# Existence of a minimizer to the particle number-Casimir functional for the
Einstein-Vlasov system

Håkan Andréasson
Mathematical Sciences
Chalmers University of Technology
Göteborg University
S-41296 Göteborg, Sweden
email<EMAIL_ADDRESS>

Markus Kunze
Mathematisches Institut
Universität Köln
Weyertal 86-90
D-50931 Köln, Germany
email<EMAIL_ADDRESS>
###### Abstract

In 2001 Wolansky [13] introduced a particle number-Casimir functional for the
Einstein-Vlasov system. Two open questions are associated with this
functional. First, a meaningful variational problem should be formulated and
the existence of a minimizer to this problem should be established. The second
issue is to show that a minimizer, for some choice of the parameters, is a
static solution of the Einstein-Vlasov system. In the present work we solve
the first problem by proving the existence of a minimizer to the particle
number-Casimir functional. On the technical side, it is a main achievement
that we are able to bypass the non-compactness of minimizing sequences by new
arguments in both $v$-space and $x$-space, which might have several further
applications. We note that such compactness results for the Einstein-Vlasov
system have been absent in the literature, whereas similar results have been
known in the Newtonian case. We also provide arguments which give strong
support that minimizers corresponding to small masses are static solutions of
the Einstein-Vlasov system. Furthermore, our analysis leads us to propose a
new stability criterion for static solutions: We conjecture that a static
solution for which the Casimir-binding energy is positive is stable for mass-
preserving perturbations.

##  1 Introduction

The Einstein-Vlasov system is a PDE system from Mathematical General
Relativity, where Einstein’s field equations are used jointly with a kinetic
description of the matter to model the time evolution of a collection of non-
colliding objects (‘particles’); the objects to be described in astrophysics
include stars, galaxies or galaxy clusters. The system simplifies
considerably, if some symmetry assumptions are added, like spherical symmetry.

It is well-known that there exists a large number of spherically symmetric
static (i.e., time-independent) solutions to the Einstein-Vlasov system [12,
2]. Whether or not a subclass of these solutions can be obtained as minimizers
of a variational problem, as should be expected from general physics
principles, has however remained an open question. The main reason why this is
of interest is related to stability. In the Newtonian analogue, i.e., for the
Vlasov-Poisson system, stability of a class of static solutions was first
shown by Guo and Rein as a consequence of their result that the solutions are
minimizers of an energy-Casimir functional [10]. Many quite general results on
stability for the Vlasov-Poisson system have since then been obtained, see
[11, 12] and the references therein.

For the Einstein-Vlasov system it was claimed by Wolansky in [13] that a class
of static solutions of the Einstein-Vlasov system are minimizers to a particle
number-Casimir functional. However, the proof in [13] contains serious flaws
as observed in [4], and hence the problem was still open. It naturally splits
into two parts: Firstly, one needs to formulate a variational problem and to
show that a minimizer to this variational problem does exist. The second part
is to prove that a minimizer, for some choices of the parameters, is a static
solution of the Einstein-Vlasov system.

The main purpose of the present work concerns part one, i.e., we set up a
variational problem and establish that a minimizer to this problem exists. To
show existence of a minimizer to a variational problem often requires that
compactness properties of the minimizing sequence can be derived. Indeed, this
is central in the Vlasov-Poisson case [10], but in the relativistic case such
compactness results have so far been absent in the literature. An essential
part of the present paper is about obtaining such compactness results for
minimizing sequences. Moreover, we provide arguments which give strong support
that a class of minimizers are static solutions; we are however not able to
fully prove this. Furthermore, our existence proof leads us to a new criterion
which we conjecture to be sufficient for stability.

We now specify in detail the problem we solve and we also include some more
comments on the Newtonian case, since there are some fundamental differences
between the relativistic and the non-relativistic systems. Let the Casimir
function be defined as

$\chi(f)=\frac{k}{k+1}\,f^{1+\frac{1}{k}}$

for $k\in]0,2]$. The functional we are dealing with is

${\cal
D}(f)=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(\chi(f)-\alpha
f)\,dx\,dv,$ (1.1)

where $\alpha>0$, $\lambda=\lambda_{f}$ is given by

$e^{-2\lambda}=1-\frac{2m}{r}$

and $m=m_{f}$ is the mass function defined as

$m(r)=\int_{|x|\leq r}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv.$

We refer to this functional as the particle number-Casimir functional
(introduced by Wolansky [13] and also used in [5]), since
$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,f\,dx\,dv$ represents
the particle number. We point out that $\mathcal{D}(f)$ is a conserved
quantity along solutions $f=f(t)$ of the full Einstein-Vlasov system. From now
on we fix $\alpha=1$, but let us nevertheless explain the reason for including
a general $\alpha$ in (1.1). The goal is to show that a minimizer of
$\mathcal{D}$ over a certain set ${\cal A}$ of admissible functions is a
static solution that has the form

$f_{0}(r,w,\ell)=\Big{(}1-Ce^{\mu_{0}(r)}\sqrt{1+w^{2}+\ell^{2}/r^{2}}\Big{)}_{+}^{k},$
(1.2)

where $r=|x|$, $w=\frac{x\cdot v}{r}$ and $\ell=|x\wedge v|$ are the spherical
coordinates and $e^{\mu_{0}(r)}$ is one of the metric coefficients. A family
of static solutions of this form can be parametrized by the central red shift
[6, 9]. If the mass and the radius of the spatial support of the solutions for
such a sequence is plotted (the ‘mass-radius diagram’), then a spiral is
obtained [6]. A solution of the form (1.2) can thus not have an arbitrary
large mass. However, if $f_{0}$ is a static solution with induced mass $M$,
then another static solution can be generated by applying the scaling
$\tilde{f}(x,v)=\gamma^{2}f_{0}(\gamma x,v)$, see Lemma 5.1 ahead. The induced
mass of $\tilde{f}$ is $M/\gamma$. Hence, if we want to find static solutions
with large masses as minimizers, then the corresponding particle number-
Casimir functional (1.1) needs to be modified, i.e., one needs to have a
different value than $\alpha=1$. This will result in a slightly different form
of $f_{0}$ which depends on $\alpha$, cf. [13]. For minimizing the functional
${\cal D}$, it is however sufficient to deal with just one $\alpha$, which we
take to be $\alpha=1$.

We will consider functions $f=f(x,v)\geq 0$ of $x,v\in\mathbb{R}^{3}$ that are
spherically symmetric, and our goal is to minimize ${\cal D}$ over an
appropriate class of functions ${\cal A}$. For $f\in{\cal A}$ we will require
the mass constraint

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=M.$
(1.3)

We note that also $M$ is a conserved quantity of the Einstein-Vlasov system.
This constraint will not be sufficient to enforce the mass function $m=m_{f}$
to have $2m/r$ stay away from $1$. Without a restriction on $m/r$, the
functional can be shown to be unbounded from below. This is in stark contrast
to the Newtonian case, where the corresponding functional is bounded from
below and a global minimizer exists. For this reason we will need to add
another condition, which we choose to be of the form
$\rho(r)=\rho_{f}(r)\leq\sigma_{0}$ for $r\in[0,\infty[$, where

$\rho(x)=\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$ (1.4)

denotes the density induced by $f$, and $\rho(x)=\rho(r)$ by abuse of
notation. Here another important technical difference to the Vlasov-Poisson
case arises: For Einstein-Vlasov (1.4) and (1.3) only gives $\rho\in
L^{1}_{x}(\mathbb{R}^{3})$, whereas the density $\rho_{{\rm
VP}}(x)=\int_{\mathbb{R}^{3}}f(x,v)\,dv$ for Vlasov-Poisson is know to satisfy
$\rho_{{\rm VP}}\in L^{5/3}_{x}(\mathbb{R}^{3})$.

To summarize, we consider the convex set ${\cal A}$ given by

$\displaystyle{\cal A}={\cal A}_{M,\sigma_{0}}$ $\displaystyle=$
$\displaystyle\Big{\\{}f=f(x,v)\geq 0\,\,\mbox{is measurable and spherically
symmetric},$ (1.5) $\displaystyle\hskip
10.00002pt\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=M,\,\,\rho(r)\leq\sigma_{0}\,\,\mbox{for}\,\,r\in[0,\infty[\Big{\\}},$

where $M>0$ and $\beta\in]0,\frac{1}{2}[$ are fixed, and for technical reasons

$0<\sigma_{0}\leq\min\Big{\\{}1,\frac{3\beta^{3}}{4\pi M^{2}}\Big{\\}}.$ (1.6)

It is shown in Lemma 9.2 below that for $f\in\mathcal{A}$ it holds that
$m/r\leq\beta$. Let us point out that the formulation of the variational
problem by Wolansky in [13] is different from ours. In fact, Wolansky
eventually considers another functional (defined in terms of a suitable
Lagrangian). However, in both formulations a restriction on $m/r$ is
necessary. This restriction introduces additional ‘boundaries’ of the set
${\cal A}$, and the minimization problem is turned into an obstacle problem,
which is one reason for the flaws in [13].

We want to minimize $\mathcal{D}$ over the set $\mathcal{A}$ and we thus
define

$I=\inf_{f\in{\cal A}}{\cal D}(f).$ (1.7)

It is not difficult to establish that $I<0$; see Lemma 4.2 ahead. The main
purpose of this paper is to show that $I$ is attained, in the sense that there
exists a function $f_{0}\in{\cal A}$ such that ${\cal D}(f_{0})=I$. This will
be achieved under a further assumption, which we call the Casimir-binding-
energy condition (CBEC).

###### Definition 1.1

We say that the Casimir-binding-energy condition (CBEC) holds, if

$I=\inf_{f\in{\cal A}}{\cal D}(f)<-M.$

Analytically we are able to show that (CBEC) holds for all $k\in]0,2]$, cf.
Lemma 4.2. The reason for the terminology ‘Casimir-binding-energy’ condition
is due to the definition of the binding energy as

$E_{b}(f)=M_{0}-M=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,f\,dx\,dv-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f\,dx\,dv.$

As mentioned above, $M_{0}$ is a conserved quantity called the particle
number, or rest mass. We thus find it natural to call the quantity

$E_{Cb}(f)=-\mathcal{D}(f)-M=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(f-\chi(f))\,dx\,dv-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f\,dx\,dv$

the Casimir-binding energy. We note that both the binding energy and the
Casimir-binding energy are conserved quantities. Moreover, due to $f\geq 0$ we
have $E_{b}(f)\geq E_{Cb}(f)$, so that the (CBEC) condition implies that a
minimizer $f_{0}$ (which we show that exists) has the property that
$E_{b}(f_{0})>0$. It is known that perturbations of static solutions for which
$E_{b}(f)>0$ have better stability properties than in the case where
$E_{b}(f)<0$, although the condition $E_{b}(f)>0$ is certainly not sufficient
for stability [6, 9]. In the literature, the strong binding energy hypothesis
[14] claims that the first local maximum of the binding energy along a
sequence of static solutions signals the onset of instability. This claim has
recently been investigated carefully using numerical simulations, and
counterexamples were found, see [9]. There is however strong support that the
weak binding energy hypothesis holds, which says that static solutions up to
the first local maximum of the binding energy along a sequence are stable. To
decide whether or not a steady state has the property that it is situated
before the first local maximum of the binding energy along a sequence requires
that such a sequence of static solutions can be generated. We propose a more
direct criterion that can be straightforwardly checked, without the need of
having a sequence of static solutions. The class of perturbations that we
address have the property that they leave the total mass unchanged, and the
perturbed density function $f$ off a static solution $f_{0}$ is also required
to be non-negative. We call this class mass-preserving perturbations.

Conjecture: Let $f_{0}$ be an isotropic static solution of the form (1.2). If
$\mathcal{D}(f_{0})<-M$, where $M$ is the induced mass of $f_{0}$, then
$f_{0}$ is stable for mass-preserving perturbations.

In Section 8 a more detailed discussion of the (CBEC) is given and some
comments about the stated conjecture are also provided.

Let us end this introduction by commenting about the relation between
stability and the existence of a minimizer. As mentioned before, for the
Vlasov-Poisson system stability is a consequence of the fact that a static
solution is obtained as a minimizer of an energy-Casimir functional over a
certain set (which we simply denote by $\mathcal{A}_{{\rm VP}}$ without being
more specific). A perturbation $f$ of a static solution $f_{0}$ will stay in
the set $\mathcal{A}_{{\rm VP}}$ as it evolves, and this fact is crucial for
deducing stability. In the relativistic case, a perturbation of a static
solution will not necessarily remain in the set $\mathcal{A}$ due to the
constraint $\rho\leq\sigma_{0}$. Hence, if a static solution is a minimizer of
the particle number-Casimir functional, it is necessary to control the
evolution of a perturbation of the static solution in a much more precise
fashion to make a conclusion about stability. This is highly non-trivial and
again points at the fundamental difference between the Newtonian and the
relativistic case. Nevertheless, this provides us with another method to
approach the non-linear stability problem.

This paper is organized as follows. In the next section the main results and
some ideas of the proofs are outlined. A new method to obtain minimizing
sequences that decay in $v$-space is presented in Section 3, and it relies on
a rearrangement argument. This property is crucial for deriving further
features of the limiting function $f_{0}$ of the minimizing sequence, see
Section 4. By using the Casimir-binding-energy condition (CBEC), the compact
support of $f_{0}$ in $x$-space can be proved in Section 5. In addition, the
existence of a minimizer $f_{0}\in\mathcal{A}$ is shown in this section. A
variational inequality satisfied by $f_{0}$ is derived in Section 6. As a
consequence of this inequality we obtain that the minimizer has compact
support also in $v$-space. In Section 7, a connection to the results in [5] is
made by proving that the minimizer necessarily takes a certain form, of which
a static solution is a special case. A particularly transparent way to state
the variational inequality is also derived, which clarifies under which
conditions a minimizer is a static solution. An argument is also given which
gives strong support to the belief that the minimizer is a static solution,
provided that the mass $M$ is sufficiently small. A discussion of the Casimir-
binding-energy condition (CBEC) is contained in Section 8, as well as a
discussion of the conjecture on stability that we propose. Section 9 has some
technical results, in particular, it is shown that the ‘Casimir part’ of the
functional $\mathcal{D}$ is convex for $k\in]0,2]$.

##  2 Main results and ideas of proofs

The minimization problem (1.7), (1.5), (1.1) is quite non-standard. Certainly
a minimizing sequence $(f_{n})\subset{\cal A}$ will be bounded in
$L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$, and we can extract a
subsequence (that is not relabeled) such that

$f_{n}\rightharpoonup
f_{0}\,\,\mbox{in}\,\,L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ (2.1)

as $n\to\infty$ for an appropriate limit function $f_{0}\in
L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$. However, it is far from
obvious why we should have $f_{0}\in{\cal A}$ and ${\cal D}(f_{0})=I$. There
is also no direct concentration compactness principle available that could be
of help at this point.

In this paper we will first introduce a new technique that improves minimizing
sequence by mass rearrangement in $v$-space. This procedure should be thought
of as two ‘machines’, with the feature that you stick in a function with
certain properties, then turn the crank, and out comes another function that
has better properties. The first machine modifies $f$ into $\hat{f}$ and adds
the condition $\hat{f}(x,v)\leq 1$, and in addition we have
$\rho_{\hat{f}}=\rho_{f}$ as well as ${\cal D}(\hat{f})\leq{\cal D}(f)$. Then
the second machine takes $\hat{f}$ into some $\tilde{f}$ so that
$\rho_{\tilde{f}}=\rho_{\hat{f}}$ and ${\cal D}(\tilde{f})\leq{\cal
D}(\hat{f})$ as well as $\tilde{f}(x,v)\leq 1$. The newly gained property of
$\tilde{f}$ is

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq
P_{0}+1,$ (2.2)

for a certain $P_{0}>0$ depending only on $k$, $M$ and $\beta$. If we apply
this procedure along the minimizing sequence, we thus may suppose that
additionally $f_{n}(x,v)\leq 1$ holds, as well as (2.2) for all $f_{n}$ (with
uniform constants). It turns out that (2.2) is sufficient, together with the
weak convergence (2.1), to prove the following facts about the limit function
$f_{0}\in L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$.

###### Lemma 2.1

The following assertions hold:

  * •

$0\leq f_{0}\leq 1$;

  * •

$f_{0}$ is spherically symmetric;

  * •

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq P_{0}+1$;

  * •

$m_{f_{n}}(r)\to m_{f_{0}}(r)$ as $n\to\infty$ for $r\in[0,\infty[$, in
particular $m_{f_{0}}(r)/r\leq\beta$ for $r\in]0,\infty[$;

  * •

$\rho_{f_{0}}(r)\leq\sigma_{0}$ for $r\in[0,\infty[$;

  * •

$\lambda_{f_{n}}(r)\to\lambda_{f_{0}}(r)$ as $n\to\infty$ for
$r\in[0,\infty[$.

While Lemma 2.1 is satisfying, and (2.2) is very useful to control the loss of
mass in $v$-space, we need an additional input to make sure that also no mass
is lost in $x$-space, in the sense that
$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv=M$
for the limit function $f_{0}$. Observe that we already know that
$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv=\int_{\mathbb{R}^{3}}\int_{|v|\leq
P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv+{\cal O}(P^{-1/4})$, which serves as a
substitute for compact support in $v$. To close the argument, we will be able
to determine the potential loss of mass in $x$-space by comparison to a
function $h\in{\cal A}$ such that ${\cal D}(h)<0$ and $|D(h)|>M$; such a
function $h$ must exist due to (CBEC).

Lastly, it is also possible to prove that in fact $f_{0}$ must have compact
support, both in $x$ and in $v$. Hence we arrive at the following main result
of this paper.

###### Theorem 2.2

Let $k\in]0,2]$. Then (CBEC) is verified and there is a function $f_{0}\in
L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ that is a minimizer, i.e., we
have $f_{0}\in{\cal A}$ and ${\cal D}(f_{0})=I$. Furthermore, $f_{0}\leq 1$,
and there are $R_{0},S_{0}>0$ such that $f_{0}(x,v)=0$ for a.e. $|x|\geq
R_{0}$, $v\in\mathbb{R}^{3}$ and for a.e. $x\in\mathbb{R}^{3}$, $|v|\geq
S_{0}$.

Up to the compact support property in $v$-space, the proof of this theorem is
completed in Section 5, cf. Theorem 5.10. To establish that the minimizer has
compact support in $v$-space we make use of the variational inequality
satisfied by $f_{0}$, as outlined in Section 6. The compact support property
in $v$-space is proven in Corollary 6.3.

The important question whether or not the minimizer from Theorem 2.2 is a
static solution of the Einstein-Vlasov system is considered in Section 7. By
making use of the variational inequality satisfied by $f_{0}$ and the results
in [5], we will find that

$f_{0}(r,w,\ell)=\Big{(}1-\tilde{\varepsilon}(r)\sqrt{1+w^{2}+\ell^{2}/r^{2}}\Big{)}_{+}^{k}$

for the function
$\tilde{\varepsilon}(r)=(G^{\prime})^{-1}(-4\pi\kappa\rho_{0}(r))$, where
$\kappa=\frac{k+1}{16\pi^{2}}$ and

$G(\varepsilon)=\int_{0}^{\infty}\xi^{2}\,{(1-\varepsilon\sqrt{1+\xi^{2}})}_{+}^{k+1}\,d\xi,\textrm{for
}\varepsilon\in\mathbb{R}.$

This information can be used to rewrite the variational inequality as

$\int_{0}^{\infty}\frac{d}{dr}\Big{(}e^{\lambda_{0}+\mu_{0}}(m_{g}-m_{0})\Big{)}\,U\,dr\geq
0.$ (2.3)

Here $g\in\mathcal{A}$ is free to be taken such that ${\rm supp}\,g\subset{\rm
supp}\,f_{0}$, and $\mu_{0}$ and $\lambda_{0}$ are the induced metric
functions for the minimizer $f_{0}$. In addition,

$U(r)=-\left\\{\begin{array}[]{c@{\quad:\quad}l}e^{-\mu_{0}(r)}\tilde{\varepsilon}(r)&r\in[0,R_{0}]\\\
0&r\in]R_{0},\infty[\end{array}\right.,$

where $R_{0}$ denotes the end of the $r$-support of $f_{0}$. It turns out that
if $U(r)={\rm const.}$, then $f_{0}$ is a static solution. Hence, if the
inequality in (2.3) could be turned into an equality, the claim would follow
from integration by parts, if $U$ was differentiable. If the minimizer stays
away from the boundary specified in the set $\mathcal{A}$, i.e., if the
induced density satisfies $\rho_{0}(r)<\sigma_{0}$, then all of this can be
made rigorous. We give an argument that for small mass $M$, the induced
density $\rho_{0}$ must at least partially stay away from of the boundary. Up
to such technical issues, we thus provide strong support that a minimizer
corresponding to a small mass is a static solution of the Einstein-Vlasov
system.

##  3 Rearrangement in $v$-space

Throughout this section we fix $M>0$ and $\beta\in]0,\frac{1}{2}[$. Let

$\displaystyle\tilde{{\cal A}}_{M,\beta}$ $\displaystyle=$
$\displaystyle\Big{\\{}f=f(x,v)\geq 0\,\,\mbox{is measurable and spherically
symmetric},$ $\displaystyle\hskip
10.00002pt\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=M,\,\,\frac{m(r)}{r}\leq\beta\,\,\mbox{for}\,\,r\in]0,\infty[,\,\,\rho(r)\leq
1\,\,\mbox{for}\,\,r\in[0,\infty[\Big{\\}}.$

Since it will be needed below, we remark that the function

$\phi(s)=\chi(s)-s=\frac{k}{k+1}\,s^{1+1/k}-s,\quad s\in[0,\infty[,$ (3.2)

has its zeros at $s=0$ and $s=(\frac{k+1}{k})^{k}$, attains its negative
minimal value $\phi(1)=-\frac{1}{k+1}$ at $s=1$ and decreases in $[0,1]$.

We improve a given $f\in\tilde{{\cal A}}_{M,\beta}$ (to be thought of as a
member of a minimizing sequence for ${\cal D}$) in several steps.

###### Lemma 3.1

Let $f=f(x,v)\geq 0$ be spherically symmetric and so that $f\in\tilde{{\cal
A}}_{M,\beta}$, i.e.,

  * (a)

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=M$;

  * (b)

$m(r)/r\leq\beta$ for $r\in]0,\infty[$;

  * (c)

$\rho(r)\leq 1$ for $r\in[0,\infty[$.

Then there is a spherically symmetric function $\hat{f}=\hat{f}(x,v)\geq 0$
such that

  * •

$\rho_{\hat{f}}=\rho_{f}$, in particular (a)-(c) holds for $\hat{f}$ and
$\hat{f}\in\tilde{{\cal A}}_{M,\beta}$;

  * •

${\cal D}(\hat{f})\leq{\cal D}(f)$;

  * •

$\hat{f}(x,v)\leq 1$ for $(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}$.

Proof : To construct $\hat{f}$, we define the sets

$E(x)=\\{v\in\mathbb{R}^{3}:f(x,v)>1\\}$

for $x\in\mathbb{R}^{3}$, and also

$\rho_{D}(x)=\int_{E(x)}\sqrt{1+|v|^{2}}\,(f(x,v)-1)\,dv.$

Then $0\leq\rho_{D}(x)\leq\rho(x)\leq 1$. Let
$H=\\{v\in\mathbb{R}^{3}:a<|v|<b\\}$ for $a=\sqrt[3]{\frac{1}{4\pi}}$ and
$b=\sqrt[3]{\frac{25}{4\pi}}$, and moreover

$H_{<}(x)=\Big{\\{}v\in H:f(x,v)<\frac{1}{2}\Big{\\}},\quad
H_{>}(x)=\Big{\\{}v\in H:f(x,v)\geq\frac{1}{2}\Big{\\}}$

for $x\in\mathbb{R}^{3}$. We have $|H|=(4\pi/3)(b^{3}-a^{3})=8$ for the
Lebesgue measure of $H$, and since $H=H_{<}(x)\cup H_{>}(x)$ is a disjoint
union, also $|H_{<}(x)|+|H_{>}(x)|=8$. We claim that $|H_{>}(x)|<4$. In the
opposite case $|H_{>}(x)|\geq 4$ we would obtain

$1\geq\rho(x)\geq\int_{H_{>}(x)}\sqrt{1+|v|^{2}}\,f(x,v)\,dv\geq\frac{1}{2}\int_{H_{>}(x)}\sqrt{1+|v|^{2}}\,dv\geq\frac{1}{2}\,|H_{>}(x)|\geq
2,$

which is a contradiction. From $|H_{>}(x)|<4$ it follows that $|H_{<}(x)|>4$.
Noting that $E(x)$ and $H_{<}(x)$ are disjoint, we define

$\hat{f}(x,v)=\left\\{\begin{array}[]{c@{\quad:\quad}c}1&v\in
E(x)\\\\[4.30554pt]
\frac{\rho_{D}(x)}{\sqrt{1+|v|^{2}}\,|H_{<}(x)|}+f(x,v)&v\in
H_{<}(x)\\\\[4.30554pt] f(x,v)&v\in\mathbb{R}^{3}\setminus(E(x)\cup
H_{<}(x))\end{array}\right..$ (3.3)

To see that $\rho_{\hat{f}}=\rho_{f}$, we calculate

$\displaystyle\rho_{f}(x)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$
$\displaystyle=$
$\displaystyle\int_{E(x)}\sqrt{1+|v|^{2}}\,f(x,v)\,dv+\int_{H_{<}(x)}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$
$\displaystyle+\,\int_{\mathbb{R}^{3}\setminus(E(x)\cup
H_{<}(x))}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$ $\displaystyle=$
$\displaystyle\int_{E(x)}\sqrt{1+|v|^{2}}\,(f(x,v)-1)\,dv+\int_{E(x)}\sqrt{1+|v|^{2}}\,dv$
$\displaystyle+\,\int_{H_{<}(x)}\sqrt{1+|v|^{2}}\,f(x,v)\,dv+\int_{\mathbb{R}^{3}\setminus(E(x)\cup
H_{<}(x))}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$ $\displaystyle=$
$\displaystyle\rho_{D}(x)+\int_{E(x)}\sqrt{1+|v|^{2}}\,dv$
$\displaystyle+\,\int_{H_{<}(x)}\sqrt{1+|v|^{2}}\,f(x,v)\,dv+\int_{\mathbb{R}^{3}\setminus(E(x)\cup
H_{<}(x))}\sqrt{1+|v|^{2}}\,f(x,v)\,dv.$

On the other hand,

$\displaystyle\rho_{\hat{f}}(x)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,\hat{f}(x,v)\,dv$
$\displaystyle=$
$\displaystyle\int_{E(x)}\sqrt{1+|v|^{2}}\,dv+\int_{H_{<}(x)}\sqrt{1+|v|^{2}}\,\Big{(}\frac{\rho_{D}(x)}{\sqrt{1+|v|^{2}}\,|H_{<}(x)|}+f(x,v)\Big{)}\,dv$
$\displaystyle+\,\int_{\mathbb{R}^{3}\setminus(E(x)\cup
H_{<}(x))}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$ $\displaystyle=$
$\displaystyle\int_{E(x)}\sqrt{1+|v|^{2}}\,dv+\rho_{D}(x)+\int_{H_{<}(x)}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$
$\displaystyle+\,\int_{\mathbb{R}^{3}\setminus(E(x)\cup
H_{<}(x))}\sqrt{1+|v|^{2}}\,f(x,v)\,dv,$

so that both expressions agree. Next we argue that ${\cal D}(\hat{f})\leq{\cal
D}(f)$ holds. Since $m_{\hat{f}}=m_{f}$, and hence
$\lambda_{\hat{f}}=\lambda_{f}$, we need to establish the relation

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f}}\,(\chi(\hat{f})-\hat{f})\,dx\,dv\leq\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f}}\,(\chi(f)-f)\,dx\,dv.$

Due to (3.3), this comes down to showing that

$\int_{\mathbb{R}^{3}}dx\,e^{\lambda_{f}}\int_{E(x)\cup
H_{<}(x)}dv\,(\chi(\hat{f})-\hat{f})\leq\int_{\mathbb{R}^{3}}dx\,e^{\lambda_{f}}\int_{E(x)\cup
H_{<}(x)}dv\,(\chi(f)-f).$ (3.4)

By definition, if $v\in H_{<}(x)$, then $f(x,v)<1/2$ and also
$\delta(x,v)=\frac{\rho_{D}(x)}{\sqrt{1+|v|^{2}}\,|H_{<}(x)|}\leq\frac{1}{|H_{<}(x)|}<1/4$.
Since $1/4+1/2\leq 1$, we deduce that
$\phi(\hat{f}(x,v))=\phi(\delta(x,v)+f(x,v))\leq\phi(f(x,v))$. Furthermore, if
$v\in E(x)$, then $\phi(\hat{f}(x,v))=\phi(1)\leq\phi(f(x,v))$, which implies
(3.4). Due to $1/4+1/2\leq 1$ and $f(x,v)\leq 1$ for $v\not\in E(x)$, we also
have $\hat{f}(x,v)\leq 1$ for $(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}$.
$\Box$

Define $P_{0}\geq 10$ by

$P_{0}=\max\Big{\\{}10,\Big{(}\frac{1+M}{4\pi}\Big{)}^{4/3},\frac{64(k+1)^{2}}{1-2\beta},\frac{256(k+1)^{2}M^{4}}{(1-2\beta)^{2}}\Big{\\}}.$
(3.5)

###### Lemma 3.2

Let $f=f(x,v)\geq 0$ be spherically symmetric and so that $f\in\tilde{{\cal
A}}_{M,\beta}$, i.e.,

  * (a)

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=M$;

  * (b)

$m(r)/r\leq\beta$ for $r\in]0,\infty[$;

  * (c)

$\rho(r)\leq 1$ for $r\in[0,\infty[$.

In addition, suppose that

  * (d)

$f(x,v)\leq 1$ for $(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}$;

  * (e)

there is a $P\geq P_{0}$ such that

$P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq P}\sqrt{1+|v|^{2}}\,f\,dx\,dv\geq
1.$

Then there is a spherically symmetric function $\tilde{f}=\tilde{f}(x,v)\geq
0$ such that

  * •

$\rho_{\tilde{f}}=\rho_{f}$, in particular (a)-(c) holds for $\tilde{f}$ and
$\tilde{f}\in\tilde{{\cal A}}_{M,\beta}$;

  * •

${\cal D}(\tilde{f})\leq{\cal D}(f)$;

  * •

${\rm
supp}(\tilde{f})\subset\\{(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}:|v|\leq
P+1\\}$;

  * •

$\tilde{f}(x,v)\leq 1$ for $(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}$;

  * •

we have

$P^{1/4}\int_{\mathbb{R}^{3}}\int_{P\leq|v|\leq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv\leq 1.$ (3.6)

Proof : We decompose

$\displaystyle M$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv$
$\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{3}}\int_{|v|\leq
P}\sqrt{1+|v|^{2}}\,f\,dx\,dv+\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}\sqrt{1+|v|^{2}}\,f\,dx\,dv$ $\displaystyle=:$ $\displaystyle M_{<}+M_{>}.$

Note that we have $M_{>}>0$, since otherwise (e) could not hold. Next we
introduce the numbers

$N=\bigg{[}\frac{(1+M)P}{M_{>}}\bigg{]}\in\mathbb{N},\quad\delta=\frac{1}{N},\quad
q=\sqrt{P},$

along with the strips

$S_{j}=\Big{\\{}(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}:\frac{q}{2}+j\delta
q\leq|v|\leq\frac{q}{2}+(j+1)\delta q\Big{\\}},\quad j=0,\ldots,N-1;$

then

$\bigcup_{j=0}^{N-1}S_{j}=\Big{\\{}(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}:\frac{q}{2}\leq|v|\leq\frac{3q}{2}\Big{\\}}.$
(3.7)

Observe that $M_{<}\leq M$ yields $\frac{(1+M)P}{M_{>}}\geq P\geq P_{0}\geq
10$, so at least $N\geq 9$. Furthermore,

$N=\bigg{[}\frac{(1+M)P}{M_{>}}\bigg{]}\geq\frac{(1+M)P}{M_{>}}-1\geq\frac{(1+M)P}{2M_{>}}$
(3.8)

is found. We also split up

$\displaystyle{\cal D}(f)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv$
$\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{3}}\int_{|v|\leq
P}e^{\lambda}\,(\chi(f)-f)\,dx\,dv+\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}e^{\lambda}\,(\chi(f)-f)\,dx\,dv$ $\displaystyle=:$ $\displaystyle{\cal
D}_{<}+{\cal D}_{>}.$

Since $f(x,v)\leq 1$ pointwise,

$\chi(f(x,v))-f(x,v)=\phi(f(x,v))\leq 0$ (3.9)

pointwise; recall (3.2). In particular, ${\cal D}_{<}\leq 0$ and ${\cal
D}_{>}\leq 0$. Due to $m(r)/r\leq\beta$ we obtain $e^{-2\lambda}=1-2m/r\geq
1-2\beta$, so that

$1\leq e^{\lambda}\leq\frac{1}{\sqrt{1-2\beta}}=c_{\beta}.$ (3.10)

It follows that

$\displaystyle{\cal D}_{>}$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}e^{\lambda}\,(\chi(f)-f)\,dx\,dv\geq-\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}e^{\lambda}\,f\,dx\,dv\geq-c_{\beta}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}f\,dx\,dv$ (3.11) $\displaystyle\geq$
$\displaystyle-\frac{c_{\beta}}{\sqrt{1+P^{2}}}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}\sqrt{1+|v|^{2}}\,f\,dx\,dv=-\frac{c_{\beta}}{\sqrt{1+P^{2}}}\,M_{>}\,.$

Furthermore,

$\displaystyle|{\cal D}_{<}|$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{|v|\leq
P}e^{\lambda}\,(f-\chi(f))\,dx\,dv\leq\int_{\mathbb{R}^{3}}\int_{|v|\leq
P}e^{\lambda}\,f\,dx\,dv\leq c_{\beta}\int_{\mathbb{R}^{3}}\int_{|v|\leq
P}f\,dx\,dv$ (3.12) $\displaystyle\leq$ $\displaystyle c_{\beta}M_{<}\leq
c_{\beta}M.$

Now we assert that there is an $i\in\\{0,\ldots,N-1\\}$ such that

$\int\int_{S_{i}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv\geq\frac{{\cal D}_{<}}{N}.$
(3.13)

Otherwise we had

$\int\int_{S_{j}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv<\frac{{\cal D}_{<}}{N}$

for all $j\in\\{0,\ldots,N-1\\}$, which upon summation and on account of (3.7)
would yield

$\int\int_{\frac{q}{2}\leq|v|\leq\frac{3q}{2}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv<{\cal
D}_{<}.$

From $3q/2=3\sqrt{P}/2\leq P$ in conjunction with (3.9), it follows that

${\cal F}_{<}=\int\int_{|v|\leq
P}e^{\lambda}\,(\chi(f)-f)\,dx\,dv\leq\int\int_{\frac{q}{2}\leq|v|\leq\frac{3q}{2}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv<{\cal
D}_{<},$

which is a contradiction and completes the proof of (3.13).

Let

$\rho_{>}(x)=\int_{|v|\geq
P}\sqrt{1+|v|^{2}}\,f(x,v)\,dv,\quad\rho_{i}(x)=\int_{\bar{S}_{i}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv,$

where

$\bar{S}_{i}=\Big{\\{}v\in\mathbb{R}^{3}:\frac{q}{2}+i\delta
q\leq|v|\leq\frac{q}{2}+(i+1)\delta q\Big{\\}}$

denotes the $v$-section of $S_{i}$. Then both $\rho_{>}$ and $\rho_{i}$ are
radial, and also $\bar{S}_{i}\subset\\{v\in\mathbb{R}^{3}:|v|\leq
3q/2\\}\leq\\{v\in\mathbb{R}^{3}:|v|\leq P\\}$. We claim:

  * (A)

There is a $\xi_{1}\in]P,P+1[$ such that

$\int_{P\leq|v|\leq\xi_{1}}\sqrt{1+|v|^{2}}\,dv=1.$

  * (B)

There is a $\xi_{2}\in]\frac{q}{2}+i\delta q,\frac{q}{2}+(i+1)\delta q[$ such
that

$\int_{\frac{q}{2}+i\delta q\leq|v|\leq\xi_{2}}\sqrt{1+|v|^{2}}\,dv=1.$

To establish (A),
$\varphi_{1}(\xi)=\int_{P\leq|v|\leq\xi}\sqrt{1+|v|^{2}}\,dv$ for
$\xi\in[P,P+1]$ has $\varphi_{1}(P)=0$ and

$\displaystyle\varphi_{1}(P+1)$ $\displaystyle=$
$\displaystyle\int_{P\leq|v|\leq
P+1}\sqrt{1+|v|^{2}}\,dv=4\pi\int_{P}^{P+1}u^{2}\sqrt{1+u^{2}}\,du\geq
4\pi\int_{P}^{P+1}u^{3}\,du$ $\displaystyle=$
$\displaystyle\pi((P+1)^{4}-P^{4})\geq 4\pi P>1.$

For (B), the argument is similar if we use the function
$\varphi_{2}(\xi)=\int_{\frac{q}{2}+i\delta
q\leq|v|\leq\xi}\sqrt{1+|v|^{2}}\,dv$ for $\xi\in[\frac{q}{2}+i\delta
q,\frac{q}{2}+(i+1)\delta q]$, which has $\varphi_{2}(\frac{q}{2}+i\delta
q)=0$ and

$\displaystyle\varphi_{2}\Big{(}\frac{q}{2}+(i+1)\delta q\Big{)}$
$\displaystyle=$ $\displaystyle\int_{\frac{q}{2}+i\delta
q\leq|v|\leq\frac{q}{2}+(i+1)\delta q}\sqrt{1+|v|^{2}}\,dv$
$\displaystyle\geq$ $\displaystyle 4\pi\int_{\frac{q}{2}+i\delta
q}^{\frac{q}{2}+(i+1)\delta
q}u^{3}\,du=\pi\,q^{4}\Big{(}(1+2(i+1)\delta)^{4}-(1+2i\delta)^{4}\Big{)}$
$\displaystyle=$
$\displaystyle\pi\,q^{4}\Big{(}8\delta+24[(i+1)^{2}-i^{2}]\delta^{2}+32[(i+1)^{3}-i^{3}]\delta^{3}+16[(i+1)^{4}-i^{4}]\delta^{4}\Big{)}$
$\displaystyle\geq$ $\displaystyle 8\pi\,q^{4}\delta=8\pi\,\frac{P^{2}}{N}\geq
8\pi\,\frac{P^{2}M_{>}}{(1+M)P}\geq 8\pi\,\frac{P^{3/4}}{1+M}>1,$

where we have used that $P^{1/4}M_{>}\geq 1$ is verified by (e), along with
the definition of $P_{0}$. Thus (A) and (B) have been established, and we can
put

$h_{1}(x,v)={\bf 1}_{[P,\,\xi_{1}]}(|v|)\rho_{i}(x),\quad h_{2}(x,v)={\bf
1}_{[\frac{q}{2}+i\delta q,\,\xi_{2}]}(|v|)\rho_{>}(x).$

Define $\tilde{f}$ by means of

$\tilde{f}(x,v)=\left\\{\begin{array}[]{c@{\quad:\quad}c}0&|v|\geq
P+1\\\\[4.30554pt] h_{1}(x,v)&P\leq|v|\leq P+1\\\\[4.30554pt]
f(x,v)&(x,v)\in\\{(x,v):|v|\leq P\\}\setminus S_{i}\\\\[4.30554pt]
h_{2}(x,v)&(x,v)\in S_{i}\end{array}\right..$

To see that $\tilde{f}$ has the desired properties, we begin by calculating
its density $\rho_{\tilde{f}}$. Here we have, by (A) and (B) above,

$\displaystyle\rho_{\tilde{f}}(x)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,\tilde{f}(x,v)\,dv$
$\displaystyle=$ $\displaystyle\int_{P\leq|v|\leq
P+1}\sqrt{1+|v|^{2}}\,h_{1}(x,v)\,dv+\int_{\\{|v|\leq
P\\}\setminus\bar{S}_{i}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$
$\displaystyle+\,\int_{\bar{S}_{i}}\sqrt{1+|v|^{2}}\,h_{2}(x,v)\,dv$
$\displaystyle=$
$\displaystyle\rho_{i}(x)\int_{P\leq|v|\leq\xi_{1}}\sqrt{1+|v|^{2}}\,dv+\int_{\\{|v|\leq
P\\}\setminus\bar{S}_{i}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$
$\displaystyle+\rho_{>}(x)\int_{\frac{q}{2}+i\delta
q\leq|v|\leq\xi_{2}}\sqrt{1+|v|^{2}}\,dv$ $\displaystyle=$
$\displaystyle\int_{\bar{S}_{i}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv+\int_{\\{|v|\leq
P\\}\setminus\bar{S}_{i}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$
$\displaystyle+\int_{|v|\geq P}\sqrt{1+|v|^{2}}\,f(x,v)\,dv$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f(x,v)\,dv=\rho_{f}(x),$

so that in particular $m_{\tilde{f}}=m_{f}$ and consequently
$\lambda_{\tilde{f}}=\lambda_{f}$. We can also write

$\displaystyle{\cal D}(\tilde{f})$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{\tilde{f}}}\,(\chi(\tilde{f})-\tilde{f})\,dx\,dv=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f}}\,(\chi(\tilde{f})-\tilde{f})\,dx\,dv$
(3.14) $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{3}}\int_{P\leq|v|\leq
P+1}e^{\lambda_{f}}\,(\chi(h_{1})-h_{1})\,dx\,dv+\int_{\mathbb{R}^{3}}\int_{\\{|v|\leq
P\\}\setminus\bar{S}_{i}}e^{\lambda_{f}}\,(\chi(f)-f)\,dx\,dv$
$\displaystyle+\int_{\mathbb{R}^{3}}\int_{\bar{S}_{i}}e^{\lambda_{f}}\,(\chi(h_{2})-h_{2})\,dx\,dv.$

For the contribution on $\\{P\leq|v|\leq P+1\\}$ we note that
$0\leq\rho_{i}(x)\leq\rho(x)\leq 1$ yields $0\leq h_{1}(x,v)\leq 1$, and
accordingly $\chi(h_{1}(x,v))-h_{1}(x,v)=\phi(h_{1}(x,v))\leq 0$. Thus

$\int_{\mathbb{R}^{3}}\int_{P\leq|v|\leq
P+1}e^{\lambda_{f}}\,(\chi(h_{1})-h_{1})\,dx\,dv\leq 0.$ (3.15)

For the part on $\\{|v|\leq P\\}\setminus\bar{S}_{i}$ we use (3.13) to get

$\displaystyle\int_{\mathbb{R}^{3}}\int_{\\{|v|\leq
P\\}\setminus\bar{S}_{i}}e^{\lambda_{f}}\,(\chi(f)-f)\,dx\,dv$ (3.16)
$\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{3}}\int_{|v|\leq
P}e^{\lambda_{f}}\,(\chi(f)-f)\,dx\,dv-\int_{\mathbb{R}^{3}}\int_{\bar{S}_{i}}e^{\lambda_{f}}\,(\chi(f)-f)\,dx\,dv$
$\displaystyle=$ $\displaystyle{\cal
D}_{<}-\int\int_{S_{i}}e^{\lambda_{f}}\,(\chi(f)-f)\,dx\,dv$
$\displaystyle\leq$ $\displaystyle{\cal D}_{<}-\frac{{\cal D}_{<}}{N}.$

Lastly, as for $h_{1}$ we obtain $0\leq h_{2}(x,v)\leq 1$, so that

$h_{2}-\chi(h_{2})=h_{2}\Big{(}1-\frac{k}{k+1}\,h_{2}^{1/k}\Big{)}\geq\frac{1}{k+1}\,h_{2}.$
(3.17)

Hence (3.10), the definition of $h_{2}$ and (B) above yields

$\displaystyle\int_{\mathbb{R}^{3}}\int_{\bar{S}_{i}}e^{\lambda_{f}}\,(h_{2}-\chi(h_{2}))\,dx\,dv$
$\displaystyle\geq$
$\displaystyle\frac{1}{k+1}\int_{\mathbb{R}^{3}}\int_{\bar{S}_{i}}e^{\lambda_{f}}\,h_{2}\,dx\,dv\geq\frac{1}{k+1}\int_{\mathbb{R}^{3}}\int_{\bar{S}_{i}}h_{2}\,dx\,dv$
(3.18) $\displaystyle=$
$\displaystyle\frac{1}{k+1}\Big{(}\int_{\mathbb{R}^{3}}\rho_{>}(x)\,dx\Big{)}\Big{(}\int_{\bar{S}_{i}}{\bf
1}_{[\frac{q}{2}+i\delta q,\,\xi_{2}]}(|v|)\,dv\Big{)}$ $\displaystyle=$
$\displaystyle\frac{1}{k+1}\,M_{>}\Big{(}\int_{\frac{q}{2}+i\delta
q\leq|v|\leq\xi_{2}}\,\frac{\sqrt{1+|v|^{2}}}{\sqrt{1+|v|^{2}}}\,dv\Big{)}$
$\displaystyle\geq$
$\displaystyle\frac{1}{k+1}\,\frac{M_{>}}{\sqrt{1+\frac{9q^{2}}{4}}}\Big{(}\int_{\frac{q}{2}+i\delta
q\leq|v|\leq\xi_{2}}\,\sqrt{1+|v|^{2}}\,dv\Big{)}$ $\displaystyle=$
$\displaystyle\frac{1}{k+1}\,\frac{M_{>}}{\sqrt{1+\frac{9q^{2}}{4}}}.$

Now if we put (3.15), (3.16) and (3.18) into (3.14), it follows that

${\cal D}(\tilde{f})\leq{\cal D}_{<}-\frac{{\cal
D}_{<}}{N}-\frac{1}{k+1}\,\frac{M_{>}}{\sqrt{1+\frac{9q^{2}}{4}}}.$

Since $q^{2}=P$ and $1+9P/4\leq 4P$, we deduce from (3.8), (3.12) and (3.11)
that

$\displaystyle{\cal D}(\tilde{f})$ $\displaystyle\leq$ $\displaystyle{\cal
D}_{<}+\frac{2|{\cal
D}_{<}|M_{>}}{(1+M)P}-\frac{1}{2}\,\frac{1}{k+1}\,\frac{M_{>}}{\sqrt{P}}$
$\displaystyle=$ $\displaystyle{\cal D}_{<}+\frac{2|{\cal
D}_{<}|M_{>}}{(1+M)P}-\frac{1}{4}\,\frac{1}{k+1}\,\frac{M_{>}}{\sqrt{P}}-\frac{1}{4}\,\frac{1}{k+1}\,\frac{M_{>}}{\sqrt{P}}$
$\displaystyle\leq$ $\displaystyle{\cal D}_{<}+{\cal
D}_{>}+\frac{M_{>}}{\sqrt{P}}\bigg{[}\frac{2c_{\beta}}{\sqrt{P}}-\frac{1}{4}\,\frac{1}{k+1}\bigg{]}+\frac{M_{>}}{\sqrt{P}}\bigg{[}\frac{c_{\beta}\sqrt{P}}{\sqrt{1+P^{2}}}-\frac{1}{4}\,\frac{1}{k+1}\bigg{]}$
$\displaystyle\leq$ $\displaystyle{\cal D}_{<}+{\cal D}_{>}={\cal D}(f),$

where we have also used the definition (3.5) of $P_{0}$. The asserted support
property of $\tilde{f}$ is just by definition, and so is $\tilde{f}(x,v)\leq
1$, since $h_{1}(x,v)\leq 1$, $h_{2}(x,v)\leq 1$ as well as $f(x,v)\leq 1$.
Finally to establish (3.6), we get from (A), and the definitions of $\rho_{i}$
and $\bar{S}_{i}$,

$\displaystyle\int_{\mathbb{R}^{3}}\int_{P\leq|v|\leq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{P\leq|v|\leq
P+1}\sqrt{1+|v|^{2}}\,h_{1}\,dx\,dv$ $\displaystyle=$
$\displaystyle\bigg{(}\int_{P\leq|v|\leq\xi_{1}}\sqrt{1+|v|^{2}}\,dv\bigg{)}\bigg{(}\int_{\mathbb{R}^{3}}\rho_{i}(x)\,dx\bigg{)}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\bar{S}_{i}}\sqrt{1+|v|^{2}}\,f\,dx\,dv$
$\displaystyle=$ $\displaystyle\int\int_{S_{i}}\sqrt{1+|v|^{2}}\,f\,dx\,dv.$

Recalling that $f(x,v)\leq 1$, the analogous argument as is (3.17) shows that
$f-\chi(f)\geq\frac{1}{k+1}\,f$. Furthermore, on $S_{i}$ we have $|v|\leq
3q/2$. By means of (3.10), (3.13), (3.12) and (3.8) we may therefore bound

$\displaystyle P^{1/4}\int_{\mathbb{R}^{3}}\int_{P\leq|v|\leq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$ $\displaystyle\leq$
$\displaystyle(k+1)\sqrt{1+\frac{9q^{2}}{4}}\,P^{1/4}\int\int_{S_{i}}(f-\chi(f))\,dx\,dv$
$\displaystyle\leq$
$\displaystyle(k+1)\sqrt{1+\frac{9q^{2}}{4}}\,P^{1/4}\int\int_{S_{i}}e^{\lambda}(f-\chi(f))\,dx\,dv$
$\displaystyle\leq$
$\displaystyle(k+1)\sqrt{1+\frac{9q^{2}}{4}}\,P^{1/4}\,\frac{|{\cal
D}_{<}|}{N}$ $\displaystyle\leq$
$\displaystyle(k+1)\sqrt{1+\frac{9q^{2}}{4}}\,P^{1/4}\,\frac{c_{\beta}M}{N}$
$\displaystyle\leq$
$\displaystyle(k+1)\sqrt{1+\frac{9P}{4}}\,P^{1/4}\,\frac{2c_{\beta}MM_{>}}{(1+M)P}$
$\displaystyle\leq$ $\displaystyle(k+1)\,\frac{4c_{\beta}M}{P^{1/4}}$
$\displaystyle\leq$ $\displaystyle 1,$

and this completes the proof. $\Box$

###### Corollary 3.3

Let $f=f(x,v)\geq 0$ be spherically symmetric and so that $f\in\tilde{{\cal
A}}_{M,\beta}$, i.e.,

  * (a)

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=M$;

  * (b)

$m(r)/r\leq\beta$ for $r\in]0,\infty[$;

  * (c)

$\rho(r)\leq 1$ for $r\in[0,\infty[$.

Then there is a spherically symmetric function $\tilde{f}=\tilde{f}(x,v)\geq
0$ such that

  * •

$\rho_{\tilde{f}}=\rho_{f}$, in particular (a)-(c) holds for $\tilde{f}$ and
$\tilde{f}\in\tilde{{\cal A}}_{M,\beta}$;

  * •

${\cal D}(\tilde{f})\leq{\cal D}(f)$;

  * •

$\tilde{f}(x,v)\leq 1$ for $(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}$;

  * •

we have

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq
P_{0}+1.$ (3.19)

Proof : From $f$ we can pass to $\hat{f}$ by applying Lemma 3.1, and hence we
may suppose w.l.o.g. that already $f$ has the additional property that

$f(x,v)\leq 1\quad\mbox{for}\quad(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}.$
(3.20)

Now we need to distinguish a few cases. Case 1: We have

$P_{0}^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P_{0}}\sqrt{1+|v|^{2}}\,f\,dx\,dv\geq 1.$

Then (a)-(e) from Lemma 3.2 are satisfied for $f$ and $P_{0}$. Thus there is
$\tilde{f}$ such that $\rho_{\tilde{f}}=\rho_{f}$, ${\cal
D}(\tilde{f})\leq{\cal D}(f)$, ${\rm
supp}(\tilde{f})\subset\\{(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}:|v|\leq
P_{0}+1\\}$ and $\tilde{f}(x,v)\leq 1$ pointwise; no further property of
$\tilde{f}$ will be needed here. To prove (3.19), let $P\geq P_{0}+1$. If
$|v|\geq P$, then $\tilde{f}(x,v)=0$ and hence

$P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv=0.$

Case 2: We have

$P_{0}^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P_{0}}\sqrt{1+|v|^{2}}\,f\,dx\,dv<1$

and

$P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq P}\sqrt{1+|v|^{2}}\,f\,dx\,dv\leq
1,\quad P\geq P_{0},$

holds for $f$. Since

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f\,dx\,dv\leq\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}\sqrt{1+|v|^{2}}\,f\,dx\,dv$

we may simply take $\tilde{f}=f$ in this case. Case 3: We have

$P_{0}^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P_{0}}\sqrt{1+|v|^{2}}\,f\,dx\,dv<1$

and there is a $P_{1}\geq P_{0}$ such that

$P_{1}^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P_{1}}\sqrt{1+|v|^{2}}\,f\,dx\,dv>1$

is verified. Then $P_{1}>P_{0}$, and since the function
$\varphi(P)=P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P}\sqrt{1+|v|^{2}}\,f\,dx\,dv$ is continuous on $[P_{0},P_{1}]$, there is a
first $\hat{P}\in]P_{0},P_{1}[$ such that

$\hat{P}^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq\hat{P}}\sqrt{1+|v|^{2}}\,f\,dx\,dv=1.$

In particular,

$P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq P}\sqrt{1+|v|^{2}}\,f\,dx\,dv\leq
1,\quad P\in[P_{0},\hat{P}].$ (3.21)

For $f$ and $\hat{P}$ we can apply Lemma 3.2 to obtain a density $\tilde{f}$
such that $\rho_{\tilde{f}}=\rho_{f}$, ${\cal D}(\tilde{f})\leq{\cal D}(f)$,
${\rm
supp}(\tilde{f})\subset\\{(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}:|v|\leq\hat{P}+1\\}$
and $\tilde{f}(x,v)\leq 1$ pointwise. In addition, we can use that

$\hat{P}^{1/4}\int_{\mathbb{R}^{3}}\int_{\hat{P}\leq|v|\leq\hat{P}+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv\leq
1.$ (3.22)

To show (3.19), let $P\geq P_{0}+1$. If $P\geq\hat{P}$ and $|v|\geq P+1$, then
$|v|\geq\hat{P}+1$, and thus the integral in question vanishes by the support
properties of $\tilde{f}$. Hence we may assume that $P_{0}+1\leq P\leq\hat{P}$
holds. First we will consider the case where $P+1\leq\hat{P}$, so that we have
$P_{0}\leq P<P+1\leq\hat{P}<\hat{P}+1$. We write

$\displaystyle P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$ (3.23) $\displaystyle=$
$\displaystyle
P^{1/4}\int_{\mathbb{R}^{3}}\int_{P+1\leq|v|\leq\hat{P}+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$
$\displaystyle=$ $\displaystyle
P^{1/4}\int_{\mathbb{R}^{3}}\int_{P+1\leq|v|\leq\hat{P}}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv+P^{1/4}\int_{\mathbb{R}^{3}}\int_{\hat{P}\leq|v|\leq\hat{P}+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv\qquad$

Now

$P^{1/4}\int_{\mathbb{R}^{3}}\int_{P+1\leq|v|\leq\hat{P}}\sqrt{1+|v|^{2}}\,f\,dx\,dv\leq
P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq P}\sqrt{1+|v|^{2}}\,f\,dx\,dv\leq 1$

by (3.21), as $P\leq\hat{P}$. The second term in (3.23) is also bounded by
$1$, since $P^{1/4}\leq\hat{P}^{1/4}$ and we may use (3.22). It remains to
consider the case where $P+1\geq\hat{P}$, so that $P_{0}\leq P\leq\hat{P}\leq
P+1\leq\hat{P}+1$. Then

$\displaystyle P^{1/4}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$ $\displaystyle=$ $\displaystyle
P^{1/4}\int_{\mathbb{R}^{3}}\int_{P+1\leq|v|\leq\hat{P}+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$
$\displaystyle\leq$
$\displaystyle\hat{P}^{1/4}\int_{\mathbb{R}^{3}}\int_{\hat{P}\leq|v|\leq\hat{P}+1}\sqrt{1+|v|^{2}}\,\tilde{f}\,dx\,dv$
$\displaystyle\leq$ $\displaystyle 1,$

once again by (3.22). $\Box$

##  4 Properties of the limit function

###### Remark 4.1

If (1.6) holds, then ${\cal A}_{M,\sigma_{0}}\subset\tilde{{\cal
A}}_{M,\beta}$, where $\tilde{{\cal A}}_{M,\beta}$ is from (3). To see this,
$\rho(r)\leq\sigma_{0}\leq 1$ for $r\in[0,\infty[$ by the definition of
$\sigma_{0}$. Recalling $\sigma_{M}=\frac{3}{32\pi M^{2}}$, also Lemma 9.2
ahead shows that

$\frac{2m(r)}{r}\leq\bigg{(}\frac{\sigma_{0}}{\sigma_{M}}\bigg{)}^{1/3}\leq\bigg{(}\frac{96\pi\beta^{3}M^{2}}{12\pi
M^{2}}\bigg{)}^{1/3}=2\beta$

for $r\in[0,\infty[$, and hence $m(r)/r\leq\beta$ for $r\in[0,\infty[$.

We recall that

$I=\inf_{f\in{\cal A}}{\cal D}(f).$

###### Lemma 4.2

If $f\in{\cal A}$, then ${\cal D}(f)\geq-c_{\beta}M$ for
$c_{\beta}=\frac{1}{\sqrt{1-2\beta}}$; in particular, $I$ is well-defined.
Furthermore,

$I<0.$

In addition, for every exponent $k>0$ there exists a function $h\in{\cal A}$
such that ${\cal D}(h)<0$ and $|{\cal D}(h)|>M$.

Proof : For $f\in{\cal A}$ we have $1\leq e^{\lambda}\leq c_{\beta}$, as in
(3.10), and also recall Remark 4.1. Therefore

${\cal
D}(f)=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv\geq-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,f\,dx\,dv\geq-
c_{\beta}M.$

To show that $I$ is negative, consider

$\tilde{f}=A\,{\bf 1}_{[0,a]}(|x|)\,{\bf 1}_{[0,b]}(|v|)$

for $A\in]0,1]$ and $a,b>0$. Then

$\displaystyle\tilde{\rho}(r)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,\tilde{f}\,dv=A\,{\bf
1}_{[0,a]}(r)\,\theta(b),$ $\displaystyle\theta(b)$ $\displaystyle=$
$\displaystyle
4\pi\int_{0}^{b}s^{2}\sqrt{1+s^{2}}\,ds=\frac{\pi}{2}\Big{(}b\sqrt{1+b^{2}}(1+2b^{2})-\arcsin
b\Big{)}.$

This gives the mass function

$\tilde{m}(r)=4\pi\int_{0}^{r}s^{2}\tilde{\rho}(s)\,ds=\left\\{\begin{array}[]{c@{\quad:\quad}c}\frac{4\pi}{3}\,A\,\theta(b)\,r^{3}&r\leq
a\\\\[4.30554pt] \frac{4\pi}{3}\,A\,\theta(b)\,a^{3}&r\geq
a\end{array}\right.,$

and hence the condition $\tilde{f}\in{\cal A}$ is equivalent to

$\frac{4\pi}{3}\,A\,\theta(b)\,a^{3}=M\quad\mbox{and}\quad
A\,\theta(b)\leq\sigma_{0}\leq\min\Big{\\{}1,\frac{3\beta^{3}}{4\pi
M^{2}}\Big{\\}}.$ (4.1)

In addition, we obtain

$e^{\tilde{\lambda}}=\frac{1}{\sqrt{1-\frac{2\tilde{m}(r)}{r}}}=\left\\{\begin{array}[]{c@{\quad:\quad}c}\frac{1}{\sqrt{1-\frac{8\pi}{3}\,A\,\theta(b)\,r^{2}}}&r\leq
a\\\\[8.61108pt]
\frac{1}{\sqrt{1-\frac{8\pi}{3}\,A\,\theta(b)\,\frac{a^{3}}{r}}}&r\geq
a\end{array}\right..$

Since $\tilde{f}=A$ on its support and
$\chi(A)-A=-A(1-\frac{k}{k+1}\,A^{\frac{1}{k}})<0$ due to $A\leq 1$, this
yields

$\displaystyle{\cal D}(\tilde{f})$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\tilde{\lambda}}(\chi(\tilde{f})-\tilde{f})\,dx\,dv$
(4.2) $\displaystyle=$
$\displaystyle-A\Big{(}1-\frac{k}{k+1}\,A^{\frac{1}{k}}\Big{)}\,\frac{16\pi^{2}}{3}\,b^{3}\int_{0}^{a}\frac{r^{2}}{\sqrt{1-\frac{8\pi}{3}\,A\,\theta(b)\,r^{2}}}\,dr<0.$

Note that here explicitly, for $r\in[0,a]$,

$\frac{2\tilde{m}(r)}{r}=\frac{8\pi}{3}\,A\,\theta(b)\,r^{2}\leq\frac{8\pi}{3}\,A\,\theta(b)\,a^{2}\leq
2\beta<1$

for $r\leq a$, as predicted by Lemma 9.2, using that (4.1) yields
$\theta(b)=\frac{3M}{4\pi Aa^{3}}$ and $\frac{3M}{4\pi
a^{3}}=A\theta(b)\leq\frac{3\beta^{3}}{4\pi M^{2}}$, whence

$\frac{8\pi}{3}\,A\,\theta(b)\,a^{2}=\frac{2M}{a}\leq 2\beta.$

The condition (4.1) is easy to satisfy: We will fix $A=A(k)$ depending only on
$k$ below. Then first take $b>0$ small enough such that
$A\theta(b)\leq\sigma_{0}$ holds and thereafter put $a=(\frac{3M}{4\pi
A\theta(b)})^{1/3}$. In this case $\tilde{f}\in{\cal A}$ and ${\cal
D}(\tilde{f})<0$ by (4.2), so that in particular $I<0$.

For the last assertion, the integral (4.2) can be evaluated explicitly. It is
found that, with $c=(\frac{8\pi}{3}\,A\theta(b))^{1/2}$, one has

${\cal
D}(\tilde{f})=-A\Big{(}1-\frac{k}{k+1}\,A^{\frac{1}{k}}\Big{)}\,\frac{8\pi^{2}}{3}\,\frac{b^{3}}{c^{3}}\,\Big{(}\arcsin(ca)-ca\sqrt{1-c^{2}a^{2}}\Big{)}.$

Thus if (4.1) holds, the condition $|{\cal D}(\tilde{f})|>M$ is equivalent to

$4\Big{(}1-\frac{k}{k+1}\,A^{\frac{1}{k}}\Big{)}\,\frac{\arcsin(ca)-ca\sqrt{1-c^{2}a^{2}}}{c^{3}a^{3}}>\frac{b\sqrt{1+b^{2}}(1+2b^{2})-\arcsin
b}{b^{3}}.$ (4.3)

As $x\to 0^{+}$, the expansions are

$\displaystyle\arcsin(x)-x\sqrt{1-x^{2}}=\frac{2}{3}\,x^{3}+{\cal O}(x^{5}),$
$\displaystyle x\sqrt{1+x^{2}}(1+2x^{2})-\arcsin x=\frac{7}{3}\,x^{3}+{\cal
O}(x^{5}),$

and we already know from above that we can take $b\to 0$ and
$a=(\frac{3M}{4\pi A\theta(b)})^{1/3}$ to make sure that $\tilde{f}\in{\cal
A}$. From the expansions we deduce as $b\to 0$

$\displaystyle\theta(b)$ $\displaystyle=$
$\displaystyle\frac{\pi}{2}\,\Big{(}\frac{7}{3}\,b^{3}+{\cal
O}(b^{5})\Big{)}=\frac{7\pi}{6}\,b^{3}+{\cal O}(b^{5}),$ $\displaystyle c$
$\displaystyle=$
$\displaystyle\sqrt{\frac{8\pi}{3}\,A\theta(b)}=\sqrt{A}\,\frac{2\sqrt{7}\pi}{3}\,b^{3/2}\,(1+{\cal
O}(b^{2})),$ $\displaystyle a$ $\displaystyle=$
$\displaystyle\Big{(}\frac{3M}{4\pi
A\theta(b)}\Big{)}^{1/3}=A^{-1/3}\,\Big{(}\frac{9M}{14\pi^{2}}\Big{)}^{1/3}\,\frac{1}{b}\,(1+{\cal
O}(b^{2})),$ $\displaystyle ca$ $\displaystyle=$ $\displaystyle
A^{1/6}\,\frac{2\sqrt{7}\pi}{3}\,\Big{(}\frac{9M}{14\pi^{2}}\Big{)}^{1/3}b^{1/2}\,(1+{\cal
O}(b^{2})),$

so that also $ca\to 0$ as $b\to 0$. Therefore we may use the expansions to
infer

$\frac{\arcsin(ca)-ca\sqrt{1-c^{2}a^{2}}}{c^{3}a^{3}}=\frac{2}{3}+{\cal
O}(b),\quad\frac{b\sqrt{1+b^{2}}(1+2b^{2})-\arcsin b}{b^{3}}=\frac{7}{3}+{\cal
O}(b^{2}).$

Thus if

$4\Big{(}1-\frac{k}{k+1}\,A^{\frac{1}{k}}\Big{)}\,\cdot\frac{2}{3}>\frac{7}{3}$
(4.4)

holds, then (4.3) will be verified for $b>0$ small enough; the condition (4.4)
reads as

$\frac{k}{k+1}\,A^{\frac{1}{k}}<\frac{1}{8},$

and it can be satisfied, if $k>0$ is fixed and $A=A(k)>0$ is taken to be
sufficiently small; for instance $A(k)=(\frac{1}{8})^{k}$ will do. $\Box$

Let $(f_{n})\subset{\cal A}$ be a minimizing sequence, i.e.

$\lim_{n\to\infty}{\cal D}(f_{n})=I.$ (4.5)

For $M>0$ and $\beta\in]0,\frac{1}{2}[$ given, let $P_{0}>0$ be determined by
(3.5). Since ${\cal A}={\cal A}_{M,\sigma_{0}}\subset\tilde{{\cal
A}}_{M,\beta}$ by Remark 4.1, the results from Section 3 are applicable, and
in particular every $f_{n}$ satisfies (a)-(c) from Corollary 3.3. Hence there
is $\tilde{f}_{n}\in\tilde{{\cal A}}_{M,\beta}$ such that
$0\leq\tilde{f}_{n}\leq 1$, $\rho_{\tilde{f}_{n}}=\rho_{f_{n}}$, ${\cal
D}(\tilde{f}_{n})\leq{\cal D}(f_{n})$, and

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,\tilde{f}_{n}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq
P_{0}+1.$ (4.6)

Since $\rho_{\tilde{f}_{n}}=\rho_{f_{n}}$, also $\tilde{f}_{n}\in{\cal A}$.
Therefore ${\cal D}(\tilde{f}_{n})\geq I$, and thus $\lim_{n\to\infty}{\cal
H}(\tilde{f}_{n})=I$, which shows that $(\tilde{f}_{n})$ is a minimizing
sequence, too. Thus we may simply drop the tilde and assume that the original
sequence $(f_{n})$ possesses the property (4.6), and moreover $0\leq f_{n}\leq
1$ for every $n\in\mathbb{N}$.

###### Lemma 4.3

$(f_{n})\subset L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ is bounded.

Proof : By (3.10),

$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\chi(f_{n})\,dx\,dv$
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{n}}}\chi(f_{n})\,dx\,dv$
$\displaystyle=$ $\displaystyle{\cal
D}(f_{n})+\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{n}}}\,f_{n}\,dx\,dv$
$\displaystyle\leq$ $\displaystyle{\cal D}(f_{n})+c_{\beta}M$

together with (4.5) yields the claim. $\Box$

As a consequence of Lemma 4.3 we may extract a subsequence (not relabeled)
such that

$f_{n}\rightharpoonup
f_{0}\,\,\mbox{in}\,\,L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ (4.7)

as $n\to\infty$ for an appropriate limit function $f_{0}\in
L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$. Now we are ready for the proof
of Lemma 2.1, the statement of which we repeat for convenience.

###### Lemma 4.4

The following assertions hold:

  * •

$0\leq f_{0}\leq 1$;

  * •

$f_{0}$ is spherically symmetric;

  * •

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq P_{0}+1$;

  * •

$m_{f_{n}}(r)\to m_{f_{0}}(r)$ as $n\to\infty$ for $r\in[0,\infty[$, in
particular $m_{f_{0}}(r)/r\leq\beta$ for $r\in]0,\infty[$;

  * •

$\rho_{f_{0}}(r)\leq\sigma_{0}$ for $r\in[0,\infty[$;

  * •

$\lambda_{f_{n}}(r)\to\lambda_{f_{0}}(r)$ as $n\to\infty$ for
$r\in[0,\infty[$.

Proof : If $U\subset\mathbb{R}^{3}\times\mathbb{R}^{3}$ is measurable and of
finite measure, then

$0\leq\int\int_{U}f_{n}\,dx\,dv\to\int\int_{U}f_{0}\,dx\,dv,\quad n\to\infty,$

which yields $f_{0}\geq 0$ a.e. Similarly,

$0\leq\int\int_{U}(1-f_{n})\,dx\,dv\to\int\int_{U}(1-f_{0})\,dx\,dv,\quad
n\to\infty,$

for each such $U$ implies that $f_{0}\leq 1$ a.e. Let $A\in{\rm SO}(3)$ and
$X,V\subset\mathbb{R}^{3}$ be measurable and of finite measure. We have

$\displaystyle\int_{X}\int_{V}f_{0}(x,v)\,dx\,dv$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}\int_{X}\int_{V}f_{n}(x,v)\,dx\,dv=\lim_{n\to\infty}\int_{X}\int_{V}f_{n}(Ax,Av)\,dx\,dv$
$\displaystyle=$
$\displaystyle\lim_{n\to\infty}\int_{AX}\int_{AV}f_{n}(x,v)\,dx\,dv=\int_{AX}\int_{AV}f_{0}(x,v)\,dx\,dv$
$\displaystyle=$ $\displaystyle\int_{X}\int_{V}f_{0}(Ax,Av)\,dx\,dv$

as $n\to\infty$, and it follows that $f_{0}$ is spherically symmetric.

If $P\geq P_{0}+1$, $R>0$ and $S>P+1$, then by (4.6) for $n\in\mathbb{N}$

$\frac{2}{P^{1/4}}\geq\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv\geq\int_{|x|\leq R}\int_{P+1\leq|v|\leq
S}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv.$

Since ${\bf 1}_{|x|\leq R}\sqrt{1+|v|^{2}}\,{\bf 1}_{P+1\leq|v|\leq S}\in
L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})^{\ast}$, we can pass to the
limit $n\to\infty$ and get

$\frac{2}{P^{1/4}}\geq\int_{|x|\leq R}\int_{P+1\leq|v|\leq
S}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv$

for every $R>0$ and $S>P+1$. Taking the limits $R,S\to\infty$, we arrive at

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq P_{0}+1.$
(4.8)

Next fix $0\leq r_{1}<r_{2}$. We claim that

$\lim_{n\to\infty}\int_{r_{1}<|x|<r_{2}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=\int_{r_{1}<|x|<r_{2}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv.$
(4.9)

To establish (4.9), let $(P_{N})$ be a sequence satisfying $P_{N}\geq P_{0}+1$
and $\lim_{N\to\infty}P_{N}=\infty$. For $n,N\in\mathbb{N}$ we deduce from
(4.6) and (4.8)

$\displaystyle\bigg{|}\int_{r_{1}<|x|<r_{2}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{n}\,dx\,dv-\int_{r_{1}<|x|<r_{2}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{0}\,dx\,dv\bigg{|}$
$\displaystyle=$ $\displaystyle\bigg{|}\int_{r_{1}<|x|<r_{2}}\int_{|v|\leq
P_{N}+1}\sqrt{1+|v|^{2}}f_{n}\,dx\,dv+\int_{r_{1}<|x|<r_{2}}\int_{|v|\geq
P_{N}+1}\sqrt{1+|v|^{2}}f_{n}\,dx\,dv$
$\displaystyle\quad-\,\int_{r_{1}<|x|<r_{2}}\int_{|v|\leq
P_{N}+1}\sqrt{1+|v|^{2}}f_{0}\,dx\,dv-\int_{r_{1}<|x|<r_{2}}\int_{|v|\geq
P_{N}+1}\sqrt{1+|v|^{2}}f_{0}\,dx\,dv\bigg{|}$ $\displaystyle\leq$
$\displaystyle\bigg{|}\int_{r_{1}<|x|<r_{2}}\int_{|v|\leq
P_{N}+1}\sqrt{1+|v|^{2}}f_{n}\,dx\,dv-\int_{r_{1}<|x|<r_{2}}\int_{|v|\leq
P_{N}+1}\sqrt{1+|v|^{2}}f_{0}\,dx\,dv\bigg{|}+\frac{4}{P_{N}^{1/4}}$
$\displaystyle=:$ $\displaystyle\delta_{n,N}+\frac{4}{P_{N}^{1/4}}.$

Let $\varepsilon>0$ be given. Then there is a $N_{0}\in\mathbb{N}$ such that
$4P_{N_{0}}^{-1/4}\leq\varepsilon$. As $\lim_{n\to\infty}\delta_{n,N_{0}}=0$,
we find an $n_{0}\in\mathbb{N}$ so that $\delta_{n,N_{0}}\leq\varepsilon$ for
$n\geq n_{0}$, whence altogether we obtain $|\ldots|\leq 2\varepsilon$ for
$n\geq n_{0}$, which completes the proof of (4.9). Taking $r_{1}=0$ and
$r_{2}=r$, we deduce that in particular
$\lim_{n\to\infty}m_{f_{n}}(r)=m_{f_{0}}(r)$ holds as $n\to\infty$, for
$r\in[0,\infty[$. In addition, for $r>0$ and $\varepsilon>0$ we get from (4.9)
and $f_{n}\in{\cal A}$

$\displaystyle 4\pi\int_{r}^{r+\varepsilon}s^{2}\rho_{f_{0}}(s)\,ds$
$\displaystyle=$
$\displaystyle\int_{r<|x|<r+\varepsilon}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv$
$\displaystyle=$
$\displaystyle\lim_{n\to\infty}\int_{r<|x|<r+\varepsilon}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv$
$\displaystyle=$
$\displaystyle\lim_{n\to\infty}\int_{r<|x|<r+\varepsilon}\rho_{f_{n}}(x)\,dx$
$\displaystyle\leq$
$\displaystyle\frac{4\pi\sigma_{0}}{3}\,((r+\varepsilon)^{3}-r^{3})=\frac{4\pi\sigma_{0}}{3}\,(3r^{2}\varepsilon+3r\varepsilon^{2}+\varepsilon^{3}).$

Therefore, dividing by $\varepsilon>0$ and taking the limit $\varepsilon\to
0$, we see that for a.e. $r>0$ we have $4\pi r^{2}\rho_{f_{0}}(r)\leq
4\pi\sigma_{0}r^{2}$, which means that $\rho_{f_{0}}(r)\leq\sigma_{0}$.

Lastly, to show that $\lambda_{f_{n}}(r)\to\lambda_{f_{0}}(r)$ pointwise, note
first that

$e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}=\frac{2}{r}\,\frac{e^{2\lambda_{f_{n}}}e^{2\lambda_{f_{0}}}}{e^{\lambda_{f_{n}}}+e^{\lambda_{f_{0}}}}\,(m_{f_{n}}-m_{f_{0}}),$
(4.10)

as is verified by a direct calculation. From above we know
$\lim_{n\to\infty}m_{f_{n}}(r)=m_{f_{0}}(r)$ for every $r>0$. Since, owing to
(3.10), $1\leq e^{2\lambda_{f_{n}}}\leq c_{\beta}^{2}$ and $1\leq
e^{2\lambda_{f_{0}}}\leq c_{\beta}^{2}$, the claim follows. $\Box$

##  5 The Casimir-binding-energy condition and existence of a minimizer

We continue to consider

${\cal
D}(f)=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv,$
(5.1)

which is (1.1).

First we study the scaling properties of the system; also see [9]. Henceforth
we will assume that $k\in]0,2]$.

###### Lemma 5.1

Suppose that $f=f(x,v)\geq 0$ is measurable and spherically symmetric such
that $\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f\,dx\,dv=M$.
For $\gamma>0$ define

$f_{\gamma}(x,v)=\gamma^{2}f(\gamma x,v).$

Then

$M_{\gamma}=\gamma^{-1}M,\quad\rho_{\gamma}(r)=\gamma^{2}\rho(\gamma r),\quad
m_{\gamma}(r)=\gamma^{-1}m(\gamma r),\quad\lambda_{\gamma}(r)=\lambda(\gamma
r),$

and if furthermore $\gamma\leq 1$, then also

${\cal D}(f_{\gamma})\leq\gamma^{-1}\,{\cal D}(f).$

Proof : By changing variables as $\tilde{x}=\gamma x$,
$d\tilde{x}=\gamma^{3}dx$, we calculate

$M_{\gamma}=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{\gamma}\,dx\,dv=\gamma^{2}\gamma^{-3}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f(\tilde{x},v)\,d\tilde{x}\,dv=\gamma^{-1}M.$

Moreover,

$\rho_{\gamma}(r)=\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{\gamma}(x,v)\,dv=\gamma^{2}\rho(\gamma
r)$

yields

$m_{\gamma}(r)=4\pi\int_{0}^{r}s^{2}\rho_{\gamma}(s)\,ds=4\pi\gamma^{2}\int_{0}^{r}s^{2}\rho(\gamma
s)\,ds=4\pi\gamma^{-1}\int_{0}^{\gamma
r}\tau^{2}\rho(\tau)\,d\tau=\gamma^{-1}m(\gamma r).$

This in turn leads to

$e^{-2\lambda_{\gamma}(r)}=1-\frac{2m_{\gamma}(r)}{r}=1-\frac{m(\gamma
r)}{\gamma r}=e^{-2\lambda(\gamma r)},$

so that $\lambda_{\gamma}(r)=\lambda(\gamma r)$. It follows that

$\displaystyle{\cal D}(f_{\gamma})$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{\gamma}}\,(\chi(f_{\gamma})-f_{\gamma})\,dx\,dv$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda(\gamma
r)}\,\Big{(}\frac{k}{k+1}\,\gamma^{2(1+\frac{1}{k})}\,f(\gamma
x,v)^{1+\frac{1}{k}}-\gamma^{2}f(\gamma x,v)\Big{)}\,dx\,dv$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda(\tilde{r})}\,\Big{(}\frac{k}{k+1}\,\gamma^{\frac{2}{k}-1}\,f(\tilde{x},v)^{1+\frac{1}{k}}-\gamma^{-1}f(\tilde{x},v)\Big{)}\,d\tilde{x}\,dv$
$\displaystyle=$ $\displaystyle{\cal
D}(f)+\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda(\tilde{r})}\,\Big{(}\frac{k}{k+1}\,[\gamma^{\frac{2}{k}-1}-1]\,f(\tilde{x},v)^{1+\frac{1}{k}}-[\gamma^{-1}-1]f(\tilde{x},v)\Big{)}\,d\tilde{x}\,dv.$

Since $k\in]0,2]$ we have $\frac{2}{k}-1\geq 0$, and hence
$\gamma^{\frac{2}{k}-1}-1\leq 0$ due to $\gamma\leq 1$. Therefore we obtain

$\displaystyle{\cal D}(f_{\gamma})$ $\displaystyle\leq$ $\displaystyle{\cal
D}(f)+(\gamma^{-1}-1)\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(-f)\,dx\,dv$
$\displaystyle\leq$ $\displaystyle{\cal
D}(f)+(\gamma^{-1}-1)\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda}\,(\chi(f)-f)\,dx\,dv$
$\displaystyle=$ $\displaystyle\gamma^{-1}\,{\cal D}(f),$

as claimed. $\Box$

###### Lemma 5.2

Let $f\in{\cal A}$ and $\varepsilon,\eta,R>0$ be such that $\frac{2M}{R}\leq
1-\frac{1}{(1+\eta)^{2}}$,

$f\leq 1,\quad{\cal D}(f)\leq I+\varepsilon\quad\mbox{and}\quad\Delta
M=\int_{|x|\geq R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv\in]0,M[$

are satisfied. If $f_{1}={\bf 1}_{\\{|x|\leq R\\}}f$, then

$\frac{\Delta M}{M-\Delta M}\,|{\cal D}(f_{1})|\leq(1+\eta)\Delta
M+\varepsilon.$

Proof : Let $f_{2}={\bf 1}_{\\{|x|\geq R\\}}f$. Since $f\leq 1$ we have
$\chi(f)-f\leq 0$ pointwise, recall (3.2). Therefore in particular
$\chi(f_{1})-f_{1}\leq 0$ pointwise and $\chi(f_{2})-f_{2}\leq 0$ pointwise.
If $r\leq R$, then

$\displaystyle m_{f_{1}}(r)$ $\displaystyle=$ $\displaystyle
4\pi\int_{0}^{r}s^{2}\rho_{f_{1}}(s)\,ds=\int_{|x|\leq
r}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{1}(x,v)\,dx\,dv$ (5.2)
$\displaystyle=$ $\displaystyle\int_{|x|\leq
r}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f(x,v)\,dx\,dv=m_{f}(r),$

and hence also

$e^{\lambda_{f_{1}}(r)}=e^{\lambda_{f}(r)},\quad r\leq R.$

This gives

$\displaystyle|{\cal D}(f)|$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f}(r)}\,(f-\chi(f))\,dx\,dv$
$\displaystyle=$ $\displaystyle\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f}(r)}\,(f-\chi(f))\,dx\,dv+\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f}(r)}\,(f-\chi(f))\,dx\,dv$
$\displaystyle=$ $\displaystyle\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{1}}(r)}\,(f_{1}-\chi(f_{1}))\,dx\,dv+\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f}(r)}\,(f-\chi(f))\,dx\,dv$
$\displaystyle=$ $\displaystyle|{\cal D}(f_{1})|+\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f}(r)}\,(f-\chi(f))\,dx\,dv.$

If $r\geq R$, then $\frac{2M}{R}\leq 1-\frac{1}{(1+\eta)^{2}}$ leads to

$e^{\lambda_{f}(r)}=\frac{1}{\sqrt{1-\frac{2m_{f}(r)}{r}}}\leq\frac{1}{\sqrt{1-\frac{2M}{R}}}\leq
1+\eta.$

It follows that

$\displaystyle\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f}(r)}\,(f-\chi(f))\,dx\,dv$
$\displaystyle\leq$ $\displaystyle(1+\eta)\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}f\,dx\,dv$ $\displaystyle\leq$
$\displaystyle(1+\eta)\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f\,dx\,dv\leq(1+\eta)\Delta M.$

As a consequence,

$|{\cal D}(f)|\leq|{\cal D}(f_{1})|+(1+\eta)\Delta M.$ (5.3)

Next we apply the scaling property from Lemma 5.1 to $f_{1}$, $M$ replaced by
the total mass $M-\Delta M$ of $f_{1}$ and $\gamma=\frac{M-\Delta
M}{M}\in]0,1[$ to obtain a spherically symmetric function $f_{1,\gamma}\geq 0$
such that

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{1,\gamma}\,dx\,dv=\gamma^{-1}(M-\Delta
M)=M$

and also

${\cal D}(f_{1,\gamma})\leq\frac{M}{M-\Delta M}\,{\cal D}(f_{1})\leq 0.$

From $\rho_{f_{1,\gamma}}(r)=\gamma^{2}\rho_{f_{1}}(\gamma
r)\leq\rho_{f_{1}}(\gamma r)\leq\sigma_{0}$ we deduce that
$f_{1,\gamma}\in{\cal A}$, so that

${\cal D}(f)-\varepsilon\leq I\leq{\cal D}(f_{1,\gamma})\leq\frac{M}{M-\Delta
M}\,{\cal D}(f_{1}).$

Taking minus signs, this can be rewritten as

$\frac{M}{M-\Delta M}\,|{\cal D}(f_{1})|\leq|{\cal D}(f)|+\varepsilon.$

Using (5.3), we arrive at

$\frac{\Delta M}{M-\Delta M}\,|{\cal D}(f_{1})|\leq(1+\eta)\Delta
M+\varepsilon,$

as was to be shown. $\Box$

Now we turn to the Casimir-binding-energy condition (CBEC), as stated in
Definition 1.1.

###### Remark 5.3

(a) Observe that (CBEC) holds if and only if there is a function $h\in{\cal
A}$ such that ${\cal D}(h)<0$ and

$|{\cal D}(h)|>M.$

(b) According to (a) and Lemma 4.2, (CBEC) is verified in particular for
$k\in]0,2]$.

###### Corollary 5.4

Let $h$ be as in Remark 5.3(a) and $\delta>0$ such that $|{\cal D}(h)|\geq
M+\delta$. Next define

$\eta=\min\Big{\\{}1,\frac{\delta}{6M}\Big{\\}}$

and take

$R\geq\max\Big{\\{}\frac{8M}{3},\frac{2M}{1-\frac{1}{(1+\frac{\delta}{6M})^{2}}}\Big{\\}}.$
(5.4)

If $g\in{\cal A}$ satisfies

$g\leq 1,\quad{\cal D}(g)\leq I+\eta(\Delta M)\quad\mbox{and}\quad\Delta
M\in]0,M[,$

where

$\Delta M=\int_{|x|\geq R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,g\,dx\,dv,$

then

$\Delta M\geq\frac{\delta}{18}.$

In particular, if $(f_{n})\subset{\cal A}$ is a minimizing sequence such that
$f_{n}\leq 1$ and

$\lim_{n\to\infty}{(\Delta M)}_{n}=\varepsilon_{0}$

for some $\varepsilon_{0}\in]0,M[$, where

${(\Delta M)}_{n}=\int_{|x|\geq
R_{n}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv\quad\mbox{with}\quad
R_{n}\geq\max\Big{\\{}\frac{8M}{3},\frac{2M}{1-\frac{1}{(1+\frac{\delta}{6M})^{2}}}\Big{\\}},$
(5.5)

then

$\varepsilon_{0}\geq\frac{\delta}{18}.$

Proof : We can apply Lemma 5.2 with $f=g$ and $\varepsilon=\eta(\Delta M)$,
since $\frac{2M}{R}\leq 1-\frac{1}{(1+\eta)^{2}}$. It follows that

$\frac{\Delta M}{M-\Delta M}\,|{\cal D}(g_{1})|\leq(1+\eta)\Delta
M+\varepsilon=(1+2\eta)\Delta M$

for $g_{1}={\bf 1}_{\\{|x|\leq R\\}}\,g$, and hence

$|{\cal D}(g_{1})|\leq(M-\Delta M)(1+2\eta).$

Now we distinguish two cases. Case 1: $|{\cal D}(h)|\leq|{\cal
D}(g_{1})|+\frac{\delta}{6}$. Then

$\displaystyle M+\delta$ $\displaystyle\leq$
$\displaystyle|\mathcal{D}(h)|\leq|{\cal
D}(g_{1})|+\frac{\delta}{6}\leq(M-\Delta M)(1+2\eta)+\frac{\delta}{6}$
$\displaystyle\leq$ $\displaystyle M(1+2\eta)+\frac{\delta}{6}\leq
M+\frac{\delta}{2},$

which is a contradiction. Case 2: $|{\cal D}(h)|\geq|{\cal
D}(g_{1})|+\frac{\delta}{6}$. If $r\leq R$, then

$m_{g_{1}}(r)=4\pi\int_{0}^{r}s^{2}\rho_{g_{1}}(s)\,ds=\int_{|x|\leq
r}\int_{\mathbb{R}^{3}}g_{1}(x,v)\,dx\,dv=\int_{|x|\leq
r}\int_{\mathbb{R}^{3}}g(x,v)\,dx\,dv=m_{g}(r),$

and hence also $\lambda_{g_{1}}(r)=\lambda_{g}(r)$ for $r\leq R$. Therefore we
obtain

$\displaystyle{\cal D}(h)$ $\displaystyle\leq$ $\displaystyle{\cal
D}(g_{1})-\frac{\delta}{6}$ $\displaystyle=$ $\displaystyle\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{g_{1}}}(\chi(g_{1})-g_{1})\,dx\,dv-\frac{\delta}{6}$
$\displaystyle=$ $\displaystyle\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{g}}(\chi(g)-g)\,dx\,dv-\frac{\delta}{6}$
$\displaystyle=$ $\displaystyle{\cal D}(g)+\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{g}}(g-\chi(g))\,dx\,dv-\frac{\delta}{6}$
$\displaystyle\leq$ $\displaystyle{\cal
D}(g)+\frac{1}{\sqrt{1-\frac{2M}{R}}}\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}g\,dx\,dv-\frac{\delta}{6}$ $\displaystyle\leq$
$\displaystyle{\cal D}(g)+(1+\eta)\Delta M-\frac{\delta}{6}$
$\displaystyle\leq$ $\displaystyle I+\eta(\Delta M)+(1+\eta)\Delta
M-\frac{\delta}{6}$ $\displaystyle=$ $\displaystyle I+(1+2\eta)\Delta
M-\frac{\delta}{6}$ $\displaystyle\leq$ $\displaystyle I+3\Delta
M-\frac{\delta}{6}.$

Since $h\in{\cal A}$, we have $I\leq{\cal D}(h)$, which implies that $\Delta
M\geq\frac{\delta}{18}$.

For the second part, $R_{n}$ satisfies (5.4) by hypothesis. Due to the fact
that $(f_{n})$ is a minimizing sequence, we have $\lim_{n\to\infty}{\cal
D}(f_{n})=I$. Then ${(\Delta M)}_{n}\to\varepsilon_{0}>0$ shows that ${\cal
D}(f_{n})\leq I+\eta{(\Delta M)}_{n}$ is verified for $n$ large enough. In
addition, $\varepsilon_{0}\in]0,M[$ yields that also ${(\Delta
M)}_{n}\in]0,M[$ for $n$ sufficiently large. Thus the first part applies to
deduce that ${(\Delta M)}_{n}\geq\frac{\delta}{18}$ for large $n$, so that
$\varepsilon_{0}\geq\frac{\delta}{18}$ is found in the limit $n\to\infty$.
$\Box$

The last part of the previous result can be sharpened as follows.

###### Corollary 5.5

Let $h$ be as in Remark 5.3(a) and $\delta>0$ such that $|{\cal D}(h)|\geq
M+\delta$. Let $(R_{n})$ be a sequence such that

$R_{n}\geq\max\Big{\\{}\frac{8M}{3},\frac{2M}{1-\frac{1}{(1+\frac{\delta}{6M})^{2}}}\Big{\\}},\quad
n\in\mathbb{N}.$ (5.6)

Then there is no minimizing sequence $(f_{n})\subset{\cal A}$ such that
$f_{n}\leq 1$ and

$\lim_{n\to\infty}{(\Delta M)}_{n}\in]0,M[,$

where

${(\Delta M)}_{n}=\int_{|x|\geq
R_{n}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv.$

Proof : Since $\delta>0$ has the property that $|{\cal D}(h)|\geq M+\delta$,
there is no loss of generality in assuming that $\delta<20M$, as otherwise
$\delta$ could be replaced by a smaller number. Assuming the contrary of what
is stated, let $\varepsilon_{0}=\lim_{n\to\infty}{(\Delta M)}_{n}\in]0,M[$. We
distinguish two cases. Case 1: $\varepsilon_{0}\leq\frac{\delta}{20}$. From
Corollary 5.4 we however deduce that
$\frac{\delta}{18}\leq\varepsilon_{0}\leq\frac{\delta}{20}$, which is a
contradiction. Case 2: $\varepsilon_{0}>\frac{\delta}{20}$. Then we may assume
that

${(\Delta M)}_{n}=\int_{|x|\geq
R_{n}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv>\frac{\delta}{20},\quad
n\in\mathbb{N}.$

For every fixed $n\in\mathbb{N}$ we consider the continuous function
$\kappa_{n}(R)=\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv$. Then
$\kappa_{n}(R_{n})>\frac{\delta}{20}$ and $\lim_{R\to\infty}\kappa_{n}(R)=0$.
Hence there is a $\tilde{R}_{n}>R_{n}$ such that
$\kappa_{n}(\tilde{R}_{n})=\frac{\delta}{20}$. For

${(\widetilde{\Delta
M})}_{n}=\int_{|x|\geq\tilde{R}_{n}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv$

we then get ${(\widetilde{\Delta
M})}_{n}=\kappa_{n}(\tilde{R}_{n})=\frac{\delta}{20}$ for $n\in\mathbb{N}$.
Furthermore, since $\tilde{R}_{n}>R_{n}$, also $\tilde{R}_{n}$ satisfies
(5.5). As $\frac{\delta}{20}<M$ by the remark at the beginning of the
argument, we can once again apply Corollary 5.4 to yield
$\frac{\delta}{20}\geq\frac{\delta}{18}$, which is impossible. $\Box$

From now on we take $(f_{n})\subset{\cal A}$ to be a minimizing sequence for
${\cal D}$ with the additional property (4.6), and $0\leq f_{n}\leq 1$ for
$n\in\mathbb{N}$, that led to the weak limit $f_{0}$; see the paragraphs
before and following Lemma 4.3.

###### Lemma 5.6

For every $R>0$ we have

$\lim_{n\to\infty}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv$ (5.7)

and

$\lim_{n\to\infty}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\varphi\,f_{n}\,dx\,dv=\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\varphi\,f_{0}\,dx\,dv$ (5.8)

for every bounded and continuous function $\varphi=\varphi(x,v)\geq 0$.

Proof : For $n\in\mathbb{N}$ and $P\geq P_{0}+1$ we use (4.6) and Lemma 4.4 to
bound

$\displaystyle\bigg{|}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv-\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\bigg{|}$
$\displaystyle=$ $\displaystyle\bigg{|}\int_{|x|\leq R}\int_{|v|\leq
P+1}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv+\int_{|x|\leq R}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv$ $\displaystyle-\,\int_{|x|\leq
R}\int_{|v|\leq P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv-\int_{|x|\leq
R}\int_{|v|\geq P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\bigg{|}$
$\displaystyle\leq$ $\displaystyle\bigg{|}\int_{|x|\leq R}\int_{|v|\leq
P+1}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv-\int_{|x|\leq R}\int_{|v|\leq
P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\bigg{|}+\frac{4}{P^{1/4}}.$

Given $\varepsilon>0$ choose $P\geq P_{0}+1$ large enough so that
$4P^{-{1/4}}\leq\varepsilon/2$. Since $|\ldots|\to 0$ as $n\to\infty$ for the
first term on the right-hand side due to the weak convergence, we find
$n_{0}\in\mathbb{N}$ with the property that $|\ldots|\leq\varepsilon/2$ for
$n\geq n_{0}$, which establishes (5.7).

To show (5.8), one can proceed in an analogous way, using that this time

$\displaystyle\bigg{|}\int_{|x|\leq R}\int_{|v|\geq
P+1}\varphi\,f_{n}\,dx\,dv-\int_{|x|\leq R}\int_{|v|\geq
P+1}\varphi\,f_{0}\,dx\,dv\bigg{|}$ $\displaystyle\leq$
$\displaystyle\int_{|x|\leq R}\int_{|v|\geq
P+1}\varphi\,f_{n}\,dx\,dv+\int_{|x|\leq R}\int_{|v|\geq
P+1}\varphi\,f_{0}\,dx\,dv$ $\displaystyle\leq$
$\displaystyle{\|\varphi\|}_{\infty}\int_{|x|\leq R}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv+{\|\varphi\|}_{\infty}\int_{|x|\leq
R}\int_{|v|\geq P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv$ $\displaystyle\leq$
$\displaystyle\frac{4{\|\varphi\|}_{\infty}}{P^{1/4}}$

for the error term. For the other term, the weak convergence may still be
used, owing to the fact that ${\bf 1}_{\\{|x|\leq R\\}}{\bf 1}_{\\{|v|\leq
P+1\\}}\varphi\in{L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})}^{\ast}$.
$\Box$

Now we are going to show that (CBEC) implies that $f_{0}$ has compact support
in $x$.

###### Lemma 5.7

Let $f_{0}\in L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ be the limiting
function as obtained in Lemma 4.4. Then there exists a $R_{0}>0$ such that

$\int_{\mathbb{R}^{3}}\int_{|x|\geq R_{0}}\sqrt{1+|v|^{2}}f_{0}\,dx\,dv=0,$

which means that $f_{0}(x,v)=0$ for a.e. $|x|\geq R_{0}$ and
$v\in\mathbb{R}^{3}$.

Proof : Let $\kappa(R)=\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv$ and assume that
$\kappa(R)>0$ for all $R>0$. Let $h$ be as in Remark 5.3(a) and $\delta>0$
such that $|{\cal D}(h)|\geq M+\delta$. Since $\lim_{R\to\infty}\kappa(R)=0$,
there is

$R_{0}\geq\max\Big{\\{}\frac{8M}{3},\frac{2M}{1-\frac{1}{(1+\frac{\delta}{6M})^{2}}}\Big{\\}}$

so that

$0<\varepsilon_{0}:=\kappa(R_{0})\leq\frac{M}{2}.$

In particular, $\varepsilon_{0}\in]0,M[$, and in the present case, the
relation (5.7) can be rewritten as

$\lim_{n\to\infty}\int_{|x|\leq
R_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=M-\varepsilon_{0}.$

As $f_{n}$ has mass $M$, this in turn implies

$\lim_{n\to\infty}\int_{|x|\geq
R_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=\varepsilon_{0}.$

Since $(f_{n})$ is a minimizing sequence, we obtain a contradiction to
Corollary 5.5, upon taking $R_{n}=R_{0}$. $\Box$

###### Lemma 5.8

Let $f_{0}\in L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ be the limiting
function as obtained in Lemma 4.4. Then $f_{0}\neq 0$ and $f_{0}\in{\cal A}$.

Proof : First we are going to show that $f_{0}\neq 0$. Otherwise we would get

$\lim_{n\to\infty}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=0$

for every $R>0$ from (5.7). Let $h$ be as in Remark 5.3(a) and $\delta>0$ such
that $|{\cal D}(h)|\geq M+\delta$. In particular, $h\in{\cal A}$ implies that
$I\leq{\cal D}(h)\leq-M-\delta$. Fix

$R=\frac{2M}{1-\frac{1}{(1+\frac{\delta}{2M})^{2}}}>2M.$

If $n\in\mathbb{N}$, then by (3.10)

$\displaystyle|{\cal D}(f_{n})|$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{n}}\,(f_{n}-\chi(f_{n}))\,dx\,dv\leq\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{n}}\,f_{n}\,dx\,dv$
$\displaystyle\leq$ $\displaystyle c_{\beta}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv+\frac{1}{\sqrt{1-\frac{2M}{R}}}\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}f_{n}\,dx\,dv$ $\displaystyle\leq$ $\displaystyle
c_{\beta}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv+\frac{1}{\sqrt{1-\frac{2M}{R}}}\,M$
$\displaystyle=$ $\displaystyle c_{\beta}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv+\Big{(}1+\frac{\delta}{2M}\Big{)}M.$

Passing to the limit $n\to\infty$, we deduce

$M+\delta\leq|I|\leq M+\frac{\delta}{2},$

which is a contradiction.

In order to establish that $f_{0}\in{\cal A}$, due to Lemma 4.4 the only thing
that remains to be verified is that $f_{0}$ has mass $M$. Suppose that
$m=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{0}\,dx\,dv<M$.
Due to $f_{0}\neq 0$, then in fact $m\in]0,M[$. Let $R_{0}>0$ be such that
$f_{0}(x,v)=0$ for a.e. $|x|\geq R_{0}$ and $v\in\mathbb{R}^{3}$; this is in
accordance with Lemma 5.7. By (5.7) it follows that for every $R\geq R_{0}$ we
have

$\displaystyle\lim_{n\to\infty}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv$ $\displaystyle=$
$\displaystyle\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv$ $\displaystyle=$
$\displaystyle\int_{|x|\leq
R_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv=m.$

Since every $f_{n}$ has mass $M$, we deduce that, for every $R\geq R_{0}$,

$\lim_{n\to\infty}\int_{|x|\geq
R}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=\varepsilon_{0}$

for $\varepsilon_{0}=M-m\in]0,M[$. In particular, if $\tilde{R}_{0}\geq R_{0}$
satisfies

$\tilde{R}_{0}\geq\max\Big{\\{}\frac{8M}{3},\frac{2M}{1-\frac{1}{(1+\frac{\delta}{6M})^{2}}}\Big{\\}},$

then

$\lim_{n\to\infty}\int_{|x|\geq\tilde{R}_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv=\varepsilon_{0}.$

Since $(f_{n})$ is a minimizing sequence, we may now apply Corollary 5.5 with
$R_{n}=\tilde{R}_{0}$ to get a contradiction. $\Box$

###### Lemma 5.9

Let $f_{0}\in L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ be the limiting
function as obtained in Lemma 4.4. Then

$\lim_{n\to\infty}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{n}}}f_{n}\,dx\,dv=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{0}\,dx\,dv,$

up to taking a further subsequence.

Proof : First we will show that

$\lim_{n\to\infty}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,f_{n}\,dx\,dv=0.$
(5.9)

For, let $R>0$ and $P\geq P_{0}+1$. Then due to $f_{n}\leq 1$, by (4.10),
(3.10) and (4.6) [remember that $\tilde{f}_{n}$ had been renamed to $f_{n}$],

$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,f_{n}\,dx\,dv$
(5.10) $\displaystyle=$ $\displaystyle\int_{|x|\leq R}\int_{|v|\leq
P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,f_{n}\,dx\,dv+\int_{|x|\geq
R}\int_{|v|\leq P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,f_{n}\,dx\,dv$
$\displaystyle+\,\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,f_{n}\,dx\,dv$
$\displaystyle\leq$ $\displaystyle\int_{|x|\leq R}\int_{|v|\leq
P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,dx\,dv+\int_{|x|\geq
R}\int_{|v|\leq
P+1}\frac{2}{r}\,\frac{e^{2\lambda_{f_{n}}}e^{2\lambda_{f_{0}}}}{e^{\lambda_{f_{n}}}+e^{\lambda_{f_{0}}}}\,|m_{f_{n}}-m_{f_{0}}|\,f_{n}\,dx\,dv$
$\displaystyle+\,2c_{\beta}\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{n}\,dx\,dv$ $\displaystyle\leq$
$\displaystyle\int_{|x|\leq R}\int_{|v|\leq
P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,dx\,dv+\frac{2M}{R}\,c_{\beta}^{4}\int_{|x|\geq
R}\int_{|v|\leq P+1}f_{n}\,dx\,dv+\frac{4c_{\beta}}{P^{1/4}}$
$\displaystyle\leq$ $\displaystyle\int_{|x|\leq R}\int_{|v|\leq
P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,dx\,dv+\frac{2M^{2}}{R}\,c_{\beta}^{4}+\frac{4c_{\beta}}{P^{1/4}}.$

Given $\varepsilon>0$ we fix $R>0$ and $P\geq P_{0}+1$ so large that the last
two terms add up to something less than $\varepsilon/2$. The dominated
convergence theorem implies that

$\lim_{n\to\infty}\int_{|x|\leq R}\int_{|v|\leq
P+1}|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\,dx\,dv=0.$ (5.11)

In fact, we have $e^{\lambda_{f_{n}}(r)}\to e^{\lambda_{f_{0}}(r)}$ as
$n\to\infty$ for $r\in[0,\infty[$ pointwise by Lemma 4.4 and moreover
$|e^{\lambda_{f_{n}}}-e^{\lambda_{f_{0}}}|\leq 2c_{\beta}$ by (3.10). Thus
(5.11) holds, and accordingly we may choose $n_{0}\in\mathbb{N}$ such that the
first term in (5.10) is less that $\varepsilon/2$ for every $n\geq n_{0}$.
This completes the argument for verifying (5.9).

Therefore in order to prove the lemma, it remains to show that

$\lim_{n\to\infty}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n}\,dx\,dv=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{0}\,dx\,dv.$
(5.12)

From (5.8) and (3.10) we deduce that

$\lim_{n\to\infty}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n}\,dx\,dv=\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{0}\,dx\,dv$

holds for every $R>0$. Choose $R_{0}>0$ such that $f_{0}(x,v)=0$ for a.e.
$|x|\geq R_{0}$ and $v\in\mathbb{R}^{3}$; this is possible owing to Lemma 5.7.
If $R\geq R_{0}$, then

$\lim_{n\to\infty}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n}\,dx\,dv=\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{0}\,dx\,dv=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{0}\,dx\,dv=:\theta_{0},$

and $\theta_{0}>0$. Let $(\varepsilon_{m})$ be a sequence such that
$\varepsilon_{m}\downarrow 0$ as $m\to\infty$. Hence for every
$m\in\mathbb{N}$ and $R\geq R_{0}$ there is $n(m,R)\in\mathbb{N}$ such that

$\bigg{|}\int_{|x|\leq
R}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n}\,dx\,dv-\theta_{0}\bigg{|}\leq\varepsilon_{m},\quad
n\geq n(m,R).$

Taking $R=m$ (for $m$ sufficiently large, say $m\geq m_{0}$) and defining
$n_{m}=n(m,m)+m$, we then obtain

$\bigg{|}\int_{|x|\leq
m}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m}}\,dx\,dv-\theta_{0}\bigg{|}\leq\varepsilon_{m},\quad
m\geq m_{0},$

as well as $n_{m}\to\infty$ as $m\to\infty$. Now we need to distinguish two
cases. Case 1: There is a further subsequence $(f_{n_{m_{k}}})$ of
$(f_{n_{m}})$ such that

$\lim_{k\to\infty}\int_{|x|\geq
m_{k}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m_{k}}}\,dx\,dv=0.$

Then

$\displaystyle\bigg{|}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m_{k}}}\,dx\,dv-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{0}\,dx\,dv\bigg{|}$
$\displaystyle\leq$ $\displaystyle\int_{|x|\geq
m_{k}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m_{k}}}\,dx\,dv+\bigg{|}\int_{|x|\leq
m_{k}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m_{k}}}\,dx\,dv-\theta_{0}\bigg{|}$
$\displaystyle\leq$ $\displaystyle\int_{|x|\geq
m_{k}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m_{k}}}\,dx\,dv+\varepsilon_{m_{k}}\to
0,\quad k\to\infty,$

which yields the assertion (5.12) along the specified subsequence. Case 2:
There is a $\delta_{0}>0$ such that

$\int_{|x|\geq
m}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{0}}}f_{n_{m}}\,dx\,dv\geq\delta_{0},\quad
m\geq m_{0}.$

Then (3.10) shows that

$\int_{|x|\geq
m}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{n_{m}}\,dx\,dv\geq\delta_{1},\quad
m\geq m_{0},$

where $\delta_{1}=c_{\beta}^{-1}\delta_{0}>0$. Since every term on the left-
hand side is bounded from above by the mass, there is a further subsequence
(not relabelled) such that

$\varepsilon_{0}=\lim_{m\to\infty}\int_{|x|\geq
m}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{n_{m}}\,dx\,dv\in[\delta_{1},M]$

does exist. Now $\varepsilon_{0}=M$ is impossible, since then

$\displaystyle\int_{|x|\leq
R_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{n_{m}}\,dx\,dv$
$\displaystyle\leq$ $\displaystyle\int_{|x|\leq
m}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{n_{m}}\,dx\,dv$ $\displaystyle=$
$\displaystyle M-\int_{|x|\geq
m}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}f_{n_{m}}\,dx\,dv\to
M-\varepsilon_{0}=0$

as $m\to\infty$, but

$\lim_{m\to\infty}\int_{|x|\leq
R_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{n_{m}}\,dx\,dv=\int_{|x|\leq
R_{0}}\int_{\mathbb{R}^{3}}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv=M$

due to (5.7) and the support properties of $f_{0}$. Altogether, we have seen
that $\varepsilon_{0}\in]0,M[$. Since also ${(f_{n_{m}})}_{m\in\mathbb{N}}$ is
a minimizing sequence and $R_{m}=m$ will satisfy (5.6) for large $m$,
Corollary 5.5 yields a contradiction. $\Box$

The following theorem summarizes the foregoing results. Up to the compact
support property of $f_{0}$ in $v$ that is shown later (up to now we only know
(5.13)), it contains all the statements of the main theorem, Theorem 2.2.

###### Theorem 5.10

Let $k\in]0,2]$. Then (CBEC) is verified. Suppose that $f_{0}\in
L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$ is the limiting function as
obtained in Lemma 4.4. Then $f_{0}$ is a minimizer, i.e., $f_{0}\in{\cal A}$
and ${\cal D}(f_{0})=I$. Furthermore, $f_{0}\leq 1$,

$\int_{\mathbb{R}^{3}}\int_{|v|\geq
P+1}\sqrt{1+|v|^{2}}\,f_{0}\,dx\,dv\leq\frac{2}{P^{1/4}},\quad P\geq P_{0}+1,$
(5.13)

and there is $R_{0}>0$ such that $f_{0}(x,v)=0$ for a.e. $|x|\geq R_{0}$ and
$v\in\mathbb{R}^{3}$.

Proof : From Remark 5.3(b) we know that (CBEC) holds, and by Lemma 5.8 we have
$f_{0}\in{\cal A}$. We write the functional as

${\cal
D}(f)=\Psi(f)-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f}}f\,dx\,dv,\quad\Psi(f)=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f}}\chi(f)\,dx\,dv,$

and consider ${\cal D}(f_{n})$, disregarding the fact that we would have to
pass to subsequences in what follows. To deal with the term $\Psi(f_{n})$, we
are going to use Lemma 9.1, which asserts that $\Psi$ is convex. Since
$(f_{n})\subset{\cal A}$ and also ${\cal A}$ is convex, it follows from
Mazur’s Theorem [7, Thm. 3.9] and (4.7) that
$\Psi(f_{0})\leq\liminf_{n\to\infty}\Psi(f_{n})$. Hence by Lemma 5.9

$\displaystyle{\cal D}(f_{0})$ $\displaystyle=$
$\displaystyle\Psi(f_{0})-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{0}}f_{0}\,dx\,dv$
$\displaystyle\leq$
$\displaystyle\liminf_{n\to\infty}\bigg{(}\Psi(f_{n})-\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{f_{n}}}f_{n}\,dx\,dv\bigg{)}=\liminf_{n\to\infty}{\cal
D}(f_{n})=I.$

Since $f_{0}\in{\cal A}$, also $I\leq{\cal D}(f_{0})$, which yields ${\cal
D}(f_{0})=I$. For the stated additional properties of $f_{0}$, see Lemma 4.4
and Lemma 5.7. $\Box$

##  6 A variational inequality and compact support in $v$-space

From Lemma 5.7 we already know that $f_{0}(x,v)=0$ for a.e. $|x|\geq R_{0}$
and $v\in\mathbb{R}^{3}$. Also $0\leq f_{0}\leq 1$ a.e. by construction.
Denote $\rho_{0}=\rho_{f_{0}}$, $m_{0}=m_{f_{0}}$,
$\lambda_{0}=\lambda_{f_{0}}$, etc.

###### Lemma 6.1

If $g\in{\cal A}$, then

$\int\int\Big{[}e^{\lambda_{0}}(\chi^{\prime}(f_{0})-1)(g-f_{0})+e^{3\lambda_{0}}\,(\chi(f_{0})-f_{0})\,\frac{m_{g}-m_{0}}{r}\Big{]}\,dx\,dv\geq
0$ (6.1)

Proof : Let $g\in{\cal A}$. From the convexity of ${\cal A}$ and the
minimality of $f_{0}$ we can infer that $\varphi(t)={\cal
D}(tg+(1-t)f_{0})\geq{\cal D}(f_{0})=\varphi(0)$ for $t\in[0,1]$. Therefore by
definition of ${\cal D}$

$\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{0}}\,(\chi(f_{0})-f_{0})\,dx\,dv\leq\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{tg+(1-t)f_{0}}}\,\Big{(}\chi(tg+(1-t)f_{0})-f_{0}-t(g-f_{0})\Big{)}\,dx\,dv$
(6.2)

for $t\in[0,1]$. To begin with, we recall from (3.10) that $1\leq
e^{\lambda_{g}}\leq c_{\beta}$ for all $g\in{\cal A}$. Also

$\Big{|}\frac{m_{g}(r)}{r}-\frac{m_{0}(r)}{r}\Big{|}=\frac{4\pi}{r}\,\Big{|}\int_{0}^{r}s^{2}\,(\rho_{g}(s)-\rho_{0}(s))\,ds\Big{|}.$

Hence if $0<r\leq 1$, then

$\Big{|}\frac{m_{g}(r)}{r}-\frac{m_{0}(r)}{r}\Big{|}\leq\frac{4\pi}{r}\,\Big{(}\int_{0}^{r}s^{2}\,(\rho_{g}(s)+\rho_{0}(s))\,ds\Big{)}\leq\frac{8\pi\sigma_{0}}{3r}\,r^{3}\leq\frac{8\pi}{3}\,\sigma_{0}.$

On the other hand, if $r\geq 1$, then

$\Big{|}\frac{m_{g}(r)}{r}-\frac{m_{0}(r)}{r}\Big{|}\leq\frac{4\pi}{r}\,\int_{0}^{r}s^{2}\,(\rho_{g}(s)+\rho_{0}(s))\,ds\leq
2M.$

Thus we get

$2e^{2\lambda_{0}(r)}\Big{|}\frac{m_{g}(r)-m_{0}(r)}{r}\Big{|}\leq
2c_{g}^{2}\,\Big{(}\frac{8\pi}{3}\,\sigma_{0}+2M\Big{)}=:C_{0}$ (6.3)

in any case. Henceforth we restrict the range of $t$ to
$t\in[0,1]\cap[0,\frac{1}{2C_{0}}]$. Since $m_{f}$ is linear in $f$, we can
write pointwise a.e.

$\displaystyle e^{-2\lambda_{tg+(1-t)f_{0}}}$ $\displaystyle=$ $\displaystyle
1-\frac{2m_{tg+(1-t)f_{0}}}{r}=1-\frac{2m_{0}}{r}-2t\,\frac{m_{g}-m_{0}}{r}=e^{-2\lambda_{0}}-2t\,\frac{m_{g}-m_{0}}{r}$
$\displaystyle=$ $\displaystyle
e^{-2\lambda_{0}}\Big{(}1-2t\,e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r}\Big{)}.$

In view of (6.3) and $0\leq t\leq\frac{1}{2C_{0}}$, it follows that

$e^{\lambda_{tg+(1-t)f_{0}}}=e^{\lambda_{0}}\frac{1}{(1-2t\,e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r})^{1/2}}.$

We have

$\frac{1}{(1-x)^{1/2}}=1+\frac{1}{2}\,x+\psi(x)x^{2},\quad|\psi(x)|\leq
C,\quad x\in\Big{[}0,\frac{1}{2}\Big{]}.$

Therefore we may continue

$\displaystyle e^{\lambda_{tg+(1-t)f_{0}}}$ $\displaystyle=$ $\displaystyle
e^{\lambda_{0}}\Big{(}1+t\,e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r}+\psi\Big{(}2t\,e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r}\Big{)}\Big{[}2t\,e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r}\Big{]}^{2}\Big{)}$
$\displaystyle=$ $\displaystyle
e^{\lambda_{0}}\Big{(}1+t\,e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r}+t^{2}\Psi\Big{)},$

where $|\Psi(t,r)|\leq C$ is uniformly bounded. Next,

$\displaystyle\chi(tg+(1-t)f_{0})$ $\displaystyle=$
$\displaystyle\chi(f_{0})+t\chi^{\prime}(f_{0})(g-f_{0})+\int_{0}^{t(g-f_{0})}\Big{(}\chi^{\prime}(f_{0}+s)-\chi^{\prime}(f_{0})\Big{)}\,ds$
$\displaystyle=$
$\displaystyle\chi(f_{0})+t\chi^{\prime}(f_{0})(g-f_{0})+t(g-f_{0})\,\Phi,$
$\displaystyle\Phi(t,x,v)$ $\displaystyle=$
$\displaystyle\frac{1}{t}\int_{0}^{t}\Big{(}\chi^{\prime}(f_{0}(x,v)+\tau(g(x,v)-f_{0}(x,v)))-\chi^{\prime}(f_{0}(x,v))\Big{)}\,d\tau.$

Going back to (6.2) and using the previous expansions, we get for
$t\in]0,\frac{1}{2C_{0}}]$, after some simplification,

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{0}}\bigg{(}\Big{[}(\chi^{\prime}(f_{0})-1)(g-f_{0})+e^{2\lambda_{0}}(\chi(f_{0})-f_{0})\,\frac{m_{g}-m_{0}}{r}+(g-f_{0})\Phi\Big{]}$
$\displaystyle\hskip
60.00009pt+\,t\Big{[}e^{2\lambda_{0}}\frac{m_{g}-m_{0}}{r}\,(g-f_{0})(\chi^{\prime}(f_{0})-1+\Phi)+(\chi(f_{0})-f_{0})\Psi\Big{]}$
$\displaystyle\hskip
60.00009pt+\,t^{2}\Big{[}\chi^{\prime}(f_{0})(g-f_{0})\Psi+(g-f_{0})\Psi\,\Phi-(g-f_{0})\Psi\Big{]}\bigg{)}\,dx\,dv.$

It remains to show that the error terms tend to zero as $t\to 0$. For, we will
only prove that

$I(t)=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}e^{\lambda_{0}}(g-f_{0})\Phi\,dx\,dv\to
0,\quad t\to 0,$ (6.4)

since the other terms are similar or easier. For a.e.
$(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}$ we have $\Phi(t,x,v)\to 0$ as
$t\to 0$, since $\chi^{\prime}$ is continuous on $[0,\infty[$, $0\leq
f_{0}(x,v)\leq 1$ a.e. and $0\leq g(x,v)<\infty$ a.e.; also note that
$f_{0}(x,v)+\tau(g(x,v)-f_{0}(x,v))=(1-\tau)f_{0}(x,v)+\tau g(x,v)\geq 0$. No
matter if $k\leq 1$ or $k\geq 1$, there is a constant $C_{k}>0$ such that
$\chi^{\prime}(s_{1}+s_{2})=(s_{1}+s_{2})^{1/k}\leq
C_{k}(s_{1}^{1/k}+s_{2}^{1/k})$ for $s_{1},s_{2}\geq 0$. Therefore pointwise
a.e.

$\displaystyle|\Phi(t,x,v)|$ $\displaystyle\leq$
$\displaystyle\frac{1}{t}\int_{0}^{t}\Big{(}C_{k}(1-\tau)^{1/k}f_{0}(x,v)^{1/k}+C_{k}\tau^{1/k}g(x,v)^{1/k}+f_{0}(x,v)^{1/k}\Big{)}\,d\tau$
$\displaystyle\leq$ $\displaystyle(C_{k}+1)(f_{0}(x,v)^{1/k}+g(x,v)^{1/k}),$

and hence

$e^{\lambda_{0}}\,|g(x,v)-f_{0}(x,v)|\,|\Phi(t,x,v)|\leq
c_{\beta}(C_{k}+1)(f_{0}(x,v)+g(x,v))\,(f_{0}(x,v)^{1/k}+g(x,v)^{1/k}).$

Since $f_{0},g\in L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$, the right-
hand side is integrable, by Hölder’s inequality. Therefore the dominated
convergence theorem shows that (6.4) is valid. $\Box$

###### Remark 6.2

If $\rho_{g}=\rho_{0}$ and $g\geq 0$, then $g\in{\cal A}$ is for free, and due
to $m_{g}=m_{0}$ the second term in (6.1) drops out. Thus we have

$\rho_{g}=\rho_{0}\quad\Longrightarrow\quad\int\int_{{\rm
supp}\,f_{0}}e^{\lambda_{0}}(1-\chi^{\prime}(f_{0}))(f_{0}-g)\,dx\,dv\geq\int\int_{({\rm
supp}\,f_{0})^{c}}e^{\lambda_{0}}g\,dx\,dv.$ (6.5)

This may be used to prove that $f_{0}$ has compact support in $v$. Taking
together Theorem 5.10 and Corollary 6.3, this in particular completes the
proof of Theorem 2.2.

###### Corollary 6.3

In the setting of Theorem 5.10 the following holds for the minimizer $f_{0}$.
There is a $S_{0}>0$ such that $f_{0}(x,v)=0$ for a.e. $x\in\mathbb{R}^{3}$
and $|v|\geq S_{0}$.

Proof : We consider the boxes

$B=\\{(x,v):|x|\in[r_{1},r_{2}],|v|\in[s_{1},s_{2}]\\},\quad\hat{B}=\\{(x,v):|x|\in[r_{1},r_{2}],|v|\in[\hat{s}_{1},\hat{s}_{2}]\\},$
(6.6)

for $r_{2}>r_{1}>0$, $s_{2}>s_{1}>0$ and $\hat{s}_{2}>\hat{s}_{1}>0$ such that
$B\cap\hat{B}=\emptyset$ and $B$ is below $\hat{B}$. Define

$g(x,v)=\left\\{\begin{array}[]{c@{\quad:\quad}c}f_{0}(x,v)+b(x,v)&(x,v)\in
B\\\\[4.30554pt] \frac{1}{2}f_{0}(x,v)&(x,v)\in\hat{B}\\\\[4.30554pt]
f_{0}(x,v)&(x,v)\in\mathbb{R}^{6}\setminus(B\cup\hat{B})\end{array}\right.,$

where

$b(x,v)=\bigg{(}\frac{1}{2}\int_{\hat{s}_{1}\leq|\tilde{v}|\leq\hat{s}_{2}}\sqrt{1+|\tilde{v}|^{2}}\,f_{0}(x,\tilde{v})\,d\tilde{v}\bigg{)}\,\frac{1}{4\pi(s_{2}-s_{1})}\,\frac{1}{|v|^{2}\sqrt{1+|v|^{2}}}.$

Then $\rho_{g}(r)=\rho_{0}(r)$ for $r\in[0,\infty[\setminus[r_{1},r_{2}]$. If
$r\in[r_{1},r_{2}]$ and since $B$ is below $\hat{B}$, we get

$\displaystyle\rho_{g}(r)$ $\displaystyle=$ $\displaystyle\int_{0\leq|v|\leq
s_{1}}\sqrt{1+|v|^{2}}\,f_{0}(x,v)\,dv+\int_{s_{1}\leq|v|\leq
s_{2}}\sqrt{1+|v|^{2}}\,(f_{0}(x,v)+b(x,v))\,dv$
$\displaystyle+\int_{s_{2}\leq|v|\leq\hat{s}_{1}}\sqrt{1+|v|^{2}}\,f_{0}(x,v)\,dv+\frac{1}{2}\int_{\hat{s}_{1}\leq|v|\leq\hat{s}_{2}}\sqrt{1+|v|^{2}}\,f_{0}(x,v)\,dv$
$\displaystyle+\int_{|v|\geq\hat{s}_{2}}\sqrt{1+|v|^{2}}\,f_{0}(x,v)\,dv$
$\displaystyle=$ $\displaystyle\rho_{f_{0}}(r)+\int_{s_{1}\leq|v|\leq
s_{2}}\sqrt{1+|v|^{2}}\,b(x,v)\,dv-\frac{1}{2}\int_{\hat{s}_{1}\leq|v|\leq\hat{s}_{2}}\sqrt{1+|v|^{2}}\,f_{0}(x,v)\,dv$
$\displaystyle=$
$\displaystyle\rho_{f_{0}}(r)+\bigg{(}\frac{1}{2}\int_{\hat{s}_{1}\leq|\tilde{v}|\leq\hat{s}_{2}}\sqrt{1+|\tilde{v}|^{2}}\,f_{0}(x,\tilde{v})\,d\tilde{v}\bigg{)}\,\frac{4\pi}{4\pi(s_{2}-s_{1})}\int_{s_{1}}^{s_{2}}\frac{u^{2}\sqrt{1+u^{2}}}{u^{2}\sqrt{1+u^{2}}}\,\,du$
$\displaystyle-\,\frac{1}{2}\int_{\hat{s}_{1}\leq|v|\leq\hat{s}_{2}}\sqrt{1+|v|^{2}}\,f_{0}(x,v)\,dv$
$\displaystyle=$ $\displaystyle\rho_{0}(r).$

Thus we have seen that $\rho_{g}=\rho_{0}$ and (6.5) applies. It follows that

$\int\int_{{\rm
supp}\,f_{0}}e^{\lambda_{0}}(1-\chi^{\prime}(f_{0}))(f_{0}-g)\,dx\,dv\geq\int\int_{({\rm
supp}\,f_{0})^{c}}e^{\lambda_{0}}g\,dx\,dv,$

which leads to

$\displaystyle\frac{1}{2}\int\int_{\hat{B}}e^{\lambda_{0}}(1-\chi^{\prime}(f_{0}))f_{0}\,dx\,dv$
$\displaystyle=$ $\displaystyle\frac{1}{2}\int\int_{{\rm
supp}\,f_{0}\cap\hat{B}}e^{\lambda_{0}}(1-\chi^{\prime}(f_{0}))f_{0}\,dx\,dv$
$\displaystyle\geq$ $\displaystyle\int\int_{{\rm supp}\,f_{0}\cap
B}e^{\lambda_{0}}(1-\chi^{\prime}(f_{0}))\,b\,dx\,dv+\int\int_{({\rm
supp}\,f_{0})^{c}\cap B}e^{\lambda_{0}}\,b\,dx\,dv$ $\displaystyle=$
$\displaystyle\int\int_{B}e^{\lambda_{0}}(1-\chi^{\prime}(f_{0}))\,b\,dx\,dv$
$\displaystyle=$ $\displaystyle\int_{r_{1}\leq|x|\leq
r_{2}}dx\,e^{\lambda_{0}(r)}\int_{s_{1}\leq|v|\leq
s_{2}}dv\,(1-\chi^{\prime}(f_{0}(x,v)))\,\bigg{(}\frac{1}{2}\int_{\hat{s}_{1}\leq|\tilde{v}|\leq\hat{s}_{2}}\sqrt{1+|\tilde{v}|^{2}}\,f_{0}(x,\tilde{v})\,d\tilde{v}\bigg{)}\,$
$\displaystyle\hskip
120.00018pt\times\frac{1}{4\pi(s_{2}-s_{1})}\,\frac{1}{|v|^{2}\sqrt{1+|v|^{2}}}$
$\displaystyle=$ $\displaystyle\frac{1}{2}\int_{r_{1}\leq|x|\leq
r_{2}}dx\,e^{\lambda_{0}(r)}\int_{\hat{s}_{1}\leq|v|\leq\hat{s}_{2}}dv\,\sqrt{1+|v|^{2}}\,f_{0}(x,v)\int_{s_{1}\leq|\tilde{v}|\leq
s_{2}}d\tilde{v}\,(1-\chi^{\prime}(f_{0}(x,\tilde{v})))\,$
$\displaystyle\hskip
120.00018pt\times\frac{1}{4\pi(s_{2}-s_{1})}\,\frac{1}{|\tilde{v}|^{2}\sqrt{1+|\tilde{v}|^{2}}}.$

This can be re-expressed as

$\displaystyle\int\int_{\hat{B}}dx\,dv\,e^{\lambda_{0}(r)}(1-\chi^{\prime}(f_{0}(x,v)))f_{0}(x,v)$
$\displaystyle\geq$
$\displaystyle\frac{1}{4\pi(s_{2}-s_{1})}\int\int_{\hat{B}}\,dx\,dv\,e^{\lambda_{0}(r)}\,\sqrt{1+|v|^{2}}\,f_{0}(x,v)$
$\displaystyle\hskip 120.00018pt\int_{s_{1}\leq|\tilde{v}|\leq
s_{2}}d\tilde{v}\,(1-\chi^{\prime}(f_{0}(x,\tilde{v})))\,\frac{1}{|\tilde{v}|^{2}\sqrt{1+|\tilde{v}|^{2}}}.$

This relation is verified for all boxes $\hat{B}$ above $B$ as in (6.6) that
do not intersect. Next recall from $f_{0}\in{\cal A}$ and Theorem 5.10 that
$\rho_{0}(r)\leq\sigma_{0}$ and $f_{0}\leq 1$ a.e. Take

$N\geq\max\Big{\\{}\sqrt[3]{\frac{3\sigma_{0}}{2\pi}},\sqrt[3]{\frac{3\sigma_{0}}{2\pi}\,2^{k}}\Big{\\}}.$

To begin with, for a.e. $x\in\mathbb{R}^{3}$,

$\displaystyle\int_{|\tilde{v}|\leq
N}d\tilde{v}\,(1-\chi^{\prime}(f_{0}(x,\tilde{v})))\,\frac{1}{|\tilde{v}|^{2}\sqrt{1+|\tilde{v}|^{2}}}$
(6.9) $\displaystyle\geq$
$\displaystyle\frac{1}{N^{2}\sqrt{1+N^{2}}}\int_{|\tilde{v}|\leq
N}d\tilde{v}\,(1-f_{0}(x,\tilde{v})^{1/k})$ $\displaystyle=$
$\displaystyle\frac{1}{N^{2}\sqrt{1+N^{2}}}\bigg{(}\frac{4\pi}{3}\,N^{3}-\int_{|\tilde{v}|\leq
N}d\tilde{v}\,f_{0}(x,\tilde{v})^{1/k}\bigg{)}.$

If $k\leq 1$, then $f_{0}^{1/k}\leq f_{0}$, and hence

$\displaystyle\int_{|\tilde{v}|\leq
N}d\tilde{v}\,(1-\chi^{\prime}(f_{0}(x,\tilde{v})))\,\frac{1}{|\tilde{v}|^{2}\sqrt{1+|\tilde{v}|^{2}}}$
$\displaystyle\geq$
$\displaystyle\frac{1}{N^{2}\sqrt{1+N^{2}}}\bigg{(}\frac{4\pi}{3}\,N^{3}-\rho_{f_{0}}(r)\bigg{)}$
$\displaystyle\geq$
$\displaystyle\frac{1}{N^{2}\sqrt{1+N^{2}}}\bigg{(}\frac{4\pi}{3}\,N^{3}-\sigma_{0}\bigg{)}$
$\displaystyle\geq$ $\displaystyle\frac{2\pi}{3}\,\frac{N}{\sqrt{1+N^{2}}}.$

On the other hand, if $k\geq 1$ and $f_{0}(x,\tilde{v})^{1/k}\geq 1/2$, then

$2^{k-1}f_{0}(x,\tilde{v})f_{0}(x,\tilde{v})^{-1/k}=(2f_{0}(x,\tilde{v})^{1/k})^{k-1}\geq
1$

yields $f_{0}(x,\tilde{v})^{1/k}\leq 2^{k-1}f_{0}(x,\tilde{v})$. It follows
that

$\displaystyle\int_{|\tilde{v}|\leq N}d\tilde{v}\,f_{0}(x,\tilde{v})^{1/k}$
$\displaystyle=$ $\displaystyle\int_{\\{|\tilde{v}|\leq
N,\,f_{0}(x,\tilde{v})^{1/k}\leq
1/2\\}}d\tilde{v}\,f_{0}(x,\tilde{v})^{1/k}+\int_{\\{|\tilde{v}|\leq
N,\,f_{0}(x,\tilde{v})^{1/k}\geq 1/2\\}}d\tilde{v}\,f_{0}(x,\tilde{v})^{1/k}$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\,\frac{4\pi}{3}\,N^{3}+2^{k-1}\int_{\\{|\tilde{v}|\leq
N,\,f_{0}(x,\tilde{v})^{1/k}\geq 1/2\\}}d\tilde{v}\,f_{0}(x,\tilde{v})$
$\displaystyle\leq$ $\displaystyle\frac{2\pi}{3}\,N^{3}+2^{k-1}\rho_{0}(r).$

Thus in the case $k\geq 1$ we deduce from (6.9) and the choice of $N$ that

$\displaystyle\int_{|\tilde{v}|\leq
N}d\tilde{v}\,(1-\chi^{\prime}(f_{0}(x,\tilde{v})))\,\frac{1}{|\tilde{v}|^{2}\sqrt{1+|\tilde{v}|^{2}}}$
$\displaystyle\geq$
$\displaystyle\frac{1}{N^{2}\sqrt{1+N^{2}}}\bigg{(}\frac{2\pi}{3}\,N^{3}-2^{k-1}\rho_{f_{0}}(r)\bigg{)}$
$\displaystyle\geq$
$\displaystyle\frac{1}{N^{2}\sqrt{1+N^{2}}}\bigg{(}\frac{2\pi}{3}\,N^{3}-2^{k-1}\sigma_{0}\bigg{)}$
$\displaystyle\geq$ $\displaystyle\frac{\pi}{3}\,\frac{N}{\sqrt{1+N^{2}}}.$

Therefore we have seen that in all cases the estimate

$\int_{|\tilde{v}|\leq
N}d\tilde{v}\,(1-\chi^{\prime}(f_{0}(x,\tilde{v})))\,\frac{1}{|\tilde{v}|^{2}\sqrt{1+|\tilde{v}|^{2}}}\geq\frac{\pi}{3}\,\frac{N}{\sqrt{1+N^{2}}}$

is verified. Thus taking $s_{1}=0$ and $s_{2}=N$ in (6), we therefore obtain
for any box $\hat{B}$ that lies above the line $\\{|v|=N\\}$ the estimate

$\displaystyle
12\sqrt{1+N^{2}}\int\int_{\hat{B}}dx\,dv\,e^{\lambda_{0}(r)}(1-\chi^{\prime}(f_{0}(x,v)))f_{0}(x,v)$
$\displaystyle\geq$
$\displaystyle\int\int_{\hat{B}}\,dx\,dv\,e^{\lambda_{0}(r)}\,\sqrt{1+|v|^{2}}\,f_{0}(x,v).$

Since this holds for all such boxes $\hat{B}$, we must have pointwise a.e. in
the set $\\{(x,v)\in\mathbb{R}^{3}\times\mathbb{R}^{3}:|v|\geq N\\}$ the
relation

$\sqrt{1+|v|^{2}}\,f_{0}(x,v)\leq
12\sqrt{1+N^{2}}\,(1-\chi^{\prime}(f_{0}(x,v)))f_{0}(x,v)\leq
12\sqrt{1+N^{2}}\,f_{0}(x,v),$

and we can in turn infer that $|v|\leq 12\sqrt{1+N^{2}}$ on ${\rm
supp}\,f_{0}$. $\Box$

##  7 When is the minimizer a static solution?

We recall some results from [5] (with $\alpha=1$ there), which are stated in
the notation that we are using in the present paper. Let $r>0$ be fixed and
put

$\Xi_{r}=\Big{\\{}\psi=\psi(w,\ell)\in
L^{1+1/k}(\mathbb{R}\times[0,\infty[):\psi\geq
0,\psi(w,\ell)=0\,\,\mbox{for}\,\,\sqrt{1+w^{2}+\ell^{2}/r^{2}}\geq\sqrt{1+S_{0}^{2}}\Big{\\}}.$

As compared to [5], the number ‘$2$’ in the definition of $\Xi_{r}$ has been
replaced by $\sqrt{1+S_{0}^{2}}$, and as remarked in [5], the results still
hold with such a modification. Now we consider the functionals

$\hat{H}(\psi)=\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,(\chi(\psi)-\psi),\quad\hat{F}(\psi,r,a)=\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\sqrt{1+w^{2}+\ell^{2}/r^{2}}\,\psi-\frac{r^{2}}{\pi}\,a,$
(7.1)

for $r>0$, functions $\psi=\psi(w,\ell)\in\Xi_{r}$ and $a>0$; here
$d\ell^{2}=2\ell\,d\ell$. The next lemma is [5, Lemma 5.1].

###### Lemma 7.1

For $r>0$ and $a>0$ there is a unique function
$\psi=\psi(w,\ell;r,a)\in\Xi_{r}$ such that

$\inf_{\psi\in\Xi_{r}}\,\\{\hat{H}(\psi):\hat{F}(\psi,r,a)=0\\}=\hat{H}(\psi(\cdot,\cdot;r,a))$
(7.2)

and

$\hat{F}(\psi(\cdot,\cdot;r,a),r,a)=0.$ (7.3)

For $r>0$ and $a=0$ this unique function is $\psi(w,\ell;r,0)=0$.

###### Corollary 7.2

We have

$\psi(\cdot,\cdot;r,\rho_{0}(r))=f_{0}(r,\cdot,\cdot)$ (7.4)

for a.e. $r\geq 0$.

Proof : Since

$4\pi^{2}\int_{0}^{\infty}dr\int_{\mathbb{R}}dw\int_{0}^{\infty}d\ell^{2}\,f_{0}^{1+1/k}=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}f_{0}^{1+1/k}\,dx\,dv<\infty,$

we have
$\int_{\mathbb{R}}dw\int_{0}^{\infty}d\ell^{2}\,f_{0}(r,\cdot,\cdot)^{1+1/k}<\infty$
for a.e. $r\geq 0$, so that $0\leq f_{0}(r,\cdot,\cdot)\in
L^{1+1/k}(\mathbb{R}\times[0,\infty[)$. Also, by Corollary 6.3, the support
condition in $\Xi_{r}$ is satisfied, and hence
$f_{0}(r,\cdot,\cdot)\in\Xi_{r}$. Next observe that

$\hat{F}(f_{0}(r,\cdot,\cdot),r,\rho_{0}(r))=\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\sqrt{1+w^{2}+\ell^{2}/r^{2}}\,f_{0}(r,w,\ell)-\frac{r^{2}}{\pi}\,\rho_{0}(r)=0$

due to $dv=\frac{\pi}{r^{2}}\,d\ell^{2}\,dw$. It follows from the minimizing
property of the function $\psi$ that

$\hat{H}(\psi(\cdot,\cdot;r,\rho_{0}(r)))\leq\hat{H}(f_{0}(r,\cdot,\cdot)).$
(7.5)

Let the spherically symmetric function $g\geq 0$ be defined by

$g(r,w,\ell)=\psi(w,\ell;r,\rho_{0}(r))\geq 0.$

Then $\rho_{g}=\rho_{0}\leq\sigma_{0}$ by the constraint $\hat{F}=0$. In
particular, $g$ has total mass $M$. Next, (7.5) says that

$\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,(\chi(g(r,\cdot,\cdot))-g(r,\cdot,\cdot))\leq\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,(\chi(f_{0}(r,\cdot,\cdot))-f_{0}(r,\cdot,\cdot)),$

and upon integration $\int_{0}^{\infty}dr$ it follows that

$\displaystyle\frac{k}{k+1}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}g^{1+1/k}\,dx\,dv$
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}g\,dx\,dv+\frac{k}{k+1}\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}f_{0}^{1+1/k}\,dx\,dv$
$\displaystyle\leq$ $\displaystyle
M+\frac{k}{k+1}\|f_{0}\|_{L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})}^{1+1/k}<\infty.$

As a consequence, $g\in L^{1+1/k}(\mathbb{R}^{3}\times\mathbb{R}^{3})$, which
shows that $g\in{\cal A}$. Since $f_{0}$ is a minimizer, and $m_{g}=m_{0}$
yields $\lambda_{g}=\lambda_{0}$,

$\displaystyle 0$ $\displaystyle\geq$ $\displaystyle{\cal D}(f_{0})-{\cal
D}(g)=4\pi^{2}\int_{0}^{\infty}dr\,e^{\lambda_{0}}\int_{\mathbb{R}}dw\int_{0}^{\infty}d\ell^{2}\,[(\chi(f_{0})-f_{0})-(\chi(g)-g)]$
$\displaystyle=$ $\displaystyle
4\pi^{2}\int_{0}^{\infty}dr\,e^{\lambda_{0}}[\hat{H}(f_{0}(r,\cdot,\cdot))-\hat{H}(g(r,\cdot,\cdot))]\geq
0,$

where in the last step (7.5) has been used. Hence
$\hat{H}(f_{0}(r,\cdot,\cdot))=\hat{H}(g(r,\cdot,\cdot))$ for a.e. $r\geq 0$,
and from the uniqueness assertion in Lemma 7.1 we infer that (7.4) holds.
$\Box$

Now we can follow the steps in [5] and arrive at [5, equ. (5.15)] to deduce
that

$f_{0}(r,w,\ell)=\Big{(}1-\tilde{\varepsilon}(r)\sqrt{1+w^{2}+\ell^{2}/r^{2}}\Big{)}_{+}^{k}$
(7.6)

for $\tilde{\varepsilon}(r)=(G^{\prime})^{-1}(-4\pi\kappa\rho_{0}(r))$, where
$\kappa=\frac{k+1}{16\pi^{2}}$ and

$G(\varepsilon)=\int_{0}^{\infty}\xi^{2}\,{(1-\varepsilon\sqrt{1+\xi^{2}})}_{+}^{k+1}\,d\xi,\quad\varepsilon\in\mathbb{R};$

see [5, Section 7.1] for the properties of $G$ and its derivatives. It is
important to note that the fact that $\rho_{\ast}$ from [5] satisfies the
(reformulated) Euler-Lagrange equation enters only in the end, where it is
shown that the function $\tilde{\varepsilon}$ is proportional to
$e^{\mu_{\ast}}$.

This allows the following reformulation of (6.1).

###### Corollary 7.3

If $g\in{\cal A}$, then

$\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,e^{\lambda_{0}}\bigg{[}(\chi^{\prime}(f_{0})-1)(g-f_{0})+e^{2\lambda_{0}}\,f_{0}\frac{d}{dw}\,[(\chi^{\prime}(f_{0})-1)w]\,\frac{m_{g}-m_{0}}{r}\bigg{]}\geq
0.$ (7.7)

Proof : From Lemma 5.7 and Corollary 6.3 we know that $f_{0}(r,w,\ell)=0$ for
$r\geq R_{0}$ or if $1+w^{2}+\ell^{2}/r^{2}\geq 1+S_{0}^{2}$. If $(r,\ell)$
are fixed, then (7.6) implies that the support of $f_{0}(r,\cdot,\ell)$ is
given by the (symmetric) interval $I_{r,\ell}=\\{w\in\mathbb{R}:|w|^{2}\leq
1/\tilde{\varepsilon}(r)^{2}-1-\ell^{2}/r^{2}\\}$, and
$f_{0}(r,w,\ell)=(1-\tilde{\varepsilon}(r)\sqrt{1+w^{2}+\ell^{2}/r^{2}})^{k}$
on $I_{r,\ell}$, which is a differentiable function in $w$. Therefore we may
integrate by parts in $w$ to obtain

$\displaystyle\int_{\mathbb{R}^{3}}(\chi(f_{0})-f_{0})\,dv$ $\displaystyle=$
$\displaystyle\frac{\pi}{r^{2}}\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,(\chi(f_{0})-f_{0})=\frac{\pi}{r^{2}}\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}}dw\,(\chi(f_{0})-f_{0})\,\frac{d}{dw}\,w$
$\displaystyle=$
$\displaystyle-\frac{\pi}{r^{2}}\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}}dw\,(\chi^{\prime}(f_{0})-1)\,\frac{\partial
f_{0}}{\partial w}\,w$ $\displaystyle=$
$\displaystyle\frac{\pi}{r^{2}}\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}}dw\,\frac{d}{dw}\,[(\chi^{\prime}(f_{0})-1)w]\,f_{0}$
$\displaystyle=$
$\displaystyle\frac{\pi}{r^{2}}\int_{0}^{\infty}d\beta\int_{\mathbb{R}}dw\,\frac{d}{dw}\,[(\chi^{\prime}(f_{0})-1)w]\,f_{0}.$

Hence (7.7) follows from (6.1) and $dx\,dv=4\pi^{2}dr\,d\ell^{2}\,dw$. $\Box$

The relation (7.7) can also be restated in a slimmer form.

###### Corollary 7.4

Let $I_{r,\ell}=\\{w\in\mathbb{R}:|w|^{2}\leq
1/\tilde{\varepsilon}(r)^{2}-1-\ell^{2}/r^{2}\\}$ and

$U(r)=-\left\\{\begin{array}[]{c@{\quad:\quad}l}e^{-\mu_{0}(r)}\tilde{\varepsilon}(r)&r\in[0,R_{0}]\\\
0&r\in]R_{0},\infty[\end{array}\right..$

If $g\in{\cal A}$, then

$\int_{0}^{\infty}\frac{d}{dr}\Big{(}e^{\lambda_{0}+\mu_{0}}(m_{g}-m_{0})\Big{)}\,U\,dr\geq
4\pi^{2}\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}(1-\tilde{\varepsilon}\,\sqrt{1+|v|^{2}})g.$

In particular, if $g\in{\cal A}$ and ${\rm supp}\,g\subset{\rm supp}\,f_{0}$,
then

$\int_{0}^{\infty}\frac{d}{dr}\Big{(}e^{\lambda_{0}+\mu_{0}}(m_{g}-m_{0})\Big{)}\,U\,dr\geq
0.$ (7.8)

Proof : Define the functions

$a=\chi^{\prime}(f_{0})-1,\quad{\cal
E}=\sqrt{1+|v|^{2}}=\sqrt{1+w^{2}+\ell^{2}/r^{2}},\quad b=\frac{a}{{\cal
E}}=\frac{\chi^{\prime}(f_{0})-1}{\sqrt{1+w^{2}+\ell^{2}/r^{2}}}.$

We write

$\displaystyle\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,e^{\lambda_{0}}\bigg{[}(\chi^{\prime}(f_{0})-1)(g-f_{0})+e^{2\lambda_{0}}\,f_{0}\frac{d}{dw}\,[(\chi^{\prime}(f_{0})-1)w]\,\frac{m_{g}-m_{0}}{r}\bigg{]}$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,e^{\lambda_{0}}\bigg{[}b\,{\cal
E}(g-f_{0})+e^{2\lambda_{0}}\,f_{0}\frac{d}{dw}\,(b{\cal
E}w)\,\frac{m_{g}-m_{0}}{r}\bigg{]}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}}dw\,e^{\lambda_{0}}\bigg{[}b\,{\cal
E}(g-f_{0})+e^{2\lambda_{0}}\,f_{0}\frac{d}{dw}\,(b{\cal
E}¸w)\,\frac{m_{g}-m_{0}}{r}\bigg{]}$
$\displaystyle+\,\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}b\,{\cal
E}g.$

On the interval $I_{r,\ell}$ we have
$f_{0}(r,w,\ell)=(1-\tilde{\varepsilon}(r)\sqrt{1+w^{2}+\ell^{2}/r^{2}})^{k}$,
whence $\chi^{\prime}(s)=s^{1/k}$ leads to
$b(r,w,\ell)=-\tilde{\varepsilon}(r)$, which is independent of $w$. On the
other hand, on $I_{r,\ell}^{c}$ we have $b(r,w,\ell)=-1/{\cal E}$. This gives

$\displaystyle\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,e^{\lambda_{0}}\bigg{[}(\chi^{\prime}(f_{0})-1)(g-f_{0})+e^{2\lambda_{0}}\,f_{0}\frac{d}{dw}\,[(\chi^{\prime}(f_{0})-1)w]\,\frac{m_{g}-m_{0}}{r}\bigg{]}$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}}dw\,e^{\lambda_{0}}\tilde{\varepsilon}\bigg{[}{\cal
E}(g-f_{0})+e^{2\lambda_{0}}\,f_{0}\frac{d}{dw}\,({\cal
E}w)\,\frac{m_{g}-m_{0}}{r}\bigg{]}$
$\displaystyle-\,\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}\,g$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,e^{\lambda_{0}}\tilde{\varepsilon}\,{\cal
E}(g-f_{0})+\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}\tilde{\varepsilon}\,{\cal
E}g$
$\displaystyle-\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,e^{3\lambda_{0}}\,\tilde{\varepsilon}\,f_{0}\frac{d}{dw}\,({\cal
E}w)\,\frac{m_{g}-m_{0}}{r}$
$\displaystyle-\,\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}\,g$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}dr\,e^{\lambda_{0}+\mu_{0}}\,U\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,{\cal
E}(g-f_{0})$
$\displaystyle+\int_{0}^{\infty}dr\,e^{3\lambda_{0}+\mu_{0}}\,U\,\frac{m_{g}-m_{0}}{r}\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,f_{0}\Big{(}{\cal
E}+\frac{w^{2}}{{\cal E}}\Big{)}$
$\displaystyle-\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}(1-\tilde{\varepsilon}\,{\cal
E})g$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}dr\,e^{\lambda_{0}+\mu_{0}}\,U\,\frac{r^{2}}{\pi}\,(\rho_{g}-\rho_{0})+\int_{0}^{\infty}dr\,e^{3\lambda_{0}+\mu_{0}}\,U\,\frac{m_{g}-m_{0}}{r}\,\frac{r^{2}}{\pi}(\rho_{0}+p_{0})$
$\displaystyle-\int_{0}^{\infty}dr\int_{0}^{\infty}d\ell^{2}\int_{I_{r,\ell}^{c}}dw\,e^{\lambda_{0}}(1-\tilde{\varepsilon}\,{\cal
E})g.$

Next note that, by definition of $\lambda_{0}$ and $\mu_{0}$,

$\displaystyle\frac{d}{dr}\Big{(}e^{\lambda_{0}+\mu_{0}}(m_{g}-m_{0})\Big{)}$
$\displaystyle=$ $\displaystyle 4\pi
r^{2}\,e^{\lambda_{0}+\mu_{0}}(\rho_{g}-\rho_{0})+e^{\lambda_{0}+\mu_{0}}(\lambda^{\prime}_{0}+\mu^{\prime}_{0})(m_{g}-m_{0})$
$\displaystyle=$ $\displaystyle 4\pi
r^{2}\,e^{\lambda_{0}+\mu_{0}}(\rho_{g}-\rho_{0})+4\pi
r^{2}\,e^{\lambda_{0}+\mu_{0}}e^{2\lambda_{0}}\,\frac{m_{g}-m_{0}}{r}\,(\rho_{0}+p_{0}),$

which yields the claim, if we use Corollary 7.3. $\Box$

Clearly, if it can be shown that $U(r)={\rm const.}$, then the minimizer
$f_{0}$ is a static solution having the form

$f_{0}(r,w,\ell)=\Big{(}1-Ce^{\mu_{0}(r)}\sqrt{1+w^{2}+\ell^{2}/r^{2}}\Big{)}_{+}^{k}.$
(7.9)

The following lemma implies that if the induced density stays away from the
boundary in the set $\mathcal{A}$ and if $f_{0}>0$ on a strip, then $f_{0}$
takes the form (7.9) in this domain.

###### Lemma 7.5

Suppose that $\rho_{0}(r)\leq\sigma_{0}-\delta_{0}$ for some $\delta_{0}>0$.
If

$f_{0}(r,w,\ell)\geq\varepsilon_{0}\quad\mbox{a.e.~{}in}\quad[r_{1},r_{2}]\times[w_{1},w_{2}]\times[\ell_{1},\ell_{2}]$

for some $\varepsilon_{0}>0$, $0<r_{1}<r_{2}<R_{0}$, $w_{1}\leq w_{2}$ and
$0\leq\ell_{1}\leq\ell_{2}$, then

$U(r)={\rm const.}\quad\mbox{a.e.~{}in}\quad[r_{1},r_{2}].$

Proof : For $A>0$ consider

$g=f_{0}+\frac{A}{\sqrt{1+|v|^{2}}}\,{\bf 1}_{\\{|v|\leq
1\\}}\,\frac{\eta^{\prime}(r)}{4\pi r^{2}}\,{\bf 1}_{[w_{1},w_{2}]}(w)\,{\bf
1}_{[\ell_{1},\ell_{2}]}(\ell),$

where $\eta\in C^{1}([0,\infty[)$ is such that $\eta(r)=0$ outside of
$[r_{1},r_{2}]$. Since $r_{1}>0$ we have ${\|\frac{\eta^{\prime}(r)}{4\pi
r^{2}}\|}_{L^{\infty}_{r}}\leq C_{0}$. If $(r,w,\ell)\in
Q=[r_{1},r_{2}]\times[w_{1},w_{2}]\times[\ell_{1},\ell_{2}]$, then
$f_{0}(r,w,\ell)\geq\varepsilon_{0}$. Thus if
$|A|\leq\frac{\varepsilon_{0}}{2C_{0}}$, then $g(r,w,\ell)\geq 0$ in $Q$. On
the other hand, for $(r,w,\ell)\not\in Q$ we get
$g(r,w,\ell)=f_{0}(r,w,\ell)\geq 0$. Integrating against $\sqrt{1+|v|^{2}}$ in
$v$, we obtain

$\rho_{g}(r)=\rho_{0}(r)+AC_{1}\,\frac{\eta^{\prime}(r)}{4\pi r^{2}}$

for $C_{1}=\int{\bf 1}_{\\{|v|\leq 1\\}}\,{\bf 1}_{[w_{1},w_{2}]}(w)\,{\bf
1}_{[\ell_{1},\ell_{2}]}(\ell)\,dv$. Hence if
$|A|\leq\frac{\delta_{0}}{C_{0}C_{1}}$, then $\rho_{g}(r)\leq\sigma_{0}$ for
all $r$. For the mass functions it follows

$m_{g}(r)=m_{0}(r)+AC_{1}\,\int_{0}^{r}\eta^{\prime}(s)\,ds=m_{0}(r)+AC_{1}\eta(r),$

and in particular $\eta(r)=0$ for $r\geq r_{2}$ implies that $g$ has total
mass $M$. In summary, we have shown that $g\in{\cal A}$. If
$f_{0}(r,w,\ell)=0$, then $(r,w,\ell)\not\in Q$ and
$g(r,w,\ell)=f_{0}(r,w,\ell)=0$. Thus (7.8) is applicable and yields

$\int_{0}^{\infty}\frac{d}{dr}(e^{\lambda_{0}+\mu_{0}}\eta)\,U\,dr\geq 0.$

Since $\eta$ can be replaced by $-\eta$, it follows that

$\int_{0}^{\infty}\frac{d}{dr}(e^{\lambda_{0}+\mu_{0}}\eta)\,U\,dr=0$ (7.10)

for all $C^{1}$-function $\eta$ that vanish outside of $[r_{1},r_{2}]$. Let
$\varphi\in C_{0}^{\infty}(]r_{1},r_{2}[)$ be a test function. By extending
$\varphi$ by zero outside of $[r_{1},r_{2}]$, we consider $\varphi$ to be
defined on $[0,\infty[$. Then $\eta=e^{-(\lambda_{0}+\mu_{0})}\varphi$ is a
$C^{1}$-function that vanishes outside of $[r_{1},r_{2}]$. From (7.10) we
therefore deduce

$\int_{r_{1}}^{r_{2}}\varphi^{\prime}\,U\,dr=0$

for all $\varphi\in C_{0}^{\infty}(]r_{1},r_{2}[)$, so that $U(r)={\rm
const.}$ a.e. in $[r_{1},r_{2}]$. $\Box$

Let

$R_{0}=\inf\\{R>0:f_{0}(r,w,\ell)=0\,\,\mbox{for
a.e.}\,\,r\in[R,\infty[,w\in\mathbb{R},\ell\in[0,\infty[\\}.$

Next we show a further support property of $f_{0}$.

###### Lemma 7.6

Suppose that $\rho_{0}(r)\leq\sigma_{0}-\delta_{0}$ for some $\delta_{0}>0$.
Then there is no interval $]a,b[$ such that $0<a<b<R_{0}$ and
$f_{0}(r,w,\ell)=0$ for all a.e.
$(r,w,\ell)\in]a,b[\times\mathbb{R}\times[0,\infty[$.

Proof : Assume, on the contrary, that there is such an interval $]a,b[$. By
moving $b$ closer to $a$ if necessary, we may suppose that
$\delta=b-a\leq\min\\{\frac{1}{2}R_{0},((\frac{\sigma_{0}}{\sigma_{0}-\delta_{0}})^{1/2}-1)a\\}$.
Define $f_{0}^{-}:=f_{0}\,\textbf{1}_{[0,a]}(|x|)$ and
$f_{0}^{+}:=f_{0}\,\textbf{1}_{[b,R_{0}]}(|x|)$ to obtain
$f_{0}=f_{0}^{-}+f_{0}^{+}$. For $r\in[a,\infty[$ define
$f^{+}(r,w,\ell)=f_{0}^{+}(r+\delta,w,\ell)$, i.e.,
$f^{+}=f_{0}\textbf{1}_{[a,R_{0}-\delta]}(|x|)$, and furthermore
$f=f_{0}^{-}+f^{+}$. We also put

$F_{0}(r)=\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\sqrt{1+w^{2}+\ell^{2}/r^{2}}\,f_{0}(r,w,\ell)$

and

$G_{0}(r)=\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\Big{(}\chi(f_{0}(r,w,\ell))-f_{0}(r,w,\ell)\Big{)};$

the functions $F$ and $G$ are defined analogously by replacing $f_{0}$ by $f$.
Note that

$\rho_{0}(r)=\frac{\pi}{r^{2}}\,F_{0}(r),\quad\rho(r)=\frac{\pi}{r^{2}}\,F(r),$
(7.11)

if $\rho$ denotes the density as induced by $f$. If $r\in[0,a]$, then
$f_{0}^{+}(r,\cdot,\cdot)=f^{+}(r,\cdot,\cdot)=0$ a.e. and hence
$F_{0}(r)=F(r)$ as well as $G_{0}(r)=G(r)$. If $r\in[a,\infty[$, then
$f_{0}^{-}(r,\cdot,\cdot)=0$ a.e., so that $F_{0}(r+\delta)=F(r)$ and
$G_{0}(r+\delta)=G(r)$ by definition. For the mass functions, if $r\in[0,a]$,
then $m_{0}(r)=m(r)$ by (7.11). If $r\in[a,\infty[$, then

$\displaystyle m(r)$ $\displaystyle=$ $\displaystyle
m(a)+4\pi^{2}\int_{a}^{r}F(s)\,ds=m_{0}(a)+4\pi^{2}\int_{a}^{r}F_{0}(s+\delta)\,ds$
$\displaystyle=$ $\displaystyle
m_{0}(b)+4\pi^{2}\int_{b}^{r+\delta}F_{0}(s)\,ds=m_{0}(r+\delta).$

Here we used that $m_{0}(a)=m_{0}(b)$ since $f_{0}=0$ on $]a,b[$. In
particular, the induced total masses of $f_{0}$ and $f$ are the same.
Moreover,

$\rho(r)=\frac{\pi}{r^{2}}\,F(r)=\left\\{\begin{array}[]{c@{\quad:\quad}l}\frac{\pi}{r^{2}}\,F_{0}(r)&r\in[0,a]\\\\[4.30554pt]
\frac{\pi}{r^{2}}\,F_{0}(r+\delta)&r\in[a,R_{0}-\delta]\\\\[4.30554pt]
0&r\in[R_{0}-\delta,\infty[\end{array}\right.=\left\\{\begin{array}[]{c@{\quad:\quad}l}\rho_{0}(r)&r\in[0,a]\\\\[4.30554pt]
\frac{(r+\delta)^{2}}{r^{2}}\,\rho_{0}(r+\delta)&r\in[a,R_{0}-\delta]\\\\[4.30554pt]
0&r\in[R_{0}-\delta,\infty[\end{array}\right..$

If $r\geq a$, then
$\frac{(r+\delta)^{2}}{r^{2}}\rho_{0}(r+\delta)\leq\sigma_{0}$ by the choice
of $\delta$ and the assumption on $\rho_{0}$. Therefore we obtain
$\rho(r)\leq\sigma_{0}$ for all $r\in[0,\infty[$, and in turn we have shown
that $f\in{\cal A}$. Lastly, we observe that

$\displaystyle{\cal D}(f)$ $\displaystyle=$ $\displaystyle
4\pi^{2}\int_{0}^{R_{0}-\delta}\frac{G(r)}{\sqrt{1-\frac{2m(r)}{r}}}\,dr$
$\displaystyle=$ $\displaystyle
4\pi^{2}\int_{0}^{a}\frac{G_{0}(r)}{\sqrt{1-\frac{2m_{0}(r)}{r}}}\,dr+4\pi^{2}\int_{a}^{R_{0}-\delta}\frac{G(r)}{\sqrt{1-\frac{2m(r)}{r}}}\,dr=:T_{1}+T_{2}.$

To estimate $T_{2}$, we recall from Theorem 5.10 that $f_{0}\leq 1$, which
implies that $G_{0}$ is non-positive. It follows that

$\displaystyle T_{2}$ $\displaystyle=$ $\displaystyle
4\pi^{2}\int_{a}^{R_{0}-\delta}\frac{G_{0}(r+\delta)}{\sqrt{1-\frac{2m_{0}(r+\delta)}{r}}}\,dr\leq
4\pi^{2}\int_{a}^{R_{0}-\delta}\frac{G_{0}(r+\delta)}{\sqrt{1-\frac{2m_{0}(r+\delta)}{r+\delta}}}\,dr$
(7.12) $\displaystyle=$ $\displaystyle
4\pi^{2}\int_{b}^{R_{0}}\frac{G_{0}(r)}{\sqrt{1-\frac{2m_{0}(r)}{r}}}\,dr.$

Actually if we had equality in (7.12), then $G_{0}(r+\delta)m_{0}(r+\delta)=0$
a.e. in $[a,R_{0}-\delta]$, which means that $G_{0}(r)m_{0}(r)=0$ a.e. in
$[b,R_{0}]$. If there is a $r_{1}\in[b,R_{0}]$ such that $m_{0}(r_{1})=0$,
then there is also a last $r_{0}\in[b,R_{0}]$ with this property; recall that
$m_{0}$ is continuous. From $m_{0}(r_{0})=0$ it then follows that $f_{0}=0$
a.e. in $[0,r_{0}]\times\mathbb{R}\times[0,\infty[$. Now $r_{0}=R_{0}$ is
impossible, since this would yield $f_{0}=0$ a.e., contradicting $I<0$ from
Lemma 4.2. Therefore we have $r_{0}<R_{0}$ and thus $G_{0}(r)=0$ for a.e.
$r\in[r_{0},R_{0}]$, which by virtue of $\chi(f_{0})-f_{0}\leq 0$ in turn
yields $f_{0}=0$ a.e. in $[r_{0},R_{0}]$, and yet a further contradiction,
this time to the definition of $R_{0}$. As a consequence, $m_{0}(r)\neq 0$ for
$r\in[b,R_{0}]$, so that $G_{0}(r)=0$ for a.e. $r\in[b,R_{0}]$ and hence
$f_{0}=0$ a.e. in $[b,R_{0}]\times\mathbb{R}\times[0,\infty[$, which is again
impossible by the definition of $R_{0}$. Thus indeed
$T_{2}<4\pi^{2}\int_{b}^{R_{0}}(1-\frac{2m_{0}(r)}{r})^{-1/2}\,G_{0}(r)\,dr$
in (7.12). Hence we obtain

${\cal
D}(f)<4\pi^{2}\int_{0}^{a}\frac{G_{0}(r)}{\sqrt{1-\frac{2m_{0}(r)}{r}}}\,dr+4\pi^{2}\int_{b}^{R_{0}}\frac{G_{0}(r)}{\sqrt{1-\frac{2m_{0}(r)}{r}}}\,dr={\cal
D}(f_{0}).$

Since $f_{0}$ is a minimizer, this is a contradiction. $\Box$

In the following lemma we show that there are minimizers that are not constant
$\rho_{0}(r)=\sigma_{0}$ on the entire interval $[0,M/\beta]$, if the mass $M$
is small. This indicates that in this case one ultimately might be able to
prove that $\rho_{0}(r)$ stays below $\sigma_{0}$ on the whole interval, which
is in accordance with what we have seen numerically.

###### Lemma 7.7

Let $\beta$ be sufficiently small so that $e^{\lambda}<21/20$. Let
$M=\sqrt{3\beta^{3}/4\pi}$ so that $\sigma_{0}=1$, cf. (1.6). If
$\rho_{0}=\sigma_{0}=1$ on $[0,r_{\ast}]$, then $r_{\ast}<M/\beta$.

Proof : Assume that $\rho_{0}=\sigma_{0}=1$ on $[0,M/\beta]$. Let $K=12/25$.
Since $f_{0}\leq 1$ we have

$\int_{|v|\leq K}\sqrt{1+|v|^{2}}f_{0}\,dv\leq\int_{|v|\leq
K}\sqrt{1+|v|^{2}}\,dv\leq\frac{\sigma_{0}}{2}=\frac{1}{2},$

which implies that

$M_{<}=4\pi\int_{0}^{M/\beta}r^{2}\int_{|v|\leq
K}\sqrt{1+|v|^{2}}f_{0}\,dv\,dr\leq\frac{M}{2}.$

Clearly, we have that $M_{>}=M-M_{<}$. From (3.11) we get

$|\mathcal{D}_{>}|\leq\frac{21}{20}\,\frac{M_{>}}{\sqrt{1+K^{2}}}=\frac{21}{20}\,\frac{(M-M_{<})}{\sqrt{1+K^{2}}}.$

In addition, from (3.12) we have

$|\mathcal{D}_{<}|\leq\frac{21}{20}\,M_{<}\leq\frac{21}{40}\,M.$

Hence we obtain

$|\mathcal{D}|\leq\frac{21}{20}\,M_{<}+\frac{21}{20}\,\frac{(M-M_{<})}{\sqrt{1+K^{2}}}\leq\frac{21}{20}\,\frac{M}{\sqrt{1+K^{2}}}+\frac{21}{40}\,M\Big{(}1-\frac{1}{\sqrt{1+K^{2}}}\Big{)}<0.9983\,M<M,$

which contradicts (CBEC). $\Box$

###### Remark 7.8

In the argument above we used that $f_{0}\leq 1$. In the case of small mass,
one can deduce that $f_{0}\ll 1$, at least on a large set of its support. This
fact can be used to improve the result, so that $\rho<\sigma_{0}$ on a large
part of the interval, but since we anyway have not been able to prove this on
the entire interval we have settled for the result above.

##  8 On the Casimir-binding-energy condition (CBEC)

In this section we discuss the relevance of the Casimir-binding-energy
condition (CBEC) for static solutions, we find strong evidence that static
solutions that are not too relativistic will satisfy this condition. We also
comment on the stability conjecture proposed in the introduction.

Let us start by deriving a couple of general results for static solutions.

###### Lemma 8.1

Suppose that $f_{0}$ is a static solution of the form (7.9), i.e.,
$f_{0}=(1-e^{\mu_{0}(r)-\mu_{0}(R_{0})}{\cal E})_{+}^{k}$ for ${\cal
E}=\sqrt{1+|v|^{2}}=\sqrt{1+w^{2}+\ell^{2}/r^{2}}$. Then

${\cal D}(f_{0})=-4\pi
e^{-\mu_{0}(R_{0})}\int_{0}^{R_{0}}r^{2}(p_{0}+\rho_{0})\,e^{\mu_{0}+\lambda_{0}}\,dr.$

Proof : From the proof of Corollary 7.3 it follows that

$\displaystyle\int_{\mathbb{R}^{3}}(\chi(f_{0})-f_{0})\,dv$ $\displaystyle=$
$\displaystyle\frac{\pi}{r^{2}}\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\frac{d}{dw}\,[(\chi^{\prime}(f_{0})-1)w]\,f_{0}$
$\displaystyle=$
$\displaystyle-\frac{\pi}{r^{2}}\,e^{\mu_{0}(r)-\mu_{0}(R_{0})}\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\frac{d}{dw}\,[{\cal
E}w]\,f_{0}$ $\displaystyle=$
$\displaystyle-\frac{\pi}{r^{2}}\,e^{\mu_{0}(r)-\mu_{0}(R_{0})}\int_{0}^{\infty}d\ell^{2}\int_{\mathbb{R}}dw\,\Big{(}\frac{w^{2}}{\cal
E}+{\cal E}\Big{)}\,f_{0}$ $\displaystyle=$
$\displaystyle-(p_{0}+\rho_{0})\,e^{\mu_{0}(r)-\mu_{0}(R_{0})}.$

The claim follows by inserting this expression into the formula for ${\cal
D}$. $\Box$

###### Lemma 8.2

Suppose that $f_{0}=(1-e^{\mu_{0}(r)-\mu_{0}(R_{0})}{\cal E})_{+}^{k}$. Then

$\frac{1}{2}\,e^{-\mu_{0}(R_{0})}M\leq|{\cal D}(f_{0})|\leq
e^{-\mu_{0}(R_{0})}M.$

Proof : From [1, equ. (10)] it follows that

$4\pi\int_{0}^{R_{0}}r^{2}(\rho_{0}+p_{0}+q_{0})e^{\mu_{0}+\lambda_{0}}\,dr=M.$
(8.1)

Here $q_{0}$ denotes twice the tangential pressure. From this relation and
Lemma 8.1 we can derive bounds for $|{\cal D}(f_{0})|$. We immediately have

$|{\cal D}(f_{0})|\leq e^{-\mu_{0}(R_{0})}M,$

and furthermore, since $\rho_{0}\geq p_{0}+q_{0}$,

$\displaystyle|{\cal D}(f_{0})|$ $\displaystyle\geq$ $\displaystyle 4\pi
e^{-\mu_{0}(R_{0})}\int_{0}^{R_{0}}r^{2}\rho_{0}\,e^{\mu_{0}+\lambda_{0}}\,dr$
$\displaystyle\geq$ $\displaystyle 2\pi
e^{-\mu_{0}(R_{0})}\int_{0}^{R_{0}}r^{2}(\rho_{0}+p_{0}+q_{0})\,e^{\mu_{0}+\lambda_{0}}\,dr$
$\displaystyle=$ $\displaystyle\frac{1}{2}\,e^{-\mu_{0}(R_{0})}M,$

which completes the argument. $\Box$

We note that in our case the pressure is isotropic and we have $q_{0}=2p_{0}$,
so that

$4\pi\int_{0}^{R_{0}}r^{2}(\rho_{0}+3p_{0})\,e^{\mu_{0}+\lambda_{0}}\,dr=M.$

From Lemma 8.2 it first seems that highly relativistic solutions for which

$e^{-\mu_{0}(R_{0})}=\frac{1}{1-\frac{2M}{R_{0}}}$

is large should be good candidates to satisfy the (CBEC). Note that
$2M/R_{0}<8/9$ for any static solution of the Einstein-Vlasov system, cf. [3],
whereas for isotropic solutions an analytic bound is not known, but numerical
simulations indicate that $2M/R_{0}<1/2$ in that case. We now argue why it is
misleading that highly relativistic solutions are the best candidates. For,
let us consider highly relativistic isotropic solutions where the extreme case
$2M/R_{0}=1/2$ is attained. Then Lemma 8.2 gives the lower bound

$|{\cal D}(f_{0})|\geq\frac{1}{2}\,e^{-\mu_{0}(R_{0})}M=\frac{M}{\sqrt{2}}.$

On the other hand, for static solutions that are not very relativistic, but
close to being Newtonian, we have $p_{0}\ll\rho_{0}$, so that

$\displaystyle|{\cal D}(f_{0})|$ $\displaystyle=$ $\displaystyle 4\pi
e^{-\mu_{0}(R_{0})}\int_{0}^{R_{0}}r^{2}(p_{0}+\rho_{0})\,e^{\mu_{0}+\lambda_{0}}\,dr$
$\displaystyle\simeq
e^{-\mu_{0}(R_{0})}4\pi\int_{0}^{R_{0}}r^{2}(\rho_{0}+3p_{0})\,e^{\mu_{0}+\lambda_{0}}\,dr=e^{-\mu_{0}(R_{0})}M>M.$

Hence a static solution that is not too relativistic is a good candidate for
satisfying the (CBEC). Now, it is straightforward to numerically check (CBEC)
on particular static solutions, and indeed we find that solutions that are not
too relativistic do verify the condition.

In the introduction we conjecture that isotropic static solutions which
satisfy the (CBEC) are stable with respect to mass-preserving perturbations.
An interesting observation is that when $k\ll 1$, the Casimir-binding energy
$E_{Cb}$ and the binding energy $E_{b}$ almost coincide. If the conjecture is
true it indicates that when $k=0$, the sign of $E_{b}$ alone determines
whether a static solution is stable or unstable. The first author
investigated, together with Gerhard Rein, stability of static shells in [6].
(Note that the conjecture we pose concerns isotropic solutions rather than
shells, which are anisotropic, but a comparison is nevertheless relevant.) In
[6] the perturbations had the form $f=Af_{0}$, where $A<1$ or $A>1$, which we
here call amplitude-perturbations. From the results in Table 1 in [6] it
follows that the sign of $E_{b}$ does not determine stability in the case of
amplitude-perturbations; if $A>1$ there are several cases when the static
solution is unstable even if $E_{b}>0$. (The binding energy in [6] is in fact
the fractional binding energy but the sign is the same as of $E_{b}$.) Now,
the case $A>1$ does not preserve the mass and such a perturbation is outside
the scope of our conjecture. In a very limited study we have investigated the
case $k=0$ for shells and we do find numerical support that for mass-
preserving perturbations the sign of $E_{b}$ alone determines the stability
properties. An interesting aspect of this is that in the numerical study [8],
another type of perturbations were considered which are called dynamically
accessible perturbations. The results in [8] indicate that the stability
properties for dynamically accessible perturbations and amplitude-
perturbations are similar in the following sense: if an amplitude-perturbation
with $A>1$ is unstable then a dynamically accessible perturbation is also
unstable if the perturbation is such that the particles of the perturbation
initially go inward, i.e. the steady state is pushed towards collapse rather
than towards dispersion. In our situation with mass-preserving perturbations
we have therefore tried perturbations of both types, i.e. perturbations that
push towards collapse and towards dispersion. In both cases we find numerical
support that a shell with $k=0$, and such that $E_{b}>0$, is stable in
accordance with our conjecture. This indicates an essential difference between
mass-preserving perturbations and dynamically accessible perturbations. A more
extensive numerical study will be carried out in future work to fully
understand this intriguing issue.

##  9 Some technical results

###  9.1 A convexity property

The following observation is a crucial ingredient of the minimization