The problem we
consider is linear, in that the background is fixed. This constitutes a first
step towards understanding the more complicated nonlinear coupled problem,
where one also considers the effect of the energy-momentum tensor of the
solution to the Klein-Gordon equation on the Einstein equation. This nonlinear
coupled problem is much more complicated, and usually requires, as a first
step, a detailed understanding of our simpler linear problem.

There are several motivations behind the interest in this question. Firstly,
one may consider the linear wave equations as a proxy for the Einstein
equations, with the ultimate goal of understanding the qualitative behaviour
of solutions to the Einstein equations by slicing spacetime into spacelike
hypersurfaces. After this first step, one may then proceed to consider
linearised Einstein equations (which can be reduced to tensor wave-like linear
equations), and finally, the full nonlinear Einstein equations. With the
addition of a positive cosmological constant to the Einstein field equations,
the expectation is that the resulting accelerated expansion has a dominating
effect on the decay of solutions. Precise estimates on solutions may then
prove useful in formulating and proving cosmic no-hair theorems (e.g. [3],
[7]).

The wave equation $\square_{g}\phi=0$ in expanding cosmological spacetimes
$(M,g)$ has been amply studied in the literature, see for example [5], [8],
[10], [22], and the references therein. It is a natural question to also study
the Klein-Gordon equation $\square_{g}\phi-m^{2}\phi=0$, the degenerate
version of which, when $m^{2}=0$, is the wave equation. For example, in [22,
§6], also the case of the Klein-Gordon equation in the Schwarzschild-de Sitter
spacetime is considered. In [20], the asymptotic behaviour of the solutions to
the Klein-Gordon equation near the Big Bang singularity is studied, while we
investigate the asymptotics of the Klein-Gordon equation in the far future in
the case of the de Sitter universe in flat FLRW form, and in the cosmological
region of the Reissner-Nordström-de Sitter solution. Recently, in [11], among
other things, decay estimates for the solutions to the Klein-Gordon equation
were obtained in de Sitter models (see in particular, Corollary 2.1 and the
less obvious Proposition 3.1). However, these results are proved via Fourier
transformation (reminiscent of our mode calculation in Appendix A) and do not
seem to be as sharp as our Theorem 3.1.

The wave equation in the de Sitter spacetime having flat $3$-dimensional
spatial sections was considered in Rendall [19]. There it was shown that the
time derivative $\partial_{t}\phi=:\dot{\phi}$ decays at least as
$e^{-Ht}=(a(t))^{-1}$, where $H=\sqrt{\Lambda/3}$ is the Hubble constant, and
$\Lambda>0$ is the cosmological constant. Moreover, it was conjectured that
the decay is of the order $e^{-2Ht}=(a(t))^{-2}$. The almost-exact conjectured
decay rate of $|\dot{\phi}|\lesssim(a(t))^{-2+\delta}$ (where $\delta>0$ can
be chosen arbitrarily at the outset) follows as a corollary of a result shown
recently [8, Remark 1.1]. We improve this result, to obtaining full conformity
with Rendall’s conjecture, in our result Theorem 2.3 below.

Finally, from the pure mathematical perspective, analysis of linear wave
equations on Lorentzian manifolds is a natural topic of study within the realm
of hyperbolic partial differential equations and differential geometry; see
for example [2], [24, §7, Chap.2].

A naive heuristic indication of the effect of the accelerated expansion on the
decay of the solution, based on physical energy considerations, can be
obtained as follows. Considering an expanding FLRW model with flat
$n$-dimensional spatial sections of radius $a(t)$, we have on the one hand
that the energy density of a solution $\phi$ of the Klein-Gordon equation is
of the order of $m^{2}\phi^{2}$. On the other hand, if the wavelength of the
particles associated with $\phi$ follows the expansion, then it is
proportional to $a(t)$, and so the energy varies as

$E^{2}\sim m^{2}+p^{2}\sim A+\frac{B}{(a(t))^{2}},$

where $A,B>0$ are constants. Thus $\displaystyle
m^{2}\phi^{2}(a(t))^{n}\;\propto\;\left(A+\frac{B}{(a(t))^{2}}\right),$ giving

$m^{2}\phi^{2}\sim(a(t))^{-n}\left(A+\frac{B}{(a(t))^{2}}\right).$

As $\dot{a}\geq 0$ (expanding FLRW spacetime), the term
$A+\frac{B}{(a(t))^{2}}$ approaches a finite positive value, and so one may
expect

$\phi\sim(a(t))^{-\frac{n}{2}}.$

We will find out that in fact things are much more complicated: this decay
rate is valid only for $|m|\geq\frac{n}{2}$. In order to obtain precise
conjectures on the expected decay, we will consider Fourier modes for
spatially-periodic solutions to the Klein-Gordon equation, or equivalently,
consider the expanding de Sitter universe in flat FLRW form with toroidal
spatial sections. This exercise already demonstrates that the underlying decay
mechanism is the cosmological expansion, as opposed to dispersion. The Fourier
mode analysis, which is peripheral to the rest of the paper, is relegated to
Appendix A.

In the cosmological region of the Reissner-Nordström-de Sitter spacetimes, the
expanding region is foliated by spacelike hypersurfaces of ‘constant $r$’. One
expects the decay rate with respect to $r$, for the solution to the Klein-
Gordon equation, in the cosmological region of the Reissner-Nordström-de
Sitter spacetime, to be the same as the one for the de Sitter universe in flat
FLRW form, when $e^{t}$ is replaced by $r$. We show that this expectation is
correct, and a suitable modification of the technique used in the case of the
de Sitter universe in flat FLRW form, does enable one to obtain the expected
decay rates also for the case of the Reissner-Nordström-de Sitter spacetime.

Our main results are as follows:

  * •

Theorem 2.3 considers the $m=0$ case (wave equation), and we obtain a decay
estimate on $\partial_{t}{\phi}$, improving a corollary of [8, Theorem 1], and
proving the aforementioned Rendall’s conjecture.

  * •

Theorem 5.3 improves [8, Theorem 2], and we obtain a decay estimate on
$\partial_{r}\phi$, using a similar method to the one we use for proving
Rendall’s conjecture.

  * •

Theorem 3.1 gives the decay rate of the solutions $\phi$ to the Klein-Gordon
equation in the de Sitter universe in flat FLRW form.

  * •

Theorem 4.2 gives the decay rate of the solutions $\phi$ to the Klein-Gordon
equation in the cosmological region of the RNdS model.

Theorems 2.3, 3.1, 4.2, 5.3 are stated and proved in Sections 2, 3, 4, 5,
respectively. The Fourier mode analysis for spatially-periodic solutions to
the Klein-Gordon equation is given in Appendix A, while Appendix B contains a
technical lemma which is needed in the proof of Theorem 3.1. Finally, in
Appendix C, we establish the sharpness of the bound of the $|m|=\frac{n}{2}$
case of Theorem 3.1.

###  1.1. Relation of our results to previous work

Our decay rates for the Klein-Gordon equation solutions in the case of the de
Sitter universe can be retrieved from the article [25] by setting $x=e^{-t}$,
$Y=\mathbb{R}^{n}$ therein. However, the methods used are entirely different:
our proof in this case is more explicit, and more elementary (relying on
energy methods, rather than technical tools from microlocal analysis of
partial differential operators).

In the article [13], the Klein-Gordon equation is studied in the Nariai
spacetime using energy methods, and en route it is also established that
solutions of the Klein-Gordon equation decay exponentially in the de Sitter
case (with spherical spatial sections). However, the decay rates are not given
explicitly.

The article [9] contains a general discussion of redshift estimates, which we
use to prove our results in the context of the Reissner-Nordström-de Sitter
spacetime. Similar estimates are used in the article [22] to study the wave
equation in the Schwarzschild-de Sitter spacetime, of which the Reissner-
Nordström-de Sitter spacetime is a perturbation for large radius.
Nevertheless, we do not appeal to these results, and instead of extracting
what we need from these sources, we give a less technical, self-contained
derivation for the convenience of the reader in §4.5. Here we follow [8]
(where a similar derivation was given for the wave equation).

Acknowledgements: We thank Pedro Girão for suggesting the idea behind the
proofs of Theorems 2.2 and 5.3. JN was partially supported by FCT/Portugal
through UID/MAT/04459/2013 and grant (GPSEinstein) PTDC/MAT-ANA/1275/2014.

##  2\. Decay in the de Sitter universe in flat FLRW form; $m=0$

In [8, Theorem 1], the following result was shown:

######  Theorem 2.1.

Suppose that

  * •

$\delta>0$,

  * •

$I\subset\mathbb{R}$ is an open interval of the form $(t_{*},+\infty)$,
$t_{0}\in I$,

  * •

$a(\cdot)\in C^{1}(I)$ with $\dot{a}(t)\geq 0$ for $t\geq t_{0}$, and
$\epsilon>0$ is such that

$\phantom{aaaaaaaaaaaaaaaa}\displaystyle\int_{t_{0}}^{\infty}\frac{1}{(a(t))^{\epsilon}}dt<+\infty,$

  * •

$n\geq 2$,

  * •

$(M,g)$ is an expanding FLRW spacetime with flat $n$-dimensional

sections, given by $I\times\mathbb{R}^{n}$, with the metric

$g=-dt^{2}+(a(t))^{2}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right),$ (1)

  * •

$k>\frac{n}{2}+2$, $\;\;\phi_{0}\in H^{k}(\mathbb{R}^{n})$, $\;\;\phi_{1}\in
H^{k-1}(\mathbb{R}^{n})$, and

  * •

$\phi$ is a smooth solution to the Cauchy problem

$\left\\{\begin{array}[]{rcll}\square_{g}\phi&=&0,&\;\;\;\;(t\geq
t_{0},\;\mathbf{x}\in\mathbb{R}^{n}),\\\
\phi(t_{0},\mathbf{x})&=&\phi_{0}(\mathbf{x})&\;\;\;\;(\mathbf{x}\in\mathbb{R}^{n}),\\\
\partial_{t}\phi(t_{0},\mathbf{x})&=&\phi_{1}(\mathbf{x})&\;\;\;\;(\mathbf{x}\in\mathbb{R}^{n}).\end{array}\right.$

Then

$\forall t\geq
t_{0},\;\;\;\;\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left(a(t)\right)^{-2+\epsilon+\delta}.$

Here, the symbol $\lesssim$ is used to mean that there exists a constant
$C(\delta)$, independent of $\epsilon$, such that

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\leq
C(\delta)\left(a(t)\right)^{-2+\epsilon+\delta}.$

We also use the standard notation $H^{k}(\mathbb{R}^{n})$ for the Sobolev
space,

$\|\phi\|^{2}_{H^{k}(\mathbb{R}^{n})}:=\int_{\mathbb{R}^{n}}\sum_{|{\bm{\alpha}}|\leq
k}(\partial_{\bm{\alpha}}\phi)^{2}d^{n}\mathbf{x}<+\infty\;\;\textrm{ for
}\phi\in H^{k}(\mathbb{R}^{n}),$

where

$\displaystyle{\bm{\alpha}}$ $\displaystyle=$
$\displaystyle(\alpha_{1},\cdots,\alpha_{n})\in\mathbb{N}_{0}^{n},\quad\mathbb{N}_{0}=\\{0,1,2,3,\cdots\\},$
$\displaystyle|{\bm{\alpha}}|$ $\displaystyle:=$
$\displaystyle\alpha_{1}+\cdots+\alpha_{n},\textrm{ and}$
$\displaystyle\partial_{\bm{\alpha}}$ $\displaystyle=$
$\displaystyle(\partial_{x_{1}})^{\alpha_{1}}\cdots(\partial_{x_{n}})^{\alpha_{n}};$

see for example [26, p.249] or [24, Chap. 4].

######  Remark 2.2 (Smoothness assumption on the solution $\phi$).

In Theorem 2.1 (and later also in Theorems 2.3, 3.1, 4.2, 5.1, 5.3), we will
assume, for the sake of simplicity of exposition, that the solution $\phi$ to
the wave/Klein-Gordon equation is smooth. However, these theorems are also
true without this assumption. To see this, we note that for non-smooth
solutions with initial data in $H^{k}\times H^{k-1}$, we can approximate the
initial data by smooth functions in $H^{k}\times H^{k-1}$, prove the bounds
for the $H^{k}$ norms of the corresponding solutions, and then take limits.
Since the solution of the problem with rough initial data is in
$C^{0}(I,H^{k})\cap\;\\!C^{1}(I,H^{k-1})$, these bounds will continue to be
true in the limit, and we can then use the Sobolev embedding theorem. This
enables one to drop the smoothness assumption.

In Theorem 2.1 above, in particular, if

$a(t)=e^{Ht},$

where $H$ is the Hubble constant, then since $\epsilon>0$ can be taken to be
arbitrarily small, we obtain

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim(a(t))^{-2+\delta}=e^{-(2-\delta)Ht},$

and this is in agreement with Rendall’s conjecture up to the small quantity
$\delta>0$. We will show below that in fact one gets the exact rate
$(a(t))^{-2}$ when $n>2$. There is no loss of generality in assuming that
$H=1$. Our result is the following.

######  Theorem 2.3.

Suppose that

  * •

$I\subset\mathbb{R}$ is an open interval of the form $(t_{*},+\infty)$,
$t_{0}\in I$,

  * •

$n>2$,

  * •

$(M,g)$ is the expanding de Sitter universe in flat FLRW form, with flat
$n$-dimensional sections, given by $I\times\mathbb{R}^{n}$, with the metric

$g=-dt^{2}+e^{2t}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right),$

  * •

$k>\frac{n}{2}+2$, $\;\;\phi_{0}\in H^{k}(\mathbb{R}^{n})$, $\;\;\phi_{1}\in
H^{k-1}(\mathbb{R}^{n})$, and

  * •

$\phi$ is a smooth solution to the Cauchy problem

$\left\\{\begin{array}[]{rcll}\square_{g}\phi&=&0,&\;\;\;\;(t\geq
t_{0},\;\mathbf{x}\in\mathbb{R}^{n}),\\\
\phi(t_{0},\mathbf{x})&=&\phi_{0}(\mathbf{x})&\;\;\;\;(\mathbf{x}\in\mathbb{R}^{n}),\\\
\partial_{t}\phi(t_{0},\mathbf{x})&=&\phi_{1}(\mathbf{x})&\;\;\;\;(\mathbf{x}\in\mathbb{R}^{n}).\end{array}\right.$

Then

$\forall t\geq
t_{0},\;\;\;\;\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left(a(t)\right)^{-2}.$

###### Proof.

We proceed in several steps.

Step 1: Bound on $\Delta\phi$.

We will follow the preliminary steps of the proof of [8, Theorem 1] in order
to obtain a bound on $\Delta\phi$, which will be needed in the proof of our
Theorem 2.3. We repeat this preliminary step here from [8, §2.2] for the sake
of completeness and for the convenience of the reader.

For a vector field $X=X^{\mu}\partial_{\mu}$, it can be shown that

$\nabla_{\mu}X^{\mu}=\frac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}\;\\!X^{\mu}),$

where $g:=\det[g_{\mu\nu}]$ is the determinant of the matrix $[g_{\mu\nu}]$
describing the metric in the chart. Then it follows that

$\square_{g}\phi=\nabla_{\mu}(\partial^{\mu}\phi)=\frac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}\;\\!\partial^{\mu}\phi).$

Thus $\square_{g}\phi=0$ can be rewritten as
$\partial_{\mu}(\sqrt{-g}\;\\!\partial^{\mu}\phi)=0$. With the metric for the
de Sitter universe in flat FLRW form given by

$g=-dt^{2}+(a(t))^{2}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right),$

the wave equation can be rewritten as

$\partial_{\mu}(a^{n}\partial^{\mu}\phi)=0,$

that is,

$-\ddot{\phi}-\frac{n\dot{a}}{a}\dot{\phi}+\frac{1}{a^{2}}\delta^{ij}\partial_{i}\partial_{j}\phi=0.$

We recall (see e.g. [26, Appendix E]) that the energy-momentum tensor for the
wave equation is

$T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi-\frac{g_{\mu\nu}}{2}\partial_{\alpha}\phi\partial^{\alpha}\phi.$
(2)

Then it can be shown that

$\nabla_{\mu}T^{\mu\nu}=0.$

From (2), have in particular that

$T_{00}=\frac{1}{2}\left(\dot{\phi}^{2}+a^{-2}\delta^{ij}\partial_{i}\phi\partial_{j}\phi\right).$

Define the vector field

$X=a^{2-n}\frac{\partial}{\partial t}.$

Then $X$ is future-pointing ($g(X,\partial_{t})<0$) and causal ($X$ is time-
like since $g(X,X)<0$). We form the current $J$, given by

$J_{\mu}=T_{\mu\nu}X^{\nu}.$

Then it can be shown that

$J=(X\cdot\phi)\textrm{grad }\phi-\frac{1}{2}g(\textrm{grad }\phi,\textrm{grad
}\phi)X.$

(Here $X\cdot\phi$ means the application of the vector field $X$ on $\phi$.)
It follows that $g(J,J)\leq 0$, so that $J$ is causal. Also, $J$ is past-
pointing. To see this, we choose $E_{1},\cdots,E_{n}$ orthogonal and spacelike
such that $\\{X,E_{1},\cdots,E_{n}\\}$ forms an orthogonal basis in each
tangent space. Then expressing

$\textrm{grad }\phi=c^{0}X+c^{1}E_{1}+\cdots c^{n}E_{n},$

we obtain

$\displaystyle g(J,X)\\!\\!$ $\displaystyle\\!\\!=\\!\\!$
$\displaystyle\\!\\!\\!\left(g(X,\textrm{grad
}\phi)\right)^{2}-\frac{1}{2}g(\textrm{grad }\phi,\textrm{grad }\phi)\cdot
g(X,X)$ $\displaystyle\\!\\!=\\!\\!$
$\displaystyle\\!\\!\\!\frac{(c^{0})^{2}}{2}\\!\left(g(X,X)\right)^{2}\\!-\\!\frac{1}{2}\left((c^{1})^{2}g(E_{1},E_{1})\\!+\\!\cdots\\!+\\!(c^{n})^{2}g(E_{n},E_{n})\right)\cdot
g(X,X)\\!\geq\\!0.$

Set

$N=\frac{\partial}{\partial t},$

the future unit normal vector field. We define the energy $E$ by

$E(t)=\int_{\\{t\\}\times\mathbb{R}^{n}}J_{\mu}N^{\mu}=\int_{\mathbb{R}^{n}}a^{2}T_{00}d^{n}\mathbf{x}=\int_{\mathbb{R}^{n}}\frac{1}{2}\left(a^{2}\dot{\phi}^{2}+\delta^{ij}\partial_{i}\partial_{j}\phi\right)d^{n}\mathbf{x}.$

The deformation tensor $\Pi$ associated with the multiplier $X$ is

$\Pi=\frac{1}{2}\mathcal{L}_{X}g=-dt\mathcal{L}_{X}dt+\dot{a}a^{3-n}\delta_{ij}dx^{i}dx^{j}.$

It can be shown that

$\mathcal{L}_{X}dt=(2-n)\dot{a}a^{1-n}dt.$

Thus

$\Pi=(n-2)\dot{a}a^{1-n}dt^{2}+\dot{a}a^{3-n}\delta_{ij}dx^{i}dx^{j}.$

Claim: $\nabla_{\mu}J^{\mu}=T^{\mu\nu}\Pi_{\mu\nu}.$

We have

$\displaystyle\Pi_{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\mathcal{L}_{X}g)(\partial_{\mu},\partial_{\nu})=\frac{1}{2}\left(\mathcal{L}_{X}(g_{\mu\nu})-g(\mathcal{L}_{X}\partial_{\mu},\partial_{\nu})-g(\partial_{\mu},\mathcal{L}_{X}\partial_{\nu})\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(X(g_{\mu\nu})-g([X,\partial_{\mu}],\partial_{\nu})-g(\partial_{\mu},[X,\partial_{\nu}])\right).$

But by the definition of the Levi-Civita connection $\nabla$,

$\displaystyle g(\nabla_{\partial_{\mu}}\partial_{\nu},X)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\partial_{\mu}(g(\partial_{\nu},X))+\partial_{\nu}(g(\partial_{\mu},X))-X(g(\partial_{\mu},\partial_{\nu}))\right.$
$\displaystyle\left.+g([X,\partial_{\mu}],\partial_{\nu})+g([X,\partial_{\nu}],\partial_{\mu})+g(X,[\partial_{\mu},\partial_{\nu}])\right)\phantom{\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\partial_{\mu}(g(\partial_{\nu},X))+\partial_{\nu}(g(\partial_{\mu},X))\right)$
$\displaystyle-\frac{1}{2}\left(X(g_{\mu\nu})-g([X,\partial_{\mu}],\partial_{\nu})-g(\partial_{\mu},[X,\partial_{\nu}])\right).$

So

$\displaystyle\Pi_{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(X(g_{\mu\nu})-g([X,\partial_{\mu}],\partial_{\nu})-g(\partial_{\mu},[X,\partial_{\nu}])\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\partial_{\mu}(g(\partial_{\nu},X))+\partial_{\nu}(g(\partial_{\mu},X))\right)-g(\nabla_{\mu}\partial_{\nu},X).$

Writing $X=k\partial_{0}$, where $k(t):=(a(t))^{2-n}$, we have

$\Pi_{\mu\nu}=\frac{1}{2}\left(\partial_{\mu}(kg_{\nu
0})+\partial_{\nu}(kg_{\mu
0})\right)-g(\nabla_{\mu}\partial_{\nu},k\partial_{0})=-\dot{k}\delta_{\mu
0}\delta_{\nu 0}+k\Gamma_{\mu\nu}^{0}.$

We have

$\nabla_{\mu}X_{\nu}=\nabla_{\mu}(-k\delta_{\nu 0})=-\dot{k}\delta_{\mu
0}\delta_{\nu 0}+k\Gamma_{\mu\nu}^{0}.$

So $\Pi_{\mu\nu}=\nabla_{\mu}X_{\nu}$, and consequently,

$\nabla_{\mu}J^{\mu}=\nabla_{\mu}(T^{\mu\nu}X_{\nu})=0+T^{\mu\nu}\nabla_{\mu}X_{\nu}=T^{\mu\nu}\Pi_{\mu\nu}.$

This completes the proof of our claim that
$\nabla_{\mu}J^{\mu}=T^{\mu\nu}\Pi_{\mu\nu}.$

So the ‘bulk term’ is

$\displaystyle\nabla_{\mu}J^{\mu}=T^{\mu\nu}\Pi_{\mu\nu}$ $\displaystyle=$
$\displaystyle(n-2)\dot{a}a^{1-n}\dot{\phi}^{2}+\frac{n-2}{2}\dot{a}a^{1-n}\partial_{\alpha}\phi\partial^{\alpha}\phi$
$\displaystyle+\dot{a}a^{-1-n}\delta^{ij}\partial_{i}\phi\partial_{j}\phi-\frac{n}{2}\dot{a}a^{1-n}\partial_{\alpha}\phi\partial^{\alpha}\phi$
$\displaystyle=$ $\displaystyle(n-1)\dot{a}a^{1-n}\dot{\phi}^{2}\geq
0.\phantom{\frac{1}{2}}$

For each $R>0$, define the set

$B_{0}:=\\{(t_{0},\mathbf{x})\in
I\times\mathbb{R}^{n}:\langle\mathbf{x},\mathbf{x}\rangle_{\scriptscriptstyle\mathbb{R}^{n}}\leq
R^{2}\\}.$

The future domain of dependence of $B_{0}$ is the set

$D^{+}(B_{0}):=\left\\{p\in M\;\Big{|}\;\begin{array}[]{ll}\textrm{Every past
inextendible causal curve}\\\ \textrm{through }p\textrm{ intersects
}B_{0}.\end{array}\right\\}.$

(Here by a causal curve, we mean one whose tangent vector at each point is a
causal vector. A curve $c:(a,b)\rightarrow M$ which is smooth and future
directed111That is, $\dot{c}$ is future-pointing. is called past inextendible
if $\lim\limits_{t\rightarrow a}c(t)$ does not exist.)

Let $t_{1}>t_{0}$. We will now apply the divergence theorem to the region

$\mathcal{R}:=D^{+}(B_{0})\;\\!\cap\;\\!\\{(t,\mathbf{x})\in M:t\leq
t_{1}\\}.$

For preliminaries on the divergence theorem in the context of a time-oriented
Lorentzian manifold, we refer the reader to [26, Appendix B]. We have

$\int_{\mathcal{R}}(\nabla_{\mu}J^{\mu})\epsilon=\int_{\partial\mathcal{R}}J\intprod\epsilon,$

where $\partial\mathcal{R}$ denotes the boundary of $\mathcal{R}$, $\epsilon$
is the volume form on $M$ induced by $g$, and denotes contraction in the first
index.

Since $J$ is past-pointing, the boundary integral over the null portion $C$ of
the boundary $\partial\mathcal{R}$ is nonpositive. Also, because
$\nabla_{\mu}J^{\mu}$ is nonnegative, we have that the volume integral over
$\mathcal{R}$ is nonnegative. This gives an inequality on the two boundary
integrals, one over $B_{0}$, and the other over
$B_{1}:=D^{+}(B_{0})\cap\\{t=t_{1}\\}$, as follows:

$\int_{B_{0}}\frac{1}{2}(a^{2}\dot{\phi}^{2}+\delta^{ij}\partial_{i}\phi\partial_{j}\phi)d^{n}\mathbf{x}\geq\int_{B_{1}}\frac{1}{2}(a^{2}\dot{\phi}^{2}+\delta^{ij}\partial_{i}\phi\partial_{j}\phi)d^{n}\mathbf{x}.$

Passing the limit $R\rightarrow\infty$ yields $E(t_{0})\geq E(t_{1}).$ As the
choice of $t_{1}>t_{0}$ was arbitrary, we have

$\forall t\geq t_{0},\;\;E(t)\leq E(t_{0})<\infty.$

The finiteness of $E(t_{0})$ follows from our assumption that $\phi_{0}\in
H^{k}(\mathbb{R}^{n})$ and $\phi_{1}\in H^{k-1}(\mathbb{R}^{n})$ for a $k$
satisfying $k>\frac{n}{2}+2\geq 1$. From here, it follows that for all $t\geq
t_{0}$,

$\displaystyle\int_{\mathbb{R}^{n}}\dot{\phi}^{2}d^{n}\mathbf{x}\lesssim\frac{1}{a^{2}},\;\;\textrm{
and}$
$\displaystyle\int_{\mathbb{R}^{n}}\delta^{ij}\partial_{i}\phi\partial_{j}\phi
d^{n}\mathbf{x}\lesssim 1.$

But since each partial derivative $\partial_{i}\phi$ is also a solution of the
wave equation, and as $k\geq 2$, we obtain, by applying the above to the
partial derivatives $\partial_{i}\phi$, that also

$\int_{\mathbb{R}^{n}}(\Delta\phi)^{2}d^{n}\mathbf{x}\lesssim 1.$

In fact, since $k>\frac{n}{2}+2$, we also obtain that for a
$k^{\prime}>\frac{n}{2}$,

$\|\Delta\phi\|_{H^{k^{\prime}}(\mathbb{R}^{n})}\lesssim 1.$

Finally, by the Sobolev inequality (see e.g. [14, (7.30), p.158]), we obtain

$\|\Delta\phi\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim 1.$ (3)

This completes Step 1 of the proof of Theorem 2.3.

Step 2: The wave equation in conformal coordinates.

The key point of departure from the earlier derivation of the estimates from
[8] is the usage of ‘conformal coordinates’, which renders the wave equation
in a form where it becomes possible to integrate, leaving essentially just the
time derivative of $\phi$ with other terms (e.g. $\Delta\phi$) for which we
have a known bound. An application of the triangle inequality will then
deliver the desired bound.

Define

$\tau=\int_{t_{0}}^{t}\frac{1}{a(s)}ds.$

Then

$\displaystyle\frac{d\tau}{dt}=\frac{1}{a(t)}\;\;\textrm{ and
}\;\;a(t)\displaystyle\frac{d}{dt}=\frac{d}{d\tau}.$

With a slight abuse of notation, we write $a(\tau):=a(t(\tau))$. Then
$dt=a(\tau)d\tau$. So

$g=-dt^{2}+(a(t))^{2}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right)=(a(\tau))^{2}\left(-d\tau^{2}+\delta_{ij}dx^{i}dx^{j}\right).$

The wave equation $\square_{g}\phi=0$ can be rewritten as
$\partial_{\mu}(\sqrt{-g}\;\\!\partial^{\mu}\phi)=0$, which becomes

$\partial_{\mu}(a^{n+1}\partial^{\mu}\phi)=0.$

Separating the partial derivative operators with respect to the $\tau$ and
$\mathbf{x}$ coordinates, we obtain the wave equation in conformal coordinates

$\partial_{\tau}(a^{n-1}\partial_{\tau}\phi)=a^{n-1}\Delta\phi,$

where $\Delta$ is the usual Laplacian on $\mathbb{R}^{n}$. This completes Step
2 of the proof of Theorem 2.3.

Step 3: $n>2$ and $a(t)=e^{t}$.

We have

$\tau=\int_{t_{0}}^{t}\frac{1}{e^{s}}ds=e^{-t_{0}}-\frac{1}{e^{t}}=e^{-t_{0}}-\frac{1}{a},$
(4)

and so

$a(\tau)=\displaystyle\frac{1}{e^{-t_{0}}-\tau}.$

We note that $\tau\in[0,e^{-t_{0}})$. Also,

$\displaystyle a(\tau=0)=\frac{1}{e^{-t_{0}}}=e^{t_{0}}=a(t=t_{0}).$

Integrating

$\partial_{\tau}(a^{n-1}\partial_{\tau}\phi)=a^{n-1}\Delta\phi$

from $\tau=0$ to $\tau$, we obtain

$a^{n-1}\partial_{\tau}\phi-a(t_{0})^{n-1}\left.\partial_{\tau}\phi\right|_{\tau=0}=\int_{0}^{\tau}\Delta\phi\frac{1}{(e^{-t_{0}}-\tau)^{n-1}}d\tau,$

and so

$a^{n-1}a\partial_{t}\phi=a(t_{0})^{n-1}a(t_{0})\left.\partial_{t}\phi\right|_{t=t_{0}}+\int_{0}^{\tau}\Delta\phi\frac{1}{(e^{-t_{0}}-\tau)^{n-1}}d\tau,$

that is,

$\partial_{t}\phi=(a(t))^{-n}\left(a(t_{0})^{n}\phi_{1}+\int_{0}^{\tau}\Delta\phi\frac{1}{(e^{-t_{0}}-\tau)^{n-1}}d\tau\right).$

Hence, using the bound from (3), namely
$\|\Delta\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\leq C$ for all $t\geq
t_{0}$, we obtain

$\displaystyle\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\phantom{\Big{(}\frac{a(t_{0})}{a(t)}\Big{)}^{n-2}}$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!\\!(a(t))^{-n}\\!\left(a(t_{0})^{n}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}+\int_{0}^{\tau}\|\Delta\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\frac{1}{(e^{-t_{0}}-\tau)^{n-1}}d\tau\right)$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!(a(t))^{-n}\\!\left(a(t_{0})^{n}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}\\!+\\!\frac{C}{n-2}\left((e^{-t_{0}}\\!-\\!\tau)^{2-n}\\!-(e^{-t_{0}})^{2-n}\right)\\!\right)$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!(a(t))^{-n}\\!\left(a(t_{0})^{n}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}\\!+\\!\frac{C}{n-2}\left((a(t))^{n-2}\\!-(a(t_{0}))^{n-2}\right)\\!\right)$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!\\!(a(t))^{-n}(a(t))^{n-2}\left(\\!\frac{a(t_{0})^{n}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}}{(a(t))^{n-2}}\\!+\\!\frac{C}{n-2}\left(1\\!-\\!\Big{(}\frac{a(t_{0})}{a(t)}\Big{)}^{n-2}\right)\\!\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{(a(t))^{2}}\left(\frac{a(t_{0})^{n}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}}{(a(t_{0}))^{n-2}}\\!+\\!\frac{C}{n-2}\left(1\\!-\\!0\right)\\!\right).$

Hence

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\leq\frac{1}{(a(t))^{2}}\left((a(t_{0}))^{2}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{n})}\\!+\\!\frac{C}{n-2}\\!\right),$

and so

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim(a(t))^{-2}.$

This completes the proof of Theorem 2.3. ∎

######  Remark 2.4.

The case when $n=2$ and $a(t)=e^{t}$:

Integrating

$\partial_{\tau}(a\partial_{\tau}\phi)=a\Delta\phi$

from $\tau=0$ to $\tau$, we obtain

$a\partial_{\tau}\phi-a(t_{0})\left.\partial_{\tau}\phi\right|_{\tau=0}=\int_{0}^{\tau}\Delta\phi\frac{1}{e^{-t_{0}}-\tau}d\tau,$

and so

$\partial_{t}\phi=(a(t))^{-2}\left(a(t_{0})^{2}\phi_{1}+\int_{0}^{\tau}\Delta\phi\frac{1}{e^{-t_{0}}-\tau}d\tau\right).$

Hence

$\displaystyle\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{2})}\phantom{\frac{a(t_{0})}{a(t)}\bigg{)}^{n-2}}$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!\\!(a(t))^{-2}\\!\left(\\!a(t_{0})^{2}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{2})}+\int_{0}^{\tau}\|\Delta\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{2})}\frac{1}{e^{-t_{0}}-\tau}d\tau\right)\phantom{\frac{a(t_{0})}{a(t)}\bigg{)}^{n-2}}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!(a(t))^{-2}\\!\left(\\!a(t_{0})^{2}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{2})}\\!+\\!C\left(-\log(e^{-t_{0}}\\!-\\!\tau)\Big{|}_{0}^{\tau}\right)\\!\right)\phantom{\frac{a(t_{0})}{a(t)}\bigg{)}^{n-2}}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!(a(t))^{-2}(\log
a(t))\\!\bigg{(}\\!\frac{a(t_{0})^{2}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{2})}}{\log
a(t)}\\!+\\!C\Big{(}1-\frac{\log a(t_{0})}{\log a(t)}\Big{)}\\!\bigg{)}$
$\displaystyle\leq$ $\displaystyle\\!\\!\\!\\!(a(t))^{-2}(\log
a(t))\\!\bigg{(}\\!\frac{a(t_{0})^{2}\|\phi_{1}\|_{L^{\infty}(\mathbb{R}^{2})}}{t_{0}}\\!+\\!C\\!\bigg{)},$

and so

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim(a(t))^{-2}\log
a(t).$

This can be viewed as an improvement to [8, Theorem 1] in the special case
when $a(t)=e^{t}$ and $n=2$, since

$\log a(t)=t\lesssim e^{\delta t}=1+\delta t+\cdots.$

######  Remark 2.5.

The case when $a(t)=t^{p}$, $p\geq 1$:

One can prove an analogue of Theorem 2.3 when $a(t)=t^{p}$ as well. In this
case, the $\epsilon$ from Theorem 2.1 can be chosen to be any number
satisfying

$\epsilon>\frac{1}{p},$

and so Theorem 2.1 gives the decay estimate

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left(a(t)\right)^{-2+\frac{1}{p}+\delta}=t^{-(2p-1-\delta^{\prime})},$

where $\delta^{\prime}>0$ can be chosen arbitrarily. We can improve this to
the following:

$\|\partial_{t}\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left(a(t)\right)^{-2+\frac{1}{p}}=t^{-(2p-1)}.$

The proof is the same, mutatis mutandis, as that of Theorem 2.3.

######  Remark 2.6.

Using a similar method, one can also obtain an improvement to [8, Theorem 2].
But we will postpone this discussion until after Section 4, since we will need
some preliminaries about the RNdS spacetime, which will be established in
Section 4.

##  3\. Decay in the de Sitter universe in flat FLRW form

The Klein-Gordon equation is $\square_{g}\phi-m^{2}\phi=0$, that is,

$\frac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}\;\partial^{\mu}\phi)-m^{2}\phi=0.$

In the case of the de Sitter universe in flat FLRW form, we obtain

$-\ddot{\phi}-\frac{n\dot{a}}{a}\dot{\phi}+\frac{1}{a^{2}}\delta^{ij}\partial_{i}\partial_{j}\phi-m^{2}\phi=0.$
(5)

In this section, we will prove Theorem 3.1. We arrive at the guesses for the
specific estimates given in Theorem 3.1 below, based on an analysis using
Fourier modes, assuming spatially periodic solutions. This Fourier mode
analysis is given in Appendix A.

######  Theorem 3.1.

Suppose that

  * •

$I\subset\mathbb{R}$ is an open interval of the form $(t_{*},+\infty)$,
$t_{0}\in I$,

  * •

$m\in\mathbb{R}$,

  * •

$n>2$,

  * •

$(M,g)$ is the expanding de Sitter universe in flat FLRW form, with flat
$n$-dimensional sections, given by $I\times\mathbb{R}^{n}$, with the metric

$g=-dt^{2}+e^{2t}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right),$

  * •

$k>\frac{n}{2}+2$, $\;\;\phi_{0}\in H^{k}(\mathbb{R}^{n})$, $\;\;\phi_{1}\in
H^{k-1}(\mathbb{R}^{n})$, and

  * •

$\phi$ is a smooth solution to the Cauchy problem

$\left\\{\begin{array}[]{rcll}\square_{g}\phi-m^{2}\phi&=&0,&\quad(t\geq
t_{0},\;\mathbf{x}\in\mathbb{R}^{n}),\\\
\phi(t_{0},\mathbf{x})&=&\phi_{0}(\mathbf{x})&\quad(\mathbf{x}\in\mathbb{R}^{n}),\\\
\partial_{t}\phi(t_{0},\mathbf{x})&=&\phi_{1}(\mathbf{x})&\quad(\mathbf{x}\in\mathbb{R}^{n}).\end{array}\right.$

Then for all $t\geq t_{0}$, we have

$\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left\\{\begin{array}[]{ll}a^{-\frac{n}{2}}&\textrm{if
}\;\;|m|>\frac{n}{2},\phantom{a^{\sqrt{\frac{n^{2}}{4}-m^{2}}}}\\\
a^{-\frac{n}{2}}\log a&\textrm{if
}\;\;|m|=\frac{n}{2},\phantom{a^{\sqrt{\frac{n^{2}}{4}-m^{2}}}}\\\
a^{-\frac{n}{2}+\sqrt{\frac{n^{2}}{4}-m^{2}}}&\textrm{if
}\;\;|m|<\frac{n}{2}.\end{array}\right.$

######  Remark 3.2.

We recall that the conformally invariant wave equation in $n+1$ dimensions is

$\left(\square_{g}-\frac{n-1}{4n}R_{g}\right)\phi=0,$

where $R_{g}$ is the scalar curvature of the metric $g$; see for instance
[26]. If $g$ is a FLRW metric with flat $n$-dimensional spatial sections,
having the form given by (1), then

$R_{g}=\frac{2n\ddot{a}}{a}+\frac{n(n-1)\dot{a}^{2}}{a^{2}}.$

Thus in de Sitter space in flat FLRW form, the conformally invariant wave
equation can be interpreted as a Klein-Gordon equation, with the mass
parameter satisfying $m^{2}=\frac{n^{2}-1}{4}$. From [8, Appendix B], we have

$\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim a^{\frac{1-n}{2}},$
(6)

which follows from using the fact that the $L^{\infty}$-norm of
$\psi(\cdot,t)$, defined by $\phi=a^{1-\frac{n+1}{2}}\psi$ (see [8, eq.
(178)]), is uniformly bounded with respect to $t$. The estimate (6) is in
complete agreement with the result of our Theorem 3.1 above, since the
relation $\frac{n^{2}}{4}-m^{2}=\frac{1}{4}$ reduces our bound
$a^{-\frac{n}{2}+\sqrt{\frac{n^{2}}{4}-m^{2}}}$ precisely to
$a^{\frac{1-n}{2}}$.

###  3.1. Preliminary energy function and estimates

Define the energy-momentum tensor $T$ by

$T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}g_{\mu\nu}(\partial_{\alpha}\phi\partial^{\alpha}\phi+m^{2}\phi^{2}).$

Then $\nabla_{\mu}T^{\mu\nu}=0$. Also, in particular,

$T_{00}=\frac{1}{2}\left(\dot{\phi}^{2}+\frac{1}{a^{2}}|\nabla\phi|^{2}+m^{2}\phi^{2}\right)=T^{00}.$

Set

$X=a^{-n}\frac{\partial}{\partial t}.$

Then $X$ is time-like and hence causal, and $X$ is future pointing.

Define $J$ by

$J^{\mu}=T^{\mu\nu}X_{\nu}.$

Then $J$ is causal and past-pointing.

Let $N=\partial_{t}$. Define the energy $E$ by

$E(t)=\int_{\\{t\\}\times\mathbb{R}^{n}}J_{\mu}N^{\mu}=\int_{\mathbb{R}^{n}}\frac{1}{2}\left(\dot{\phi}^{2}+\frac{1}{a^{2}}|\nabla\phi|^{2}+m^{2}\phi^{2}\right)d^{n}\mathbf{x}.$

Define

$\Pi=\frac{1}{2}\mathcal{L}_{X}g=-dt\mathcal{L}_{X}dt+a^{-n+1}\dot{a}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right).$

As

$\mathcal{L}_{X}dt=-na^{-n-1}\dot{a}dt,$

we have

$\Pi=na^{-n-1}\dot{a}dt^{2}+a^{-n+1}\dot{a}\left((dx^{1})^{2}+\cdots+(dx^{n})^{2}\right).$

Hence

$\nabla_{\mu}J^{\mu}=T^{\mu\nu}\Pi_{\mu\nu}=\frac{a^{-n-1}\dot{a}}{2}\left(2n\dot{\phi}^{2}+\frac{2}{a^{2}}|\nabla\phi|^{2}\right)\geq
0.$

For $R>0$, define

$B_{0}:=\\{(t_{0},\mathbf{x})\in
I\times\mathbb{R}^{n}:\langle\mathbf{x},\mathbf{x}\rangle_{\scriptscriptstyle\mathbb{R}^{n}}\leq
R^{2}\\}.$

The future domain of dependence of $B_{0}$ is denoted by $D^{+}(B_{0})$.

Let $t_{1}>t_{0}$. We will now apply the divergence theorem to the region

$\mathcal{R}:=D^{+}(B_{0})\;\\!\cap\;\\!\\{(t,\mathbf{x})\in M:t\leq
t_{1}\\}.$

We have

$\int_{\mathcal{R}}(\nabla_{\mu}J^{\mu})\epsilon=\int_{\partial\mathcal{R}}J\intprod\epsilon.$

Using

  * •

$\nabla_{\mu}J^{\mu}\geq 0$, and

  * •

the fact that the boundary contribution on $C$, the null portion of
$\partial\mathcal{R}$, is nonpositive (since $J$ is causal and past-pointing),

we obtain the inequality

$\int_{B_{0}}\\!\frac{1}{2}\\!\left(\dot{\phi}^{2}\\!+\\!\frac{1}{a^{2}}|\nabla\phi|^{2}\\!+\\!m^{2}\phi^{2}\right)\\!d^{n}\mathbf{x}\;\\!\geq\;\\!\int_{B_{1}}\\!\frac{1}{2}\\!\left(\dot{\phi}^{2}\\!+\\!\frac{1}{a^{2}}|\nabla\phi|^{2}\\!+\\!m^{2}\phi^{2}\right)\\!d^{n}\mathbf{x}.$

Passing the limit $R\rightarrow\infty$ yields $E(t_{1})\leq E(t_{0})<+\infty$.
As $t_{1}>t_{0}$ was arbitrary, we obtain

$\forall t\geq
t_{0},\;\;E(t)=\int_{\mathbb{R}^{n}}\frac{1}{2}\left(\dot{\phi}^{2}+\frac{1}{a^{2}}|\nabla\phi|^{2}+m^{2}\phi^{2}\right)d^{n}\mathbf{x}\leq
E(t_{0})\leq\infty.$

From here, it follows that for all $t\geq t_{0}$,

$\displaystyle\int_{\mathbb{R}^{n}}\dot{\phi}^{2}d^{n}\mathbf{x}\lesssim 1,$
$\displaystyle\int_{\mathbb{R}^{n}}|\nabla\phi|^{2}d^{n}\mathbf{x}\lesssim
a^{2},\;\;\textrm{ and }$
$\displaystyle\int_{\mathbb{R}^{n}}\phi^{2}d^{n}\mathbf{x}\lesssim
1\;\;\;\;(\textrm{if }m\neq 0).$

But since each partial derivative
$(\partial_{x^{1}})^{i_{1}}\cdots(\partial_{x^{n}})^{i_{n}}\phi$ is also a
solution of the Klein-Gordon equation, it follows from $\phi_{0}\in
H^{k}(\mathbb{R}^{n})$ and $\phi_{1}\in H^{k-1}(R^{n})$ for a
$k>\frac{n}{2}+2$, that also $\phi(t,\cdot)\in H^{k}(\mathbb{R}^{n})$ and
$\partial_{t}\phi(t,\cdot)\in H^{k-1}(\mathbb{R}^{n})$, and moreover

$\displaystyle\|\dot{\phi}\|_{H^{k^{\prime}}(\mathbb{R}^{n})}$
$\displaystyle\lesssim$ $\displaystyle 1,$
$\displaystyle\|\partial_{i}\phi\|_{H^{k^{\prime}}(\mathbb{R}^{n})}$
$\displaystyle\lesssim$ $\displaystyle a,\;\;\textrm{ and }$
$\displaystyle\|\phi\|_{H^{k^{\prime}}(\mathbb{R}^{n})}$
$\displaystyle\lesssim$ $\displaystyle 1\;\;\;\;(\textrm{if }m\neq 0),$

where $k^{\prime}:=k-1$.

###  3.2. The auxiliary function $\psi$ and its PDE

Motivated by the decay rate we anticipate for $\phi$, we define the auxiliary
function $\psi$ by

$\psi:=a^{\kappa}\phi,$

where

$\kappa:=\left\\{\begin{array}[]{ll}\frac{n}{2}&\textrm{if
}\;|m|\geq\frac{n}{2},\\\ \frac{n}{2}-\sqrt{\frac{n^{2}}{4}-m^{2}}&\textrm{if
}\;|m|\leq\frac{n}{2}.\end{array}\right.$

Then, using (5), it can be shown that $\psi$ satisfies the equation

$\ddot{\psi}+(\kappa^{2}-n\kappa+m^{2})\psi+(n-2\kappa)\dot{\psi}-\frac{1}{a^{2}}\Delta\psi=0.$
(7)

###  3.3. The case $|m|>\frac{n}{2}$

We have $\kappa=\frac{n}{2}$, so that $n-2\kappa=0$, while

$\kappa^{2}-n\kappa+m^{2}=m^{2}-\frac{n^{2}}{4},$

and thus (7) becomes

$\ddot{\psi}-\frac{1}{a^{2}}\Delta\psi+\left(m^{2}-\frac{n^{2}}{4}\right)\psi=0.$

We note that if $\phi\in H^{\ell}(\mathbb{R}^{n})$ and $\dot{\phi}\in
H^{\ell-1}(\mathbb{R}^{n})$ for some $\ell$, then $\psi\in
H^{\ell}(\mathbb{R}^{n})$ too, and also

$\dot{\psi}=\frac{n}{2}a^{\frac{n}{2}-1}\dot{a}\phi+a^{\frac{n}{2}}\dot{\phi}\in
H^{\ell-1}(\mathbb{R}^{n}).$

Define the new energy $\mathcal{E}$, associated with the $\psi$-evolution, by

$\mathcal{E}(t):=\frac{1}{2}\int_{\mathbb{R}^{n}}\left(\dot{\psi}^{2}+\frac{1}{a^{2}}|\nabla\psi|^{2}+\Big{(}m^{2}-\frac{n^{2}}{4}\Big{)}\psi^{2}\right)d^{n}\mathbf{x}.$

Then using the fact that $a=e^{t}=\dot{a}>0$, and also equation (7), we obtain

$\displaystyle\mathcal{E}^{\prime}(t)$ $\displaystyle=$
$\displaystyle\\!\\!\\!\int_{\mathbb{R}^{n}}\left(\dot{\psi}\ddot{\psi}-\frac{a\dot{a}}{a^{4}}|\nabla\psi|^{2}+\frac{1}{a^{2}}\langle\nabla\psi,\nabla\dot{\psi}\rangle+\Big{(}m^{2}-\frac{n^{2}}{4}\Big{)}\psi\dot{\psi}\right)d^{n}\mathbf{x}$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!\int_{\mathbb{R}^{n}}\left(\dot{\psi}\ddot{\psi}+\frac{1}{a^{2}}\langle\nabla\psi,\nabla\dot{\psi}\rangle+\Big{(}m^{2}-\frac{n^{2}}{4}\Big{)}\psi\dot{\psi}\right)d^{n}\mathbf{x}$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!\int_{\mathbb{R}^{n}}\left(\dot{\psi}\Big{(}\frac{1}{a^{2}}\Delta\psi-\Big{(}m^{2}-\frac{n^{2}}{4}\Big{)}\psi\Big{)}+\frac{1}{a^{2}}\langle\nabla\psi,\nabla\dot{\psi}\rangle+\Big{(}m^{2}-\frac{n^{2}}{4}\Big{)}\psi\dot{\psi}\right)d^{n}\mathbf{x}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\frac{1}{a^{2}}\int_{\mathbb{R}^{n}}\left(\dot{\psi}\Delta\psi+\langle\nabla\psi,\nabla\dot{\psi}\rangle\right)d^{n}\mathbf{x}=\frac{1}{a^{2}}\int_{\mathbb{R}^{n}}\nabla\cdot(\dot{\psi}\nabla\psi)d^{n}\mathbf{x}.$

For a fixed $t$, and for a ball $B(\mathbf{0},r)\subset\mathbb{R}^{n}$, where
$r>0$, it follows from the divergence theorem (since $\dot{\psi}$ and
$\nabla{\psi}$ are smooth), that

$\int_{B(\mathbf{0},r)}\nabla\cdot(\dot{\psi}\nabla\psi)\;\\!d^{n}\mathbf{x}=\int_{\partial
B(\mathbf{0},r)}\dot{\psi}\;\\!\langle\nabla\psi,\mathbf{n}\rangle\;\\!d\sigma_{r},$

where $d\sigma_{r}$ is the surface area measure on the sphere $S_{r}=\partial
B(\mathbf{0},r)$, and $\mathbf{n}$ is the outward-pointing unit normal. The
right hand side surface integral tends to $0$ as $r\rightarrow+\infty$, by an
application of Lemma 7.1, given in Appendix B.

So for $t\geq t_{0}$, we have $\mathcal{E}^{\prime}(t)\leq 0$, which yields
$\mathcal{E}(t)\leq\mathcal{E}(t_{0})$. In particular, for all $t\geq t_{0}$,
$\|\psi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\lesssim C$, that is,
$\|a^{\frac{n}{2}}\phi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\lesssim C$, and
so222We note that to reach this conclusion, we used Lemma 7.1, for which we
need $\dot{\psi}(t,\cdot),\nabla\psi(t,\cdot)\in H^{1}(\mathbb{R}^{n})$, which
means that the initial conditions for $\phi$ must be such that $\phi_{0}\in
H^{2}(\mathbb{R}^{n})$ and $\phi_{1}\in H^{1}(\mathbb{R}^{n})$.

$\|\phi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\lesssim a^{-\frac{n}{2}}.$ (8)

Then with enough regularity on $\phi_{0},\phi_{1}$ at the outset, that is, if
$\phi_{0}\in H^{k}(\mathbb{R}^{n})$ and $\phi_{1}\in H^{k-1}(\mathbb{R}^{n})$
for a $k>\frac{n}{2}+2$, and by considering
$(\partial_{x^{1}})^{i_{1}}\cdots(\partial_{x^{n}})^{i_{n}}\phi$ as a solution
to the Klein-Gordon equation, we arrive at333Note that in order to use the
estimate (8), for
$D\phi:=(\partial_{x^{1}})^{i_{1}}\cdots(\partial_{x^{n}})^{i_{n}}\phi$
replacing $\phi$, where $|(i_{1},\cdots,i_{n})|=:k^{\prime}$, we must ensure
that the initial conditions for $D\phi$, namely
$(D\phi(t_{0},\cdot),D\dot{\phi}(t_{0},\cdot))$ is in
$(H^{2}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n}))$, which is guaranteed if the
initial condition for $\phi$, namely $(\phi_{0},\phi_{1})$ is in
$(H^{k}(\mathbb{R}^{n}),H^{k-1}(\mathbb{R}^{n}))$, with $k-k^{\prime}=2$.

$\|\phi(t,\cdot)\|_{H^{k^{\prime}}(\mathbb{R}^{n})}\lesssim a^{-\frac{n}{2}},$

where $k^{\prime}:=k-2$. As $k^{\prime}=k-2>\frac{n}{2}$, we have, using the
Sobolev inequality, that

$\forall t\geq
t_{0},\;\;\;\;\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim
a^{-\frac{n}{2}}.$

This completes the proof of Theorem 3.1 in the case when $|m|>\frac{n}{2}$.

###  3.4. The case $|m|<\frac{n}{2}$

We have

$\kappa=\displaystyle\frac{n}{2}-\sqrt{\frac{n^{2}}{4}-m^{2}},\quad\;\;n-2\kappa=2\displaystyle\sqrt{\frac{n^{2}}{4}-m^{2}}\;>0,$

and $\kappa^{2}-n\kappa+m^{2}=0$. Equation (7) becomes

$\ddot{\psi}+2\Big{(}\sqrt{\frac{n^{2}}{4}-m^{2}}\Big{)}\;\dot{\psi}-\frac{1}{a^{2}}\Delta\psi=0.$

Defining
$\displaystyle\widetilde{\mathcal{E}}(t):=\frac{1}{2}\int_{\mathbb{R}^{n}}\left(\dot{\psi}^{2}+\frac{1}{a^{2}}|\nabla\psi|^{2}\right)d^{n}\mathbf{x},$
we obtain

$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\widetilde{\mathcal{E}}^{\prime}(t)\\!\\!\\!\\!\\!$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}^{n}}\left(\dot{\psi}\ddot{\psi}-\frac{\dot{a}}{a^{3}}|\nabla\psi|^{2}+\frac{1}{a^{2}}\langle\nabla\psi,\nabla\dot{\psi}\rangle\right)d^{n}\mathbf{x}\phantom{aa}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!\int_{\mathbb{R}^{n}}\\!\\!\left(\dot{\psi}\Big{(}\\!-\\!2\Big{(}\sqrt{\frac{n^{2}}{4}\\!-\\!m^{2}}\Big{)}\;\dot{\psi}\\!+\\!\frac{1}{a^{2}}\Delta\psi\Big{)}\\!-\\!\frac{\dot{a}}{a^{3}}|\nabla\psi|^{2}\\!+\\!\frac{1}{a^{2}}\langle\nabla\psi,\\!\nabla\dot{\psi}\rangle\\!\right)d^{n}\mathbf{x}\phantom{aaa}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!\\!-2\Big{(}\sqrt{\frac{n^{2}}{4}\\!-\\!m^{2}}\Big{)}\int_{\mathbb{R}^{n}}\dot{\psi}^{2}d^{n}\mathbf{x}\\!-\\!\frac{\dot{a}}{a^{3}}\int_{\mathbb{R}^{n}}|\nabla\psi|^{2}d^{n}\mathbf{x}.$

Using $a=e^{t}=\dot{a}$, we obtain

$\displaystyle\widetilde{\mathcal{E}}^{\prime}(t)$ $\displaystyle=$
$\displaystyle-4\Big{(}\sqrt{\frac{n^{2}}{4}-m^{2}}\Big{)}\;\frac{1}{2}\int_{\mathbb{R}^{n}}\dot{\psi}^{2}d^{n}\mathbf{x}-2\frac{1}{2}\int_{\mathbb{R}^{n}}\frac{1}{a^{2}}|\nabla\psi|^{2}d^{n}\mathbf{x}$
$\displaystyle\leq$
$\displaystyle-\min\left\\{4\Big{(}\sqrt{\frac{n^{2}}{4}-m^{2}}\Big{)},\;2\right\\}\cdot\frac{1}{2}\int_{\mathbb{R}^{n}}\left(\dot{\psi}^{2}+\frac{1}{a^{2}}|\nabla\psi|^{2}\right)d^{n}\mathbf{x}\phantom{aaa}$
$\displaystyle=$
$\displaystyle-\;\theta\cdot\widetilde{\mathcal{E}}(t),\phantom{\frac{1}{2}\int_{\mathbb{R}^{n}}\left(\frac{1}{a^{2}}|\nabla\psi|^{2}\right)}$

where

$\theta:=\min\left\\{4\Big{(}\sqrt{\frac{n^{2}}{4}-m^{2}}\Big{)}\;,\;2\right\\}>0.$

So
$\widetilde{\mathcal{E}}^{\prime}(t)+\theta\cdot\widetilde{\mathcal{E}}(t)\leq
0$. Multiplying throughout by $e^{\theta t}>0$, we obtain

$\frac{d}{dt}\left(e^{\theta t}\cdot\widetilde{\mathcal{E}}(t)\right)\leq 0.$

Integrating from $t_{0}$ to $t$ yields

$e^{\theta t}\cdot\widetilde{\mathcal{E}}(t)\leq e^{\theta
t_{0}}\cdot\widetilde{\mathcal{E}}(t_{0}),$

that is, $\widetilde{\mathcal{E}}(t)\lesssim e^{-\theta t}$. In particular,

$\|\dot{\psi}(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\leq\sqrt{2\;\\!\widetilde{\mathcal{E}}(t)}\lesssim
e^{-\frac{\theta}{2}t}.$

We have
$\displaystyle\psi(t,\mathbf{x})=\psi(t_{0},\mathbf{x})+\int_{t_{0}}^{t}(\partial_{t}\psi)(s,\mathbf{x})ds,$
and so

$\displaystyle\|\psi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}$ $\displaystyle\leq$
$\displaystyle\|\psi(t_{0},\cdot)\|_{L^{2}(\mathbb{R}^{n})}+\int_{t_{0}}^{t}\|(\partial_{t}\psi)(s,\cdot)\|_{L^{2}(\mathbb{R}^{n})}ds,$
$\displaystyle\lesssim$ $\displaystyle
A+\int_{t_{0}}^{t}Be^{-\frac{\theta}{2}s}ds=A+B\frac{e^{-\frac{\theta}{2}t_{0}}-e^{-\frac{\theta}{2}t}}{\theta/2}\lesssim
C.$

Thus for all $t\geq t_{0}$, we have

$\|\phi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}=a^{-\kappa}\|\psi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\lesssim
a^{-\kappa}.$

By considering
$(\partial_{x^{1}})^{i_{1}}\cdots(\partial_{x^{n}})^{i_{n}}\phi$ and using the
Sobolev inequality, we have

$\forall t\geq
t_{0},\;\;\;\;\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim
a^{-\kappa}=a^{-(\frac{n}{2}-\sqrt{\frac{n^{2}}{4}-m^{2}}\;\\!)}.$

This completes the proof of Theorem 3.1 in the case when $|m|<\frac{n}{2}$.

###  3.5. The case $|m|=\frac{n}{2}$

We have $\kappa=\frac{n}{2}$, and equation (7) becomes
$\displaystyle\ddot{\psi}-\frac{1}{a^{2}}\Delta\psi=0.$

Defining the same energy as we used earlier in the case when
$|m|<\frac{n}{2}$,

$\widetilde{\mathcal{E}}(t):=\frac{1}{2}\int_{\mathbb{R}^{n}}\left(\dot{\psi}^{2}+\frac{1}{a^{2}}|\nabla\psi|^{2}\right)d^{n}\mathbf{x},$

we obtain

$\displaystyle\widetilde{\mathcal{E}}^{\prime}(t)\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\\!\int_{\mathbb{R}^{n}}\left(\dot{\psi}\ddot{\psi}-\frac{\dot{a}}{a^{3}}|\nabla\psi|^{2}+\frac{1}{a^{2}}\langle\nabla\psi,\nabla\dot{\psi}\rangle\right)d^{n}\mathbf{x}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!\int_{\mathbb{R}^{n}}\\!\\!\left(\dot{\psi}\frac{1}{a^{2}}\Delta\psi\\!-\\!\frac{\dot{a}}{a^{3}}|\nabla\psi|^{2}\\!+\\!\frac{1}{a^{2}}\langle\nabla\psi,\\!\nabla\dot{\psi}\rangle\\!\right)d^{n}\mathbf{x}\\!=\\!-\frac{\dot{a}}{a^{3}}\int_{\mathbb{R}^{n}}\\!\\!|\nabla\psi|^{2}d^{n}\mathbf{x}\leq
0.$

So $\widetilde{\mathcal{E}}(t)\leq\widetilde{\mathcal{E}}(t_{0})$ for $t\geq
t_{0}$. In particular,
$\|\dot{\psi}(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\lesssim B$ for $t\geq t_{0}$.
Again,

$\displaystyle\psi(t,\mathbf{x})=\psi(t_{0},\mathbf{x})+\int_{t_{0}}^{t}(\partial_{t}\psi)(s,\mathbf{x})ds,$

gives

$\displaystyle\|\psi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}$ $\displaystyle\leq$
$\displaystyle\|\psi(t_{0},\cdot)\|_{L^{2}(\mathbb{R}^{n})}+\int_{t_{0}}^{t}\|(\partial_{t}\psi)(s,\cdot)\|_{L^{2}(\mathbb{R}^{n})}ds,$
$\displaystyle\lesssim$ $\displaystyle A^{\prime}+\int_{t_{0}}^{t}Bds\lesssim
A+Bt\lesssim\log a.$

Thus for all $t\geq t_{0}$, we have
$\|\phi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}=a^{-\kappa}\|\psi(t,\cdot)\|_{L^{2}(\mathbb{R}^{n})}\lesssim
a^{-\kappa}\log a.$ Hence (by considering
$(\partial_{x^{1}})^{i_{1}}\cdots(\partial_{x^{n}})^{i_{n}}\phi$ and using the
Sobolev inequality)

$\forall t\geq
t_{0},\;\;\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim
a^{-\frac{n}{2}}\log a.$ (9)

(One can show that this bound is sharp; see Appendix C.)

This completes the proof of Theorem 3.1.

##  4\. Decay in the cosmological region of the $\mathrm{RNdS}$ spacetime

The Reissner-Nordström-de Sitter (RNdS) spacetime $(M,g)$ is a solution to the
Einstein-Maxwell equations with a positive cosmological constant, and it
represents a pair444We note that there is no solution analogous to RNdS but
with only one black hole. This is analogous to (but much more complicated
than, and still not fully understood) the fact that one cannot have a single
electric charge on a spherical universe (Gauss’s law requires that the total
charge must be zero). In fact, the fundamental solution of the Laplace
equation on the sphere gives a unit positive charge at some point and a unit
negative charge at the antipodal point. One can have more than two black
holes, for instance the so-called Kastor-Traschen solution [17]. of antipodal
charged black holes in a spherical555“Spherical” here means that the Cauchy
hypersurface (that is, “space”) is an $n$-sphere. universe which is undergoing
accelerated expansion. The Reissner-Nordström-de Sitter metric in $n+1$
dimensions is given by

$g=-\frac{1}{V}dr^{2}+Vdt^{2}+r^{2}d\Omega^{2},$

where

$V=\displaystyle r^{2}+\frac{2M}{r^{n-2}}-\frac{e^{2}}{r^{n-1}}-1,$

and $d\Omega^{2}$ is the unit round metric on $S^{n-1}$. The constants $M$ and
$e$ are proportional to the mass and the charge, respectively, of the black
holes, and the cosmological constant is chosen to be

$\Lambda=\frac{n(n-1)}{2}$

by an appropriate choice of units.

Consider the polynomial

$p(r):=r^{n-1}V(r)=r^{n+1}-r^{n-1}+2Mr-e^{2}.$

As $p(0)=-e^{2}<0$ and as $p(r)\stackrel{{\scriptstyle
r\rightarrow\infty}}{{\longrightarrow}}\infty$, it follows that $p$ will have
a real root in $(0,+\infty)$, and the largest real root of $p$, which we
denote by $r_{c}$, must be positive. If $r>r_{c}$, then clearly $p(r)>0$, and
so also $V(r)>0$.

It can also be seen that $p$ has at most three distinct positive roots.
Suppose, on the contrary, that $p$ has more than three distinct positive
roots: $r_{1}<r_{2}<r_{3}<r_{4}$. Applying Rolle’s theorem to $p$ on
$[r_{i},r_{i+1}]$ ($i=1,2,3$), we conclude that $p^{\prime}$ must have three
distinct roots $r_{i}^{\prime}\in(r_{i},r_{i+1})$ ($i=1,2,3$). Applying
Rolle’s theorem to $p^{\prime}$ on $[r_{i}^{\prime},r_{i+1}^{\prime}]$
($i=1,2$), we conclude that $p^{\prime\prime}$ must have two distinct roots
$r_{i}^{\prime\prime}\in(r_{i}^{\prime},r_{i+1}^{\prime})$ ($i=1,2$). But

$p^{\prime\prime}=r^{n-3}n(n+1)\Big{(}r^{2}-\frac{(n-1)(n-2)}{n(n+1)}\Big{)},$

which has only one positive root, a contradiction.

The ‘subextremality’ assumption on the RNdS spacetime made in Theorem 3.1,
refers to a nondegeneracy of the positive roots of $p$: we assume that there
are exactly three positive roots, $r_{-},r_{+}$ and $r_{c}$, and

$0<r_{-}<r_{+}<r_{c}.$

These describe the event horizon $r=r_{+}$, and the Cauchy ‘inner’ horizon
$\;r=r_{-}$. It can be seen that the subextremality condition then implies
$p^{\prime}(r_{c})>0$. (Indeed, $p^{\prime}(r_{c})$ cannot be negative, as
otherwise $p$ would acquire a root larger than $r_{c}$ since
$p(r)\stackrel{{\scriptstyle r\rightarrow\infty}}{{\longrightarrow}}\infty$.
Also, if $p^{\prime}(r_{c})=0$, then Rolle’s theorem implies again that
$p^{\prime}$ would have three positive roots, ones in $(r_{-},r_{+})$ and
$(r_{+},r_{c})$, and one at $r_{c}$, which is impossible, as we had seen
above.) $p^{\prime}(r_{c})>0$ implies that $V^{\prime}(r_{c})>0$. We will also
assume that

$V^{\prime\prime}(r_{c})>0.$

Our assumptions have the following consequence, which will be used in our
proof of Theorem 4.2.

######  Lemma 4.1 (Global redshift).

$V^{\prime}(r)>0$ for all $r\geq r_{c}$.

###### Proof.

We have

$V^{\prime}(r)=\frac{rp^{\prime}(r)-(n-1)p(r)}{r^{n}}=\frac{2r^{n+1}+2(3-n)Mr+(n-1)e^{2}}{r^{n}}=:\frac{q(r)}{r^{n}}.$

As $V^{\prime}(r_{c})>0$, we have $q(r_{c})>0$. Also
$V^{\prime\prime}(r_{c})>0$ and so $V^{\prime}$ is increasing near $r_{c}$.
But then $q(r)=r^{n}V^{\prime}(r)$ is also increasing near $r_{c}$, and in
particular, $q^{\prime}(r_{c})\geq 0$. Let us suppose that there exists an
$r_{*}>r_{c}$ such that $V^{\prime}(r_{*})=0$, and let $r_{*}$ be the smallest
such root. Then $q(r_{*})=0$ too. We note that

$q^{\prime}=2(n+1)r^{n}+2(3-n)M,$

and so $q^{\prime}$ can have only one nonnegative root, namely
$\left(\frac{(n-3)}{n+1}M\right)^{\frac{1}{n}}\geq 0.$

  * $1^{\circ}$

$r_{*}$ is a repeated root of $q$. Then $q^{\prime}(r_{*})=0$.

If in addition $q^{\prime}(r_{c})=0$, then we arrive at a contradiction, since
$q^{\prime}$ then has two positive roots (at $r_{c}$ and at $r_{*}$), which is
impossible.

If $q^{\prime}(r_{c})>0$, then we arrive at a contradiction as follows. As $q$
is increasing near $r_{c}$, and since $q(r_{c})>0=q(r_{*})$, it follows by the
intermediate value theorem that there is some $r_{c}^{\prime}\in(r_{c},r_{*})$
such that $q(r_{c}^{\prime})=q(r_{c})$. But by Rolle’s theorem applied to $q$
on $[r_{c},r_{c}^{\prime}]$, there must exist an
$r_{*}^{\prime}\in(r_{c},r_{c}^{\prime})$ such that
$q^{\prime}(r_{*}^{\prime})=0$. Again $q^{\prime}$ acquires two zeros (at
$r_{*}$ and at $r_{*}^{\prime}$), which is impossible.

  * $2^{\circ}$

$r_{*}$ is a simple root of $q$. But as $q(r)\stackrel{{\scriptstyle
r\rightarrow\infty}}{{\longrightarrow}}\infty$, it follows that there must be
at least one more root $r_{**}>r_{*}$ of $q$. By Rolle’s theorem applied to
$q$ on $[r_{*},r_{**}]$, it follows that $q^{\prime}(r_{**}^{\prime})=0$ for
some $r_{**}^{\prime}\in(r_{*},r_{**})$.

If in addition $q^{\prime}(r_{c})=0$, then we arrive at a contradiction, since
$q^{\prime}$ then has two positive roots (at $r_{c}$ and at
$r_{**}^{\prime}$), which is impossible.

If $q^{\prime}(r_{c})>0$, then, as in the last paragraph of $1^{\circ}$ above,
there exists an $r_{*}^{\prime}\in(r_{c},r_{c}^{\prime})\subset(r_{c},r_{*})$
such that $q^{\prime}(r_{*}^{\prime})=0$. Thus $q^{\prime}$ again gets two
positive roots (at $r_{*}^{\prime}$ and at $r_{**}^{\prime}$), which is
impossible.

This shows that our assumption the $V^{\prime}$ is zero beyond $r_{c}$ is
incorrect. ∎

The hypersurfaces of constant $r$ are spacelike cylinders with a future-
pointing unit normal vector field

$N=V^{\frac{1}{2}}\frac{\partial}{\partial r},$

and volume element

$dV_{n}=V^{\frac{1}{2}}r^{n-1}dtd\Omega.$

The global structure of a maximal spherically symmetric extension of this
metric can be depicted by a conformal Penrose diagram shown below, repeated
periodically; see for example [7].

Figure 1. Conformal diagram of the Reissner-Nordström-de Sitter spacetime.

We are interested in the behaviour of the solution to the Klein-Gordon
equation in the cosmological region $\mathcal{R}_{5}$ of this spacetime (see
Figure 1), bounded by the cosmological horizon branches
$\mathcal{C}\mathcal{H}_{1}^{+}$, $\mathcal{C}\mathcal{H}_{2}^{+}$, the future
null infinity ${\mathscr{I}}^{+}$, and the point $i^{+}$. In particular, we
want to obtain estimates for the decay rate of $\phi$ as $r\rightarrow\infty$.
We guess the decay rates simply by substituting $r$ instead of $e^{t}$ in the
estimates we had obtained for the decay rate of $\phi$ with respect to $t$ in
the case of the de Sitter universe in flat FLRW form from the previous Section
3.

We will prove the following result.

######  Theorem 4.2.

Suppose that

  * •

$\epsilon>0$,

  * •

$m\in\mathbb{R}$,

  * •

$M>0$,

  * •

$e>0$,

  * •

$n>2$,

  * •

$(M,g)$ is the $(n+1)$-dimensional subextremal Reissner-Nordström-de Sitter
solution given by the metric

$g=-\frac{1}{V}dr^{2}+Vdt^{2}+r^{2}d\Omega^{2},$

where

$V=r^{2}+\frac{2M}{r^{n-2}}-\frac{e^{2}}{r^{n-1}}-1,$

and $d\Omega^{2}$ is the metric of the unit $(n-1)$-dimensional sphere
$S^{n-1}$,

  * •

$k>\frac{n}{2}+2$, and

  * •

$\phi$ is a smooth solution to $\square_{g}\phi-m^{2}\phi=0$ such that

$\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{1}^{+})}<+\infty\;\;\;\;\textrm{ and
}\;\;\;\;\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{2}^{+})}<+\infty,$

where
$\mathcal{C}\mathcal{H}_{1}^{+}\simeq\mathcal{C}\mathcal{H}_{2}^{+}\simeq\mathbb{R}\times
S^{n-1}$ are the two components of the future cosmological horizon,
parameterised by the flow parameter of the global Killing vector field
$\frac{\partial}{\partial t}$.

Then there exists a $r_{0}$ large enough so that for all $r\geq r_{0}$,

$\|\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}\lesssim\left\\{\begin{array}[]{ll}r^{-\frac{n}{2}+\epsilon}&\textrm{if
}\;\;|m|>\frac{n}{2},\phantom{a^{\sqrt{\frac{n^{2}}{4}-m^{2}}}}\\\
r^{-\frac{n}{2}+\sqrt{\frac{n^{2}}{4}-m^{2}}\;\\!+\epsilon}&\textrm{if
}\;\;|m|\leq\frac{n}{2}.\end{array}\right.$

###  4.1. Preliminary energy function

For a $\phi$ defined in the cosmological region $\mathcal{R}_{5}$, we define

$\phi^{\prime}:=\frac{\partial\phi}{\partial r}\;\;\textrm{ and
}\;\;\dot{\phi}:=\frac{\partial\phi}{\partial t}.$

We will also use the following notation:

$\begin{array}[]{ccl}{\mathring{\not{\nabla}}}\phi&&\textrm{gradient of
}\phi\textrm{ on }S^{n-1}\textrm{ with respect to the unit round metric},\\\
|{\mathring{\not{\nabla}}}\phi|&&\textrm{norm with respect to the unit round
metric},\\\ {\mathring{\not{\Delta}}}\phi&&\textrm{Laplacian of }\phi\textrm{
on }S^{n-1}\textrm{ with respect to the unit round metric},\\\
{\mathring{\not{g}}}&&\textrm{determinant of the unit round
metric}.\phantom{S^{n-1}{\mathring{\not{\nabla}}}\phi}\end{array}$

Suppose that $\phi$ satisfies the Klein-Gordon equation
$\square_{g}\phi-m^{2}\phi=0$. Recall that the energy-momentum tensor
associated with $\phi$ is given by

$T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}g_{\mu\nu}(\partial_{\alpha}\partial^{\alpha}\phi+m^{2}\phi^{2}).$

Thus

$\displaystyle T(N,N)$ $\displaystyle=$ $\displaystyle\left(\phi^{\prime
2}-\frac{1}{2}\frac{(-1)}{V}\Big{(}\phi^{\prime
2}(-V)+\dot{\phi}^{2}\frac{1}{V}+\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}+m^{2}\phi^{2}\Big{)}\right)V$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left(V\phi^{\prime
2}+\frac{1}{V}\dot{\phi}^{2}+\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}+m^{2}\phi^{2}\right).$

Define

$X:=\frac{V^{\frac{1}{2}}}{r^{n-1}}N=\frac{V^{\frac{1}{2}}}{r^{n-1}}V^{\frac{1}{2}}\frac{\partial}{\partial
r}=\frac{V}{r^{n-1}}\frac{\partial}{\partial r}.$

We define the energy

$\displaystyle E(r)$ $\displaystyle:=$ $\displaystyle\int_{\mathbb{R}\times
S^{n-1}}T(X,N)dV_{n}\phantom{\frac{1}{2}\int_{\mathbb{R}\times S^{n-1}}}$
$\displaystyle=$ $\displaystyle\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\left(V^{2}\phi^{\prime
2}+\dot{\phi}^{2}+\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}+m^{2}V\phi^{2}\right)dtd\Omega.$

###  4.2. The auxiliary function $\psi$ and its PDE

The Klein-Gordon equation $\square_{g}\phi-m^{2}\phi=0$ can be rewritten as:

$\displaystyle\frac{1}{\sqrt{-g}}\;\\!\partial_{\mu}(\sqrt{-g}\;\partial^{\mu}\phi)-m^{2}\phi=0,$
$\displaystyle\Leftrightarrow$
$\displaystyle\frac{1}{r^{n-1}\sqrt{{\mathring{\not{g}}}}}\;\\!\partial_{\mu}\left(r^{n-1}\sqrt{{\mathring{\not{g}}}}\;g^{\mu\nu}\partial_{\nu}\phi\right)-m^{2}\phi=0.$

This becomes

$\frac{1}{r^{n-1}\\!\sqrt{{\mathring{\not{g}}}}}\;\\!\bigg{(}\partial_{r}\Big{(}r^{n-1}\\!\sqrt{{\mathring{\not{g}}}}\;(-V)\partial_{r}\phi\Big{)}+\partial_{t}\Big{(}r^{n-1}\\!\sqrt{{\mathring{\not{g}}}}\;\frac{1}{V}\partial_{t}\phi\Big{)}+\frac{r^{n-1}\\!\sqrt{{\mathring{\not{g}}}}}{r^{2}}\;{\mathring{\not{\Delta}}}\phi\bigg{)}-m^{2}\phi=0$

that is,

$\displaystyle-(V\phi^{\prime})^{\prime}-\frac{(n-1)}{r}V\phi^{\prime}+\frac{\ddot{\phi}}{V}+\frac{1}{r^{2}}{\mathring{\not{\Delta}}}\phi-m^{2}\phi=0,$
$\displaystyle\Leftrightarrow$
$\displaystyle\phi^{\prime\prime}+\frac{(n-1)}{r}\phi^{\prime}+\frac{V^{\prime}}{V}\phi^{\prime}-\frac{\ddot{\phi}}{V^{2}}-\frac{1}{r^{2}V}{\mathring{\not{\Delta}}}\phi+\frac{m^{2}}{V}\phi=0.$

Define

$\psi:=r^{\kappa}\phi,$

where

$\kappa=\left\\{\begin{array}[]{cl}\displaystyle\frac{n}{2}&\textrm{ if
}|m|\geq\displaystyle\frac{n}{2},\\\\[8.5359pt]
\displaystyle\frac{n}{2}-\sqrt{\frac{n^{2}}{4}-m^{2}}&\textrm{ if
}|m|\leq\displaystyle\frac{n}{2}.\end{array}\right.$

Then, using the PDE for $\phi$, it can be shown that

$\psi^{\prime\prime}+\left(\frac{V^{\prime}}{V}+\frac{n-1}{r}-\frac{2\kappa}{r}\right)\psi^{\prime}-\frac{\ddot{\psi}}{V^{2}}-\frac{1}{r^{2}V}{\mathring{\not{\Delta}}}\psi+\theta\psi=0,$
(10)

where

$\theta:=\frac{m^{2}}{V}+\frac{\kappa}{r}\left(\frac{1}{r}-\frac{V^{\prime}}{V}\right)-\frac{\kappa}{r^{2}}(n-1-\kappa).$

###  4.3. The case $|m|>\frac{n}{2}$

Then $\kappa=\frac{n}{2}$, and (10) becomes

$\psi^{\prime\prime}+\left(\frac{V^{\prime}}{V}-\frac{1}{r}\right)\psi^{\prime}-\frac{\ddot{\psi}}{V^{2}}-\frac{1}{r^{2}V}{\mathring{\not{\Delta}}}\psi+\theta\psi=0,$
(11)

where

$\theta:=-\frac{n}{2}\left(\frac{n}{2}-1\right)\frac{1}{r^{2}}+\frac{m^{2}}{V}+\frac{n}{2r}\left(\frac{1}{r}-\frac{V^{\prime}}{V}\right).$

We will use an energy function to obtain the required decay of $\psi$ for
large $r$, and in order to do so, we will need to keep careful track of the
limiting behaviour of the various functions appearing in the expression for
$\theta$ and the coefficients of the PDE (11). We will do this step-by-step in
a sequence of lemmas.

######  Lemma 4.3.

Given any $\epsilon>0$, there exists an $r_{0}$ large enough so that for all
$r\geq r_{0}$,

$\frac{2+\epsilon}{r}\geq\frac{V^{\prime}}{V}\geq\frac{2-\epsilon}{r}.$

###### Proof.

This follows immediately from

$\lim_{r\rightarrow\infty}r\frac{V^{\prime}}{V}\\!=\\!\lim_{r\rightarrow\infty}r\cdot\frac{2r-\frac{2(n-2)M}{r^{n-1}}+\frac{e^{2}(n-1)}{r^{n}}}{r^{2}+\frac{2M}{r^{n-2}}-\frac{e^{2}}{r^{n-1}}-1}\\!=\\!\lim_{r\rightarrow\infty}\frac{2-\frac{2(n-2)M}{r^{n}}+\frac{e^{2}(n-1)}{r^{n+1}}}{1+\frac{2M}{r^{n}}-\frac{e^{2}}{r^{n+1}}-\frac{1}{r^{2}}}\\!=\\!2.$

∎

######  Lemma 4.4.

There exists $r_{0}$ large enough so that for $r\geq r_{0}$, we have
$\theta>0$.

(We note that the proof uses the fact that $|m|>\frac{n}{2}$, and so this
result is specific to this subsection.)

###### Proof.

We have

$\lim_{r\rightarrow\infty}\frac{r^{2}}{V}=\lim_{r\rightarrow\infty}\frac{1}{1+\frac{2M}{r^{n}}-\frac{e^{2}}{r^{n+1}}-\frac{1}{r^{2}}}=1,$

and so there exists a $r_{0}^{\prime}$ such that

$\displaystyle\frac{r^{2}}{V}\geq 1-\epsilon$

for $r\geq r_{0}^{\prime}$. Also, by the previous lemma, there exists a
$r_{0}>r_{0}^{\prime}$ such that

$\frac{V^{\prime}}{V}\leq\frac{2+\epsilon}{r}$

for all $r\geq r_{0}$. Then we have for $r>r_{0}$ that

$\displaystyle\theta$ $\displaystyle=$
$\displaystyle\frac{1}{r^{2}}\left(-\frac{n}{2}\left(\frac{n}{2}-1\right)+m^{2}\frac{r^{2}}{V}+\frac{n}{2}\Big{(}1-\frac{V^{\prime}}{V}r\Big{)}\right)$
$\displaystyle\geq$
$\displaystyle\frac{1}{r^{2}}\left(-\frac{n^{2}}{4}+\frac{n}{2}+m^{2}(1-\epsilon)+\frac{n}{2}\Big{(}1-\frac{(2+\epsilon)}{r}r\Big{)}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{r^{2}}\left(\delta-\epsilon\Big{(}\delta+\frac{n}{2}+\frac{n^{2}}{4}\Big{)}\right),$

where $\delta:=m^{2}-\frac{n^{2}}{4}>0.$

Taking $\epsilon$ at the outset small enough so as to satisfy
$0<\epsilon<\frac{\delta}{\delta+\frac{n}{2}+\frac{n^{2}}{4}},$ we see that
$\theta>0$ for $r\geq r_{0}$. ∎

Define the energy

$\mathcal{E}(r):=\frac{1}{2}\int_{\mathbb{R}\times S^{n-1}}\left(\psi^{\prime
2}+\frac{1}{V^{2}}\dot{\psi}^{2}+\frac{1}{r^{2}V}|{\mathring{\not{\nabla}}}\psi|^{2}+\theta\psi^{2}\right)dtd\Omega.$

(We assume for the moment that this is finite for a sufficiently large
$r_{0}$. Later on, in the subsection on redshift estimates, we will see how
our initial finiteness of Sobolev norms of $\phi$ on the two branches
$\mathcal{C}\mathcal{H}_{1}^{+}$, $\mathcal{C}\mathcal{H}_{2}^{+}$ of the
cosmological horizon guarantees this.)

We now proceed to find an expression for $\mathcal{E}^{\prime}(r)$, and to
simplify it, we will use (11), and the divergence theorem, to get rid of the
terms involving $\ddot{\psi}$ and ${\mathring{\not{\Delta}}}\psi$, the
spherical Laplacian of $\psi$:

$\displaystyle\mathcal{E}^{\prime}(r)\\!\\!\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(\psi^{\prime}\psi^{\prime\prime}+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\frac{1}{V^{2}}\dot{\psi}\dot{\psi}^{\prime}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}\right.$
$\displaystyle\left.\phantom{aaaaaaaa}+\frac{1}{r^{2}V}\langle{\mathring{\not{\nabla}}}\psi,({\mathring{\not{\nabla}}}\psi)^{\prime}\rangle+\frac{\theta^{\prime}}{2}\psi^{2}+\theta\psi\psi^{\prime}\right)dtd\Omega$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(\psi^{\prime}\bigg{(}-\Big{(}\frac{V^{\prime}}{V}-\frac{1}{r}\Big{)}\psi^{\prime}+\frac{1}{V^{2}}\ddot{\psi}+\frac{1}{r^{2}V}{\mathring{\not{\Delta}}}\psi-\cancel{\theta\psi}\bigg{)}\right.$
$\displaystyle\phantom{aaaaaaaa}+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\frac{1}{V^{2}}\dot{\psi}\dot{\psi}^{\prime}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}$
$\displaystyle\left.\phantom{aaaaaaaa}+\frac{1}{r^{2}V}\langle{\mathring{\not{\nabla}}}\psi,({\mathring{\not{\nabla}}}\psi)^{\prime}\rangle+\frac{\theta^{\prime}}{2}\psi^{2}+\cancel{\theta\psi\psi^{\prime}}\right)dtd\Omega$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(-\Big{(}\frac{V^{\prime}}{V}\\!-\\!\frac{1}{r}\Big{)}\psi^{\prime
2}+\cancel{\frac{1}{V^{2}}\ddot{\psi}\psi^{\prime}}+\cancel{\frac{1}{V^{2}}\dot{\psi}\dot{\psi}^{\prime}}+\\!\frac{1}{r^{2}V}\left(\bcancel{\psi^{\prime}{\mathring{\not{\Delta}}}\psi}+\bcancel{\langle{\mathring{\not{\nabla}}}\psi,({\mathring{\not{\nabla}}}\psi)^{\prime}\rangle}\right)\right.$
$\displaystyle\phantom{aaaaaaaa}\left.+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}+\frac{\theta^{\prime}}{2}\psi^{2}\right)dtd\Omega.$

We note that in the above, getting rid of the spherical Laplacian by using the
divergence theorem is allowed because the compact sphere $S^{n-1}$ has no
boundary. For the second time derivative, however, there is a boundary at
infinity (with two connected components), namely

$\lim_{t\rightarrow+\infty}\int_{S^{n-1}}\dot{\psi}\psi^{\prime}d\Omega-\lim_{t\rightarrow-\infty}\int_{S^{n-1}}\dot{\psi}\psi^{\prime}d\Omega,$

which can be seen to be equal to $0$, by Lemma 7.3 from Appendix B. Thus

$\mathcal{E}^{\prime}(r)=\int_{\mathbb{R}\times
S^{n-1}}\left(-\Big{(}\frac{V^{\prime}}{V}-\frac{1}{r}\Big{)}\psi^{\prime
2}+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}+\frac{\theta^{\prime}}{2}\psi^{2}\right)dtd\Omega.$

Let $\epsilon>0$ be given. Then there exists an $r_{0}$ large enough such
that:

(a)
$\displaystyle\frac{V^{\prime}}{V}-\frac{1}{r}\geq\frac{2-\epsilon}{r}-\frac{1}{r}=\frac{1-\epsilon}{r}$,

(b)
$\displaystyle\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}=-2\frac{V^{\prime}}{V}\frac{1}{V^{2}}\leq-2\frac{(2-\epsilon)}{r}\frac{1}{V^{2}}$,

(c)
$\displaystyle\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}=-\frac{1}{r^{2}V}\Big{(}\frac{2}{r}+\frac{V^{\prime}}{V}\Big{)}\leq-\frac{1}{r^{2}V}\Big{(}\frac{2}{r}+\frac{2-\epsilon}{r}\Big{)}=-\frac{1}{r^{2}V}\frac{(4-\epsilon)}{r}$,

(d)
$\displaystyle\frac{\theta^{\prime}}{\theta}\\!=\\!\frac{1}{r}\Big{(}\frac{-\frac{n}{2}(\frac{n}{2}-1)(-2)\\!-\frac{m^{2}V^{\prime}}{r^{3}V^{2}}\\!-\frac{n}{2r^{5}}(\frac{1}{r}-\frac{V^{\prime}}{V})+\frac{n}{2r^{4}}(-\frac{1}{r^{2}}-\frac{V^{\prime\prime}V-V^{\prime
2}}{V^{2}})}{-\frac{n}{2}(\frac{n}{2}-1)+\frac{m^{2}}{r^{2}V}+\frac{n}{2r^{3}}(\frac{1}{r}-\frac{V^{\prime}}{V})}\Big{)}$

$\phantom{aaa}\displaystyle\leq\frac{1}{r}(-2+\epsilon).$

Hence, using (a)-(d) above, we obtain

$\displaystyle\mathcal{E}^{\prime}(r)\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(-\Big{(}\frac{V^{\prime}}{V}-\frac{1}{r}\Big{)}\psi^{\prime
2}+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}+\frac{\theta^{\prime}}{2}\psi^{2}\\!\right)dtd\Omega$
$\displaystyle\leq$ $\displaystyle\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\left(-\frac{(1-\epsilon)}{r}\psi^{\prime
2}+\frac{1}{2}\frac{(-2)(2-\epsilon)}{rV^{2}}\dot{\psi}^{2}+\frac{1}{2}\frac{(-1)}{r^{2}V}\frac{(4-\epsilon)}{r}|{\mathring{\not{\nabla}}}\psi|^{2}\right.$
$\displaystyle\left.\phantom{aaaaaaa}+\\!\frac{1}{2}\frac{1}{r}(-2+\epsilon)\theta\psi^{2}\\!\right)dtd\Omega$
$\displaystyle=$
$\displaystyle\\!\\!\\!-\frac{1}{r}\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\left(2(1-\epsilon)\psi^{\prime
2}+2(2-\epsilon)\frac{1}{V^{2}}\dot{\psi}^{2}+(4-\epsilon)\frac{1}{r^{2}V}|{\mathring{\not{\nabla}}}\psi|^{2}\right.$
$\displaystyle\left.\phantom{\frac{1}{V^{2}}aaaaaaala}+(2-\epsilon)\theta\psi^{2}\right)dtd\Omega$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!-\frac{2(1-\epsilon)}{r}\mathcal{E}(r).$

Using Grönwall’s inequality (see e.g. [12, Appendix B(j)]), we obtain

$\mathcal{E}(r)\leq\;\mathcal{E}(r_{0})e^{\int_{r_{0}}^{r}-\frac{2(1-\epsilon)}{r}dr}\;=\mathcal{E}(r_{0})\Big{(}\frac{r}{r_{0}}\Big{)}^{-2(1-\epsilon)}\lesssim
r^{-2+2\epsilon}.$

Thus

$\int_{\mathbb{R}\times
S^{n-1}}\theta\psi^{2}dtd\Omega\;\leq\;2\mathcal{E}(r)\;\lesssim\;r^{-2+2\epsilon},$

and so

$\int_{\mathbb{R}\times
S^{n-1}}\psi^{2}dtd\Omega\;\lesssim\;\frac{r^{2\epsilon}}{r^{2}\theta}\;\lesssim\;\frac{r^{2\epsilon}}{r^{2}\frac{1}{r^{2}}}\;=\;r^{2\epsilon}.$

Hence

$\|\psi(r,\cdot)\|_{L^{2}(\mathbb{R}\times S^{n-1})}\lesssim r^{\epsilon}.$

Consequently,

$\|\phi(r,\cdot)\|_{L^{2}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-\frac{n}{2}+\epsilon}.$

Recall that $S^{n-1}$ admits $\frac{n(n-1)}{2}$ independent Killing vectors,
given by

$L_{ij}=x^{i}\frac{\partial}{\partial x^{j}}-x^{j}\frac{\partial}{\partial
x^{i}},$

for $i<j$ (under the usual embedding $S^{n-1}\subset\mathbb{R}^{n}$). As
$\frac{\partial}{\partial t}$ and $L_{ij}$ are Killing vector fields, it
follows that $\dot{\phi}$ and $L_{ij}\cdot\phi$ are also solutions to
$\square_{g}\phi-m^{2}\phi=0$. Commuting with the Killing vector fields
$\frac{\partial}{\partial t}$ and $L_{ij}$, if we assume at the moment666This
will be proved later in the subsection on redshift estimates. that at $r_{0}$
we have $\|\phi(r_{0},\cdot)\|_{H^{k}(\\{r=r_{0}\\})}<+\infty,$ then we also
obtain for all $r\geq r_{0}$ that
$\|\phi(r,\cdot)\|_{H^{k^{\prime}}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-\frac{n}{2}+\epsilon},$ where $k^{\prime}=k-2>\frac{n}{2}$. By the Sobolev
inequality777 The part of the Sobolev embedding theorem concerning inclusion
in Hölder spaces holds for a complete Riemannian manifold with a positive
injectivity radius and a bounded sectional curvature; see e.g. [16, §3.3,
Thm.3.4] or [4, Ch.2]., $\|\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}\lesssim r^{-\frac{n}{2}+\epsilon}.$

This completes the proof of Theorem 4.2 in the case when $|m|>\frac{n}{2}$
(provided we show the aforementioned finiteness of energy, which will be
carried out in Subsection 4.5 on redshift estimates).

###  4.4. The case $|m|\leq\frac{n}{2}$

Let $\epsilon^{\prime}>0$ be given. Define

$\widetilde{\mathcal{E}}(r)=\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\left(\psi^{\prime
2}+\frac{1}{V^{2}}\dot{\psi}^{2}+\frac{1}{r^{2}V}|{\mathring{\not{\nabla}}}\psi|^{2}+\frac{\epsilon^{\prime}}{r^{2}}\psi^{2}\right)dtd\Omega.$

We now proceed to find an expression for
$\widetilde{\mathcal{E}}^{\prime}(r)$, and we will simplify it using (10) and
the divergence theorem, in order to get rid of the terms involving
$\ddot{\psi}$ and the spherical Laplacian of $\psi$:

$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\widetilde{\mathcal{E}}^{\prime}(r)\phantom{\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\bm{\left(}\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\right)}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(\psi^{\prime}\psi^{\prime\prime}+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\frac{1}{V^{2}}\dot{\psi}\dot{\psi}^{\prime}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}\right.$
$\displaystyle\left.\phantom{aaaaaaaa}+\frac{1}{r^{2}V}\langle{\mathring{\not{\nabla}}}\psi,({\mathring{\not{\nabla}}}\psi)^{\prime}\rangle-\frac{\epsilon^{\prime}}{r^{3}}\psi^{2}+\frac{\epsilon^{\prime}}{r^{2}}\psi\psi^{\prime}dtd\Omega$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(\psi^{\prime}\bigg{(}-\Big{(}\frac{V^{\prime}}{V}+\frac{n-1}{r}-\frac{2\kappa}{r}\Big{)}\psi^{\prime}+\cancel{\frac{\ddot{\psi}}{V^{2}}}+\bcancel{\frac{1}{r^{2}V}{\mathring{\not{\Delta}}}\psi}-\theta\psi\bigg{)}\right.$
$\displaystyle\phantom{aaaaaaaa}+\frac{1}{2}\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\dot{\psi}^{2}+\cancel{\frac{1}{V^{2}}\dot{\psi}\dot{\psi}^{\prime}}+\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}+\bcancel{\frac{1}{r^{2}V}\langle{\mathring{\not{\nabla}}}\psi,({\mathring{\not{\nabla}}}\psi)^{\prime}\rangle}$
$\displaystyle\phantom{aaaaaa}\left.\phantom{\Bigg{(}}-\frac{\epsilon^{\prime}}{r^{3}}\psi^{2}+\frac{\epsilon^{\prime}}{r^{2}}\psi\psi^{\prime}\right)dtd\Omega$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\left(\\!-\Big{(}\frac{V^{\prime}}{V}\\!+\\!\frac{n-1}{r}\\!-\\!\frac{2\kappa}{r}\Big{)}\psi^{\prime
2}\\!+\\!\frac{1}{2}\Big{(}\frac{1}{V^{2}}\\!\Big{)}^{\prime}\dot{\psi}^{2}\\!+\\!\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\\!\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}\\!-\\!\frac{\epsilon^{\prime}}{r^{3}}\psi^{2}\\!\right)dtd\Omega$
$\displaystyle+\left(\frac{\epsilon^{\prime}}{r^{2}}-\theta\right)\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\psi\psi^{\prime}dtd\Omega.$

Again, for getting rid of the spherical Laplacian, we use the divergence
theorem, noting that the sphere $S^{n-1}$ has no boundary. For handling the
second time derivative, as before, we note that there is a boundary at
infinity (with two connected components), which can be seen to be equal to
$0$, by Lemma 7.3 from Appendix B.

Thus

$\displaystyle\widetilde{\mathcal{E}}^{\prime}(r)\\!\\!\\!\\!$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\\!\int_{\mathbb{R}\times\\!S^{n-1}}\\!\\!\\!\left(\\!-\Big{(}\frac{V^{\prime}}{V}\\!+\\!\frac{n\\!-\\!1}{r}\\!-\\!\frac{2\kappa}{r}\Big{)}\psi^{\prime
2}\\!+\\!\frac{1}{2}\Big{(}\frac{1}{V^{2}}\\!\Big{)}^{\prime}\dot{\psi}^{2}\\!+\\!\frac{1}{2}\Big{(}\frac{1}{r^{2}V}\\!\Big{)}^{\prime}|{\mathring{\not{\nabla}}}\psi|^{2}\\!\\!-\\!\frac{\epsilon^{\prime}}{r^{3}}\psi^{2}\\!\right)dtd\Omega$
$\displaystyle+\left(\frac{\epsilon^{\prime}}{r^{2}}-\theta\right)\int_{\mathbb{R}\times
S^{n-1}}\psi\psi^{\prime}dtd\Omega.$

Now there exists an $r_{0}$ large enough such that for all $r\geq r_{0}$, we
have:

  * (i)

$\displaystyle\frac{V^{\prime}}{V}+\frac{n-1}{r}-\frac{2\kappa}{r}\geq\frac{2-\epsilon^{\prime}}{r}+\frac{n-1}{r}-\frac{2\kappa}{r}=\frac{1-\epsilon^{\prime}+(n-2\kappa)}{r}\geq\frac{1-\epsilon^{\prime}}{r}$,

using $n-2\kappa\geq 0$.

  * (ii)

$\displaystyle\Big{(}\frac{1}{V^{2}}\Big{)}^{\prime}\leq-\frac{2(2-\epsilon^{\prime})}{r}\cdot\frac{1}{V^{2}}$.

  * (iii)

$\displaystyle\Big{(}\frac{1}{r^{2}V}\Big{)}^{\prime}\leq-\frac{1}{r^{2}V}\Big{(}\frac{2}{r}+\frac{2-\epsilon^{\prime}}{r}\Big{)}$.

Using (i), (ii) and (iii), it can be seen that

$\displaystyle\widetilde{\mathcal{E}}^{\prime}(r)\\!\\!\\!\\!$
$\displaystyle\\!\leq\\!$ $\displaystyle\\!\\!\\!\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\\!\left(\\!-\frac{(1\\!-\\!\epsilon^{\prime})}{r}\psi^{\prime
2}\\!-\\!\frac{1}{2}\frac{2(2\\!-\\!\epsilon^{\prime})}{rV^{2}}\dot{\psi}^{2}\\!-\\!\frac{1}{2}\frac{1}{r^{2}V}\Big{(}\frac{2}{r}\\!+\\!\frac{2\\!-\\!\epsilon^{\prime}}{r}\Big{)}|{\mathring{\not{\nabla}}}\psi|^{2}\\!-\\!\frac{\epsilon^{\prime}}{r^{3}}\psi^{2}\\!\right)dtd\Omega$
$\displaystyle+\left(\frac{\epsilon^{\prime}}{r^{2}}-\theta\right)\int_{\mathbb{R}\times\\!S^{n-1}}\psi\psi^{\prime}dtd\Omega$
$\displaystyle\\!\leq\\!$
$\displaystyle\\!\\!\\!\\!-\frac{1}{r}\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\\!\\!\\!\left(\\!2(1\\!-\\!\epsilon^{\prime})\psi^{\prime
2}\\!+\\!2(2\\!-\\!\epsilon^{\prime})\frac{1}{V^{2}}\dot{\psi}^{2}\\!+\\!(4-\epsilon^{\prime})\frac{1}{r^{2}V}|{\mathring{\not{\nabla}}}\psi|^{2}\\!+\\!\frac{2\epsilon^{\prime}}{r^{2}}\psi^{2}\\!\right)dtd\Omega$
$\displaystyle+\left(\frac{\epsilon^{\prime}}{r^{2}}-\theta\right)\int_{\mathbb{R}\times
S^{n-1}}\psi\psi^{\prime}dtd\Omega.$

Hence

$\widetilde{\mathcal{E}}^{\prime}(r)\leq-\frac{2(1-\epsilon^{\prime})}{r}\widetilde{\mathcal{E}}(r)+\Big{(}\frac{\epsilon^{\prime}}{r^{2}}-\theta\Big{)}\int_{\mathbb{R}\times
S^{n-1}}\psi\psi^{\prime}dtd\Omega.$

We have

$\displaystyle\theta$ $\displaystyle=$
$\displaystyle\frac{m^{2}}{V}+\frac{\kappa}{r}\Big{(}\frac{1}{r}-\frac{V^{\prime}}{V}\Big{)}-\frac{\kappa}{r^{2}}(n-1-\kappa)$
$\displaystyle=$
$\displaystyle\frac{1}{r^{2}}\left(\frac{m^{2}}{\frac{V}{r^{2}}}+\kappa\Big{(}1-\frac{V^{\prime}}{V}r\Big{)}-\kappa(n-1-\kappa)\right).$

As $\displaystyle\frac{V}{r^{2}}\stackrel{{\scriptstyle
r\rightarrow\infty}}{{\longrightarrow}}1$ and
$\displaystyle\frac{V^{\prime}}{V}r\stackrel{{\scriptstyle
r\rightarrow\infty}}{{\longrightarrow}}2$, it follows that

$r^{2}\theta\;\stackrel{{\scriptstyle
r\rightarrow\infty}}{{\longrightarrow}}\;\frac{m^{2}}{1}+\kappa(1-2)-\kappa(n-1-\kappa)=m^{2}-\kappa
n+\kappa^{2}=0.$

Thus, given $\epsilon^{\prime}>0$, there exists an $r_{0}$ large enough such
that for $r\geq r_{0}$, $|r^{2}\theta|<\epsilon^{\prime}$, that is,

$|\theta|<\frac{\epsilon^{\prime}}{r^{2}}.$

So

$\displaystyle\widetilde{\mathcal{E}}^{\prime}(r)$ $\displaystyle\leq$
$\displaystyle-\frac{2(1-\epsilon^{\prime})}{r}\widetilde{\mathcal{E}}(r)+\Big{(}\frac{\epsilon^{\prime}}{r^{2}}-\theta\Big{)}\int_{\mathbb{R}\times
S^{n-1}}\psi\psi^{\prime}dtd\Omega$ $\displaystyle\leq$
$\displaystyle-\frac{2(1-\epsilon^{\prime})}{r}\widetilde{\mathcal{E}}(r)+\Big{(}\frac{\epsilon^{\prime}}{r^{2}}+\frac{\epsilon^{\prime}}{r^{2}}\Big{)}\left|\int_{\mathbb{R}\times
S^{n-1}}\psi\psi^{\prime}dtd\Omega\right|.$

The Cauchy-Schwarz inequality applied to the last integral gives

$\displaystyle\left|\int_{\mathbb{R}\times
S^{n-1}}\psi\psi^{\prime}dtd\Omega\right|$ $\displaystyle\leq$
$\displaystyle\sqrt{\int_{\mathbb{R}\times
S^{n-1}}\psi^{2}dtd\Omega}\;\cdot\;\sqrt{\int_{\mathbb{R}\times
S^{n-1}}\psi^{\prime 2}dtd\Omega}$ $\displaystyle\leq$
$\displaystyle\sqrt{\frac{2r^{2}}{\epsilon^{\prime}}\widetilde{\mathcal{E}}(r)}\;\cdot\;\sqrt{2\;\\!\widetilde{\mathcal{E}}(r)}=\frac{2r}{\sqrt{\epsilon^{\prime}}}\widetilde{\mathcal{E}}(r).$

So we obtain

$\widetilde{\mathcal{E}}^{\prime}(r)\leq-\frac{2(1-\epsilon^{\prime})}{r}\widetilde{\mathcal{E}}(r)+\frac{2\epsilon^{\prime}}{r^{2}}\frac{2r}{\sqrt{\epsilon^{\prime}}}\widetilde{\mathcal{E}}(r)=(-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}})\frac{1}{r}\widetilde{\mathcal{E}}(r).$

Application of Grönwall’s inequality yields

$\widetilde{\mathcal{E}}(r)\leq\widetilde{\mathcal{E}}(r_{0})e^{\int_{r_{0}}^{r}(-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}})\frac{1}{r}dr}=\frac{\widetilde{\mathcal{E}}(r_{0})}{r_{0}^{-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}}}}r^{-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}}}.$

So

$\displaystyle\int_{\mathbb{R}\times S^{n-1}}\psi^{2}dtd\Omega$
$\displaystyle=$
$\displaystyle\frac{2r^{2}}{\epsilon^{\prime}}\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\frac{\epsilon^{\prime}}{r^{2}}\psi^{2}dtd\Omega\leq\frac{2r^{2}}{\epsilon^{\prime}}\widetilde{\mathcal{E}}(r)$
$\displaystyle\leq$
$\displaystyle\frac{2r^{2}}{\epsilon^{\prime}}\frac{\widetilde{\mathcal{E}}(r_{0})}{r_{0}^{-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}}}}r^{-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}}}=\frac{2\widetilde{\mathcal{E}}(r_{0})}{\epsilon^{\prime}r_{0}^{-2+2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}}}}r^{2\epsilon^{\prime}+4\sqrt{\epsilon^{\prime}}}.$

Thus

$\|\psi(r,\cdot)\|_{L^{2}(\mathbb{R}\times
S^{n-1})}\leq\sqrt{\frac{2\widetilde{\mathcal{E}}(r_{0})}{\epsilon^{\prime}}}\frac{1}{r_{0}^{-1+\epsilon^{\prime}+2\sqrt{\epsilon^{\prime}}}}r^{\epsilon^{\prime}+2\sqrt{\epsilon^{\prime}}},$

and so

$\|\phi(r,\cdot)\|_{L^{2}(\mathbb{R}\times
S^{n-1})}\leq\sqrt{\frac{2\widetilde{\mathcal{E}}(r_{0})}{\epsilon^{\prime}}}\frac{1}{r_{0}^{-1+\epsilon^{\prime}+2\sqrt{\epsilon^{\prime}}}}r^{-\kappa+\epsilon^{\prime}+2\sqrt{\epsilon^{\prime}}}.$

Given $\epsilon>0$, arbitrarily small, we can choose
$\epsilon^{\prime}=\epsilon^{\prime}(\epsilon)>0$ small enough so that
$\epsilon^{\prime}+2\sqrt{\epsilon^{\prime}}<\epsilon$ at the outset, so that

$\|\phi(r,\cdot)\|_{L^{2}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-\kappa+\epsilon}.$

Again assuming at the moment that at $r_{0}$ we have

$\|\phi(r_{0},\cdot)\|_{H^{k}(\\{r=r_{0}\\})}<+\infty,$

and by commuting with the Killing vector fields $\frac{\partial}{\partial t}$
and $L_{ij}$, then we also obtain for all $r\geq r_{0}$ that

$\|\phi(r,\cdot)\|_{H^{k^{\prime}}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-(\frac{n}{2}-\sqrt{\frac{n^{2}}{4}-m^{2}}\;\\!)+\epsilon},$

where $k^{\prime}=k-2>\frac{n}{2}$. By the Sobolev inequality, this yields

$\|\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-(\frac{n}{2}-\sqrt{\frac{n^{2}}{4}-m^{2}}\;\\!)+\epsilon}.$

This completes the proof of Theorem 4.2 in the case when $|m|\leq\frac{n}{2}$
(provided we show the finiteness of energy, which will be carried out in the
subsection on redshift estimates below).

###  4.5. Redshift estimates

The last step is to use redshift estimates to transfer finiteness of the
energies along the branches $\mathcal{C}\mathcal{H}_{1}^{+}$ and
$\mathcal{C}\mathcal{H}_{2}^{+}$ of the cosmological horizon to finiteness at
$r=r_{0}$, justifying the finiteness of the energies $\mathcal{E}(r_{0})$ and
$\widetilde{\mathcal{E}}(r_{0})$ assumed in the previous two subsections.

Define the new coordinate $u$ by

$u=t+\int_{r_{*}}^{r}\frac{1}{V}dr,$

where $r_{*}>r_{c}$ is arbitrary, but fixed. Then

$du=dt+\frac{1}{V}dr.$

The Reissner-Nordström-de Sitter metric can be rewritten using the coordinates
$(u,r,\cdots)$, instead of the old $(t,r,\cdots)$-coordinates, as follows

$\displaystyle g$ $\displaystyle=$
$\displaystyle-\frac{1}{V}dr^{2}+Vdt^{2}+r^{2}d\Omega^{2}=V\Big{(}-\frac{1}{V^{2}}dr^{2}+dt^{2}\Big{)}+r^{2}d\Omega^{2}$
$\displaystyle=$
$\displaystyle-V\Big{(}\frac{1}{V}dr+dt\Big{)}\Big{(}\frac{1}{V}dr-
dt\Big{)}+r^{2}d\Omega^{2}$ $\displaystyle=$ $\displaystyle-
Vdu\Big{(}-du+\frac{2}{V}dr\Big{)}+r^{2}d\Omega^{2}=Vdu^{2}-2dudr+r^{2}d\Omega^{2}.$

The matrix of the metric in the $(u,r,\cdots)$-coordinate system is

$[g_{\mu\nu}]=\left[\begin{array}[]{rr|c}V&-1&\\\ -1&0&\\\
\hline\cr&&\ast\end{array}\right].$

Since

$\det\left[\begin{array}[]{rr}V&-1\\\ -1&0\end{array}\right]=-1,$

this coordinate system extends across the cosmological horizon $r=r_{c}$
(where $V=0$). The hypersurfaces of constant $u$ are null and transverse to
the cosmological horizon. Thus only one of the branches of the cosmological
horizon, namely $\mathcal{C}\mathcal{H}_{1}^{+}$, is covered by the
$(u,r,\cdots)$-coordinates. (In order to cover the other branch
$\mathcal{C}\mathcal{H}_{2}^{+}$, where $u=-\infty$, we can introduce

$v:=-t+\int_{r_{*}}^{r}\frac{1}{V}dr,$

and use the $(v,r,\cdots)$-coordinate chart.)

We will only consider $\mathcal{C}\mathcal{H}_{1}^{+}$ in the remainder of
this subsection, since $\mathcal{C}\mathcal{H}_{2}^{+}$ can be treated in an
analogous manner.

The Killing vector field

$K=\frac{\partial}{\partial u}=\frac{\partial}{\partial t}$

is well-defined across $\mathcal{C}\mathcal{H}_{1}^{+}$, and is null on the
cosmological horizon $\mathcal{C}\mathcal{H}_{1}^{+}$, even though the
$t$-coordinate is not defined there. Consider the vector field in the
$(u,r,\cdots)$-coordinate chart given by

$Y=\left(\frac{\partial}{\partial r}\right)_{u}.$

The subscript $u$ means that the integral curves of $Y$ in the
$(u,r,\cdots)$-coordinate chart have a constant $u$-coordinate. Then we have
$du(Y)=0$ and $dr(Y)=1$, and so in the old $(t,r,\cdots)$-coordinate chart,
the vector field $Y$ can be expressed as

$Y=\frac{\partial}{\partial r}-\frac{1}{V}\frac{\partial}{\partial t}.$

Let the vector field $X$ be defined by

$X=\frac{V^{\frac{1}{2}}}{r^{n-1}}N=\frac{V}{r^{n-1}}\frac{\partial}{\partial
r}$

in the old $(t,r,\cdots)$-coordinate chart. To find the expression for $X$ in
the $(u,r,\cdots)$-coordinate chart induced basis vectors, we first find

$N=-\frac{\textrm{grad }r}{|\textrm{grad }r|}$

in the $(u,r,\cdots)$-coordinate chart induced basis vectors. Since

$\left[\begin{array}[]{rr}V&-1\\\
-1&0\end{array}\right]^{-1}=\frac{1}{-1}\left[\begin{array}[]{rr}0&1\\\
1&V\end{array}\right]=\left[\begin{array}[]{rr}0&-1\\\
-1&-V\end{array}\right],$

we have

$\langle\textrm{grad }r,\textrm{grad }r\rangle=\langle dr,dr\rangle=-V.$

If $\omega:=-\frac{1}{\sqrt{V}}dr$, then

$N=g^{\mu\nu}\omega_{\nu}=\frac{1}{\sqrt{V}}\Big{(}\frac{\partial}{\partial
u}+V\frac{\partial}{\partial r}\Big{)}.$

So

$X=\frac{\sqrt{V}}{r^{n-1}}N=\frac{1}{r^{n-1}}\Big{(}\frac{\partial}{\partial
u}+V\frac{\partial}{\partial r}\Big{)}.$

The energy

$\displaystyle E(r)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}\times
S^{n-1}}T(X,N)dV_{n}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\Big{(}V^{2}\phi^{\prime
2}+\dot{\phi}^{2}+\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}+m^{2}V\phi^{2}\Big{)}dtd\Omega$
$\displaystyle\stackrel{{\scriptstyle r\rightarrow r_{c}}}{{\longrightarrow}}$
$\displaystyle\frac{1}{2}\int\displaylimits_{\small\begin{subarray}{c}\mathbb{R}\times
S^{n-1}\\\
(\mathcal{C}\mathcal{H}_{1}^{+})\end{subarray}}(K\cdot\phi)^{2}dud\Omega+\frac{1}{2}\int\displaylimits_{\small\begin{subarray}{c}\mathbb{R}\times
S^{n-1}\\\
(\mathcal{C}\mathcal{H}_{2}^{+})\end{subarray}}(K\cdot\phi)^{2}dvd\Omega$

(since $V(r_{c})=0$). So $E(r)$ ‘loses control’ of the transverse and angular
derivatives as $r\rightarrow r_{c}$. To remedy this problem, we define a new
energy $\widetilde{E}$, by adding $Y$ to $X$, obtaining

$\widetilde{E}(r):=E(r)+\int_{\mathbb{R}\times S^{n-1}}T(Y,N)dV_{n}.$

In the old $(t,r,\cdots)$-coordinates, $N=\sqrt{V}\;\\!\partial_{r}$, and so

$\displaystyle T(Y,N)$ $\displaystyle=$ $\displaystyle
T(\partial_{r},N)-\frac{1}{V}T(\partial_{t},N)=\frac{1}{\sqrt{V}}T(N,N)-\frac{1}{\sqrt{V}}T(\partial_{t},\partial_{r})$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{V}}\left(T(N,N)-T(\partial_{t},\partial_{r})\right).$

We have

$T(\partial_{t},\partial_{r})=\dot{\phi}\;\\!\phi^{\prime}.$

So

$\displaystyle\widetilde{E}(r)\\!\\!\\!\\!$ $\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!E(r)\\!+\\!\int_{\mathbb{R}\times\\!S^{n-1}}\\!\\!\\!\left(\\!\frac{1}{\sqrt{V}}\frac{1}{2}\Big{(}V\phi^{\prime
2}\\!+\\!\frac{\dot{\phi}^{2}}{V}\\!+\\!\frac{|{\mathring{\not{\nabla}}}\phi|^{2}}{r^{2}}\\!+\\!m^{2}\phi^{2}\Big{)}\\!-\\!\frac{1}{\sqrt{V}}\dot{\phi}\phi^{\prime}\\!\right)\sqrt{V}r^{n-1}dtd\Omega$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!E(r)\\!+\\!\int_{\mathbb{R}\times
S^{n-1}}\frac{1}{2}\left(V\Big{(}\phi^{\prime}-\frac{1}{V}\dot{\phi}\Big{)}^{2}\\!+\\!\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\\!+\\!m^{2}\phi^{2}\right)r^{n-1}dtd\Omega$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!E(r)\\!+\\!\int_{\mathbb{R}\times
S^{n-1}}\frac{1}{2}\left(V(Y\cdot\phi)^{2}\\!+\\!\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\\!+\\!m^{2}\phi^{2}\right)r^{n-1}dtd\Omega.$

We now have

$\displaystyle\widetilde{E}(r_{c})$ $\displaystyle=$ $\displaystyle
E(r_{c})+\frac{r_{c}^{n-3}}{2}\int\displaylimits_{\small\begin{subarray}{c}\mathbb{R}\times
S^{n-1}\\\
(\mathcal{C}\mathcal{H}_{1}^{+})\end{subarray}}|{\mathring{\not{\nabla}}}\phi|^{2}dud\Omega+\frac{r_{c}^{n-3}}{2}\int\displaylimits_{\small\begin{subarray}{c}\mathbb{R}\times
S^{n-1}\\\
(\mathcal{C}\mathcal{H}_{2}^{+})\end{subarray}}|{\mathring{\not{\nabla}}}\phi|^{2}dvd\Omega$
$\displaystyle\phantom{E(r_{c})}+\frac{m^{2}r_{c}^{n-1}}{2}\\!\\!\\!\\!\\!\int\displaylimits_{\small\begin{subarray}{c}\mathbb{R}\times
S^{n-1}\\\
(\mathcal{C}\mathcal{H}_{1}^{+})\end{subarray}}\phi^{2}dud\Omega+\frac{m^{2}r_{c}^{n-1}}{2}\\!\\!\\!\\!\int\displaylimits_{\small\begin{subarray}{c}\mathbb{R}\times
S^{n-1}\\\ (\mathcal{C}\mathcal{H}_{2}^{+})\end{subarray}}\phi^{2}dvd\Omega,$

so that using $\widetilde{E}$ instead of $E$ allows us to regain some control
of the angular derivatives as $r\rightarrow r_{c}$. We note that
$\widetilde{E}(r_{c})$ is equivalent to

$\|\phi\|^{2}_{H^{1}(\mathcal{C}\mathcal{H}_{1}^{+})}+\|\phi\|^{2}_{H^{1}(\mathcal{C}\mathcal{H}_{2}^{+})}.$

We will now compute the deformation tensor $\Xi$ corresponding to the
multiplier $Y$. We have

  * •

$\displaystyle\Big{[}-\frac{1}{V}\partial_{t},\partial_{r}\Big{]}=-\frac{V^{\prime}}{V}\partial_{t}$,

  * •

$\displaystyle\mathcal{L}_{\partial_{r}}g=\frac{V^{\prime}}{V}dr^{2}-\frac{2}{V}dr\mathcal{L}_{\partial_{r}}dr+V^{\prime}dt^{2}+2rd\Omega^{2}=\frac{V^{\prime}}{V}dr^{2}+V^{\prime}dt^{2}+2rd\Omega^{2}$,

  * •

$\displaystyle\mathcal{L}_{-\frac{1}{V}\partial_{t}}g=-\frac{2}{V}dr\mathcal{L}_{-\frac{1}{V}\partial_{t}}dr+2Vdt\mathcal{L}_{-\frac{1}{V}\partial_{t}}dt=2Vdt\;\Big{(}-\frac{V^{\prime}}{V^{2}}\Big{)}dr$.

Hence

$\displaystyle\Xi$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mathcal{L}_{Y}g=\frac{1}{2}\mathcal{L}_{\partial_{r}-\frac{1}{V}\partial_{t}}g=\frac{1}{2}\mathcal{L}_{\partial_{r}}g+\frac{1}{2}\mathcal{L}_{-\frac{1}{V}\partial_{t}}g\phantom{\frac{1}{V^{\prime}}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{V^{\prime}}{V^{2}}dr^{2}+\frac{V^{\prime}}{2}dt^{2}+rd\Omega^{2}+\frac{V^{\prime}}{V}dtdr$
$\displaystyle=$
$\displaystyle\frac{1}{2}V^{\prime}\Big{(}dt+\frac{1}{V}dr\Big{)}^{2}+rd\Omega^{2}=\frac{1}{2}V^{\prime}du^{2}+rd\Omega^{2}.$

We have

$du(\partial_{r})=\Big{(}dt+\frac{1}{V}\Big{)}\partial_{r}=\frac{1}{V},\;\;\textrm{
and }\;\;du(\partial_{t})=1,$

and on the other hand,

$-g(Y,\partial_{r})=-g\Big{(}\partial_{r}-\frac{1}{V}\partial_{t},\;\\!\partial_{r}\Big{)}=\frac{1}{V},\textrm{
and
}-g(Y,\partial_{t})=-g\Big{(}\partial_{r}-\frac{1}{V}\partial_{t},\;\\!\partial_{t}\Big{)}=1,$

showing that

$du=-g(Y,\cdot).$

Also, we recall that

$\displaystyle T_{\mu\nu}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\phi\partial_{\nu}\phi-\frac{g_{\mu\nu}}{2}\left(\partial_{\alpha}\phi\partial^{\alpha}\phi+m^{2}\phi^{2}\right)$
$\displaystyle=$ $\displaystyle\left(d\phi\otimes
d\phi\\!-\\!\frac{1}{2}\langle d\phi,d\phi\rangle
g-\frac{1}{2}m^{2}\phi^{2}g\right)_{\mu\nu}.$

It follows that

$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!T^{\mu\nu}\Xi_{\mu\nu}\\!\\!\\!$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\Xi^{\mu\nu}T_{\mu\nu}\phantom{\frac{1}{2}}$ (12)
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\frac{1}{2}V^{\prime}(Y\cdot\phi)^{2}+\frac{1}{r^{3}}|{\mathring{\not{\nabla}}}\phi|^{2}-\frac{n-1}{2r}\langle
d\phi,d\phi\rangle-\frac{n-1}{2r}m^{2}\phi^{2}.$

We have $d\phi(\partial_{\mu})=\partial_{\mu}\phi$. So

$\displaystyle\langle d\phi,d\phi\rangle$ $\displaystyle=$ $\displaystyle
g^{\alpha\beta}(d\phi)_{\alpha}(d\phi)_{\beta}=g^{\alpha\beta}(\partial_{\alpha}\phi)(\partial_{\beta}\phi)=(\partial_{\alpha}\phi)(\partial^{\alpha}\phi)\phantom{\frac{1}{V}}$
$\displaystyle=$ $\displaystyle\phi^{\prime
2}(-V)+\dot{\phi}^{2}\frac{1}{V}+\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}.$

Also,

$\displaystyle-2(K\cdot\phi)(Y\cdot\phi)-V(Y\cdot\phi)^{2}$ $\displaystyle=$
$\displaystyle-2\dot{\phi}\Big{(}\phi^{\prime}-\frac{1}{V}\dot{\phi}\Big{)}-V\Big{(}\phi^{\prime}-\frac{1}{V}\dot{\phi}\big{)}^{2}$
$\displaystyle=$ $\displaystyle\frac{1}{V}\dot{\phi}^{2}-V\phi^{\prime 2}.$

So

$\langle
d\phi,d\phi\rangle=-2(K\cdot\phi)(Y\cdot\phi)-V(Y\cdot\phi)^{2}+\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}.$
(13)

Combining (12) and (13), we obtain

$\displaystyle T^{\mu\nu}\Xi_{\mu\nu}\\!\\!\\!\\!$ $\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\\!\Big{(}\frac{V^{\prime}}{2}\\!+\\!\frac{n\\!-\\!1}{2r}V\Big{)}(Y\\!\cdot\\!\phi)^{2}\\!+\\!\frac{n\\!-\\!1}{r}(K\\!\cdot\\!\phi)(Y\\!\cdot\\!\phi)\\!-\\!\frac{n\\!-\\!3}{2r^{3}}|{\mathring{\not{\nabla}}}\phi|^{2}\\!-\\!\frac{n\\!-\\!1}{2r}m^{2}\phi^{2}$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\\!\frac{V^{\prime}}{2}\Big{(}Y\\!\cdot\\!\phi\\!+\\!\frac{n\\!-\\!1}{rV^{\prime}}(K\\!\cdot\\!\phi)\Big{)}^{2}\\!-\\!\frac{(n\\!-\\!1)^{2}}{2r^{2}V^{\prime}}(K\\!\cdot\\!\phi)^{2}\\!-\\!\frac{n\\!-\\!3}{2r^{3}}|{\mathring{\not{\nabla}}}\phi|^{2}\\!-\\!\frac{n\\!-\\!1}{2r}m^{2}\phi^{2}.$

Now as $V^{\prime}(r)>0$ for $r\geq r_{c}$ (global redshift), it follows that
the first summand in the last expression is nonnegative, and so we obtain the
inequality

$T^{\mu\nu}\Xi_{\mu\nu}\geq-\frac{(n-1)^{2}}{2r^{2}V^{\prime}}(K\cdot\phi)^{2}-\frac{n-3}{2r^{3}}|{\mathring{\not{\nabla}}}\phi|^{2}-\frac{n-1}{2r}m^{2}\phi^{2}.$
(14)

Now suppose that $r_{0}$ is fixed. As $r^{2}V^{\prime}(r)>0$ for all
$r\in[r_{c},r_{0}]$, we have (by the extreme value theorem)

$\min_{r\in[r_{c},r_{0}]}r^{2}V^{\prime}(r)>0.$

Thus

$-\frac{(n-1)^{2}}{2r^{2}V^{\prime}}\geq-\frac{(n-1)^{2}}{\min\limits_{r\in[r_{c},r_{0}]}r^{2}V^{\prime}(r)}=:-C_{1}(r_{0}).$

Similarly, for $r\in[r_{c},r_{0}]$,
$\displaystyle\frac{1}{r^{3}}\leq\frac{1}{r_{c}^{3}}$, and so

$-\frac{n-3}{2r^{3}}\geq-\frac{n-3}{2r_{c}^{3}}=:-\widetilde{C}_{2}(r_{c}).$

Also, for $r\in[r_{c},r_{0}]$,

$-\frac{(n-1)}{2r}m^{2}\geq-\frac{(n-1)}{2r_{c}}m^{2}=:-C_{3}(r_{c}).$

Also, if we set

$\Pi:=\frac{1}{2}\mathcal{L}_{X}g,$

then from Step 1 of the proof of Theorem 5.3 (see in particular the inequality
(18) on page 18), we have for $r>r_{c}$ that

$\displaystyle T^{\mu\nu}\Pi_{\mu\nu}$ $\displaystyle\geq$
$\displaystyle-\left(1+\frac{n}{2}+\frac{e^{2}}{2r_{c}^{n-1}}\right)\frac{1}{r^{n+2}}|{\mathring{\not{\nabla}}}\phi|^{2}$
$\displaystyle\geq$
$\displaystyle-\underbrace{\left(1+\frac{n}{2}+\frac{e^{2}}{2r_{c}^{n-1}}\right)\frac{1}{r_{c}^{n+2}}}_{=:\widetilde{\widetilde{C}}_{2}(r_{c})}|{\mathring{\not{\nabla}}}\phi|^{2}.$

Set
$C_{2}(r_{c}):=\widetilde{C}_{2}(r_{c})+\widetilde{\widetilde{C}}_{2}(r_{c})$.
We have

$\displaystyle T^{\mu\nu}\Pi_{\mu\nu}\\!\\!\\!\\!$ $\displaystyle+$
$\displaystyle\\!\\!\\!T^{\mu\nu}\Xi_{\mu\nu}$ $\displaystyle\geq$
$\displaystyle-\widetilde{\widetilde{C}}_{2}(r_{c})|{\mathring{\not{\nabla}}}\phi|^{2}-C_{1}(r_{0})(K\cdot\phi)^{2}-\widetilde{C}_{2}(r_{c})|{\mathring{\not{\nabla}}}\phi|^{2}-C_{3}(r_{c})\phi^{2}$
$\displaystyle=$ $\displaystyle-
C_{1}(r_{0})(K\cdot\phi)^{2}-C_{2}(r_{c})|{\mathring{\not{\nabla}}}\phi|^{2}-C_{3}(r_{c})\phi^{2}$
$\displaystyle\geq$
$\displaystyle-\underbrace{\max\\{C_{1}(r_{0}),\;\\!C_{2}(r_{c}),\;\\!C_{3}(r_{c})\\}}_{=:C(r_{c},r_{0})>0}\cdot\left((K\cdot\phi)^{2}+|{\mathring{\not{\nabla}}}\phi|^{2}+\phi^{2}\right).$

For $r_{1}\in(r_{c},r_{0})$, and $T>0$, define
$\mathcal{D}=\\{r=r_{1}\\}\;\\!\cap\;\\!\\{-T\leq t\leq T\\}.$ We now apply
the divergence theorem, with the current $J$ corresponding to the multiplier
$X+Y$, in the region $\mathcal{T}=D^{+}(\mathcal{D})\;\\!\cap\;\\!\\{r\leq
r_{0}\\}.$ Noticing that the flux across the future null boundaries is less
than or equal to $0$, we obtain, after passing the limit $T\rightarrow\infty$,
that

$\widetilde{E}(r_{1})-\widetilde{E}(r_{0})\geq-\\!\int_{r_{1}}^{r_{0}}\\!\int_{\mathbb{R}\times
S^{n-1}}\\!\\!C(r_{0},r_{c})\left((K\cdot\phi)^{2}+|{\mathring{\not{\nabla}}}\phi|^{2}+\phi^{2}\right)r^{n-1}dtd\Omega
dr.$ (15)

But

$\displaystyle\widetilde{E}(r)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\left(V^{2}\phi^{\prime
2}+\dot{\phi}^{2}+\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}+m^{2}V\phi^{2}\right)dtd\Omega$
$\displaystyle\\!\\!\\!+\frac{1}{2}\\!\int_{\mathbb{R}\times
S^{n-1}}\left(V(Y\cdot\phi)^{2}+\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}+m^{2}\phi^{2}\right)r^{n-1}dtd\Omega.$

In particular,

$\displaystyle\int_{r_{1}}^{r_{0}}\int_{\mathbb{R}\times
S^{n-1}}\\!(K\cdot\phi)^{2}r^{n-1}dtd\Omega dr\\!$ $\displaystyle\leq$
$\displaystyle\\!\int_{r_{1}}^{r_{0}}\int_{\mathbb{R}\times
S^{n-1}}\dot{\phi}^{2}r_{0}^{n-1}dtd\Omega dr$ $\displaystyle\leq$
$\displaystyle\int_{r_{1}}^{r_{0}}2\widetilde{E}(r)r_{0}^{n-1}dr=2r_{0}^{n-1}\int_{r_{1}}^{r_{0}}\widetilde{E}(r)dr.$

Also,

$\displaystyle\int_{r_{1}}^{r_{0}}\int_{\mathbb{R}\times
S^{n-1}}|{\mathring{\not{\nabla}}}\phi|^{2}r^{n-1}dtd\Omega dr$
$\displaystyle\leq$ $\displaystyle\int_{r_{1}}^{r_{0}}\int_{\mathbb{R}\times
S^{n-1}}\frac{|{\mathring{\not{\nabla}}}\phi|^{2}}{r^{2}}r^{n-1}r^{2}dtd\Omega
dr$ $\displaystyle\leq$
$\displaystyle\int_{r_{1}}^{r_{0}}2\widetilde{E}(r)r_{0}^{2}dr=2r_{0}^{2}\int_{r_{1}}^{r_{0}}\widetilde{E}(r)dr.$

Finally,

$\displaystyle\int_{r_{1}}^{r_{0}}\int_{\mathbb{R}\times
S^{n-1}}\phi^{2}r^{n-1}dtd\Omega dr$ $\displaystyle\leq$
$\displaystyle\int_{r_{1}}^{r_{0}}\int_{\mathbb{R}\times
S^{n-1}}m^{2}\phi^{2}r^{n-1}\frac{1}{m^{2}}dtd\Omega dr$ $\displaystyle\leq$
$\displaystyle\int_{r_{1}}^{r_{0}}2\widetilde{E}(r)\frac{1}{m^{2}}dr=\frac{2}{m^{2}}\int_{r_{1}}^{r_{0}}\widetilde{E}(r)dr.$

Using the above three estimates, it follows from (15) that

$\displaystyle\widetilde{E}(r_{1})-\widetilde{E}(r_{0})$ $\displaystyle\geq$
$\displaystyle-\int_{r_{1}}^{r_{0}}C(r_{0},r_{c})\left(2r_{0}^{n-1}+2r_{0}^{2}+\frac{2}{m^{2}}\right)\widetilde{E}(r)dr$
(16) $\displaystyle=$
$\displaystyle-\int_{r_{1}}^{r_{0}}k(r_{0},r_{c})\widetilde{E}(r)dr,$

where $\displaystyle
k(r_{0},r_{c}):=C(r_{0},r_{c})\left(2r_{0}^{n-1}+2r_{0}^{2}+\frac{2}{m^{2}}\right).$

Now suppose that $r_{2}$ is such that $r_{c}<r_{1}<r_{2}<r_{0}.$

If we redo all of the above steps in order to obtain (16), but with $r_{2}$
replacing $r_{0}$, we obtain

$\widetilde{E}(r_{1})-\widetilde{E}(r_{2})\geq-\int_{r_{1}}^{r_{2}}k(r_{2},r_{c})\widetilde{E}(r)dr,$
(17)

where $\displaystyle
k(r_{2},r_{c})=C(r_{2},r_{c})\left(2r_{2}^{n-1}+2r_{2}^{2}+\frac{2}{m^{2}}\right).$

But

$\displaystyle\\!\\!k(r_{2},r_{c})\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!C(r_{2},r_{c})\left(2r_{2}^{n-1}+2r_{2}^{2}+\frac{2}{m^{2}}\right)$
$\displaystyle\leq$
$\displaystyle\\!\\!C(r_{2},r_{c})\left(2r_{0}^{n-1}+2r_{0}^{2}+\frac{2}{m^{2}}\right)$
$\displaystyle=$
$\displaystyle\\!\\!\max\\{C_{1}(r_{2}),\;\\!C_{2}(r_{c}),\;\\!C_{3}(r_{c})\\}\cdot\left(2r_{0}^{n-1}+2r_{0}^{2}+\frac{2}{m^{2}}\right).$

We have

$C_{1}(r_{2})=\frac{(n-1)^{2}}{\min\limits_{r\in[r_{c},r_{2}]}r^{2}V^{\prime}(r)}\leq\frac{(n-1)^{2}}{\min\limits_{r\in[r_{c},r_{0}]}r^{2}V^{\prime}(r)}=C_{1}(r_{0}),$

since $[r_{c},r_{2}]\subset[r_{c},r_{0}]$. Hence

$\displaystyle\\!\\!k(r_{2},r_{c})\\!\\!$ $\displaystyle\leq$
$\displaystyle\\!\\!\max\\{C_{1}(r_{2}),\;\\!C_{2}(r_{c}),\;\\!C_{3}(r_{c})\\}\\!\cdot\\!\left(2r_{0}^{n-1}+2r_{0}^{2}+\frac{2}{m^{2}}\right)$
$\displaystyle\leq$
$\displaystyle\\!\\!\max\\{C_{1}(r_{0}),\;\\!C_{2}(r_{c}),\;\\!C_{3}(r_{c})\\}\\!\cdot\\!\left(2r_{0}^{n-1}+2r_{0}^{2}+\frac{2}{m^{2}}\right)=k(r_{0},r_{c}).$

So from (17), we get

$\widetilde{E}(r_{1})-\widetilde{E}(r_{2})\geq-\int_{r_{1}}^{r_{2}}k(r_{2},r_{c})\widetilde{E}(r)dr\geq-\int_{r_{1}}^{r_{2}}k(r_{0},r_{c})\widetilde{E}(r)dr.$

Consequently, for all $r_{2}\in[r_{1},r_{0})$,
$\displaystyle\widetilde{E}(r_{2})\leq\widetilde{E}(r_{1})+\int_{r_{1}}^{r_{2}}k(r_{0},r_{c})\widetilde{E}(r)dr$.

By the integral form of Grönwall’s inequality (see e.g. [23, Thm. 1.10]), we
obtain for all $r_{2}\in[r_{1},r_{0})$ that

$\widetilde{E}(r_{2})\leq\widetilde{E}(r_{1})e^{\int_{r_{1}}^{r_{2}}k(r_{0},r_{c})dr}=\widetilde{E}(r_{1})e^{k(r_{0},r_{c})\cdot(r_{2}-r_{1})}.$

Passing the limit as $r_{2}\nearrow r_{0}$ yields

$\widetilde{E}(r_{0})\leq\widetilde{E}(r_{1})e^{k(r_{0},r_{c})\cdot(r_{0}-r_{1})},$

and this holds for all $r_{1}\in(r_{c},r_{0})$. Now passing the limit as
$r_{1}\searrow r_{c}$, we obtain

$\widetilde{E}(r_{0})\leq\widetilde{E}(r_{c})e^{k(r_{0},r_{c})\cdot(r_{0}-r_{c})}.$

Consequently,

$\displaystyle E(r_{0})\leq\widetilde{E}(r_{0})$ $\displaystyle\leq$
$\displaystyle\widetilde{E}(r_{c})e^{k(r_{0},r_{c})\cdot(r_{0}-r_{c})}$
$\displaystyle\lesssim$
$\displaystyle\widetilde{E}(r_{c})\lesssim\|\phi\|^{2}_{H^{1}(\mathcal{C}\mathcal{H}_{1}^{+})}+\|\phi\|^{2}_{H^{1}(\mathcal{C}\mathcal{H}_{2}^{+})}<+\infty.$

Commuting with the Killing vector fields $\frac{\partial}{\partial t}$ and
$L_{ij}$, we see that the hypothesis from Theorem 4.2, namely,

$\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{1}^{+})}<+\infty\;\;\;\textrm{ and
}\;\;\;\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{2}^{+})}<+\infty,$

for some $k>\frac{n}{2}+2$, yields also that

$\|\phi\|_{H^{k}(\\{r=r_{0}\\})}\lesssim\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{1}^{+})}+\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{2}^{+})}<+\infty.$

We now show that this justifies the assumption used in the previous two
subsections. For simplicity, we only consider just one of the energies

$\mathcal{E}(r)=\frac{1}{2}\int_{\mathbb{R}\times S^{n-1}}\left(\psi^{\prime
2}+\frac{1}{V^{2}}\dot{\psi}^{2}+\frac{1}{r^{2}V}|{\mathring{\not{\nabla}}}\psi|^{2}+\theta\psi^{2}\right)dtd\Omega.$

(The proof of the finiteness of $\widetilde{\mathcal{E}}(r_{0})$ is entirely
analogous.) As $\psi=r^{\kappa}\phi$, we obtain finiteness of the last
summand, namely

$\displaystyle\phantom{aaa}\int_{\mathbb{R}\times
S^{n-1}}\theta(r_{0})\left(\psi(r_{0},\cdot)\right)^{2}dtd\Omega$
$\displaystyle=$
$\displaystyle\theta(r_{0})r_{0}^{2\kappa}\int_{\mathbb{R}\times
S^{n-1}}\left(\phi(r_{0},\cdot)\right)^{2}dtd\Omega$ $\displaystyle\leq$
$\displaystyle\theta(r_{0})r_{0}^{2\kappa}\|\phi(r_{0},\cdot)\|_{H^{1}(\\{r=r_{0}\\})}^{2}<+\infty.\phantom{\int_{S^{n}}}$

We have

$\displaystyle\int_{\mathbb{R}\times
S^{n-1}}\left(\phi^{\prime}(r_{0},\cdot)\right)^{2}dtd\Omega$ $\displaystyle=$
$\displaystyle\frac{1}{(V(r_{0}))^{2}}\int_{\mathbb{R}\times
S^{n-1}}\left(V(r_{0},\cdot)\phi^{\prime}(r_{0},\cdot)\right)^{2}dtd\Omega$
$\displaystyle\leq$ $\displaystyle\frac{2E(r_{0})}{(V(r_{0}))^{2}}<+\infty.$

Since $\psi^{\prime}(r_{0},\cdot)=\kappa
r_{0}^{\kappa-1}\phi(r_{0},\cdot)+r_{0}^{\kappa}\phi^{\prime}(r_{0},\cdot)$,
and as $\phi(r_{0},\cdot)\in H^{1}(\\{r=r_{0}\\})$, we have
$\psi(r_{0},\cdot)\in L^{2}(\mathbb{R}\times S^{n-1})$, that is,
$\displaystyle\int_{\mathbb{R}\times
S^{n-1}}\left(\psi^{\prime}(r_{0},\cdot)\right)^{2}dtd\Omega<+\infty.$

We also have

$\displaystyle\int_{\mathbb{R}\times
S^{n-1}}\frac{1}{\left(V(r_{0})\right)^{2}}\left(\dot{\psi}(r_{0},\cdot)\right)^{2}dtd\Omega$
$\displaystyle=$
$\displaystyle\frac{r_{0}^{2\kappa}}{\left(V(r_{0})\right)^{2}}\int_{\mathbb{R}\times
S^{n-1}}\left(\dot{\phi}(r_{0},\cdot)\right)^{2}dtd\Omega$ $\displaystyle\leq$
$\displaystyle\frac{r_{0}^{2\kappa}}{\left(V(r_{0})\right)^{2}}\|\phi(r_{0},\cdot)\|_{H^{1}(\\{r=r_{0}\\})}^{2}<+\infty.$

Finally,

$\displaystyle\int_{\mathbb{R}\times
S^{n-1}}\frac{1}{r_{0}^{2}V(r_{0})}|{\mathring{\not{\nabla}}}\psi(r_{0},\cdot)|^{2}dtd\Omega$
$\displaystyle=$
$\displaystyle\frac{r_{0}^{2\kappa}}{r_{0}^{2}V(r_{0})}\int_{\mathbb{R}\times
S^{n-1}}|{\mathring{\not{\nabla}}}\phi(r_{0},\cdot)|^{2}dtd\Omega$
$\displaystyle\lesssim$
$\displaystyle\|\phi(r_{0},\cdot)\|_{H^{1}(\\{r=r_{0}\\})}^{2}<+\infty.\phantom{\int_{S^{n}}\frac{1}{V^{2}}}$

Thus each summand in the expression for $\mathcal{E}(r_{0})$ is finite.

This completes the proof of Theorem 4.2.

##  5\. Decay in $\mathrm{RNdS}$ when $m=0$, the wave equation

In [8, Theorem 2], the following result was shown:

######  Theorem 5.1.

Suppose that

  * •

$\delta>0$,

  * •

$M>0$,

  * •

$e\geq 0$,

  * •

$n>2$,

  * •

$(M,g)$ is the $(n+1)$-dimensional subextremal Reissner-Nordström-de Sitter
solution given by the metric

$g=-\frac{1}{V}dr^{2}+Vdt^{2}+r^{2}d\Omega^{2},$

where

$V=r^{2}+\frac{2M}{r^{n-2}}-\frac{e^{2}}{r^{n-1}}-1,$

and $d\Omega^{2}$ is the metric of the unit $(n-1)$-dimensional sphere
$S^{n-1}$,

  * •

$k>\frac{n}{2}+2$, and

  * •

$\phi$ is a smooth solution to $\square_{g}\phi=0$ such that

$\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{1}^{+})}<+\infty\;\;\;\textrm{ and
}\;\;\;\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{2}^{+})}<+\infty,$

where
$\mathcal{C}\mathcal{H}_{1}^{+}\simeq\mathcal{C}\mathcal{H}_{2}^{+}\simeq\mathbb{R}\times
S^{n-1}$ are the two components of the future cosmological horizon,
parameterised by the flow parameter of the global Killing vector field
$\frac{\partial}{\partial t}$.

Then there exists a $r_{0}$ large enough so that for all $r\geq r_{0}$,

$\|\partial_{r}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-3+\delta}.$

Using a method similar to the one we used to show Rendall’s conjecture in
Theorem 2.3, we can improve the almost-exact bound of $r^{-3+\delta}$ to
$r^{-3}$.

######  Remark 5.2.

As observed in [8, Remark 1.4], this decay rate bound of $r^{-3}$ for
$\partial_{r}\phi$ is in fact the decay rate one would expect in light of
Rendall’s conjecture. Indeed, for freely falling observers in the cosmological
region, one has

$r(\tau)\sim e^{\tau}\sim a(\tau),$

where $\tau$ is the proper time, and $a(\tau)$ is the radius of a comparable
de Sitter universe in flat FLRW form, giving

$\partial_{r}\phi\sim\frac{\partial_{\tau}\phi}{\partial_{\tau}r}\sim\frac{e^{-2\tau}}{e^{\tau}}\sim\frac{1}{r^{3}}.$

Thus our improved version of Theorem 5.1 is the following result.

######  Theorem 5.3.

Suppose that

  * •

$M>0$,

  * •

$e\geq 0$,

  * •

$n>2$,

  * •

$(M,g)$ is the $(n+1)$-dimensional subextremal Reissner-Nordström-de Sitter
solution given by the metric

$g=-\frac{1}{V}dr^{2}+Vdt^{2}+r^{2}d\Omega^{2},$

where

$V=r^{2}+\frac{2M}{r^{n-2}}-\frac{e^{2}}{r^{n-1}}-1,$

and $d\Omega^{2}$ is the metric of the unit $(n-1)$-dimensional sphere
$S^{n-1}$,

  * •

$k>\frac{n}{2}+2$, and

  * •

$\phi$ is a smooth solution to $\square_{g}\phi=0$ such that

$\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{1}^{+})}<+\infty\;\;\;\textrm{ and
}\;\;\;\|\phi\|_{H^{k}(\mathcal{C}\mathcal{H}_{2}^{+})}<+\infty,$

where
$\mathcal{C}\mathcal{H}_{1}^{+}\simeq\mathcal{C}\mathcal{H}_{2}^{+}\simeq\mathbb{R}\times
S^{n-1}$ are the two components of the future cosmological horizon,
parameterised by the flow parameter of the global Killing vector field
$\frac{\partial}{\partial t}$.

Then there exists a $r_{0}$ large enough so that for all $r\geq r_{0}$,

$\|\partial_{r}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times S^{n-1})}\lesssim
r^{-3}.$

###### Proof.

Step 1: We will first establish the following estimates: there exists an
$r_{0}$ large enough such that for all $r\geq r_{0}$,

$\displaystyle\|\ddot{\phi}(r,\cdot)\|_{L^{\infty}(\mathbb{R},S^{n-1})}\\!\\!$
$\displaystyle\lesssim$ $\displaystyle\\!\\!1,$
$\displaystyle\|{\mathring{\not{\Delta}}}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R},S^{n-1})}\\!\\!$
$\displaystyle\lesssim$ $\displaystyle\\!\\!1.$

We will follow [8, §3.2] in order to obtain the bounds above, which will be
needed in Step 2 of our proof below. We repeat this preliminary step here from
[8, §3.2] for the sake of completeness and for the convenience of the reader.

Suppose that $\phi$ satisfies the wave equation $\square_{g}\phi=0$. Recall
that the energy-momentum tensor associated with $\phi$ is given by

$T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}g_{\mu\nu}\partial_{\alpha}\partial^{\alpha}\phi.$

Thus

$\displaystyle T(N,N)$ $\displaystyle=$ $\displaystyle\left(\phi^{\prime
2}\\!-\\!\frac{1}{2}\frac{(-1)}{V}\Big{(}\phi^{\prime
2}(-V)\\!+\\!\dot{\phi}^{2}\frac{1}{V}\\!+\\!\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\Big{)}\right)V$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left(V\phi^{\prime
2}\\!+\\!\frac{1}{V}\dot{\phi}^{2}\\!+\\!\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right).$

Define $\displaystyle
X:=\frac{V^{\frac{1}{2}}}{r^{n-1}}N=\frac{V^{\frac{1}{2}}}{r^{n-1}}V^{\frac{1}{2}}\frac{\partial}{\partial
r}=\frac{V}{r^{n-1}}\frac{\partial}{\partial r}.$

The current $J$ is given by $J_{\mu}:=T_{\mu\nu}X^{\nu}.$ We define the energy

$E(r)\\!:=\\!\int_{\mathbb{R}\times
S^{n-1}}\\!T(X,N)dV_{n}\\!=\\!\frac{1}{2}\int_{\mathbb{R}\times
S^{n-1}}\\!\left(V^{2}\phi^{\prime
2}\\!+\\!\dot{\phi}^{2}\\!+\\!\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right)dtd\Omega.$

The deformation tensor $\Pi$ associated to the multiplier $X$ is given by

$\Pi=\frac{1}{2}\mathcal{L}_{X}g=-\frac{1}{V}dr\mathcal{L}_{X}dr+\frac{V}{2r^{n-1}}\frac{V^{\prime}}{V^{2}}dr^{2}+\frac{V^{\prime}V}{2r^{n-1}}dt^{2}+\frac{V}{r^{n-1}}rd\Omega^{2}.$

It can be shown that
$\displaystyle\mathcal{L}_{X}dr=\left(\frac{V^{\prime}}{r^{n-1}}-\frac{(n-1)V}{r^{n}}\right)dr.$
Thus

$\displaystyle\Pi\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!-\frac{1}{V}\left(\frac{V^{\prime}}{r^{n-1}}\\!-\\!\frac{(n-1)V}{r^{n}}\right)dr^{2}\\!+\\!\frac{V^{\prime}}{2r^{n-1}V}dr^{2}\\!+\\!\frac{V^{\prime}V}{2r^{n-1}}dt^{2}\\!+\\!\frac{V}{r^{n-1}}rd\Omega^{2}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\underbrace{\frac{V^{\prime}}{2r^{n-1}}\left(-\frac{1}{V}dr^{2}+Vdt^{2}\right)}_{=:\Pi^{(1)}}+\underbrace{\phantom{\bigg{(}}\frac{n-1}{r^{n}}dr^{2}\phantom{\bigg{)}}}_{=:\Pi^{(2)}}+\underbrace{\phantom{\bigg{(}}\frac{V}{r^{n-2}}d\Omega^{2}\phantom{\bigg{)}}}_{\Pi^{(3)}}.$

We have

$\displaystyle T^{\mu\nu}\Pi^{(1)}_{\mu\nu}\\!\\!\\!\\!$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\frac{V^{\prime}}{2r^{n-1}}\left(-\frac{1}{2}V\phi^{\prime
2}\\!-\\!\frac{1}{2V}\dot{\phi}^{2}\\!-\\!\frac{1}{2r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right.$
$\displaystyle\phantom{aaaaaa}+\left.\frac{1}{V}\left(\dot{\phi}^{2}\\!-\\!\frac{1}{2}V\Big{(}\phi^{\prime
2}(-V)+\dot{\phi}^{2}\frac{1}{V}\\!+\\!\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\Big{)}\right)\right)$
$\displaystyle=$
$\displaystyle\frac{V^{\prime}}{2r^{n-1}}\left(-\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right).$

Also,

$\displaystyle T^{\mu\nu}\Pi^{(2)}_{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{(n-1)}{2r^{n}}\left(V^{2}\phi^{\prime
2}+\dot{\phi}^{2}+\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right).$

Finally,

$\displaystyle T^{\mu\nu}\Pi^{(3)}_{\mu\nu}\\!\\!\\!$ $\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\frac{V}{r^{n-2}}\left(\frac{1}{r^{4}}|{\mathring{\not{\nabla}}}\phi|^{2}\\!-\\!\frac{(n-1)}{2r^{2}}\left(\phi^{\prime
2}(-V)\\!+\\!\dot{\phi}^{2}\frac{1}{V}\\!+\\!\frac{1}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right)\right)$
$\displaystyle\\!=\\!$
$\displaystyle\\!\\!\\!\frac{V}{r^{n+2}}|{\mathring{\not{\nabla}}}\phi|^{2}+\frac{(n-1)}{2r^{n}}\left(V^{2}\phi^{\prime
2}\\!-\\!\dot{\phi}^{2}\\!-\\!\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right).$

Consequently, the full bulk term is

$\nabla_{\mu}J^{\mu}=T^{\mu\nu}\Pi_{\mu\nu}=\frac{(n-1)}{r^{n}}V^{2}\phi^{\prime
2}+|{\mathring{\not{\nabla}}}\phi|^{2}\left(\frac{V}{r^{n+2}}-\frac{V^{\prime}}{2r^{n+1}}\right).$

Using the expression for $V$, we compute

$\frac{V}{r^{n+2}}-\frac{V^{\prime}}{2r^{n+1}}=-\frac{1}{r^{n+2}}\left(1+\frac{(n+1)e^{2}}{2r^{n-1}}-\frac{nM}{r^{n-2}}\right).$

For $r>r_{c}$, we have $V(r)>0$, and so

$\displaystyle 1+\frac{(n+1)e^{2}}{2r^{n-1}}-\frac{nM}{r^{n-2}}$
$\displaystyle=$ $\displaystyle
1-\frac{n}{2}\left(r^{2}+\frac{2M}{r^{n-2}}-\frac{e^{2}}{r^{n-1}}-1\right)-\frac{n}{2}+\frac{e^{2}}{2r^{n-1}}$
$\displaystyle=$ $\displaystyle
1-\frac{n}{2}V(r)-\frac{n}{2}+\frac{e^{2}}{2r^{n-1}}$ $\displaystyle\leq$
$\displaystyle 1-0-\frac{n}{2}+\frac{e^{2}}{2r^{n-1}}\leq
1+\frac{n}{2}+\frac{e^{2}}{2r_{c}^{n-1}}=:C.$

Hence

$\displaystyle\nabla_{\mu}J^{\mu}=T^{\mu\nu}\Pi_{\mu\nu}\geq-\frac{C}{r^{n+2}}|{\mathring{\not{\nabla}}}\phi|^{2}.$
(18)

For each $T<0$, define the set

$\mathcal{C}:=\\{r=r_{0}\\}\;\\!\cap\;\\!\\{-T\leq t\leq T\\}.$

Also, consider the region
$\mathcal{S}:=D^{+}(\mathcal{C})\;\\!\cap\;\\!\\{r\leq r_{1}\\}.$

We will apply the divergence theorem to the current $J$ on the region
$\mathcal{S}$. As the flux across the future null boundaries is nonpositive,
we have

$\displaystyle\int_{r_{0}}^{r_{1}}\int_{\mathbb{R}\times
S^{n-1}}\left(-\frac{C}{r^{n+2}}|{\mathring{\not{\nabla}}}\phi|^{2}\right)dtd\Omega
dr$ $\displaystyle\leq$
$\displaystyle\int_{\mathcal{S}}(\nabla_{\mu}J^{\mu})\epsilon=\underbrace{\int_{\begin{subarray}{c}\tiny\textrm{null}\\\
\tiny\textrm{part}\end{subarray}}J\intprod\epsilon}_{\leq
0}+\underbrace{\int_{\\{r=r_{1}\\}}J\intprod\epsilon}_{-E(r_{1})}+\underbrace{\int_{\\{r=r_{0}\\}}J\intprod\epsilon}_{E(r_{0})}.$

So $\displaystyle
E(r_{0})-E(r_{1})\geq-\int_{r_{0}}^{r_{1}}\int_{\mathbb{R}\times
S^{n-1}}\frac{C}{r^{3}}|{\mathring{\not{\nabla}}}\phi|^{2}dtd\Omega dr.$ We
have

$\int_{\mathbb{R}\times
S^{n-1}}|{\mathring{\not{\nabla}}}\phi|^{2}dtd\Omega=\int_{\mathbb{R}\times
S^{n-1}}\frac{V}{r^{2}}|{\mathring{\not{\nabla}}}\phi|^{2}\frac{r^{2}}{V}dtd\Omega\leq
2E(r)\cdot\frac{r^{2}}{V}=\frac{2r^{2}}{V}E(r).$

Since

$\lim_{r\rightarrow\infty}\frac{2r^{2}}{V}=\lim_{r\rightarrow\infty}\frac{2}{1+\frac{2M}{r^{n}}-\frac{e^{2}}{r^{n+1}}-\frac{1}{r^{2}}}=\frac{2}{1},$

there exists an $r_{0}$ large enough such that for all $r\geq r_{0}$,
$|\frac{2r^{2}}{V}-2|<1$, and in particular, $(0<)\frac{2r^{2}}{V}<3.$ Hence

$E(r_{1})\\!\leq\\!E(r_{0})\\!+\\!\int_{r_{0}}^{r_{1}}\int_{\mathbb{R}\times
S^{n-1}}\\!\left(\frac{C}{r^{3}}|{\mathring{\not{\nabla}}}\phi|^{2}\right)dtd\Omega
dr\\!\leq\\!E(r_{0})\\!+\\!\int_{r_{0}}^{r_{1}}\frac{3C}{r^{3}}E(r)dr.$

Using Grönwall’s inequality (see for e.g. [23, Thm. 1.10]), we obtain

$E(r_{1})\leq E(r_{0})e^{\int_{r_{0}}^{r_{1}}\frac{3C}{r^{3}}dr}\lesssim
C(r_{0})E(r_{0}),$

as was also noted in [8, Eq.(71)]. Thus, we have in particular that there
exists an $r_{0}$ large enough such that for all $r\geq r_{0}$,

$\int_{\mathbb{R}\times S^{n-1}}\dot{\phi}^{2}dtd\Omega\lesssim
1,\quad\textrm{ and }\quad\int_{\mathbb{R}\times
S^{n-1}}|{\mathring{\not{\nabla}}}\phi|^{2}dtd\Omega\lesssim 1.$

Commuting with the Killing vector fields $\frac{\partial}{\partial t}$ and
$L_{ij}$, we obtain (after the transferral of the finiteness of the energies
along the branches $\mathcal{C}\mathcal{H}_{1}^{+}$ and
$\mathcal{C}\mathcal{H}_{2}^{+}$ of the cosmological horizon to finiteness at
$r=r_{0}$, and an application of Sobolev’s inequality) that for all $r\geq
r_{0}$,

$\|\ddot{\phi}(r,\cdot)\|_{L^{\infty}(\mathbb{R},S^{n-1})}\lesssim
1,\quad\textrm{ and
}\quad\|{\mathring{\not{\Delta}}}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R},S^{n-1})}\lesssim
1.$

Step 2: In this step, we will write the wave equation in new coordinates which
‘equalises’ the magnitude of the coefficient weights for the $r$ and $t$
coordinates in the matrix of the metric.

To this end, we define $\displaystyle\rho=\int_{r_{0}}^{r}\frac{1}{V(r)}dr.$
Then $\displaystyle\frac{d\rho}{dr}=\frac{1}{V(r)}\textrm{ and
}V(r)\displaystyle\frac{d}{dr}=\frac{d}{d\rho}.$

With a slight abuse of notation, we write $V(\rho):=V(r(\rho)).$ We have

$\displaystyle g$ $\displaystyle=$
$\displaystyle-\frac{1}{V}dr^{2}+Vdt^{2}+r^{2}d\Omega^{2}$ $\displaystyle=$
$\displaystyle-
Vd\rho^{2}+Vdt^{2}+(r(\rho))^{2}d\Omega^{2}.\phantom{\frac{1}{V}}$

The wave equation $\square_{g}\phi=0$ can be rewritten as
$\partial_{\mu}(\sqrt{-g}\;\\!\partial^{\mu}\phi)=0$, which becomes
$\partial_{\mu}(Vr^{n-1}\partial^{\mu}\phi)=0.$ Separating the differential
operators with respect to the $\rho,t,\cdots$ coordinates, we obtain

$\partial_{\rho}(r^{n-1}\partial_{\rho}\phi)=r^{n-1}\ddot{\phi}+Vr^{n-3}{\mathring{\not{\Delta}}}\phi.$

Integrating from $\rho_{0}:=\rho(r_{0})=0$ to $\rho=\rho(r)$, we obtain

$r^{n-1}\partial_{\rho}\phi-
r_{0}^{n-1}\left.(\partial_{\rho}\phi)\right|_{\rho=\rho_{0}}=\int_{0}^{\rho}\left(r^{n-1}\ddot{\phi}+Vr^{n-3}{\mathring{\not{\Delta}}}\phi\right)d\rho,$

and so

$r^{n-1}V\partial_{r}\phi=r_{0}^{n-1}V(r_{0})\left.(\partial_{r}\phi)\right|_{r=r_{0}}+\int_{0}^{\rho}\left(r^{n-1}\ddot{\phi}+Vr^{n-3}{\mathring{\not{\Delta}}}\phi\right)d\rho,$

that is,

$\partial_{r}\phi=\Big{(}\frac{r_{0}}{r}\Big{)}^{n-1}\frac{V(r_{0})}{V(r)}\left.(\partial_{r}\phi)\right|_{r=r_{0}}+\frac{1}{r^{n-1}V}\int_{0}^{\rho}\left(r^{n-1}\ddot{\phi}+Vr^{n-3}{\mathring{\not{\Delta}}}\phi\right)d\rho.$

Hence

$\displaystyle\|\partial_{r}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}\phantom{\Big{(}\frac{a(t_{0})}{a(t)}\Big{)}^{n-2}}$
$\displaystyle\leq$
$\displaystyle\\!\\!\\!\\!\\!\Big{(}\frac{r_{0}}{r}\Big{)}^{n-1}\frac{V(r_{0})}{V(r)}\|(\partial_{r}\phi)(r_{0},\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}$
$\displaystyle\\!\\!\\!\\!\\!+\frac{1}{r^{n-1}V}\\!\int_{\rho_{0}}^{\rho}\\!\left(r^{n-1}\|\ddot{\phi}(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}\\!+\\!Vr^{n-3}\|{\mathring{\not{\Delta}}}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}\right)\\!d\rho.$

Using the fact that $V\sim r^{2}$ for $r\geq r_{0}$, with $r_{0}$ large
enough, and the estimates from Step 1 above, we obtain

$\displaystyle\|\partial_{r}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}$ $\displaystyle\lesssim$
$\displaystyle\frac{A}{r^{n+1}}+\frac{B}{r^{n+1}}\int_{\rho_{0}}^{\rho}\left(r(\rho)\right)^{n-1}d\rho\phantom{\frac{1}{V}}$
$\displaystyle\lesssim$
$\displaystyle\frac{A}{r^{n+1}}+\frac{B}{r^{n+1}}\int_{r_{0}}^{r}r^{n-1}\frac{1}{V(r)}dr$
$\displaystyle\lesssim$
$\displaystyle\frac{A}{r^{n+1}}+\frac{B^{\prime}}{r^{n+1}}\int_{r_{0}}^{r}r^{n-3}dr.$

Recalling that $n>2$, we have

$\|\partial_{r}\phi(r,\cdot)\|_{L^{\infty}(\mathbb{R}\times
S^{n-1})}\lesssim\frac{A}{r^{n+1}}+\frac{B^{\prime}}{r^{n+1}}\frac{1}{(n-2)}(r^{n-2}-r_{0}^{n-2})\lesssim\frac{1}{r^{3}}.$

This completes the proof of Theorem 5.3. ∎

##  6\. Appendix A: Fourier modes (de Sitter in flat FLRW form)

In this appendix, we give the details of the Fourier modal analysis that
motivates the specific estimates given in Theorem 3.1, starting with spatially
periodic solutions to the Klein-Gordon equation.

Let $\mathbb{T}^{n}=\mathbb{R}^{n}/(2\pi\mathbb{Z})^{n}$. Suppose that the
‘spatially periodic’
$\phi:\mathbb{R}\times\mathbb{T}^{n}\rightarrow\mathbb{R}$ satisfies the
Klein-Gordon equation (5). Writing

$\phi=\sum_{\mathbf{k}\in\mathbb{Z}^{n}}c_{\mathbf{k}}(t)e^{i\langle\mathbf{k},\mathbf{x}\rangle},$

(5) yields

$-\ddot{c}_{\mathbf{k}}-\frac{n\dot{a}}{a}\dot{c}_{\mathbf{k}}+\frac{1}{a^{2}}\delta^{pq}ik_{p}ik_{q}c_{\mathbf{k}}-m^{2}c_{\mathbf{k}}=0,$

that is,

$\ddot{c}_{\mathbf{k}}+\frac{n\dot{a}}{a}\dot{c}_{\mathbf{k}}+\frac{k^{2}}{a^{2}}c_{\mathbf{k}}+m^{2}c_{\mathbf{k}}=0,$

where

$k^{2}:=\langle\mathbf{k},\mathbf{k}\rangle.$

So

$\displaystyle\frac{d}{dt}(a^{n}\dot{c}_{\mathbf{k}})$ $\displaystyle=$
$\displaystyle na^{n-1}\dot{a}\dot{c}_{\mathbf{k}}+a^{n}\ddot{c}_{\mathbf{k}}$
$\displaystyle=$ $\displaystyle
na^{n-1}\dot{a}\dot{c}_{\mathbf{k}}+a^{n}\left(-\frac{n\dot{a}}{a}\dot{c}_{\mathbf{k}}-\Big{(}\frac{k^{2}}{a^{2}}+m^{2}\Big{)}c_{\mathbf{k}}\right)$
$\displaystyle=$
$\displaystyle-a^{n}\left(\frac{k^{2}}{a^{2}}+m^{2}\right)c_{\mathbf{k}},$

that is,

$\frac{d}{dt}(a^{n}\dot{c}_{\mathbf{k}})+a^{n-2}(k^{2}+m^{2}a^{2})c_{\mathbf{k}}=0.$
(19)

Let

$\tau=\int\frac{1}{a(t)}dt.$

Then $\frac{d}{dt}=\frac{1}{a}\frac{d}{d\tau},$ and so (19) becomes (with
$\frac{d}{d\tau}=:{\phantom{\cdot}}^{\prime}\;$)

$\frac{1}{a}\left(a^{n}\frac{1}{a}c_{\mathbf{k}}^{\prime}\right)^{\prime}+a^{n-1}(k^{2}+m^{2}a^{2})c_{\mathbf{k}}=0,$

that is,

$(a^{n-1}c_{\mathbf{k}}^{\prime})^{\prime}+a^{n-1}(k^{2}+m^{2}a^{2})c_{\mathbf{k}}=0.$
(20)

Defining $d_{\mathbf{k}}$ by
$c_{\mathbf{k}}=:a^{-\frac{n-1}{2}}d_{\mathbf{k}},$ we have

$c_{\mathbf{k}}^{\prime}=a^{-\frac{n-1}{2}}d_{\mathbf{k}}^{\prime}-\frac{n-1}{2}a^{-\frac{n-1}{2}-1}a^{\prime}d_{\mathbf{k}}.$

(20) yields

$\left(a^{\frac{n-1}{2}}d_{\mathbf{k}}^{\prime}-\frac{n-1}{2}a^{\frac{n-1}{2}-1}a^{\prime}d_{\mathbf{k}}\right)^{\prime}+a^{\frac{n-1}{2}}(k^{2}+m^{2}a^{2})d_{\mathbf{k}}=0,$

that is,

$d_{\mathbf{k}}^{\prime\prime}+\left(k^{2}+m^{2}a^{2}-\frac{(n-1)}{2}\frac{a^{\prime\prime}}{a}-\frac{(n-1)(n-3)}{4}\Big{(}\frac{a^{\prime}}{a}\Big{)}^{2}\right)d_{\mathbf{k}}=0.$
(21)

Now if $a(t)=e^{t}$, then we may take $\tau=-e^{-t}$, so that
$-t=\log(-\tau)$, that is, $-\tau=e^{-t}$. We remark that relative to our
earlier use of conformal coordinates in (4) on page 4, we are taking
$t_{0}=+\infty$ for simplicity. We then have

$a=e^{-\log(-\tau)}=-\frac{1}{\tau},\quad\quad
a^{\prime}=\frac{1}{\tau^{2}},\quad\quad
a^{\prime\prime}=-\frac{2}{\tau^{3}}.$

Hence (21) becomes

$d_{\mathbf{k}}^{\prime\prime}\\!+\\!\left(k^{2}\\!+\\!\frac{m^{2}}{\tau^{2}}\\!-\\!\frac{n-1}{2}\Big{(}\\!-\\!\frac{2}{\tau^{3}}\Big{)}\Big{(}\\!-\\!\frac{\tau}{1}\Big{)}\\!-\\!\frac{(n-1)(n-3)}{4}\Big{(}\frac{1}{\tau^{2}}\frac{(-\tau)^{2}}{1}\Big{)}\right)d_{\mathbf{k}}\\!=\\!0,$

that is,

$d_{\mathbf{k}}^{\prime\prime}+\left(k^{2}-\frac{\mu}{\tau^{2}}\right)d_{\mathbf{k}}=0,$
(22)

where

$\mu:=n-1+\frac{(n-1)(n-3)}{4}-m^{2}.$

The general solution to this equation is888See for example [27, p.95]. For the
relevant notation, see also [27, pages 82,100,101]. given by

$C_{1}\sqrt{\tau}\;J_{\nu}(|k|\tau)+C_{2}\sqrt{\tau}\;Y_{\nu}(|k|\tau),$ (23)

where $\nu$ satisfies

$\nu^{2}=\frac{1}{4}+\mu=\frac{n^{2}}{4}-m^{2}.$

Here $J_{\nu}$ denotes the Bessel function of the first kind,

$J_{\nu}(z)=\sum_{m=0}^{\infty}\frac{(-1)^{m}}{m!\;\\!\Gamma(m+\nu+1)}\Big{(}\frac{z}{2}\Big{)}^{2m+\nu},$

and $Y_{\nu}$ is the Bessel function of the second kind,

$Y_{\nu}(z)=\frac{J_{\nu}(z)\cos(\nu\pi)-J_{-\nu}(z)}{\sin(\nu\pi)},$

where the right hand side is replaced by its limiting value if $\nu$ is an
integer. Without loss of generality, in the solution (23), we may only
consider $\nu$ such that $\textrm{Re}(\nu)\geq 0$.

We note that as $t\rightarrow\infty$, $-\tau=e^{-t}\searrow 0$, and so
$\tau\nearrow 0$. We now use the asymptotic expansions of $J_{\nu}(z)$ and
$Y_{\nu}(z)$ as $z\nearrow 0$ (see e.g. [1, 9.1.7-9]):

  * $\underline{1}^{\circ}$

If $\nu\neq 0$ (that is, $m\neq\pm\frac{n}{2}$), then as $\tau\nearrow 0$, we
have

$\displaystyle J_{\nu}(|k|\tau)$ $\displaystyle=$ $\displaystyle
C(-\tau)^{\nu}+O(|\tau|),$ $\displaystyle Y_{\nu}(|k|\tau)$ $\displaystyle=$
$\displaystyle A(-\tau)^{\nu}+B(-\tau)^{-\nu}+C(-\tau)^{2-\nu}+O(|\tau|).$

So as $\tau\nearrow 0$ or $t\rightarrow\infty$, we have

$\displaystyle d_{\mathbf{k}}$ $\displaystyle=$ $\displaystyle
A(-\tau)^{\frac{1}{2}+\nu}+B(-\tau)^{\frac{1}{2}-\nu}+C(-\tau)^{\frac{5}{2}-\nu}+O(|\tau|)$
$\displaystyle=$ $\displaystyle
Ae^{(-\frac{1}{2}-\nu)t}+Be^{(-\frac{1}{2}+\nu)t}+Ce^{(-\frac{5}{2}+\nu)t}+O(e^{-t}).$

But
$c_{\mathbf{k}}=a^{-\frac{n}{2}+\frac{1}{2}}d_{\mathbf{k}}=e^{(-\frac{n}{2}+\frac{1}{2})t}d_{\mathbf{k}}$,
and so

$c_{\mathbf{k}}=Ae^{(-\frac{n}{2}-\nu)t}+Be^{(-\frac{n}{2}+\nu)t}+Ce^{(-\frac{n}{2}-2+\nu)t}+O(e^{-\frac{n+1}{2}t})$

as $t\rightarrow+\infty$. We recall that $\textrm{Re}(\nu)\geq 0$, and so
keeping only the dominating term, we have

$|c_{\mathbf{k}}|=C^{\prime}e^{-\left(\frac{n}{2}-\textrm{Re}(\nu)\right)t}+O(e^{-\frac{n+1}{2}t})$

as $t\rightarrow+\infty$. Thus we expect $\phi$ to satisfy

$\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left\\{\begin{array}[]{ll}a^{-\frac{n}{2}}&\textrm{if
}\;|m|>\frac{n}{2},\\\
a^{-\frac{n}{2}+\sqrt{\frac{n^{2}}{4}-m^{2}}}&\textrm{if
}\;|m|<\frac{n}{2}.\end{array}\right.$

  * $\underline{2}^{\circ}$

If $\nu=0$ (that is, $m=\pm\frac{n}{2}$), then as $\tau\nearrow 0$, we have

$\displaystyle J_{\nu}(|k|\tau)$ $\displaystyle=$ $\displaystyle C+O(|\tau|),$
$\displaystyle Y_{\nu}(|k|\tau)$ $\displaystyle=$ $\displaystyle
C\log(-\tau)+O(|\tau|).$

This implies that

$d_{\mathbf{k}}=(A+Bt)e^{-\frac{1}{2}t}+O(e^{-t})$

as $t\rightarrow+\infty$. Hence

$|c_{\mathbf{k}}|=(A+Bt)e^{-\frac{n}{2}t}+O(e^{-\frac{n+1}{2}t}).$

Thus we expect $\phi$ to satisfy

$\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim a^{-\frac{n}{2}}\log
a\;\;\textrm{ if }\;m=\pm\frac{n}{2}.$

Summarising, $\phi$ is expected to have the decay

$\|\phi(t,\cdot)\|_{L^{\infty}(\mathbb{R}^{n})}\lesssim\left\\{\begin{array}[]{ll}a^{-\frac{n}{2}}&\displaystyle\textrm{if
}|m|>\frac{n}{2},\phantom{a^{\sqrt{\frac{n^{2}}{4}-m^{2}}}}\\\\[5.69046pt]
a^{-\frac{n}{2}}\log a&\displaystyle\textrm{if
}|m|=\frac{n}{2},\phantom{a^{\sqrt{\frac{n^{2}}{4}-m^{2}}}}\\\\[5.69046pt]
a^{-\frac{n}{2}+\sqrt{\frac{n^{2}}{4}-m^{2}}}&\displaystyle\textrm{if
}|m|<\frac{n}{2}.\end{array}\right.$

This motivates the decay estimates in Theorem 3.1.

##  7\. Appendix B

In this section, we prove the technical result we had used in the proof of
Theorem 3.1, in Section 3.

######  Lemma 7.1.

If $f,g\in H^{1}(\mathbb{R}^{n})$, then
$\displaystyle\lim_{r\rightarrow+\infty}\int_{S_{r}}fgd\sigma_{r}=0.$

###### Proof.

By the Cauchy-Schwarz inequality,

$\left|\int_{S_{r}}fgd\sigma_{r}\right|^{2}\leq\int_{S_{r}}|f|^{2}d\sigma_{r}\cdot\int_{S_{r}}|g|^{2}d\sigma_{r},$

and so, it is enough to show that

$\lim\limits_{r\rightarrow\infty}\int\limits_{S_{r}}|f|^{2}d\sigma_{r}=0.$

Suppose this does not hold. Then there exists an increasing sequence
$(r_{k})_{k}$ such that $r_{k}\stackrel{{\scriptstyle
k\rightarrow\infty}}{{\longrightarrow}}\infty$, and there exists an
$\epsilon>0$ such that for each $k$,

$\int_{S_{r_{k}}}|f|^{2}d\sigma_{r_{k}}>\epsilon.$

(The plan is to use the trace theorem to fatten these $S_{r_{k}}$-slices to
‘annuli’ $A_{k}$ and obtain $\|f\|_{H^{1}(A_{k})}^{2}>\widetilde{\epsilon}>0$
for all $k$, giving rise to the contradiction that

$\infty>\|f\|_{H^{1}(\mathbb{R}^{n})}^{2}>\sum_{k}\|f\|^{2}_{H^{1}(A_{k})}>\sum_{k}\tilde{\epsilon}=+\infty.$

So we will construct a subsequence $(r_{k_{m}})_{m}$ of $(r_{k})_{k}$ and a
sequence $(\delta_{m})_{m}$ of positive numbers such that
$r_{k_{1}}<r_{k_{1}}+\delta_{1}<r_{k_{2}}<r_{k_{2}}+\delta_{2}<r_{k_{3}}<\cdots$,
and such that for the ‘annuli’
$A_{m}:=\\{\mathbf{x}:r_{k_{m}}<|\mathbf{x}|<r_{k_{m}}+\delta_{m}\\}$, we have
$\|f\|_{H^{1}(A_{m})}^{2}>\widetilde{\epsilon}$. We will need to keep track of
the constants in the trace theorems on our annuli $A_{m}$, and we will use the
following [15, p.41].)

######  Theorem 7.2.

Let $\Omega$ be a bounded open subset of $\mathbb{R}^{n}$ with a Lipschitz
boundary $\Gamma$. Then for $f\in H^{1}(\Omega)$ and for all
$\epsilon\in(0,1)$,

$\|f\|^{2}_{L^{2}(\partial\Omega)}\leq\frac{\|\bm{\mu}\|_{C^{1}(\overline{\Omega})}}{\delta}\left(\epsilon^{1/2}\|\nabla
f\|_{L^{2}(\Omega)}^{2}+(1+\epsilon^{-1/2})\|f\|_{L^{2}(\Omega)}^{2}\right),$

where $\bm{\mu}\in C^{1}(\overline{\Omega},\mathbb{R}^{n})$ is such that
$\bm{\mu}\cdot\mathbf{n}\geq\delta$ on $\partial\Omega$, $\mathbf{n}$ being
the outer normal vector.

If $\Omega$ is an annulus $A=\\{\mathbf{x}:r<\|\mathbf{x}\|<R\\}$ (which is