# The limit point in the Jante’s law process has a continuous distribution

Edward Crane111School of Mathematics, University of Bristol, BS8 1TH, UK and
Stanislav Volkov222Centre for Mathematical Sciences, Lund University, Box 118
SE-22100, Lund, Sweden

###### Abstract

We study a stochastic model of consensus formation, introduced in 2015 by
Grinfeld, Volkov and Wade, who called it a _multidimensional randomized
Keynesian beauty contest_. The model was generalized by Kennerberg and Volkov,
who called their generalization the _Jante’s law process_. We consider a
version of the model where the space of possible opinions is a convex body
$\mathscr{B}$ in $\mathbb{R}^{d}$. $N$ individuals in a population each hold a
(multidimensional) opinion in $\mathscr{B}$. Repeatedly, the individual whose
opinion is furthest from the center of mass of the $N$ current opinions
chooses a new opinion, sampled uniformly at random from $\mathscr{B}$.
Kennerberg and Volkov showed that the set of opinions that are not furthest
from the center of mass converges to a random limit point. We show that the
distribution of the limit opinion is continuous, thus proving the conjecture
made after Proposition 3.2 in Grinfeld et al.

Keywords: Jante’s law process, consensus formation, Keynesian beauty contest,
rank-driven process, interacting particle system

Subject classification: 60J05, 60D05, secondary 60K35

##  1 Introduction

###  1.1 The multidimensional randomized Keynesian beauty contest

Let $N\geq 3$ and $d\geq 1$ be integers. Let $\lambda$ denote Lebesgue measure
on $\mathbb{R}^{d}$ and let $\mathscr{B}$ be a _convex body_ , i.e. a closed
convex subset of $\mathbb{R}^{d}$ with non-empty interior. In particular,
$\lambda(\mathscr{B})>0$. For the moment, while describing the model studied
in [1], we will also assume that $\mathscr{B}$ is bounded, so that
$\lambda(\mathscr{B})<\infty$, but later we will drop this assumption. We say
that a random variable is uniform on $\mathscr{B}$, i.e. it is a
$U(\mathscr{B})$ random variable, when its distribution is proportional to
$\lambda|_{\mathscr{B}}$. (This notion requires $\lambda(\mathscr{B})<\infty$,
of course.)

A discrete time Markov process $X(t)_{t=0}^{\infty}$, taking values in
$([0,1]^{d})^{N}$, called the _multidimensional randomized Keynesian beauty
contest_ was studied in [1]. In the sequel [2] a generalization of the process
was studied which the authors called the _Jante’s law process_ 333The _law of
Jante_ is literary caricature of the virtue in Scandinavian culture of not
standing out from the crowd. It appeared in Aksel Sandemose’s satirical novel
_En flyktning krysser sitt spor_ , (A fugitive crosses his tracks), published
in 1933. (see §1.4). In the present paper the process $X(t)$ will take values
in $\mathscr{B}^{N}$, so our setting is only a little more general than that
of [1]; however we prefer to retain the shorter name for the process.

If $X$ is either a finite sequence $(x_{1},\dots,x_{m})$ of $m\geq 1$ points
of $\mathbb{R}^{d}$, or a set $\\{x_{1},\dots,x_{m}\\}$ of $m\geq 1$ distinct
points of $\mathbb{R}^{d}$, then we denote by $\mu(X)$ its center of mass,
i.e.

$\mu(X)=\frac{1}{m}(x_{1}+\dots+x_{m})\,.$

We denote by $\|x-y\|=\sqrt{\sum_{i=1}^{d}(x_{i}-y_{i})^{2}}$ the Euclidean
distance between $x$ and $y$ in $\mathbb{R}^{d}$, and by

$\mathrm{B}(x,\rho)=\\{y\in{\mathbb{R}}^{d}:\ \|x-y\|<\rho\\}$

the open ball of radius $\rho$ centered at $x$.

Let $X(0)=(X_{1}(0),\dots,X_{N}(0))\in\mathscr{B}^{N}$ be a possibly random
initial state. We will always assume that a.s. the $N$ points of $X(0)$ are
distinct. An informal description of each step of the Markov chain $X(t)$ is
that a new point arrives, distributed uniformly in $\mathscr{B}$, so that now
there are $N+1$ points, and then the point among these $N+1$ that is furthest
from the new center of mass is thrown out, so that $N$ points remain.

Formally, let $\left(\zeta_{t}\right)_{t=1}^{\infty}$ be an i.i.d. sequence of
random variables uniformly distributed on $\mathscr{B}$, denoted by
$U(\mathscr{B})$, independent of the initial state $X(0)$. For each time
$t=0,1,2,\dots$ in turn, let $j(t)$ be the index of the furthest point of
$X(t)$ from $\mu(X(t))$, that is, $j(t)$ is defined by

$\|X_{j(t)}(t)-\mu(X(t))\|=\max\left\\{\|X_{i}(t)-\mu(X(t))\|:1\leq i\leq
N\right\\}\,;$

in case of a tie, choose $j(t)$ uniformly at random among the tied
indices444Any other way of breaking ties would also be acceptable, since a.s.
ties do not occur.. Define $X(t+1)$ by

$X_{i}(t+1)=\begin{cases}X_{i}(t),&\text{if
$i\in\\{1,\dots,N\\}\setminus\\{j_{t}\\}$},\\\ \zeta_{t},&\text{if
$i=j_{t}$},\end{cases}$

i.e., the point of $X(t)$ that is furthest from the center of mass of
$(X_{1}(t),\dots,X_{N}(t))$ is removed and its place is taken by the new point
$\zeta_{t}$. Note that although $X(t+1)$ and $\mu(X(t+1))$ depend continuously
on the new point $\zeta_{t}$, they depend discontinuously on $\zeta_{t-1}$.

Let $X^{\prime}(t)$ denote the core of the configuration $X(t)$, that is the
_set_ $\\{X_{i}(t):i\neq j_{t}\\}$. It has $M:=N-1$ points.

######  Definition 1.

We say that $X^{\prime}(t)$ converges to some point $\xi\in{\mathbb{R}}^{d}$
if

$\text{for all }\varepsilon>0\text{ there is }t_{\varepsilon}<\infty\text{
such that }X^{\prime}(t)\subset B(\xi,\varepsilon)\text{ for all
}t>t_{\varepsilon}\,.$

It was shown in [1] that $X^{\prime}(t)$ converges a.s. to a random limit
point $\xi$ in the above sense; additionally, in the case $d=1$, the
distribution of $\xi$ has support $[0,1]$. It was conjectured555The conjecture
is stated immediately after [1, Proposition 3.2]. that the distribution of
$\xi$ is, in fact, continuous. The purpose of this paper is to prove this
continuity conjecture in the $d$-dimensional setting described above.

A.s. for every $t\geq 0$, $X(t)$ is a sequence of $N$ distinct points, and it
will be convenient to discard from the underlying probability space the null
set where this fails. Hence, from now on, we work with a version of the
process where for each $t$ the points of $X(t)$ are surely distinct. In
particular, the core process $X^{\prime}(t)$ a.s. takes its values in the set
of subsets of $\mathscr{B}$ of cardinality $M$. Assuming $X^{\prime}(t)=S$, we
can ask what is the set of possible values of $\zeta_{t}$ which would result
in $\zeta_{t}\in X^{\prime}(t+1)$, that is, that we keep the newly sampled
point in the core while throwing out one of the original points. The answer
depends only on $S$, and not on the point $X_{j_{t}}(t)$. The set of possible
locations which could enter the core at time $t+1$ is denoted by

$\mathrm{Keep}(S;\mathscr{B})\;:=\;\\{x\in\mathscr{B}\,:\,\|x-\mu(\\{x\\}\cup
S)\|<\max(\\{\|s-\mu(\\{x\\}\cup S)\|:s\in S\\})\\}.$ (1)

For later use, we extend the definition (1) to allow $\mathscr{B}$ to be
unbounded.

Figure 1: $d=2$ and $M=3$. In this example, the core consists of points
$(0,0)$, $(10,0)$, and $(4,6)$. A newly sampled point will join the core if
and only if it lies in one of the three shaded areas; the colors indicate
which point will be removed.

Observe that in the one-dimensional case $\mathscr{B}$ is necessarily an
interval $[a,b]\subset{\mathbb{R}}$; and then $\mathrm{Keep}(S;\mathscr{B})$
is also a non-empty subinterval of $[a,b]$. If $d\geq 2$ the geometry of this
set is a little more complicated, see e.g. Figure 1. Later, in Lemma 2, we
will show that $\mathrm{Keep}(S;\mathscr{B})$ is in fact always the
intersection of $\mathscr{B}$ with a non-empty union of at most $M$ bounded
open balls which only depend on $S$. Moreover, $\mathrm{Keep}(S,\mathscr{B})$
always contains a nonempty open ball whose center is $\mu(S)$. As a result,
$\mathrm{Keep}(S;\mathscr{B})$ always has finite and positive Lebesgue
measure, so that it makes sense to sample a point uniformly from
$\mathrm{Keep}(S;\mathscr{B})$. In fact, conditional on $X^{\prime}(t)=S$ and
$X^{\prime}(t+1)\neq X^{\prime}(t)$, the new point $\zeta_{t}$ is uniformly
distributed on $\mathrm{Keep}(S;\mathscr{B})$.

###  1.2 Time-changed models

For convenience, we will work with a time-change of the process
$X^{\prime}(t)$, namely the sequence $Y(0),Y(1),Y(2),\dots$ defined by
$Y(0)=X^{\prime}(0)$ and for $n\geq 1$, $Y(n)=X^{\prime}(t_{n})$, where
$t_{1}<t_{2}<t_{3}<\dots$ is the increasing sequence of all times $t\geq 1$
for which $X^{\prime}(t)\neq X^{\prime}(t-1)$. The sequence $Y(n)_{n\geq 0}$
is itself a Markov chain. It is an instance of a $\mathscr{B}$-valued Jante’s
law process (defined below). When studying $Y(\cdot)$, there is no longer any
reason to insist that $\mathscr{B}$ be bounded, since we can describe the law
of $Y(\cdot)$ without making any reference to $U(\mathscr{B})$ random
variables. Instead, the definition involves sampling uniformly from sets of
the form $\mathrm{Keep}(S;\mathscr{B})$, which always have finite positive
Lebesgue measure, regardless of whether $\mathscr{B}$ is bounded. So in the
following definition and for the rest of the paper we allow $\mathscr{B}$ to
be unbounded.

######  Definition 2 ($\mathscr{B}$-valued Jante’s law process
$Y(n)_{n\in\mathbb{N}_{0}}$).

Let $\mathscr{B}$ be any closed convex subset of $\mathbb{R}^{d}$ with non-
empty interior. Let $Y(0)$ be any (possibly random) set of $M\geq 2$ distinct
points in $\mathscr{B}$. For each $n\geq 1$, sample a point $y_{n}$ uniformly
from $\mathrm{Keep}(Y(n-1);\mathscr{B})$, conditionally independently of
$(Y(i):0\leq i<n-1)$ given Y(n-1), and set

$\displaystyle Y(n):=\\{y_{n}\\}\cup Y(n-1)\setminus\\{r_{n}\\}\,,$ (2)

where $r_{n}$ is the (a.s. unique) point $r\in Y(n-1)$ which maximizes
$\|r-\mu(Y(n-1)\cup\\{y\\})\|$.

By definition, a $\mathscr{B}$-valued Jante’s law process is a Markov chain.
When working with $\mathscr{B}$-valued Jante’s law processes, it is convenient
to discard from our probability space the null set where for any $n$ either
there is a tie in choosing $r_{n}$, or $y_{n}\in\bigcup_{i=0}^{n-1}Y(i)$. From
the definition of $\mathrm{Keep}(S;\mathscr{B})$, it is surely the case that
$r_{n}\neq y_{n}$, so $Y(n)\neq Y(n-1)$.

By a slight modification of the proof in [1] we will show that even when
$\mathscr{B}$ is unbounded, the $\mathscr{B}$-valued Jante’s law process
$Y(\cdot)$ a.s. converges to a random limit point $\xi$, in the sense that

$(\forall\varepsilon>0)(\exists
n_{\varepsilon}<\infty)(n>n_{\varepsilon}\implies Y(n)\subset
B(\xi,\varepsilon))\,.$

Our strategy for proving that the limit point $\xi$ is a continuous random
variable is to compare $Y(\cdot)$ with an $\mathbb{R}^{d}$-valued Jante’s law
process denoted by $Z(n)_{n\in\mathbb{N}_{0}}$; we call the Markov chain
$Z(\cdot)$ the _scale-free core process_. It eliminates the complicated effect
of the boundary of $\mathscr{B}$ and has the great advantage of being
invariant in law with respect to scaling and translation.666An analogous
modification of the original process $X(t)$ was already used as a tool in [1,
§3.4]. It will be useful to have notation for the points added and removed at
each step of the scale-free core process $Z(n)$:

$Z(n)=\\{z_{n}\\}\cup(Z(n-1)\setminus\\{s_{n}\\})\,,$

Also let $z_{\infty}$ denote the a.s. limit of the process $Z(n)$, in the
sense of Definition 1.

###  1.3 Outline and results

In §2 we collect a number of basic results, including the fact that every
$\mathscr{B}$-valued Jante process almost surely converges to a random limit
(Proposition 1). We give a standalone proof because for the case where
$\mathscr{B}$ is unbounded this is not a special case of the results of [1,
2], and because the required lemmas are needed again later in the paper.

In §3 we discuss a property of convex bodies called _uniform geometry_. This
property is used later in the paper to avoid some geometric complications.
Bounded convex sets have uniform geometry. We reduce the continuity of the
distribution of the limit point of a $\mathscr{B}$-valued Jante process to the
case where $\mathscr{B}$ is bounded.

Our first continuity result is for the scale-free core process:

######  Theorem 1.

For an arbitrary deterministic initial condition $Z(0)=S$, where $S$ is any
set of $M$ distinct points in $\mathbb{R}^{d}$, the distribution $\pi_{S}$ of
$z_{\infty}$ is continuous.

Our proof of Theorem 1 is less straightforward than one might expect. The main
difficulty is to show that a.s. all of the original points are eventually
removed; we isolate this statement as Proposition 2 and prove it in §4 using a
supermartingale argument, a trick using translational symmetry, and a
compactness argument. Then in §5 we complete the proof of Theorem 1 by showing
that the distribution of the limit is a mixture of continuous distributions.
Each distribution in the mixture is seen to be continuous because it is a
convolution with a uniform distribution over a set of positive Lebesgue
measure.

In §6 we use another supermartingale argument and a coupling argument to
deduce the continuity of the distribution of the limit point for any
$\mathscr{B}$-valued Jante’s law process.

######  Theorem 2.

Let $\mathscr{B}\subset\mathbb{R}^{d}$ be any convex body, and let $S$ be any
set of $M$ distinct points of $\mathscr{B}$. Let $\xi$ be the limit point of
the $\mathscr{B}$-valued Jante’s law process $Y(\cdot)$ started at $Y(0)=S$.
Then the distribution of $\xi$ is continuous.

Of course this also shows that the distribution of the random limit $\xi$ of
the original core process $X^{\prime}(t)$ started at $X^{\prime}(0)=S$ has a
continuous distribution. Theorem 2 covers the original setting of [1] where
$\mathscr{B}=[0,1]^{d}$, so it resolves the continuous distribution conjecture
made in that paper. We remark that if $X^{\prime}(0)$ is random, then the
distribution of the limit point $\xi$ is still continuous, since it is a
mixture of such distributions. We also remark that
$\mathbb{P}(\xi\in\partial\mathscr{B})=0$ because
$\lambda(\partial\mathscr{B})=0$ for any convex body
$\mathscr{B}\subseteq\mathbb{R}^{d}$.

###  1.4 Related models

In the paper [2], the authors generalized the original model of [1] by
allowing the common distribution of the new points $\zeta_{t}$ to be an
arbitrary continuous distribution on $\mathbb{R}^{d}$. They also allowed for
more than one point to be replaced simultaneously. It was shown that if the
distribution of $\zeta_{t}$ has bounded support then a.s. a limit point $\xi$
exists. However, although the support of $\xi$ is certainly a subset of the
support of $\zeta$, it may not be the whole of the support of $\zeta$. For
example, if $\zeta_{t}$ is uniform on $[0,1]\cup[2,3]$ and a.s. all
coordinates of $X(0)$ lie in $[0,1]$, then $X^{\prime}(t)\subset[0,1]$ for all
$t$, and after passing to the subsequence of times at which
$\zeta_{t}\in[0,1]$, we recover the original process with $\zeta\sim
U([0,1])$. In particular, in this case the support of $\xi$ is $[0,1]$. In
case $\zeta$ has unbounded support, [2] shows that a.s. either $X^{\prime}(t)$
converges to a random limit point $\xi$ _or_ $X^{\prime}(t)\to\infty$, i.e.
$\inf\\{|x|:x\in X^{\prime}(t)\\}\to\infty$. However, no examples of
distributions of $\zeta_{t}$ are known to satisfy $X^{\prime}(t)\to\infty$
with positive probability. For an arbitrary common distribution of the random
variables $\zeta_{t}$, and an arbitrary distribution of $X(0)$ on
$\mathbb{R}^{d}$, such that there is a positive probability that $X(t)$
converges, one can ask whether the distribution of the limit $\xi$ conditional
on its existence is absolutely continuous with respect to the distribution of
the $\zeta_{t}$. We do not address this more general continuity problem in
this paper, although it might be possible to extend our techniques to other
cases where $\zeta_{t}$ has a continuous distribution with a sufficiently
well-behaved density.

In another paper [3], the authors studied asymptotic properties of a modified
version of Jante’s law process. In this model, called the _$p$ -contest_, at
each moment of time the point farthest from $p\mu$, where $\mu$ is the current
centre of mass, is removed and replaced by an independent $\zeta$-distributed
point; note that the case $p=1$ would correspond to the original Jante’s law
process. Finally, in [4], a local version of the process was considered. In
this version, $N$ vertices are placed on a circle, so that each vertex has
exactly two neighbours. To each vertex assign a real number, called fitness.
At each unit of time the vertex whose fitness deviates most from the average
of the fitnesses of its two immediate neighbours has its fitness replaced by a
random value drawn independently according to some distribution $\zeta$. The
authors showed that if $\zeta$ has a uniform or a discrete uniform
distribution, all fitnesses but at most one converge to the same limit. We
would also like to mention that there is a related continuous-time model,
called “Brownian bees”, which involves a fixed number $N$ of particles which
perform independent Brownian motions, and branch from time to time
simultaneously with removal of the particle most distant from the origin, thus
keeping the total number of particles constant; see [6].

##  2 Preliminaries

For any set $X\subset\mathbb{R}^{d}$ we denote by $\mathrm{Conv}(X)$ the
convex hull of $X$, i.e. the smallest convex set containing $X$. We denote by
$\partial X$ the boundary of $X$ and by $X^{\circ}$ the interior of $X$. For
any $\delta>0$ we denote by $N_{\delta}(X)$ the $\delta$-neighbourhood of $X$,
i.e. the set of points whose Euclidean distance to $X$ is less than $\delta$.

Some basic facts about continuity of $\mathbb{R}^{d}$-valued random variables
and continuity of Borel probability measures on $\mathbb{R}^{d}$ are collected
in the appendix. The main two facts that we will use are that any mixture of
continuous random variables is continuous, and that if a continuous random
variable is conditioned on an event of positive probability then the
conditioned random variable is continuous. We will see a simple argument that
uses these two facts in the proof of Lemma 10 at the end of section 3.

For any set $X=\\{x_{1},\dots,x_{M}\\}\subset\mathbb{R}^{d}$ of cardinality
$M=N-1$ we let

$\Sigma(X)=M\mu(X)=\sum_{i=1}^{M}x_{i}$

and we define the following geometric functionals of $X$:

$\displaystyle\begin{array}[]{rclrcl}F(X)&=&\displaystyle\sum_{1\leq i<j\leq
M}\|x_{i}-x_{j}\|^{2}\,=\,M\sum_{i=1}^{M}\|x_{i}-\mu(X)\|^{2}\,,&d(X)&=&\displaystyle\min_{i\neq
j}\|x_{i}-x_{j}\|\,,\\\ A(X)&=&\max_{w\in X}\|w-\mu(X)\|\,,&D(X)&=&\max_{i\neq
j}\|x_{i}-x_{j}\|\,.\end{array}$ (5)

We call $F(X)$ the moment of inertia of $X$. The functional $F$ serves as a
Lyapunov function for the Jante’s law processes $Y(\cdot)$ and $Z(\cdot)$. One
can easily verify the following inequalities among the geometric functionals:

$\displaystyle\begin{split}\frac{M}{M-1}\,A(X)&\leq D(X)\leq 2A(X),\\\
\frac{M(M-1)}{2}\,d(X)^{2}&\leq F(X)\leq\frac{M(M-1)}{2}\,D(X)^{2},\\\
\frac{M^{2}}{(M-1)}\,A(X)^{2}&\leq F(X)\leq M^{2}A(X)^{2}.\end{split}$ (6)

######  Lemma 1.

Let $X=\\{x_{1},\dots,x_{M}\\}$ be any set of $M\geq 2$ distinct points in
$\mathbb{R}^{d}$, and suppose $z\in\mathbb{R}^{d}\setminus X$. Then

$\displaystyle F(X)-F\left(\\{z\\}\cup X\setminus\\{x_{j}\\}\right)$
$\displaystyle=$
$\displaystyle(M+1)\left(\left\|x_{j}-\frac{z+\Sigma(X)}{M+1}\right\|^{2}-\left\|z-\frac{z+\Sigma(X)}{M+1}\right\|^{2}\right)$
(7) $\displaystyle=$
$\displaystyle(M-1)\left(\left\|x_{j}-\frac{\Sigma(X)-x_{j}}{M-1}\right\|^{2}-\left\|z-\frac{\Sigma(X)-x_{j}}{M-1}\right\|^{2}\right)\,.$
(8)

From (7) it follows that a point $z\in\mathscr{B}\setminus X$ belongs to
$\mathrm{Keep}(X;\mathscr{B})$ if and only if there exists $x\in X$ such that
$F(\\{z\\}\cup(X\setminus\\{x\\}))<F(X)$. Moreover, if
$z\in\mathrm{Keep}(X;\mathscr{B})$ then any choice of $j$ which maximizes
$\|x_{j}-\frac{\Sigma(X)+z}{M+1}\|$ is also a choice which minimizes
$F(\\{z\\}\cup(X\setminus\\{x_{j}\\}))$.

###### Proof.

We have

$\displaystyle F(X)-F(\\{z\\}\cup(X\setminus\\{x_{j}\\}))$ $\displaystyle=$
$\displaystyle\sum_{i:i\neq j}\|x_{j}-x_{i}\|^{2}-\sum_{i:i\neq
j}\|z-x_{i}\|^{2}$ $\displaystyle=$ $\displaystyle\sum_{i:i\neq
j}(\|x_{j}\|^{2}-\|z\|^{2}-2(x_{j}-z)\cdot x_{i})$ $\displaystyle=$
$\displaystyle(M-1)(x_{j}-z)\cdot(x_{j}+z)-2(x_{j}-z)\cdot(\Sigma(X)-x_{j})$
$\displaystyle=$
$\displaystyle(x_{j}-z)\cdot((M+1)(x_{j}+z)-2z-2\Sigma(X))\,.$

On the other hand, expanding the squared distances on the RHS of (7) as dot
products, the terms $\pm(M+1)\left\|\frac{z+\Sigma(X)}{M+1}\right\|^{2}$
cancel, and we are left with the same quantity:

$(M+1)(\|x_{j}\|^{2}-\|z\|^{2})-2(x_{j}-z)\cdot(z+\Sigma(X))=(x_{j}-z)\cdot((M+1)(x_{j}+z)-2z-2\Sigma(X))\,.$

The proof of (8) is a similar calculation. The remaining statements follow
immediately from (7). ∎

######  Corollary 1 (Essentially Lemma 2.1 in [1]).

Let $Y(\cdot)$ be a $\mathscr{B}$-valued Jante’s law process with $M\geq 2$
(where $\mathscr{B}$ may be ${\mathbb{R}}^{d}$). Then $F(Y(n))<F(Y(n-1))$ for
each $n\geq 1$.

######  Lemma 2.

Let $X=\\{x_{1},\dots,x_{M}\\}\subset\mathscr{B}$, where $M\geq 2$. Then

$\mathrm{Keep}(X;\mathscr{B})=\mathscr{B}\cap\bigcup_{j=1}^{M}B\left(\frac{\Sigma(X)-x_{j}}{M-1}\,,\,\frac{M}{M-1}\|x_{j}-\mu(X)\|\right)\,,$
(9)

###### Proof.

Let $X=\\{x_{1},\dots,x_{M}\\}$. For any point $z\in\mathbb{R}^{d}$,
$z\in\mathrm{Keep}(X;\mathbb{R}^{d})$ if and only if there exists
$j\in\\{1,\dots,M\\}$ such that

$\left\|z-\frac{z+\Sigma(X)}{M+1}\right\|<\left\|x_{j}-\frac{z+\Sigma(X)}{M+1}\right\|\,.$
(10)

Straightforward algebraic manipulation shows that each of the following three
inequalities is equivalent to (10).

$\displaystyle\begin{split}\left\|Mz-\Sigma(X)\right\|&<\left\|(M+1)x_{j}-z-\Sigma(X)\right\|\\\
\sum_{i:i\neq j}\|z-x_{i}\|^{2}&<\sum_{i:i\neq j}\|x_{j}-x_{i}\|^{2}\\\
\left\|z-\frac{\Sigma(X)-x_{j}}{M-1}\right\|&<\left\|x_{j}-\frac{\Sigma(X)-x_{j}}{M-1}\right\|=\frac{M}{M-1}\|x_{j}-\mu(X)\|\,.\end{split}$
(11)

The expression (9) follows from the last of these inequalities. ∎

######  Lemma 3.

Let $X=\\{x_{1},\dots,x_{M}\\}\subset\mathscr{B}$, where $M\geq 2$. Then

$\mathrm{Keep}(X;\mathscr{B})\subseteq\mathscr{B}\cap
B\left(\mu(X),\frac{M+1}{M-1}A(X)\right)\,.$ (12)

###### Proof.

For every $j\in\\{1,\dots,M\\}$ we have

$\left\|\mu(X)-\frac{\Sigma(X)-x_{j}}{M-1}\right\|=\frac{1}{M-1}\|x_{j}-\mu(X)\|\leq\frac{1}{M-1}A(X)\,.$
(13)

Suppose $z\in\mathrm{Keep}\left(X;\mathbb{R}^{d}\right)$ and let $j$ be such
that the equivalent inequalities (10)-(11) hold. Then by (11) we have

$\left\|z-\frac{\Sigma(X)-x_{j}}{M-1}\right\|<\frac{M}{M-1}\|x_{j}-\mu(X)\|\leq\frac{M}{M-1}A(X)\,.$
(14)

Summing the inequalities (13) and (14), and using the triangle inequality, we
obtain

$\|z-\mu(X)\|\leq\left\|z-\frac{\Sigma(X)-x_{j}}{M-1}\right\|+\left\|\mu(X)-\frac{\Sigma(X)-x_{j}}{M-1}\right\|<\frac{M+1}{M-1}\cdot
A(X)\,,$

i.e. $z\in B\left(\mu(X),\frac{M+1}{M-1}\,A(X)\right)$. ∎

######  Lemma 4.

Let $X=\\{x_{1},\dots,x_{M}\\}\subset\mathscr{B}$, where $M\geq 2$ and the
$x_{i}$ are distinct. Then

$\mathscr{B}\cap
B\left(\mu(X),A(X)\right)\subseteq\mathrm{Keep}(X;\mathscr{B})\,.$ (15)

###### Proof.

Let $x_{j}$ be a point of $X$ furthest from $\mu(X)$. Then $\mu(X)$ lies on
the line segment joining $\frac{\Sigma(X)-x_{j}}{M-1}$ and $x_{j}$, since
$\mu(X)=\left(1-\frac{1}{M}\right)\left(\frac{\Sigma(X)-x_{j}}{M-1}\right)+\frac{1}{M}x_{j}$.
We have $\|\mu(X)-x_{j}\|=A(X)$ and
$\left\|\frac{\Sigma(X)-x_{j}}{M-1}-x_{j}\right\|=\left\|\frac{M\mu(X)}{M-1}-\frac{M}{M-1}x_{j}\right\|=\frac{M}{M-1}A(X)$.
Therefore

$B(\mu(X),A(X))\subseteq
B\left(\frac{\Sigma(X)-x_{j}}{M-1},\frac{M}{M-1}\|x_{j}-\mu(X)\|\right)\,.$

Indeed, the boundary spheres of these two balls are tangent at the point
$x_{j}$. The result then follows from Lemma 2. ∎

######  Lemma 5.

Let $X=\\{x_{1},\dots,x_{M}\\}\subset\mathscr{B}$, where $M\geq 2$ and the
$x_{i}$ are distinct. For each $x_{i}\in X$ we have

$\mathscr{B}\cap
B\left(\mu(X\setminus\\{x_{i}\\}),\frac{M}{M+1}A(X)\right)\subseteq\mathrm{Keep}(X;\mathscr{B})\,.$

###### Proof.

$\mu(X\setminus\\{x_{i}\\})=\frac{\Sigma(X)-x_{i}}{M-1}$ so
$\|\mu(X)-\mu(X\setminus\\{x_{i}\\})\|=\frac{1}{M-1}\|\mu(X)-x_{i}\|$.
Therefore by Lemma 4 we have

$\mathscr{B}\cap
B\left(\mu(X\setminus\\{x_{i}\\}),A(X)-\frac{1}{M-1}\|\mu(X)-x_{i}\|\right)\subseteq\mathrm{Keep}(X;\mathscr{B})\,.$

On the other hand from Lemma 2 we have

$\mathscr{B}\cap
B\left(\mu(X\setminus\\{x_{i}\\}),\frac{M}{M-1}\|\mu(X)-x_{i}\|\right)\subseteq\mathrm{Keep}(X;\mathscr{B})\,.$

The maximum of the two radii defining these sets is at least
$\frac{M}{M+1}A(X)$. ∎

######  Lemma 6.

Let $X=\\{x_{1},\dots,x_{M}\\}\subset\mathscr{B}$, where $M\geq 2$. Suppose
$z\in B\left(\mu(X),\alpha A(X)\right)$ for some $0<\alpha<1$, and $z\not\in
X$. Let $x_{k}$ be a point of $X$ farthest from $(z+\Sigma(X))/(M+1)$. Then

$F(X)-F(\\{z\\}\cup(X\setminus\\{x_{k}\\}))\geq(1-\alpha)((M-1)\alpha+M+1)A(X)^{2}\,>\frac{1-\alpha}{M}\,F(X)\,.$
(16)

###### Proof.

Let $x_{j}$ be a point of $X$ farthest from $\mu(X)$, so that
$\|x_{j}-\mu(X)\|=A(X)$ and

$\left\|x_{j}-\frac{\Sigma(X)-x_{j}}{M-1}\right\|=\frac{M}{M-1}A(X)\,.$

By the triangle inequality,

$\left\|z-\frac{\Sigma(X)-x_{j}}{M-1}\right\|\leq\|z-\mu(X)\|+\left\|\mu(X)-\frac{\Sigma(X)-x_{j}}{M-1}\right\|\leq\alpha
A(X)+\frac{1}{M-1}A(X)\,.$

By Lemma 1 and equation (8) we have

$\displaystyle F(\\{z\\}\cup(X\setminus\\{x_{k}\\}))$ $\displaystyle\leq$
$\displaystyle F(\\{z\\}\cup(X\setminus\\{x_{j}\\}))$ $\displaystyle=$
$\displaystyle
F(X)-(M-1)\left(\left\|x_{j}-\frac{\Sigma(X)-x_{j}}{M-1}\right\|^{2}-\left\|z-\frac{\Sigma(X)-x_{j}}{M-1}\right\|^{2}\right)$
$\displaystyle\leq$ $\displaystyle
F(X)-(M-1)\left(\left(\frac{M}{M-1}\right)^{2}-\left(\alpha+\frac{1}{M-1}\right)^{2}\right)A(X)^{2}$
$\displaystyle=$ $\displaystyle F(X)-(1-\alpha)((M-1)\alpha+M+1))A(X)^{2}\,.$

The final strict inequality in (16) follows from the final inequality in
(6).777In fact one could obtain
$F(X)-F(\\{z\\}\cup(X\setminus\\{x_{k}\\}))\geq\frac{(1-\alpha)(M+1)}{M^{2}}F(X)$,
but the simpler lower bound given in (16) will suffice for the applications.

∎

The following estimate is well-known (see e.g. [5, Lemma 2.6]).

######  Lemma 7.

Let $\mathscr{B}\subset\mathbb{R}^{d}$ be a convex body. Let $x\in\mathscr{B}$
and $0<r_{1}<r_{2}$. Then

$\frac{\lambda(\mathscr{B}\cap B(x,r_{1}))}{\lambda(\mathscr{B}\cap
B(x,r_{2}))}\geq\left(\frac{r_{1}}{r_{2}}\right)^{d}\,.$

###### Proof.

Since $d$-dimensional Lebesgue measure $\lambda$ is translation-invariant, we
may assume w.l.o.g. that $x=0$. Since $\mathscr{B}$ is convex and contains
$0$, the dilation $(r_{2}/r_{1})\mathscr{B}$ contains $\mathscr{B}$, so

$\mathscr{B}\cap
B(0,r_{2})\subseteq\left(\frac{r_{2}}{r_{1}}\right)(\mathscr{B}\cap
B(0,r_{1}))\,.$

Comparing volumes yields the desired inequality. ∎

######  Corollary 2 (Adapted from Lemma 2.3 and Lemma 2.5 in [1]).

Let $Y(0),Y(1),Y(2),\dots$ be a $\mathscr{B}$-valued Jante’s law process,
where $\mathscr{B}$ is any convex subset of $\mathbb{R^{d}}$ with non-empty
interior. Let ${\cal{F}}_{n}$ be the $\sigma$-algebra generated by
$Y(i)_{i\leq n}$. Then for every $n\geq 1$,

${\mathbb{P}}\left(F(Y(n+1))-F(Y(n))<-\frac{1}{4M}F(Y(n))\,\mid\,\mathscr{F}_{n}\right)\,\geq\,4^{-d}\,.$
(17)

Consequently,

$\displaystyle\mathbb{E}(\log F(Y(n+1))-\log
F(Y(n))\,|\,{\cal{F}}_{n})\leq-\frac{1}{4M}\,4^{-d}\,.$ (18)

###### Proof.

Suppose $Y(n)=X$. Since $\mathscr{B}$ is convex, $\mu(X)\in\mathscr{B}$. The
new point $y_{n+1}$ in $Y(n+1)$ is distributed uniformly in
$\mathrm{Keep}(X;\mathscr{B})$, and by Lemmas 3 and 4 we have

$\mathscr{B}\cap
B\left(\mu(X),A(X)\right)\subseteq\mathrm{Keep}(X;\mathscr{B})\subseteq\mathscr{B}\cap
B\left(\mu(X),\frac{M+1}{M-1}A(X)\right)\,.$

Since $M\geq 2$, we have $\frac{M+1}{4(M-1)}A(X)\leq A(X)$, so

$\mathbb{P}\left(y_{n+1}\in\mathscr{B}\cap
B\left(\mu(X),\frac{M+1}{4(M-1)}A(X)\right)\right)\geq\frac{\lambda\left(\mathscr{B}\cap
B\left(\mu(X),\frac{M+1}{4(M-1)}A(X)\right)\right)}{\lambda\left(\mathscr{B}\cap
B\left(\mu(X),\frac{M+1}{M-1}A(X)\right)\right)}\,\geq 4^{-d}\,,$

where the second inequality follows from Lemma 7. When $y_{n+1}\in
B\left(\mu(X),\frac{M+1}{4(M-1)}A(X)\right)$, we may take $X=Y(n)$ and
$\alpha=\frac{M+1}{4(M-1)}$ in Lemma 6, so that inequality (16) becomes

$\displaystyle
F(Y(n))-F(Y(n+1))\;>\;\frac{3M-5}{4M(M-1)}\,F(Y(n))\;\geq\;\frac{1}{4M}\,F(Y(n))\,.$
(19)

The final consequence follows from the fact that $F(Y(n+1))/F(Y(n))\leq 1$ (by
Corollary 1) and the inequality $\log x\leq x-1$ for $0<x\leq 1$. ∎

Inequality (17) implies that a.s. $Y(n)$ converges exponentially fast.

######  Lemma 8 (essentially Lemma 2.5 in [1]).

Let $N\geq 3$. For any fixed choice of $Y(0)$, there exists $\alpha>0$ such
that, a.s., $D(Y(n))\leq e^{-\alpha n}$ for all sufficiently large $n$.

###### Proof.

From Corollary (1) and (17) it follows that
$\mathbb{E}[F(Y(n))|F(Y(0))]\leq\gamma^{n}F(Y(0))$, where
$\gamma=1-\frac{4^{1-d}}{M}<1$. Hence
$\mathbb{P}[F(Y(n)\geq\gamma^{n/2}F(Y(0))]\leq\gamma^{n/2}$, and therefore a.
s. for all sufficiently large $n$ we have $F(Y(n))<\gamma^{n/2}F(Y(0))$. The
lemma follows using the inequalities (6) to bound $D(Y(n))$ above in terms of
$\sqrt{F(Y(n))}$; we may take any $\alpha>0$ such that
$e^{-\alpha}<\gamma^{1/4}$. ∎

From the preliminary results established above it is now easy to recover the
almost sure convergence result [1, Theorem 1.1] (for $\mathscr{B}=[0,1]^{d}$)
extended to the general setting of a $\mathscr{B}$-valued Jante’s law process
$Y(n)$; we state it below as Proposition 1. For the case of a bounded convex
body $\mathscr{B}$, this is in fact a special case of [2, Theorem 2]: the
regularity assumption required by that theorem is satisfied by the uniform
distribution on $\mathscr{B}$, as an immediate consequence of Lemma 7. In the
case of unbounded $\mathscr{B}$ the exponential decay of $D(Y(n))$ ensures
that the process will not escape to infinity, a fact which was also implicitly
used in [1, §3.4].

######  Proposition 1 (essentially Theorem 1.1 in [1]).

Let $d\geq 1$ and $M=N-1\geq 2$. Let $Y(0)$ consist of $M$ distinct points in
$\mathscr{B}$. Then there exists a random $\xi\in\mathscr{B}$ such that $Y(n)$
converges to $\xi$ a.s. in the sense of Definition 1.

It will be useful to know that if $F(Y(n))$ is small, then with high
probability the $\mathscr{B}$-valued Jante’s law process never travels too far
from $\mu(Y(n))$.

######  Lemma 9.

Fix any $\epsilon>0$ and $n\in{\mathbb{N}}$. Let $\gamma=1-4^{1-d}/M$ and
$n_{0}(\epsilon)=\left\lceil
2\log_{\gamma}(\epsilon(1-\sqrt{\gamma}))\right\rceil$. Denote

${\cal
E}_{j}=\left\\{\|\mu(Y(n+j))-\mu(Y(n))\|\leq\frac{2}{M\sqrt{M-1}}\left(n_{0}(\epsilon)+\frac{1}{1-\gamma^{1/4}}\right)\sqrt{F(Y(n))}\right\\}$

Then

${\mathbb{P}}\left(\bigcap_{j=1}^{\infty}{\cal E}_{j}\mid Y(n)\right)\geq
1-\epsilon\,.$

As a result,

${\mathbb{P}}\left(\left.\|\xi-\mu(Y(n))\|\leq\frac{2}{M\sqrt{M-1}}\left(n_{0}(\epsilon)+\frac{1}{1-\gamma^{1/4}}\right)\sqrt{F(Y(n))}\,\right|\,Y(n)\right)\geq
1-\epsilon$

as well.

###### Proof.

From Lemma 3 and (6), we have for every $t\geq 1$

$\displaystyle\|\mu(Y(t))-\mu(Y(t-1))\|$ $\displaystyle=$
$\displaystyle\left\|\frac{y_{t}-r_{t}}{M}\right\|$ $\displaystyle\leq$
$\displaystyle\left(1+\frac{M+1}{M-1}\right)\frac{A(Y(t-1))}{M}=\frac{2}{M-1}A(Y(t-1))$
$\displaystyle\leq$ $\displaystyle\frac{2}{M\sqrt{M-1}}\sqrt{F(Y(t-1))}\,.$

Now note that $n_{0}(\epsilon)$ is the least natural number $k$ such that
$\sum_{i=k}^{\infty}\gamma^{i/2}\leq\epsilon$. As we showed in the proof of
Lemma 8, $\mathbb{P}(F(Y(n+i)>\gamma^{i/2}F(Y(n)))\leq\gamma^{i/2}$, so with
probability at least $1-\epsilon$ it holds for all $i\geq n_{0}(\epsilon)$
that $F(Y(n+i))\leq\gamma^{i/2}F(Y(n))$. Suppose that this event occurs. Then
since $F(Y(n+i))\leq F(Y(n))$ for all $i\geq 0$, we have

$\|\mu(Y(n+n_{0}(\epsilon)))-\mu(Y(n))\|\leq
n_{0}(\epsilon)\frac{2}{M\sqrt{M-1}}\sqrt{F(Y(n))}\,.$

We bound the remaining increments by a geometric series. For every
$j>n_{0}(\epsilon)$ we have

$\|\mu(Y(n+j))-\mu(Y(n+n_{0}(\epsilon))\|\leq\left(\sum_{i=n_{0}(\epsilon)}^{j}\gamma^{i/4}\right)\frac{2}{M\sqrt{M-1}}\sqrt{F(Y(n))}$

For all $j>n_{0}(\epsilon)$ the geometric series is bounded by
$1/(1-\gamma^{1/4})$. Applying the triangle inequality and using
$\xi=\lim_{j\to\infty}Y(n+j)$ gives the claimed bounds. ∎

##  3 Reduction to the case of uniform geometry

Following the terminology of [5, Ch. 3], a convex body $\mathscr{B}$ is said
to have _uniform geometry_ when there exists some $r_{0}>0$ such that

$b(r_{0}):=\inf_{x\in\mathscr{B}}\lambda(B(x,r_{0})\cap\mathscr{B})>0$ (20)

(see [5, Ch. 3].) Since the volume $\lambda(B(x,r_{0})\cap\mathscr{B})$
depends continuously on $x$, when $\mathscr{B}$ is compact the infimum is
achieved and is positive because $\mathscr{B}$ is the closure of its interior.
That is, every bounded convex body has uniform geometry. Examples of unbounded
convex bodies with bounded geometry are $\mathbb{R}^{d}$ itself, and any
convex body which is the intersection of finitely many closed half-spaces.
However, not every unbounded convex body has uniform geometry, see for example
[5, Example 3.12].

Suppose $\mathscr{B}$ is a convex body of uniform geometry, with $r_{0}$ and
$b(r_{0})>0$ as in (20). Let $V(d)$ denote the volume of the unit ball in
$\mathbb{R}^{d}$, and define

$c=c(\mathscr{B},r_{0}):=\frac{b(r_{0})}{V(d)r_{0}^{d}}\,.$ (21)

Note that in the case $\mathscr{B}=\mathbb{R}^{d}$ we have $c=1$, and in all
other cases we have $c\leq 1/2$. For the case $\mathscr{B}=[0,1]^{d}$ which
was studied in [1], we have $c=2^{-d}$. By Lemma 7, for all $r\in(0,r_{0})$
and all $x\in\mathscr{B}$ we have

$\frac{\lambda(\mathscr{B}\cap B(x,r))}{\lambda(B(x,r))}\geq c>0\,.$ (22)

Recalling Lemma 3 and Lemma 4, for
$X=\\{x_{1},\dots,x_{M}\\}\subset\mathscr{B}$ with $M\geq 2$ we have

$\mathscr{B}\cap
B\left(\mu(X),A(X)\right)\subseteq\mathrm{Keep}(X;\mathscr{B})\subseteq\mathscr{B}\cap
B\left(\mu(X),\frac{M+1}{M-1}A(X)\right).$

It follows that when $A(X)\leq r_{0}$,

$\frac{\lambda(\mathrm{Keep}(X;\mathscr{B}))}{\lambda\left(B\left(\mu(X),\frac{M+1}{M-1}A(X)\right)\right)}\,\geq\,\left(\frac{M-1}{M+1}\right)^{d}c\,>\,0\,.$

This means that conditional on $Y(n)=X$, we may sample from the distribution
of $y_{n+1}$ by rejection sampling with success probability bounded away from
$0$. That is, we repeatedly sample a point uniformly from
$B\left(\mu(X),\frac{M+1}{M-1}A(X)\right)$ until we obtain a sample that lies
in $\mathrm{Keep}(X;\mathscr{B})$, and for each trial the probability of
success is at least $\left(\frac{M-1}{M+1}\right)^{d}c$. This property will be
used several times. If $\mathscr{B}$ did not have uniform geometry then the
success probability for this rejection sampling procedure would not be bounded
away from $0$ as $X$ ranges over all sets with sufficiently small $A(X)$.
Fortunately, we may reduce Theorem 2 to the case where $\mathscr{B}$ is
bounded and therefore has uniform geometry, as follows.

######  Lemma 10.

Suppose the conclusion of Theorem 2 holds subject to the extra hypothesis that
$\mathscr{B}$ is bounded. Then it also holds without this hypothesis.

###### Proof.

Let $\mathscr{B}$ be an unbounded convex body and consider the
$\mathscr{B}$-valued Jante’s law process $Y(n)_{n\in\mathbb{N}_{0}}$ started
at $Y(0)=S$ where $S$ is a set of $M$ distinct points of $\mathscr{B}$. For
any $R>0$, let $\mathscr{B}_{R}=\mathscr{B}\cap B(0,R)$. Define the random
variable

$K:=\inf\\{R\in\mathbb{N}\,|\,(\forall n)(Y(n)\subset\mathscr{B}_{R})\\}\,.$

Then $K<\infty$ (and the inf is actually a min) a.s., by Lemma 8 and
Proposition 1. Let
$C=\left\lceil\frac{M+1}{M-1}\sqrt{\frac{(M-1)F(S)}{M^{2}}}\right\rceil$. Then
by Corollary 1 and (6), we have $\frac{M-1}{M+1}A(Y(n))\leq C$ for all $n\geq
0$. Suppose $R\in\mathbb{N}$ satisfies $\mathbb{P}(K=R)>0$. Let
$\tilde{Y}(\cdot)$ be the $\mathscr{B}_{R+C}$-valued Jante’s law process
started at $\tilde{Y}(0)=S$, and define
$\tilde{K}=\inf\\{R\in\mathbb{N}\,|\,(\forall
n)(\tilde{Y}(n)\subset\mathscr{B}_{R})\\}$. We claim that
$\mathbb{P}(K=R)=\mathbb{P}(\tilde{K}=R)$ and that the distribution of
$Y(\cdot)$ conditioned on $K=R$ is the same as the distribution of
$\tilde{Y}(\cdot)$ conditioned on $\tilde{K}=R$. Indeed, $Y(\cdot)$ and
$\tilde{Y}(\cdot)$ may be coupled by a _maximal coupling_ , meaning that for
each $n\geq 0$, on the event $Y(n)=\tilde{Y}(n)$, conditional on $Y(n)$ we
have $Y(n+1)=\tilde{Y}(n+1)$ with the largest possible probability. When
$Y(n)=\tilde{Y}(n)$ and
$\mathrm{Keep}(Y(n);\mathscr{B})=\mathrm{Keep}(\tilde{Y}(n);\mathscr{B}_{R+C})$,
this maximal coupling probability is $1$. On the event that $K=R$, this occurs
for every $n\geq 0$. For then $\mu(Y(n))\in\mathscr{B}_{R}$ for every $n\geq
0$ (since $\mathscr{B}_{R}$ is convex), and therefore $B(\mu(Y(n),C)\subseteq
B(0,R+C)$ and so

$\displaystyle\mathrm{Keep}(Y(n);\mathscr{B})$ $\displaystyle=$
$\displaystyle\mathscr{B}\cap\mathrm{Keep}(Y(n);\mathbb{R}^{d})$
$\displaystyle\subseteq$ $\displaystyle\mathscr{B}\cap
B\left(\mu(Y(n)),\frac{M+1}{M-1}A(Y(n))\right)$ $\displaystyle\subseteq$
$\displaystyle\mathscr{B}\cap B(\mu(Y(n)),C)$ $\displaystyle\subseteq$
$\displaystyle\mathscr{B}_{R+C}\,.$

It follows that for every $n\geq 0$

$\mathrm{Keep}(\tilde{Y}(n);\mathscr{B}_{R+C})=\mathscr{B}_{R+C}\cap\mathrm{Keep}(\tilde{Y}(n);\mathscr{B})=\mathrm{Keep}(\tilde{Y}(n);\mathscr{B})=\mathrm{Keep}(Y(n);\mathscr{B})$

as required.

Let $\tilde{\xi}$ be the limit point of $\tilde{Y}(\cdot)$. By the hypothesis
that Theorem 2 holds for bounded convex bodies, the distribution of
$\tilde{\xi}$ conditioned on $\tilde{K}=R$ is continuous: it is a continuous
distribution conditioned on an event of positive probability. Now the
distribution of $\xi$ is a mixture of these conditioned distributions over the
set of $R\in\mathbb{N}$ for which $\mathbb{P}(K=R)>0$. Therefore the
distribution of $\xi$ is continuous, as required. ∎

##  4 All original points are eventually removed, a. s.

In this section, we will study the $\mathbb{R}^{d}$-valued Jante’s law process
$Z(n)_{n\geq 0}$. In particular this is an instance of a $\mathscr{B}$-valued
Jante’s law process where $\mathscr{B}$ has uniform geometry. For some of the
lemmas in this section it takes very little extra work to make the proofs
apply to $Y(n)_{n\geq 0}$ subject to the uniform geometry assumption, and we
will state them in that generality even though we will only apply them to the
process $Z(\cdot)$. So for the whole of this section we will assume that
$Y(\cdot)$ is a $\mathscr{B}$-valued Jante’s law process where $\mathscr{B}$
is a convex body with uniform geometry and $c=c(\mathscr{B},r_{0})>0$.

Let $\tau$ be the first time when all the original points of the configuration
are removed, i.e.

$\tau=\inf\\{n>0:\ Z(n)\cap Z(0)=\emptyset\\}\,,$

with the usual convention that $\tau=\infty$ if there is no $n$ for which
$Z(n)\cap Z(0)=\emptyset$. We know from Proposition 1 that for every
$\epsilon>0$ there exists a random $n(\epsilon)$ such that for all $n\geq
n(\epsilon)$ we have $Z(n)\subset B(\xi,\epsilon)$, so if for some
deterministic choice of $Z(0)$ it happened that $\mathbb{P}(\tau=\infty)>0$
then we would have $\mathbb{P}(\xi\in Z(0))>0$, and the distribution of $\xi$
would not be continuous. Therefore Theorem 1 can only hold if it is true that
that $\tau<\infty$ a.s.. It will be useful to prove this fact separately
first.

######  Proposition 2.

$\tau<\infty$ a.s.

The rest of this section is devoted to the proof of Proposition 2.

In the case $N=3$, i.e. $M=2$, a.s. for each $n\geq 0$ the set $Z(n)$ consists
of two distinct points, which are equally likely to be removed at the next
step. Thus in this simple case $\tau-1$ is a geometric random variable, in
particular a.s. finite.

The case $M=2$ is dealt with, so from now on we will assume that $M\geq 3$.
Let $V(d)$ be the volume of the unit ball in $\mathbb{R}^{d}$.

Let ${\cal{F}}_{n}$ denote the $\sigma$-algebra generated by the process
$Y(i)_{i\leq n}$. Recall from Corollary 2 that

$\mathbb{E}(\log F(Y(n+1))-\log
F(Y(n))\,|\,{\cal{F}}_{n})\leq-\frac{4^{-1-d}}{M}\,.$ (23)

To complement this we need to bound the downward drift of the smallest inter-
point distance $d(Y(n))$.

######  Lemma 11.

Suppose that $X$ is a set of $M$ distinct points of $\mathscr{B}$ such that
$A(X)\leq r_{0}$. Then

${\mathbb{E}}(\log d(Y(n))-\log d(Y(n+1))\mid
Y(n)=X)\leq\frac{M}{cd}\left(\frac{d(X)}{A(X)}\right)^{d}\,.$

###### Proof.

Suppose $Y(n)=X$, where $A(X)\leq r_{0}$. The only way the smallest inter-
point distance can decrease is for the newly sampled point $y_{n+1}$ to lie
closer than $d(Y(n))$ to one of the existing $M$ points. By Lemma 4, together
with (22),

$\lambda(\mathrm{Keep}(X;\mathscr{B}))\geq\lambda(\mathscr{B}\cap
B(\mu(X),A(X)))\geq\frac{b(r_{0})A(X)^{d}}{r_{0}^{d}}=cV(d)A(X)^{d}\,.$

Therefore

$\displaystyle{\mathbb{E}}\left(\left.\log\frac{d(Y(n))}{d(Y(n+1))}\,\right|Y(n)=X\right)$
$\displaystyle\leq$
$\displaystyle\frac{M}{\lambda(\mathrm{Keep}(X;\mathscr{B}))}\int_{0}^{d(X)}V(d)\,d\,u^{d-1}\log\frac{d(X)}{u}\,\mathrm{d}u$
$\displaystyle\leq$
$\displaystyle\frac{M}{cd}\left(\frac{d(X)}{A(X)}\right)^{d}.$

∎

Next, for any set $X$ of $M$ distinct points in $\mathbb{R}^{d}$, define

$h(X):=\log\left(\frac{\sqrt{F(X)}}{d(X)}\right)=-\log d(X)+\frac{\log
F(X)}{2}\,.$

Note that by (6), we have

$h(X)\geq\frac{1}{2}\log\left(\frac{M(M-1)}{2}\right)\geq\frac{1}{2}\log
3>0\,.$

Let $S(d,M)$ be the space of subsets of $\mathbb{R}^{d}$ of cardinality $M$,
with the obvious topology888 i.e. the topology which it inherits as a quotient
of $(\mathbb{R}^{d})^{M}\setminus\Delta$ by the permutation action of the
symmetric group $S_{M}$, where $\Delta\subset(\mathbb{R}^{d})^{M}$ is the
subset consisting of sequences in which two or more terms coincide. Note that
$S(d,M)$ is not complete with respect to the Hausdorff metric. For any $X\in
S(d,M)$, denote by $\hat{X}$ the recentered and rescaled version of $X$,
defined by

$\hat{X}=\left\\{\frac{x-\mu(X)}{\sqrt{F(X)}}:x\in X\right\\}\,.$

We have $\mu(\hat{X})=0$, $F(\hat{X})=1$, and at the same time
$h(\hat{X})=h(X)$. For any $\rho\in[0,\infty)$, define

$\displaystyle S(d,M,\rho)$ $\displaystyle:=\\{X\in
S(d,M)\,:\,h(X)\leq\rho\\}\,,$ $\displaystyle\hat{S}(d,M,\rho)$
$\displaystyle:=\\{X\in S(d,M)\,:\;h(X)\leq\rho,\;\mu(X)=0,\;F(X)=1\\}\,.$

Then $S(d,M,\rho)$ is invariant under translation and scaling, while
$\hat{S}(d,M,\rho)$ is a compact subset of $S(d,M)$.

Our next goal is to show that for the $\mathscr{B}$-valued Jante’s law process
$Y(\cdot)$, there is a constant $\rho_{2}<\infty$ such that a.s. the rescaled
and recentered set $\widehat{Y(n)}$ visits the compact set
$\hat{S}(d,M,\rho_{2})$ infinitely often. The proof uses $h$ as a Lyapunov
function.

######  Proposition 3.

There exist $\rho_{1}=\rho_{1}(d,M)<\infty$ and
$\gamma_{i}=\gamma_{i}(d,M)>0$, $i=1,2$, such that on the event that
$h(Y_{n})\geq\rho_{1}$ and $A(Y(n))\leq r_{0}$, we have

  * (a)

${\mathbb{E}}(h(Y(n+1))-h(Y(n))\mid{\cal{F}}_{n})\leq 0$, and

  * (b)

${\mathbb{P}}(h(Y(n+1))-h(Y(n))<-\gamma_{1}\,|\,{\cal{F}}_{n})\geq\gamma_{2}$.

###### Proof.

For part (a) we use inequality (23) and Lemma 11, along with (6) to compare
$A(Y(n))$ and $\sqrt{F(Y(n))}$. This requires us to take
$\rho_{1}\geq\frac{1}{d}\left(\log\left(2^{2d+3}M^{d+2}\right)-\log(cd)\right)$.

Figure 2: A ball around the center of mass with possible holes

For part (b), consider the ball “with holes” (see Figure 2)

$B^{*}:=\mathscr{B}\cap\left(B\left(\mu(Y(n)),r\right)\setminus\bigcup_{w\in
Y(n)}B\left(w,\left(\frac{c}{2M}\right)^{1/d}r\right)\right)\,,$

where

$r:=\frac{A(Y(n))}{2}$

and $c$ is the constant defined in (21), which depends on the geometry of
$\mathscr{B}$.

Recall that $Y(n+1)\setminus Y(n)=\\{y_{n+1}\\}$. By Lemma 6 (with
$\alpha=1/2$), if $y_{n+1}\in B^{*}$, then

$F(Y(n+1))\leq\left(1-\frac{1}{2M}\right)F(Y(n))\,.$

On the other hand, using Lemmas 3 and 4 with Lemma 7,

${\mathbb{P}}\left(y_{n+1}\in
B\left(\mu(Y(n)),r\right)\right)\geq\frac{\lambda(\mathscr{B}\cap
B(\mu(Y(n)),r))}{\lambda\left(\mathscr{B}\cap
B\left(\mu(Y(n)),\frac{M+1}{M-1}A(Y(n))\right)\right)}\geq\frac{(M-1)^{d}}{2^{d}(M+1)^{d}}\,.$

At the same time, the relative volume of the “holes” in $B^{*}$ is bounded. We
have $\mu(Y(n))\in\mathscr{B}$ and $r<r_{0}$, so we may use the hypothesis of
uniform geometry to compare $\lambda(\mathscr{B}\cap
B\left(\mu(Y(n)),r\right))$ with $\lambda(B(\mu(Y(n),r)))$. We find

$\frac{\mathrm{Volume}(B^{*})}{\mathrm{Volume}(\mathscr{B}\cap
B\left(\mu(Y(n)),r\right))}\geq
1-\frac{M}{c}\left(\frac{c}{2M}\right)=\frac{1}{2}.$

Hence, ${\mathbb{P}}(z\in
B^{*})\geq\gamma_{2}:=\frac{(M-1)^{d}}{2^{d+1}(M+1)^{d}}>0$. Next, on the
event that $y_{n+1}\in B^{*}$,

$\min_{w\in
Y(n)}\|y_{n+1}-w\|\geq\left(\frac{c}{2M}\right)^{1/d}r\geq\left(\frac{c}{2M}\right)^{1/d}\frac{1}{2M}\sqrt{F(Y(n))}\,.$

by (6). Let $C:=\left(\frac{c}{2M}\right)^{1/d}\frac{1}{2M}$. Then

$d(Y(n+1))\geq\min\left\\{d(Y(n)),C\sqrt{F(Y(n))}\right\\}$

and thus

$\displaystyle h(Y(n+1))-h(Y(n))$
$\displaystyle=\frac{1}{2}\log\frac{F(Y(n+1))}{F(Y(n))}-\log d(Y(n+1))+\log
d(Y(n))$
$\displaystyle\leq-\frac{1}{4M}-\min\left\\{0,\log\frac{C\sqrt{F(Y(n))}}{d(Y(n))}\right\\}$
$\displaystyle=-\frac{1}{4M}+\max\left\\{0,+\log\frac{1/C}{h(Y(n))}\right\\}$

which equals $-\gamma_{1}:=-\frac{1}{4M}<0$ provided $h(Y(n))\geq 1/C$. We see
that both parts of the lemma hold when we take

$\rho_{1}=\max\left(\frac{1}{C},\frac{1}{d}\log\left(\frac{M^{d+2}4^{d+1}}{cd}\right)\right)\,.$

∎

Let $\rho_{2}:=\frac{M\,e^{-\rho_{1}}}{2(M-1)}$ where $\rho_{1}$ is the
constant from Proposition 3. Then define

$\tilde{S}(d,M,\rho_{2}):=\left\\{X\in S(d,M):\
d(X)\geq\rho_{2}\,D(X)\right\\}\,.$

######  Proposition 4.

The event $\\{Y(n)\in\tilde{S}(d,M,\rho_{2})\\}$ occurs infinitely often a.s.

###### Proof.

First of all, by (6),

$X\in S(d,M,\rho_{1})\implies X\in\tilde{S}(d,M,\rho_{2})\,.$

For each $m\geq 1$ let $\tau_{m}=\inf\\{n\geq m:\ Y(n)\in S(d,M,\rho_{1})\\}$.
By Proposition 3 we have that $\eta_{n;m}=h(Y(n\wedge\tau_{m}))$ is a non-
negative supermartingale for $n\geq m$, for any fixed $m$. From the
supermartingale convergence theorem it follows from that a.s. $\eta_{n;m}$
converges as $n\to\infty$. Since on $\\{\tau_{m}>n\\}$ we have
${\mathbb{P}}\left(h(Y(n+1))\leq h(Y(n)\right)-\gamma_{1})>\gamma_{2}$, where
$\gamma_{i}>0$, $i=1,2$, are from Proposition 3, we conclude that a.s.
$\tau_{m}<\infty$. As a result, $Y(n)$ returns to $S(d,M,\rho_{1})$ and hence
also to $\tilde{S}(d,M,\rho_{2})$ infinitely often, a.s. ∎

For the remainder of this section we focus on the process $Z(n)_{n\geq 0}$.

######  Lemma 12.

For every set $X$ of $M$ distinct points in $\mathbb{R}^{d}$, there exists a
finite $n_{0}$ and a possible finite trajectory of the process $Z(\cdot)$, say
$Z^{*}(0),\dots,Z^{*}(n)$, such that $Z^{*}(0)=X$, $Z^{*}(n_{0})\cap
X=\emptyset$, along the trajectory there are no ties, i.e. at each step there
is a unique legal choice of point $s_{n}^{*}$ to remove, $s_{n}^{*}\in
Z^{*}(n-1)$, and the added points $z_{1}^{*},\dots z_{n}^{*}$ are distinct and
are not in $X$.

###### Proof.

We will identify a trajectory meeting the conditions of the lemma through a
combination of probabilistic and constructive reasoning. Note that the process
$Z(\cdot)$, started at $Z(0)=X$, a.s. obeys $F(Z_{n})\to 0$, implying that
$D(Z_{n})\to 0$. Let $\tau_{A}$ be the least $n$ such that $D(Z_{n})<d(X)$.
Then $\tau_{A}<\infty$ a.s., and a.s. the points $z_{1},\dots,z_{\tau_{A}}$
are distinct and not in $X$. Therefore there exists some $m\in\mathbb{N}$ and
a trajectory $Z_{0}^{*},\dots,Z_{m}^{*}$, with $Z_{i}^{*}\setminus
Z_{i-1}^{*}=\\{z_{i}^{*}\\}$ and $Z_{i-1}^{*}\setminus
Z_{i}^{*}=\\{s_{i}^{*}\\}$ for $i=1,\dots,m$, such that $D(Z_{m}^{*})<d(X)$,
hence $|Z_{m}^{*}\cap X|\leq 1$, and the points $z_{1}^{*},\dots,z_{m}^{*}$
are distinct and not in $X$. If $Z_{m}^{*}\cap X=\emptyset$ then we are done.
So from now on assume that $Z_{m}^{*}\cap X=\\{x_{\spadesuit}\\}$.

The end of the construction falls into two cases, depending on whether
$x_{\spadesuit}$ is a furthest point in $Z_{m}^{*}$ from the point

$\mu_{0}:=\frac{M\mu(Z_{m}^{*})+x_{\spadesuit}}{M+1}.$

First, suppose we are in case 1, where $x_{\spadesuit}$ is indeed a furthest
point in $Z_{m}^{*}$ from the point $\mu_{0}$. If we had
$x_{\spadesuit}=\mu(Z_{m}^{*})$ then we would also have
$x_{\spadesuit}=\mu_{0}$ and hence all points of $Z_{m}^{*}$ would have to
coincide, which was ruled out in the construction of $Z_{m}^{*}$. Therefore
$x_{\spadesuit}\neq\mu(Z_{m}^{*})$. Let
$z_{m+1}^{*}=\epsilon\mu(Z_{m}^{*})+(1-\epsilon)x_{\spadesuit}$, for some very
small $\epsilon\in(0,1)$, chosen so that $z_{m+1}^{*}$ is not in
$\bigcup_{i=0}^{m}Z_{i}^{*}$. (This can fail only for finitely many values of
$\epsilon$.) Then $x_{\spadesuit}$ is the unique furthest point of
$Z_{m}^{*}\cup\\{z_{m+1}^{*}\\}$ from its center of mass, which is

$\mu_{\epsilon}:=\frac{(M+\epsilon)\mu(Z_{m}^{*})+(1-\epsilon)x_{\spadesuit}}{M+1}\,.$

To see this, note first that

$\|z_{m+1}^{*}-\mu_{\epsilon}\|=\frac{M}{M+1}(1-\epsilon)\|x_{\spadesuit}-\mu(Z_{m}^{*})\|>0\,,$

while

$\|x_{\spadesuit}-\mu_{\epsilon}\|=\frac{M+\epsilon}{M+1}\|x_{\spadesuit}-\mu(Z_{m}^{*})\|\,.$

Thus $\|x_{\spadesuit}-\mu_{\epsilon}\|>\|z_{m+1}^{*}-\mu_{\epsilon}\|>0$.
Secondly, note that the ball
$B(\mu_{\epsilon},\|x_{\spadesuit}-\mu_{\epsilon}\|)$ contains the ball
$B(\mu_{0},\|x_{\spadesuit}-\mu_{0}\|)$, and their boundary spheres meet only
at the point $x_{\spadesuit}$. Hence
$\|x_{\spadesuit}-\mu_{\epsilon}\|>\|z-\mu_{\epsilon}\|$ for every $z\in
Z_{m}^{*}\setminus\\{x_{\spadesuit}\\}$. Hence the point $x_{\spadesuit}$ is
the unique point which may be removed, and we obtain
$Z_{m+1}^{*}=(Z_{m}\cup\\{z_{m+1}^{*}\\})\setminus\\{x_{\spadesuit}\\}$. The
trajectory $Z_{0}^{*},\dots,Z_{m+1}^{*}$ is a permissible trajectory with no
tiebreaks and no repeated points, and $Z_{m+1}^{*}\cap X=\emptyset$, as
required.

Case 2 is the case where $x_{\spadesuit}$ is not a furthest point in
$Z_{m}^{*}$ from the point $\mu_{0}$. For the rest of the proof we assume that
we are in this case. We claim that if $v\in\mathbb{R}^{d}\setminus\\{0\\}$ is
sufficiently close to $0\in\mathbb{R}^{d}$ then from the trajectory
$Z_{0}^{*},\dots,Z_{m}^{*}$ we can obtain another permissible trajectory in
which the arriving points are all perturbed by the vector $v$. The perturbed
trajectory is $Z_{0}^{\dagger},\dots,Z_{m}^{\dagger}$, where
$Z_{0}^{\dagger}=X$ and for $i=1,\dots,m$ we have
$Z_{i}^{\dagger}=(Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\})\setminus\\{s_{i}^{\dagger}\\}$,
where $z_{i}^{\dagger}=z_{i}^{*}+v$ and $s_{i}^{\dagger}$ is a point of
$Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\}$ furthest from
$\mu(Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\})$. To say that the trajectory
$Z_{0}^{\dagger},\dots,Z_{m}^{\dagger}$ is permissible means that
$z_{i}^{\dagger}\in\mathrm{Keep}(Z_{i-1}^{\dagger};\mathbb{R}^{d})$ for each
$i=1,\dots,m$, so that $z_{i}^{\dagger}\neq s_{i}^{\dagger}$. In fact, if $v$
is sufficiently close to $0$ then the corresponding points are removed at each
step, by which we mean that
$Z_{i}^{\dagger}=Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\}\setminus\\{s_{i}^{\dagger}\\}$,
where either $s_{i}^{\dagger}=s_{i}^{*}$ (if $s_{i}\in X$), or
$s_{i}^{\dagger}=s_{i}^{*}+v$ (if $s_{i}\notin X$). To see this, note that
each point $z_{i}^{\dagger}$ or $s_{i}^{\dagger}$ depends continuously on $v$,
as does $\mu(Z_{i}^{\dagger})$. The (assumed) statement that $s_{i}^{*}$ is
the unique point of $Z_{i-1}^{*}\cup\\{z_{i}^{*}\\}$ furthest from
$\mu(Z_{i-1}^{*}\cup\\{z_{i}^{*}\\})$ is equivalent to a finite collection of
strict inequalities: for each
$z\in(Z_{i}^{*}\cup\\{z_{i}^{*}\\})\setminus\\{s_{i}^{*}\\}$,

$\|s_{i}^{*}-\mu(Z_{i}^{*}\cup\\{z_{i}^{*}\\})\|>\|z-\mu(Z_{i}^{*}\cup\\{z_{i}^{*}\\})\|\,.$

It follows that when $v\in\mathbb{R}^{d}\setminus\\{0\\}$ is sufficiently
close to $0$ then for each
$z\in(Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\})\setminus\\{s_{i}^{\dagger}\\}$,

$\|s_{i}^{\dagger}-\mu(Z_{i}^{\dagger}\cup\\{z_{i}^{\dagger}\\})\|>\|z-\mu(Z_{i}^{\dagger}\cup\\{z_{i}^{\dagger}\\})\|\,,$

i.e. $s_{i}^{\dagger}$ is the unique point of
$Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\}$ furthest from
$\mu(Z_{i-1}^{\dagger}\cup\\{z_{i}^{\dagger}\\})$ for each $i=1,\dots,m$. In
particular the trajectory $Z_{0}^{\dagger},\dots,Z_{m}^{\dagger}$ involves no
tiebreaks when $v$ is sufficiently close to $0$. By similar reasoning, since
the trajectory $Z_{0}^{*},\dots,Z_{m}^{*}$ involves no repeated points, the
same is true of the trajectory $Z_{0}^{\dagger},\dots,Z_{m}^{\dagger}$ when
$v$ is sufficiently close to $0$.

Having chosen a vector $v$ suitably close to $0$, we now extend our two
trajectories one further step, so that $Z_{m+1}^{\dagger}=Z_{m}^{*}+v$, i.e.
$Z_{m}^{\dagger}$ is the translate of $Z_{m}^{*}$ by the vector $v$. To do
this, we choose $z_{m+1}^{*}=x_{\spadesuit}-v$ and
$z_{m+1}^{\dagger}=x_{\spadesuit}+v$. Then

$Z_{m}^{\dagger}\cup\\{z_{m+1}^{\dagger}\\}=\left(Z_{m}^{*}\cup\\{z_{m+1}^{*}\\}\right)+v\,.$
(24)

Notice that in the translation which relates the two configurations, the roles
of $x_{\spadesuit}$ and the new point ($z_{m+1}^{*}$ and $z_{m+1}^{\dagger}$
respectively) are swapped. We may choose $v$ so that there is no tiebreak in
either configuration when selecting the point furthest from the center of
mass. This is because the set of vectors $v$ which would cause such a tie is
contained in a finite union of hypersurfaces of dimension less than $d$, so we
may choose $v$ as close as we like to $0$ but not in any of these
hypersurfaces. From (24), we see that there are points $s_{m+1}^{*}\in
Z_{m}^{*}\cup\\{z_{m+1}^{*}\\}$ and $s_{m+1}^{\dagger}\in
Z_{m}^{\dagger}\cup\\{z_{m+1}^{\dagger}\\}$ which are furthest from the
centers of mass $\mu(Z_{m}^{*}\cup\\{z_{m+1}^{*}\\})$ and
$\mu(Z_{m}^{\dagger}\cup\\{z_{m+1}^{\dagger}\\})$ respectively, and which are
related by $s_{m+1}^{\dagger}=s_{m+1}^{*}+v$. Hence
$Z_{m+1}^{\dagger}=Z_{m+1}^{*}+v$. Before proceeding, we must check that the
trajectories $Z_{0}^{*},\dots,Z_{m+1}^{*}$ and
$Z_{0}^{\dagger},\dots,Z_{m+1}^{\dagger}$ are permissible. This is where we
use the assumption that we are in case 2, together with the continuity of all
defined points as functions of $v$. Let $u^{*}$ be a point of $Z_{m}^{*}$
which is strictly further from $\mu_{0}$ than $x_{\spadesuit}$ is from
$\mu_{0}$, and let $u^{\dagger}=u^{*}+v$. As $v\to 0$, we have
$\mu(Z_{m}^{*}\cup\\{z_{m+1}^{*}\\})\to\mu_{0}$ and
$\mu(Z_{m}^{\dagger}\cup\\{z_{m+1}^{\dagger}\\})\to\mu_{0}$, while
$z_{m+1}^{*}\to x_{\spadesuit}$, $z_{m+1}^{\dagger}\to x_{\spadesuit}$ and
$u^{\dagger}\to u^{*}$. Therefore for $v$ close enough to $0$ we have

$\|u^{*}-\mu(Z_{m}^{*}\cup\\{z_{m+1}^{*}\\})\|>\|z_{m+1}^{*}-\mu(Z_{m}^{*}\cup\\{z_{m+1}^{*}\\})\|$

and

$\|u^{\dagger}-\mu(Z_{m}^{\dagger}\cup\\{z_{m+1}^{\dagger}\\})\|>\|z_{m+1}^{\dagger}-\mu(Z_{m}^{\dagger}\cup\\{z_{m+1}^{\dagger}\\})\|\,,$

showing that both trajectories are indeed permissible.

Finally, by the same argument as in the first paragraph of this proof, there
exists a possible trajectory $Z_{m+1}^{*},\dots,Z_{m+m^{\prime}}^{*}$ such
that $|Z_{m+1}^{*}\cap Z_{m+m^{\prime}}^{*}|\leq 1$, and having no tiebreaks,
no repeated points, and no arriving points which were already in
$\bigcup_{i=0}^{m}Z_{i}^{*}$. There are now two possibilities: either
$x_{\spadesuit}\notin Z_{m+m^{\prime}}^{*}$, in which case we are done, or
$x_{\spadesuit}\notin Z_{m+m^{\prime}}^{*}$. In the latter case, consider the
trajectory $Z_{m+1}^{\dagger},\dots,Z_{m+m^{\prime}}^{\dagger}$ defined by
$Z_{i}^{\dagger}=Z_{i}^{*}+v$ for $i=m+1,\dots,m+m^{\prime}$. It is also a
permissible trajectory, being a translate of a permissible trajectory.
Moreover, $Z_{m+1}^{\dagger}\cap
Z_{m+m^{\prime}}^{\dagger}=\\{x_{\spadesuit}+v\\}$, and in particular
$x_{\spadesuit}\notin Z_{m+m^{\prime}}^{\dagger}$, so that
$Z_{m+m^{\prime}}^{\dagger}\cap X=\emptyset$. Therefore
$Z_{0}^{\dagger},\dots,Z_{m+m^{\prime}}^{\dagger}$ is a trajectory satisfying
the conclusion of the lemma, and we are done.

∎

Recall the following definition.

######  Definition 3.

For two sets $X$ and $X^{\prime}$ the Hausdorff distance is defined as

$\displaystyle d_{H}(X,X^{\prime})$ $\displaystyle=\max\left\\{\sup_{x\in
X}\left[\inf_{a^{\prime}\in
X^{\prime}}\|x-a^{\prime}\|\right],\sup_{x^{\prime}\in
X^{\prime}}\left[\inf_{a\in X}\|x^{\prime}-a\|\right]\right\\}$
$\displaystyle=\inf\left\\{\epsilon>0:\ \displaystyle
X\subseteq\bigcup_{a^{\prime}\in
X^{\prime}}\mathrm{B}(a^{\prime},\epsilon)\text{ and
}X^{\prime}\subseteq\bigcup_{a\in X}\mathrm{B}(a,\epsilon)\right\\}.$

Note that when $X$ and $X^{\prime}$ are both finite, then
$d_{H}(X,X^{\prime})=0$ if and only if $X=X^{\prime}$.

######  Lemma 13.

For every set $X\in S(d,M)$, there is a finite $n_{0}=n_{0}(X)$, an
$\epsilon_{0}=\epsilon_{0}(X)\in(0,d(X)/2)$, and a
$\delta_{0}=\delta_{0}(X)>0$ such that if $X^{\prime}\in S(d,M)$ has Hausdorff
distance less than $\epsilon_{0}$ from $X$ then

$\mathbb{P}(Z(n_{0})\cap Z(0)=\emptyset\mid Z(0)=X^{\prime})>\delta_{0}\,.$

###### Proof.

Let $z_{1-M}^{*},\dots,z_{0}^{*}$ be any $M$ distinct points in
$\mathbb{R}^{d}$, and let $X=\\{z_{1-M}^{*},\dots,z_{0}^{*}\\}$. Take a
trajectory as provided by Lemma 12. Since at each step $1\leq n\leq n_{0}$
there is a unique choice of point $s_{n}^{*}\in Z^{*}(n-1)\cup\\{z^{*}_{n}\\}$
such that $s_{n}^{*}$ maximizes the distance from
$\mu(Z^{*}(n-1)\cup\\{z_{n}^{*}\\})$, and there are only finitely many steps,
there is an $\epsilon>0$ such that for all $1\leq n\leq n_{0}$ we have for any
$z\in Z^{*}(n)$ that

$\|s_{n}^{*}-\mu(Z^{*}(n-1)\cup\\{z_{n}^{*}\\})\|\geq\|z-\mu(Z^{*}(n-1)\cup\\{z_{n}^{*}\\})\|+4\epsilon\,.$

By making $\epsilon$ smaller if necessary, we may also assume that

$\min(\|z-x\|\,:\,z\in Z^{*}(n_{0}),\,x\in X)>2\epsilon.$

Consider any perturbed trajectory $Z^{\prime}(0),\dots,Z^{\prime}(n_{0})$
where $Z_{0}^{\prime}=\\{z_{1-M}^{\prime},\dots,z_{0}^{\prime}\\}$ and for
$1\leq n\leq n_{0}$ we have $Z^{\prime}(n)=(\\{z_{n}^{\prime}\\}\cup
Z^{\prime}(n-1))\setminus\\{s_{n}^{\prime}\\}$, and for each
$i=1-M,\dots,n_{0}$ we have $\|z_{i}^{*}-z_{i}^{\prime}\|<\epsilon$. Then one
may prove by induction over $n$ that in the perturbed trajectory, the thrown
out points $s_{n}^{\prime}$ correspond to the original points $s_{n}$ in the
sense that for each $1\leq n\leq n_{0}$, the index $j$ such that
$s_{n}^{*}=z_{j}^{*}$ also satisfies $s_{n}^{\prime}=z_{j}^{\prime}$. This is
because the perturbation moves each point by a distance less than $\epsilon$
and also moves each center of mass by a distance at most $\epsilon$. It
therefore alters corresponding distances by less than $2\epsilon$, and
differences of distances by less than $4\epsilon$.

In particular, the points of $Z^{\prime}(n_{0})$ are in one-to-one
correspondence with points of $Z^{*}(n_{0})$ so that corresponding pairs of
points are at distance less than $\epsilon$. It follows that
$Z^{\prime}(n_{0})\cap X^{\prime}=\emptyset$.

Finally, consider the random process $Z(\cdot)$ started at $Z(0)=X^{\prime}$.
The probability that each new point $z_{i}$, $i=1,\dots,n_{0}$, lies in the
ball $B(z_{i}^{*},\epsilon)$ is bounded away from $0$, uniformly over all
choices of $X^{\prime}$ that are $\epsilon$-perturbations of $X$. This is
because the volumes of the sets $\mathrm{Keep}(Z(n);\mathbb{R}^{d})$ are
bounded above _a priori_ in terms of $X$, using the fact that

$\frac{M^{2}}{M-1}A(Z_{n})^{2}\leq F(Z_{n})\leq
F(X^{\prime})\leq\frac{M(M-1)}{2}D(X^{\prime})^{2}\leq\frac{M(M-1)}{2}(D(X)+2\epsilon)^{2}$

together with Lemma 3. When this occurs, we have $Z(n_{0})\cap
Z(0)=\emptyset$. ∎

######  Lemma 14.

There exist $n_{1}=n_{1}(d,M)<\infty$ and $\delta_{1}=\delta_{1}(d,M)>0$ such
that for every $X\in\tilde{S}(d,M,\rho_{2})$,

$\mathbb{P}(Z(n+n_{1})\cap Z(n)=\emptyset\mid Z(n)=X)\geq\delta_{1}\,.$

###### Proof.

First, observe that since the process $Z(\cdot)$ is time-homogeneous and
invariant under translation and scaling, we have

$\mathbb{P}(Z(n+n_{1})\cap Z(n)=\emptyset\mid Z(n)=X)=\mathbb{P}(Z(n_{1})\cap
Z(0)=\emptyset\mid Z(0)=\hat{X})\,.$

So it suffices to prove that there exist $n_{1}<\infty$ and $\delta_{1}>0$
such that for all $X\in\hat{S}(d,M,\rho_{2})$ we have

$\mathbb{P}(Z(n_{1})\cap Z(0)=\emptyset\mid Z(0)=X)\geq\delta_{1}\,.$

This follows from Lemma 13, since for each point $X$ in the compact set
$\hat{S}(d,M,\rho_{2})$, that corollary specifies a neighbourhood $U_{X}$ of
$X$ and constants $n_{0}(X)<\infty$, $\delta_{0}(X)>0$ such that for all
$X^{\prime}\in U_{X}$,

$\mathbb{P}(Z(n_{0})\cap
Z(0)=\emptyset\mid\,Z(0)=X^{\prime})\geq\delta_{0}(X)\,.$

Since $\hat{S}(d,M,\rho_{2})$ is compact, there exists a covering of
$\hat{S}(d,M,\rho_{2})$ by some finite sequence of such neighbourhoods
$U_{X_{1}},\dots,U_{X_{k}}$, and we may take

$n_{1}=\max(n_{0}(X_{1}),\dots,n_{0}(X_{k}))<\infty$

and

$\delta_{1}=\min(\delta_{0}(X_{1}),\dots,\delta_{0}(X_{k}))>0\,.$

∎

We can now prove Proposition 2, which claims that $\tau<\infty$ a. s.

###### Proof of Proposition 2.

Starting from any configuration $Z(0)=X$ of $M$ distinct points in
$\mathbb{R}^{d}$, the process $Z(\cdot)$ a.s. eventually enters
$\tilde{S}(d,M,\rho_{2})$, by Proposition 4, say at stopping time $\tau_{1}$.
Let $n_{1}=n_{1}(d,M)$ be the constant provided by Lemma 14. Inductively
define $\tau_{k}$ for $k=2,3,\dots$, by

$\tau_{k}=\min\\{t\,:\,t\geq\tau_{k-1},\,Z(\tau_{k})\in\tilde{S}(d,M,\rho_{2})\\}\,.$

For each $k\geq 1$, conditional on $\tau_{k}$ and $Z(\tau_{k})$, in time steps
$\tau_{k}+1,\dots,\tau_{k}+n_{1}$ all the points of $Z(\tau_{k})$ are removed,
with probability at least $\delta_{1}$, regardless of the configuration
$Z(\tau_{k})$. Therefore with probability $1$ there exists a finite $k$ for
which this event occurs. We then have $\tau<\tau_{k}+n_{1}<\infty$, as
required. ∎

##  5 Proof of Theorem 1

Without loss of generality, we assume that the initial state $Z(0)$ is
deterministic. This is harmless since if for each deterministic choice of
$Z(0)$ the limit $z_{\infty}$ exists a.s. and has a continuous distribution,
then if instead $Z(0)$ is random, $z_{\infty}$ still exists a.s. and its
distribution is a mixture of continuous distributions, which is necessarily
continuous.

List the elements of $Z(0)$ in an arbitrary order as
$z_{-(N-2)},z_{-(N-3)},\dots z_{0}$. Recalling that for $n\geq 1$, $z_{n}$ is
the unique point of $Z(n)\setminus Z(n-1)$, we observe that for each $m\geq 0$
we have

$\\{z_{i}:i=-(N-2),...,m\\}\;=\;\bigcup_{n=0}^{m}Z(n)\,.$

We discarded a null set to ensure that the points of the sequence
$(z_{i})_{i=-(N_{2})}^{\infty}$ are pairwise distinct. Thus for each $n\geq
1$, the unique point $s_{n}$ of $Z(n-1)\setminus Z(n)$ is equal to
$z_{\alpha(n)}$ for a unique index $\alpha(n)$ in the range
$-(N-2)\leq\alpha(n)\leq n-1$. Moreover, the indices
$\alpha(1),\alpha(2),\alpha(3),\dots$ are pairwise distinct random variables
since each point is removed at most once.

######  Lemma 15.

Fix any $K\geq N-1$ and any deterministic sequence of distinct indices
$\mathbf{i}=(i(1),\dots,i(K))$ such that for each $n=1,\dots,K$ we have
$-(N-2)\leq i(n)\leq n-1$ and $K$ is the least positive integer such that
$\\{-(N-2),\dots,0\\}\subseteq\\{i(1),\dots,i(K)\\}$. Let
${\bar{\alpha}}=(\alpha(1),\dots,\alpha(K))$. Then the event that $\tau=K$ and
$\bar{\alpha}=\mathbf{i}$ is equivalent to the event that a certain finite
collection of linear or quadratic inequalities are satisfied, which involve
the coordinates of the random points $z_{1},\dots,z_{K}$, as well as the
coordinates of the deterministic points $z_{-(N-2)},\dots,z_{0}$.

Figure 3: Area where $(z_{1},z_{2})$ can be sampled in the example of Remark 1

######  Remark 1.

For example, take $d=1$, $N=3$, and w.l.o.g. $z_{-1}=-1$, $z_{0}=1$. Then on
$\tau=2$, $\alpha(1)=-1$, $\alpha(2)=(0)$, we have $z_{1}\in(0,1)$ and
$z_{2}\in\left(2z_{1}-1,\frac{z_{1}+1}{2}\right)$ or $z_{1}\in(1,3)$ and
$z_{2}\in\left(\frac{z_{1}+1}{2},2z_{1}-1\right)$. See Figure 3.

These inequalities define a bounded semi-algebraic set999i.e., a set defined
by a number of polynomial inequalities and equalities; in our case, a.s. these
will be just inequalities.
$\mathscr{P}_{K,\mathbf{i}}\subset\left(\mathbb{R}^{d}\right)^{K}$. (When
$d=1$, $\mathscr{P}$ is a polytope, since the defining inequalities may be
reduced to a collection of affine linear inequalities.) For some sequences
$\mathbf{i}$, $\mathscr{P}_{K,\mathbf{i}}$ may be empty or have empty
interior. Note that the boundary of $\mathscr{P}_{K,\mathbf{i}}$ is contained
in a finite union of algebraic hypersurfaces where at least one of the
defining inequalities becomes an equality. These hypersurfaces have trivial
$(Kd)$-dimensional Lebesgue measure. Thus $\mathscr{P}_{K,\mathbf{i}}$ has
positive $(Kd)$-dimensional Lebesgue measure if and only if it has non-empty
interior.

Let $E_{K,\mathbf{i}}$ be the event that $\tau=K$ and
$\bar{\alpha}=\mathbf{i}$. If $\mathbb{P}\left(E_{K,\mathbf{i}}\right)>0$ then
conditional on $E_{K,\mathbf{i}}$ the sequence $\bar{\alpha}$ is distributed
according to $U\left(\mathscr{P}_{K,\mathbf{i}}\right)$; in particular it lies
a.s. in the interior $\mathscr{P}_{K,\mathbf{i}}^{\circ}$.

Define a $d$-dimensional linear subspace $L_{K}$ of
$\left(\mathbb{R}^{d}\right)^{K}$ by

$L_{K}:=\\{(v,\dots,v)\in\mathbb{R}^{d})^{K}:v\in\mathbb{R}^{d}\\}\,.$

Consider the orthogonal projection $\pi_{K}$ of
$\left(\mathbb{R}^{d}\right)^{K}$ onto $\left(L_{K}\right)^{\perp}$. We have

$\lambda_{Kd}(\mathscr{P}_{K,\mathbf{i}})=\int_{\pi_{K}(\mathscr{P}_{K,\mathbf{i}})}\lambda_{d}(\mathscr{P}_{K,\mathbf{i}}\cap\pi_{K}^{-1}(x))\,d\lambda_{(K-1)d}(x)\,.$

We conclude that if $\lambda_{Kd}\left(\mathscr{P}_{K,\mathbf{i}}\right)>0$
then for $\lambda_{Kd}$-a.e.
$\mathbf{p}=(p_{1},\dots,p_{K})\in\mathscr{P}_{K,\mathbf{i}}$, the set

$T(\mathbf{p},K,\mathbf{i}):=\\{v\in\mathbb{R}^{d}\,:\,(p_{1}+v,\dots,p_{K}+v)\in\mathscr{P}_{K,\mathbf{i}}\\}$

has $\lambda_{d}(T(\mathbf{p},K,\mathbf{i}))>0$. Therefore we can sample a
$U(\mathscr{P}_{K,\mathbf{i}})$ random variable $\mathbf{q}$ by sampling first
a $U(\mathscr{P}_{K,\mathbf{i}})$ random variable $\mathbf{p}$, then
(conditional on $\mathbf{p}$) sampling a $U(T(\mathbf{p},K,\mathbf{i}))$
random variable $\delta$ and setting
$\mathbf{q}:=\mathbf{p}+(\underbrace{\delta,\delta,\dots,\delta}_{K\text{
times}})$.

Let $\mathscr{F}_{\tau}$ be the $\sigma$-algebra generated by the stopped
process $Z(\cdot\wedge\tau)$. Let $V$ be a random variable whose distribution
conditional on $\mathscr{F}_{\tau}$ is
$U(T((z_{1},\dots,z_{\tau}),\tau,(\alpha(1),\dots,\alpha(\tau))))$, such that
$V$ is conditionally independent of $(z_{\tau+1},z_{\tau+2},\dots)$ given
$\mathscr{F}_{\tau}$. That is to say, on the event $E_{K,\mathbf{i}}$, given
$z_{-(N-2)},\dots,z_{K}$, we let $V\sim
U(T((z_{1},\dots,z_{K}),K,\mathbf{i}))$, independent of
$(z_{K+1},z_{K+2},\dots)$. Proposition 2, and the fact that ties a.s. do not
occur, imply that a.s. $E_{K,\mathbf{i}}$ holds for exactly one
$(K,\mathbf{i})$, so $V$ is well-defined. The crucial property of $V$ is that
its conditional distribution of $V$ on $\mathscr{F}_{\tau}$ is a.s.
continuous, being the uniform distribution on a subset of $\mathbb{R}^{d}$ of
positive and finite $d$-dimensional Lebesgue measure.

Now define a new random sequence
$\left(z_{n}^{\prime}\right)_{n=-(N-2)}^{\infty}$ as follows.

$z_{n}^{\prime}=\begin{cases}z_{n},&\text{if $n\leq 0$,}\\\ z_{n}+V,&\text{if
$n\geq 1$,}\end{cases}$

Conditional on $z_{-(N-2)},\dots,z_{0}$ and on $E_{K,\mathbf{i}}$, the
sequences $\left(z_{n}\right)_{n=1}^{\tau}$ and
$\left(z_{n}^{\prime}\right)_{n=1}^{\tau}$ are identically distributed: both
are $U(\mathscr{P}_{K,\mathbf{i}})$ random variables.

Let $Z^{\prime}(0)=Z(0)$, then for $n\geq 1$ let

$Z^{\prime}(n)=\\{z_{n}^{\prime}\\}\cup\left(Z^{\prime}(n-1)\setminus\\{z_{\alpha(n)}\\}\right)\,.$

For all $n\geq\tau$ we have

$Z^{\prime}(n)=\\{z+V\,:\,z\in Z(n)\\}$ (25)

Moreover $Z^{\prime}(\cdot)$ is a Markov chain with the same distribution as
$Z(\cdot)$: by construction they have the same distribution up to the random
time $\tau$, and from time $\tau$ onwards, the transitions have the correct
distribution by equation (25) together with the translation-invariance of the
transition law of $Z(\cdot)$.

Therefore the sequence $\left(z_{n}^{\prime}\right)$ has the same distribution
as the original sequence $(z_{n})$. Since $\left(z_{n}\right)$ a.s. converges
to a limit $z_{\infty}$, it also holds that $\left(z_{n}^{\prime}\right)$ a.s.
converges; let $z_{\infty}^{\prime}$ be its limit. The distribution of
$z_{\infty}^{\prime}$ coincides with the distribution of $z_{\infty}$.

On the other hand, the distribution of $z_{\infty}^{\prime}$ is also the
mixture of its conditional distributions on $\mathscr{F}_{\tau}$. But we have

$z_{\infty}^{\prime}=z_{\infty}+V$

and $z_{\infty}$ and $V$ are conditionally independent given
$\mathscr{F}_{\tau}$. Thus the distribution of $z_{\infty}^{\prime}$ is a
mixture of continuous distributions, and it is therefore continuous. This
completes the proof of Theorem 1.

##  6 Coupling $Y(\cdot)$ and $Z(\cdot)$

In this section we complete the proof of Theorem 2. By Lemma 10 it suffices to
prove the continuity of the limit point $\xi$ for a $\mathscr{B}$-valued
Jante’s law process where $\mathscr{B}$ is bounded. So we shall assume
throughout this section that $\mathscr{B}$ is bounded, and therefore has
uniform geometry.

We will need an isoperimetric inequality for inner shells of convex bodies,
for which we have been unable to find a reference. It concerns the following
problem. Suppose you have a (possibly) hollow chocolate egg whose outer
boundary is the boundary of a convex body. If all the chocolate is within
distance $r$ of the outer boundary of the egg, what is the maximum quantity of
chocolate that can possibly be contained within a ball of radius $R$? See
Figure 4.

Figure 4: Lemma 16 bounds the quantity of chocolate in the ball of radius $R$.

######  Lemma 16.

Let $R>r>0$ and let $\mathscr{B}$ be a convex body in $\mathbb{R}^{d}$. For
every $y\in\mathbb{R}^{d}$, we have

$\lambda\left(B(y,R)\cap\mathscr{B}\cap N_{r}(\mathscr{B}^{c})\right)\leq
2rd\sqrt{d}\,V(d-1)R^{d-1}\,,$ (26)

where $V(d-1)$ is the volume of the unit ball in $\mathbb{R}^{d-1}$, with the
convention that $V(0)=1$.

###### Proof.

For $d=1$ the inequality holds with $V(0)=1$, since the set
$N_{r}\left(\mathscr{B}^{c}\right)$ is a union of at most two intervals of
length at most $r$ and therefore has length at most $2r$. Now assume that
$d\geq 2$. Let $A=\mathscr{B}\cap B(y,R)\cap N_{r}(\mathscr{B}^{c})$. Consider
any point $x\in A$. Then $\|x-z\|\leq r$ for some $z\in\partial\mathscr{B}$.
Let $H$ be a supporting hyperplane through $z$, meaning that
$H\cap\mathscr{B}^{\circ}=\emptyset$. Then (as for any hyperplane through $z$)
$H$ cuts at least one of the $d$ axis-parallel lines through $x$ at a point
whose distance from $x$ is at most $r\sqrt{d}$. It follows that there is a
point of $\partial\mathscr{B}$ that lies between $x$ and $H$ on one of the
axis-parallel lines through $x$. The situation is illustrated in Fig. 5 below.

Figure 5: The supporting hyperplane at $z$ to the green convex body
$\mathscr{B}$ cuts at least one axis-parallel line through $x$ within distance
$\sqrt{d}\|z-x\|$ of $x$.

Thus $A$ is covered by the $d$ sets $A_{1},...,A_{d}$, where

$A_{i}=\left\\{x\in A\,:\,\left(\exists
t\in\left[-r\sqrt{d},r\sqrt{d}\right]\right)\left(x+te_{i}\in\partial\mathscr{B}\right)\right\\}\,,$

where $e_{1},\dots,e_{d}$ are the standard basis vectors of $\mathbb{R}^{d}$.
Since $A_{i}\subset B(y,R)$, the orthogonal projection of $A_{i}$ onto a
hyperplane orthogonal to $e_{i}$ has $d-1$-dimensional Lebesgue measure at
most $V(d-1)R^{d-1}$. Each line parallel to $e_{i}$ that meets $B(y,R)$ meets
$A_{i}$ in at most two intervals, whose total length is at most $2r\sqrt{d}$.
Hence $\lambda(A_{i})\leq 2r\sqrt{d}V(d-1)R^{d-1}$ for each $i=1,\dots,d$.
Summing over $i$ we get

$\lambda(A)\leq\lambda\left(\bigcup_{i=1}^{d}A_{i}\right)\leq\sum_{i=1}^{d}\lambda(A_{i})\leq
2rd\sqrt{d}V(d-1)R^{d-1}\,.$

∎

Lemma 16 will suffice for our purposes, but we remark that the sharp version
of the inequality is as follows. (A proof is given in the appendix.)

######  Lemma 17.

Let $R>r>0$ and let $\mathscr{B}$ be a convex body in $\mathbb{R}^{d}$. For
every $x\in\mathbb{R}^{d}$, we have

$\lambda\left(B(x,R)\cap\mathscr{B}\cap N_{r}(\mathscr{B}^{c})\right)\leq
V(d)(R^{d}-(R-r)^{d})\,,$ (27)

where $V(d)$ is the volume of the unit ball in $\mathbb{R}^{d}$.

Let $\mathscr{B}\subset\mathbb{R}^{d}$ be a bounded convex body. For any set
$X$ of $M$ distinct points of $\mathscr{B}$ define

$\mathrm{d}_{\mathscr{B}}(X)=\mathrm{dist}(X,\mathscr{B}^{c})\,.$

and for any set $X$ of $M$ distinct points of $\mathscr{B}$ that is not
entirely contained in $\partial{\mathscr{B}}$, define

$\mathrm{d}^{\circ}_{\mathscr{B}}(X)=\mathrm{dist}(X\cap\mathscr{B}^{\circ},\mathscr{B}^{c})\,,$

For $Y(n)_{n\geq 0}$ a $\mathscr{B}$-valued Jante’s law process we define

$g_{n}:=\begin{cases}\frac{1}{2}\log(F(Y(n)))-\log\left(\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n)\right),&\text{if
}Y(n)\not\subset\partial\mathscr{B};\\\ +\infty,&\text{if
}Y(n)\subset\partial\mathscr{B}\,.\end{cases}$

######  Lemma 18.

There exists $\delta>0$ (depending on $M$, $d$, and $\mathscr{B}$) such that
for any set $X_{0}$ of $M$ distinct points of $\mathscr{B}$ and any $n_{0}$,

$\mathbb{P}(\exists n\geq
n_{0}\,:\,g_{n}\leq\log(\delta^{-1})\,|\,Y\left(n_{0}\right)=X_{0})=1\,,$

and hence almost surely for infinitely many $n$ we have

$\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))\geq\delta\sqrt{F(Y(n))}\,.$

###### Proof.

We may assume without loss of generality that $X_{0}$ has at most one point in
$\partial\mathscr{B}$. Indeed, a.s. all new points are not in
$\partial\mathscr{B}$, and a.s. $D(Y(n))\to 0$ as $n\to\infty$, so there
exists an $n\geq n_{0}$ such that $D(Y(n))<d(Y(n_{0}))$ and therefore
$Y(n)\cap Y(n_{0})$ contains at most one point; now apply the lemma with
$X_{0}=Y(n)$. It follows from these assumptions that a.s. $g_{n}\in\mathbb{R}$
for every $n\geq 0$. Likewise we may assume that
$\sqrt{\frac{M-1}{M^{2}}}\sqrt{F(X_{0})}<r_{0}$, where $r_{0}$ and $c$ are the
constants used to specify the uniform geometry of $\mathscr{B}$, so that by
Corollary 1 and inequality (6) we have $A(Y(n))<r_{0}$ for all $n\geq 0$.

We will use Lemma 16 to get an upper bound for the expectation of the positive
part of the increment of $-\log\mathrm{d}^{\circ}_{\mathscr{B}}(Y(\cdot))$.

$\mathbb{E}\left(\left.\log_{+}\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))}{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n+1))}\,\right|\,Y(n)=X\right)\,=\,\frac{\bigintss\mathbf{1}(y\in\mathrm{Keep}(X;\mathscr{B}))\,\log_{+}\frac{\textstyle{\rule[-4.0pt]{0.0pt}{10.0pt}\mathrm{d}^{\circ}_{\mathscr{B}}(X)}}{\textstyle{\rule{0.0pt}{10.0pt}\mathrm{dist}\left(y,\mathscr{B}^{c}\right)}}\,d\lambda(y)}{\lambda(\mathrm{Keep}(X;\mathscr{B}))}$
(28)

Let

$r_{1}:=\mathrm{d}^{\circ}_{\mathscr{B}}(X),\qquad R:=\frac{M+1}{M-1}A(X)\,.$

By (6) we have $R\leq\frac{M+1}{M\sqrt{M-1}}\sqrt{F(X)}$ and by Lemma 3 we
have $\mathrm{Keep}(X;\mathscr{B})\subseteq B(\mu(X),R)\cap\mathscr{B}$. The
numerator of the RHS of (28) may be rewritten as

$\displaystyle\int\int\mathbf{1}(y\in\mathrm{Keep}(X;\mathscr{B}))$
$\displaystyle\cdot\mathbf{1}\left(\mathrm{dist}\left(y,\mathscr{B}^{c}\right)\leq
r\leq\mathrm{d}^{\circ}_{\mathscr{B}}(X)\right)\,\frac{dr}{r}\,d\lambda(y)$
$\displaystyle=\int_{0}^{r_{1}}\frac{1}{r}\lambda(\mathrm{Keep}(X;\mathscr{B})\cap
N_{r}(\mathscr{B}^{c}))\,dr$
$\displaystyle\leq\int_{0}^{r_{1}}\frac{1}{r}\lambda(B(\mu(X),R)\cap\mathscr{B}\cap
N_{r}(\mathscr{B}^{c}))\,dr$
$\displaystyle\leq\int_{0}^{r_{1}}2\sqrt{d}V(d-1)R^{d-1}\,dr\,=2r_{1}d\sqrt{d}\,V(d-1)R^{d-1}\,,$

where the first equality above uses Tonelli’s theorem, the next inequality
uses Lemma 3, and the final inequality uses Lemma 16. On the other hand, by
our assumption that $A(X)<r_{0}$, we have the uniform geometry bound

$\lambda(\mathrm{Keep}(X;\mathscr{B}))\geq cV(d)A(X)^{d}\,,$

where $c$ is the constant from (21). Therefore the RHS of (28) is bounded
above by

$\displaystyle\frac{2\,\mathrm{d}^{\circ}_{\mathscr{B}}(X)\,d^{3/2}\,V(d-1)\left(\frac{M+1}{M-1}A(X)\right)^{d-1}}{cV(d)A(X)^{d}}=c_{1}\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(X)}{A(X)}\leq
c_{1}M\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(X)}{\sqrt{F(X)}}$

where $c_{1}$ is a constant depending on $d$, $M$ and $c$. Thus we have shown
that

$\mathbb{E}\left(\left.\log_{+}\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))}{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n+1))}\,\right|\,Y(n)=X\right)\,\leq\,c_{1}M\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(X)}{\sqrt{F(X)}}\,.$
(29)

By almost the same argument,

$\mathbb{P}(\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n+1))<\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))\,|\,Y(n)=X)\leq\frac{\lambda\left(B(\mu(X),R)\cap\mathscr{B}\cap
N_{r_{1}}\left(\mathscr{B}^{c}\right)\right)}{\lambda(\mathrm{Keep}(X;\mathscr{B}))}\leq
c_{1}M\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(X)}{\sqrt{F(X)}}\,.$

Define

$\delta=\frac{1}{8c_{1}M^{2}4^{d}},\qquad
L=\log(\delta^{-1})+\frac{1}{2}\log\left(1-\frac{1}{4M}\right)\,,$

(note that $L<\log(\delta^{-1})$) and the stopping time

$\tau_{g}:=\inf\\{n\geq n_{0}\,:\,g_{n}\leq\log(\delta^{-1})\\}\,.$

We aim to prove that $L\vee g_{n\wedge\tau_{g}}$ is a supermartingale. Noting
that $\log(1-1/(4M))<0$ and $\log F(Y(n+1))-\log F(Y(n))\leq 0$, inequality
(17) in Corollary 2 implies

$\displaystyle\mathbb{E}$ $\displaystyle(\max(\log(F(Y(n+1)))-\log
F(Y(n)),\,2(L-\log(\delta^{-1})))\mid\mathscr{F}_{n})$
$\displaystyle=\mathbb{E}\left(\max\left(\log(F(Y(n+1)))-\log
F(Y(n)),\,\log\left(1-\frac{1}{4M}\right)\right)\mid\mathscr{F}_{n}\right)$
$\displaystyle\leq\mathbb{E}\left(\mathbf{1}\left(\left.\log(F(Y(n+1)))-\log
F(Y(n))\leq\log\left(1-\frac{1}{4M}\right)\right)\cdot\log\left(1-\frac{1}{4M}\right)\right|\mathscr{F}_{n}\right)$
$\displaystyle=\log\left(1-\frac{1}{4M}\right)\mathbb{P}\left(\left.\\!F(Y(n+1))-F(Y(n))\leq-\frac{1}{4M}F(Y(n))\,\right|\,\mathscr{F}_{n}\right)\leq-\frac{1}{4M}4^{-d}\,.$

Combining this with (29), we find that on the event
$g_{n}\geq\log(\delta^{-1})$, we have

$\displaystyle\mathbb{E}((L\vee$ $\displaystyle g_{n+1})-(L\vee
g_{n})\,|\mathscr{F}_{n})=\mathbb{E}\left(\left(L\vee
g_{n+1}\right)-g_{n}\,|\,\mathscr{F}_{n}\right)$
$\displaystyle=\,\mathbb{E}\left(\left(L-g_{n})\vee(g_{n+1}-g_{n})\right)\,|\,\mathscr{F}_{n}\right)$
$\displaystyle\leq\,\mathbb{E}\left((L+\log\delta)\vee(g_{n+1}-g_{n})\,|\,\mathscr{F}_{n}\right)$
$\displaystyle=\,\mathbb{E}\left(\frac{1}{2}\log\left(1-\frac{1}{4M}\right)\vee(g_{n+1}-g_{n})\,|\,\mathscr{F}_{n}\right)$
$\displaystyle\leq\,\mathbb{E}\left(\left.\frac{1}{2}\log\left(1-\frac{1}{4M}\right)\vee\left(\frac{1}{2}\log\frac{F(Y(n+1))}{F(Y(n))}+\log_{+}\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))}{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n+1))}\right)\,\right|\,\mathscr{F}_{n}\right)$
$\displaystyle\leq\,\mathbb{E}\left(\left.\left(\frac{1}{2}\log\left(1-\frac{1}{4M}\right)\vee\frac{1}{2}\log\frac{F(Y(n+1))}{F(Y(n))}\right)+\log_{+}\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))}{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n+1))}\,\right|\,\mathscr{F}_{n}\right)$
$\displaystyle\leq\,-\frac{1}{8M}4^{-d}+c_{1}M\frac{\mathrm{d}^{\circ}_{\mathscr{B}}(Y(n))}{\sqrt{F(Y(n))}}=-\frac{4^{-d}}{8M}+c_{1}Me^{-g_{n}}\,\leq\,-\frac{4^{-d}}{8M}+c_{1}M\delta\,=0\,.$

(Note that we used the fact that $\max(a+b,c)\leq\max(a,b)+c$ whenever $c\geq
0$ when we went from the fifth to the sixth line.) We therefore have

$\mathbb{E}((L\vee g_{(n+1)\wedge\tau_{g}})-(L\vee
g_{n\wedge\tau_{g}})\,|\,\mathscr{F}_{n})\leq 0\,.$ (30)

Using the first part of Corollary 2, we find

$\mathbb{P}(g_{n+1}-g_{n}\leq\log(1-1/(4M)))\geq 4^{-d}-c_{1}Me^{-g_{n}}\,,$

so under the same condition that $g_{n}\geq\log(\delta^{-1})$ we also have

$\mathbb{P}\left((L\vee g_{n+1})-(L\vee
g_{n})\leq\log\left(1-1/({4M})\right)\right)\geq
4^{-d}\left(1-\frac{1}{8M}\right)>0\,.$ (31)

Inequality (30) shows that the random sequence $L\vee g_{n\wedge\tau_{g}}$ is
a supermartingale that is bounded below by $L$, so it almost surely converges,
but inequality (31) shows that it can only converge to a value less than or
equal to $\log(\delta^{-1})$. Therefore $\tau_{g}<\infty$ a.s., and indeed
a.s. $g_{n}\leq\log(\delta^{-1})$ for infinitely many $n$. ∎

Recall the definition of $\mathrm{d}_{\mathscr{B}}(\cdot)$ from before Lemma
18.

######  Lemma 19.

Let101010Think of $\delta$ small and $\Delta$ large. $\delta>0$ and
$\Delta>0$. Then there exist $n_{1}\in\mathbb{N}$ and $\epsilon>0$ (both
depending on $d,M,\delta$ and $\Delta$) such that for any $n\in\mathbb{N}$,

$\mathbb{P}\left(\frac{\mathrm{d}_{\mathscr{B}}(Y(n+n_{1}))}{\sqrt{F(Y(n+n_{1}))}}>\Delta\,\left|\,\frac{\mathrm{d}_{\mathscr{B}}(Y(n))}{\sqrt{F(Y(n))}}>\delta\right.\right)>\epsilon\,.$

###### Proof.

Before we proceed with the formal proof, let us explain the idea behind it.
The main ingredient is that there is a positive probability $\epsilon$ that
for some number $n_{1}$ of consecutive steps the new point arrives very close
to the center of of mass of the existing configuration, relative to its
diameter. When this occurs, the moment of inertia decreases by a definite
multiplicative factor at each step. As a result, after $n_{1}$ steps the
square-root of the moment of inertia will have decreased by at least a factor
$\delta/(2\Delta)$. At the same time, the convex hull of the configuration
stays inside a bounding region that grows from step to step, by a
geometrically decreasing sequence of increments. It can be arranged that the
bounding region never gets closer to $\mathscr{B}^{c}$ than
$\frac{1}{2}\mathrm{d}_{\mathscr{B}}(Y(n))$, and hence
$\mathrm{d}_{\mathscr{B}}(Y(n+n_{1}))\geq\frac{1}{2}\mathrm{d}_{\mathscr{B}}(Y(n))$.

We will combine ideas from the proofs of Corollary 2 and Lemma 9. Define

$\alpha:=\min\left(\frac{M-1}{2M(M+1)}\,,\,\frac{\delta}{48(M+1)}\right),\quad\gamma:=\sqrt{1-\frac{1}{12(M+1)}}\in(0,1)$

and

$n_{1}:=\left\lceil\frac{\log\left(\frac{\delta}{2\Delta}\right)}{\log\gamma}\right\rceil\,.$

For each $i\in\\{1,\dots,n_{1}\\}$, define the event $E_{i}$ by

$E_{i}:=\left\\{y_{n+i}\in
B\left(\mu(Y(n+i-1)),\,\alpha\sqrt{F(Y(n+i-1))}\right)\right\\}\,.$

We have

$\alpha\sqrt{F(Y(n+i-1))}\geq\alpha\frac{M}{\sqrt{M-1}}A(Y(n+i-1)),$

so

$\mathbb{P}(E_{i}\,|\,Y(n+i-1))\geq\mathbb{P}\left(y_{n+i}\in\left.B\left(\mu(Y(n+i-1)),\alpha\frac{MA(Y(n+i-1))}{\sqrt{M-1}}\right)\,\right|\,Y(n+i-1)\right)\,.$

We have $\frac{\alpha M}{\sqrt{M-1}}<1$ because
$\alpha\leq\frac{M-1}{2M(M+1)}$. Hence, using Lemmas 3, 4 and 7 (as in the
proof of Corollary 2), we have

$\mathbb{P}\left(\left.E_{i}\,\right|\,Y(n+i-1)\right)\geq\left(\frac{\alpha
M\sqrt{M-1}}{M+1}\right)^{d}\,.$

Therefore

$\mathbb{P}\left(\left.\bigcap_{i=1}^{n_{1}}E_{i}\,\right|\,Y(n)\right)\geq\left(\frac{\alpha
M\sqrt{M-1}}{M+1}\right)^{dn_{1}}=:\epsilon\,.$

Now suppose that $\mathrm{d}_{\mathscr{B}}(Y(n))\geq\delta\sqrt{F(Y(n))}$ and
suppose that the event $\bigcap_{i=1}^{n_{1}}E_{i}$ occurs. Our goal is to
show that
$\mathrm{d}_{\mathscr{B}}(Y(n+n_{1}))\geq\Delta\sqrt{F(Y(n+n_{1}))}$.

Using the condition $\alpha\leq\frac{M-1}{2M(M+1)}$ together with
$\sqrt{F(Y(n+i-1))}\leq MA(Y(n+i-1))$, see (6), we obtain for each
$i=1,\dots,n_{1}$ that

$\alpha\sqrt{F(Y(n+i-1)}\,<\,\frac{M-1}{2(M+1)}A(Y(n+i-1))\,.$

Then by inequality (19) from the proof of Corollary 2, we have

$F(Y(n+i))<\gamma^{2}\,F(Y(n+i-1))\,.$

By induction we find that for $i=1,\dots,n_{1}$,

$F(Y(n+i))<\gamma^{2i}\,F(Y(n))$

and in particular

$\sqrt{F(Y(n+n_{1}))}\,\leq\,\frac{\delta}{2\Delta}\sqrt{F(Y(n))}\,.$ (32)

For every $i=1,\dots,n_{1}$, since $\mu(Y(n+i-1))\in\mathrm{Conv}(Y(n+i-1))$,

$y_{n+i}\in N_{\alpha\sqrt{F(Y(n+i-1))}}(\mathrm{Conv}(Y(n+i-1)))\,,$

so

$\mathrm{Conv}(Y(n+i))\subseteq
N_{\alpha\gamma^{i-1}\sqrt{F(Y(n))}}(\mathrm{Conv}(Y(n+i-1)))\,.$

Since

$\sum_{i=1}^{n_{1}}\alpha\gamma^{i-1}\,\sqrt{F(Y(n))}<\beta\sqrt{F(Y(n))}\,,$

where

$\beta:=\frac{\alpha}{1-\gamma}$

it follows that

$\mathrm{Conv}(Y(n+n_{1}))\subseteq
N_{\beta\sqrt{F(Y(n))}}(\mathrm{Conv}(Y(n)))\,.$

Moreover,

$\beta\,<\,24(M+1)\,\alpha\,<\,\frac{\delta}{2}\,,$

so

$\displaystyle\mathrm{d}_{\mathscr{B}}(Y(n+n_{1}))$ $\displaystyle\geq$
$\displaystyle\mathrm{d}_{\mathscr{B}}(\mathrm{Conv}(Y(n)))-\beta\sqrt{F(Y(n))}=\mathrm{d}_{\mathscr{B}}(Y(n))-\beta\sqrt{F(Y(n))}$
(33) $\displaystyle>$ $\displaystyle(\delta-\delta/2)\sqrt{F(Y(n))}\,.$

Taking square roots of both sides of (32) and dividing by (33) we obtain

$\frac{\mathrm{d}_{\mathscr{B}}(Y(n+n_{1}))}{\sqrt{F(Y(n+n_{1}))}}>\Delta\,,$

as required. ∎

######  Lemma 20.

Let $\delta>0$. Then there exist $n_{1}^{\prime}\in\mathbb{N}$ and
$\epsilon^{\prime}>0$ (both depending on $d,M$, and $\delta$) such that for
any set $X$ of $M$ points in $\mathscr{B}$ such that
$|X\cap\partial\mathscr{B}|=1\textup{ and
}\mathrm{d}_{\mathscr{B}}^{\circ}(X)>\delta\sqrt{F(X)}$, and any
$n\in\mathbb{N}$, we have

$\mathbb{P}\left(Y(n+n_{1}^{\prime})\subset\mathscr{B}^{\circ}\,\left|Y(n)=X\right.\right)>\epsilon^{\prime}\,.$

###### Proof.

The proof is very similar to the proof of Lemma 19. We take
$\Delta=\frac{1}{\sqrt{M-1}}$ and then make the same choices of constants in
terms of $M,d,\delta$ and $\Delta$ as we did there:

$\alpha:=\min\left(\frac{M-1}{2M(M+1)}\,,\,\frac{\delta}{48(M+1)}\right),\quad\gamma:=\sqrt{1-\frac{1}{12(M+1)}}\,,$
$n_{1}^{\prime}:=\left\lceil\frac{\log\left(\frac{\delta}{2\Delta}\right)}{\log\gamma}\right\rceil\,,\quad\epsilon^{\prime}=\left(\frac{\alpha
M\sqrt{M-1}}{M+1}\right)^{dn_{1}^{\prime}}\,.$

Let $y_{b}$ be the unique point in $Y(n)\cap\partial\mathscr{B}$. This time,
the good sequence of events whose probability is at least $\epsilon$ is

$E_{i}^{\prime}:=\left\\{y_{n+i}\in
B\left(\mu(Y(n+1-i)\setminus\\{y_{b}\\}),\alpha\sqrt{F(Y(n+i-1))}\right)\right\\}\,.$

Instead of applying Lemma 4, we apply Lemma 5 to see that

$\mathscr{B}\cap
B\left(\mu\left(Y(n+i-1)\setminus\\{y_{b}\\}\right),\alpha\frac{MA(Y(n+i-1))}{\sqrt{M-1}}\right)\subseteq\mathrm{Keep}(Y(n+i-1);\mathscr{B})\,.$

This requires us to check that $\frac{\alpha M}{\sqrt{M-1}}<\frac{M}{M+1}$,
which follows from $\alpha\leq\frac{M-1}{2M(M+1)}$. It follows as before that
on the event $Y(n)\cap\partial\mathscr{B}=\\{y_{b}\\}$ we have

$\mathbb{P}\left(\left.\bigcap_{i=1}^{n_{1}^{\prime}}E_{i}^{\prime}\,\right|\,Y(n)\right)\geq\epsilon^{\prime}.$

It also follows as before that on the events
$\mathrm{d}_{\mathscr{B}}(Y(n)\setminus\\{y_{b}\\})\geq\delta$ and
$\bigcap_{i=1}^{n-1}E_{i}^{\prime}$ we have

$\sqrt{F(Y(n+n_{1}^{\prime}))}\leq\frac{\delta}{2\Delta}\sqrt{F(Y(n))}\,.$

Throughout the rest of the proof, we replace each set of the form
$\mathrm{Conv}(Y(k))$ by $\mathrm{Conv}(Y(k)\setminus\\{y_{b}\\})$. Note that
$\mu(Y(k)\setminus\\{y_{b}\\})\in\mathrm{Conv}(Y(k)\setminus\\{y_{b}\\})$ so
that on the events $E_{i}^{\prime}$ we obtain

$\mathrm{Conv}(Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\})\subseteq
N_{\beta\sqrt{F(Y(n))}}\mathrm{Conv}(Y(n)\setminus\\{y_{b}\\})$

and it follows that

$\mathrm{d}_{\mathscr{B}}(Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\})\geq\frac{\delta}{2}\sqrt{F(Y(n))}\,,$

as before, and hence

$\frac{\mathrm{d}_{\mathscr{B}}^{\circ}(Y(n+n_{1}^{\prime}))}{\sqrt{F(Y(n+n_{1}^{\prime}))}}>\Delta=\frac{1}{\sqrt{M-1}}.$
(34)

On the other hand, if $y_{b}\in Y(n+n_{1}^{\prime})$ then

$\displaystyle F(Y(n+n_{1}^{\prime}))$ $\displaystyle\geq$
$\displaystyle\sum_{x\in
Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\}}\|y_{b}-x\|^{2}$ $\displaystyle=$
$\displaystyle(M-1)\left\|y_{b}-\mu(Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\})\right\|^{2}\;+\sum_{x\in
Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\}}\left\|x-\mu(Y(n+n_{1}^{\prime})\\!\setminus\\!\\{y_{b}\\})\right\|^{2}$
$\displaystyle\geq$
$\displaystyle(M-1)\|y_{b}-\mu(Y(n+n_{1}^{\prime})\\!\setminus\\!\\{y_{b}\\})\|^{2}$
$\displaystyle\geq$
$\displaystyle(M-1)\,\left[\mathrm{dist}(\mu(Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\}),\partial\mathscr{B})\right]^{2}$
$\displaystyle\geq$
$\displaystyle(M-1)\,\left[\mathrm{d}_{\mathscr{B}}\left(\mathrm{Conv}(Y(n+n_{1}^{\prime})\setminus\\{y_{b}\\})\right)\right]^{2}$
$\displaystyle=$
$\displaystyle(M-1)\,\left[\mathrm{d}_{\mathscr{B}}^{\circ}(Y(n+n_{1}^{\prime}))\right]^{2}\,,$

contrary to (34). We have shown that on the event
$\mathrm{d}_{\mathscr{B}}(Y(n)\\!\setminus\\!\\{y_{b}\\})\geq\delta$, the
event $\bigcap_{i=1}^{n-1}E_{i}^{\prime}$ occurs with probability at least
$\epsilon^{\prime}$, and when it does, we have $y_{b}\not\in
Y(n+n_{1}^{\prime})$, as required. ∎

######  Corollary 3.

Suppose $\mathscr{B}$ is a convex body in $\mathbb{R}^{d}$ with uniform
geometry and let $Y(n)_{n\geq 0}$ be a $\mathscr{B}$-valued Jante’s law
process. There exists $\delta>0$ (depending on $d$ and $\mathscr{B}$) such
that for any set $X$ of $M$ distinct points of $\mathscr{B}$, conditional on
$Y(n_{0})=X$, a.s. there exists $n\geq n_{0}$ such that
$\mathrm{d}_{\mathscr{B}}(Y(n))>\delta\sqrt{F(Y(n))}$.

###### Proof.

Consider a $\mathscr{B}$-valued Jante’s law process started at $Y(n_{0})=X$.
Define the stopping time

$\tau_{b}=\inf\\{n\geq n_{0}\,:\,Y(n)\cap\partial\mathscr{B}=\emptyset\\}\,.$

Note that if $n\geq\tau_{b}$ then $g_{n}=\frac{1}{2}\log
F(Y(n))-\log\mathrm{d}_{\mathscr{B}}(Y(n))$. Let $\delta$ be as in Lemma 18,
so that a.s. infinitely often $g_{n}<\log(\delta^{-1})$. Note that if
$n\geq\tau_{b}$ and $g_{n}<\log(\delta^{-1})$ then
$\mathrm{d}_{\mathscr{B}}(Y(n))>\delta\sqrt{F(Y(n))}$. So it suffices to show
that a.s. $\tau_{b}<\infty$.

We define a sequence of stopping times $\left(\sigma_{i}\right)_{i\geq 0}$.
Let

$\sigma_{0}=\inf\\{n\geq n_{0}\,:\,|Y(n)\cap\partial\mathscr{B}|\leq 1\\}\,.$

Since $D(Y(n))\to 0$ a.s., and a.s. all new points are in
$\mathscr{B}^{\circ}$, we have a.s. $\sigma_{0}<\infty$. Let $n_{1}$ and
$\epsilon$ be as provided by Lemma 20. Now inductively define for
$i=1,2,3,\dots$

$\sigma_{i}=\begin{cases}\inf\left\\{n>\sigma_{i-1}\,:\,\par\mathrm{d}_{\mathscr{B}}^{\circ}(Y(n))>\delta\sqrt{F(Y(n))}\right\\}&\text{if
$i$ is odd,}\\\ \sigma_{i-1}+n_{1}&\text{if $i$ is even.}\end{cases}$

By Lemma 18, a.s. $\sigma_{i}<\infty$ for all $i\geq 1$. Then by Lemma 20, for
each $m\geq 1$, on the event
$\left|Y\left(\sigma_{2m-1}\right)\cap\partial\mathscr{B}\right|=1$ we have

$\mathbb{P}\left(Y\left(\sigma_{2m}\right)\subset\mathscr{B}^{\circ}\,|\,\mathscr{F}_{\textstyle{\sigma_{2m-1}}}\right)\,\geq\epsilon.$

Hence

$\mathbb{P}\left(Y(\sigma_{2m})\not\subset\mathscr{B}^{\circ}\right)<(1-\epsilon)^{m}\,,$

so $\tau_{b}<\infty$ a.s., as required. ∎

######  Corollary 4.

Suppose $\mathscr{B}$ is a convex body in $\mathbb{R}^{d}$ with uniform
geometry and let $Y(n)_{n\geq 0}$ be a $\mathscr{B}$-valued Jante’s law
process. Let $\Delta>0$ be given. Then for any set $X$ of $M$ distinct points
of $\mathscr{B}$, conditional on $Y(n_{0})=X$, a.s. there exists $n\geq n_{0}$
such that $\mathrm{d}_{\mathscr{B}}(Y(n))>\Delta\sqrt{F(Y(n))}$.

###### Proof.

We combine Lemma 19 with Corollary 3. Consider a $\mathscr{B}$-valued Jante’s
law process started at $Y(n_{0})=X$. Now let $n_{1}$ and $\epsilon$ be as
given by Lemma 19, applied to the $\Delta$ specified in the statement of
Corollary 4. Define a sequence of stopping times
$\left(\kappa_{i}\right)_{i\geq 0}$ as follows. $\kappa_{0}=n_{0}$, and then
inductively for $i=1,2,3,\dots$

$\kappa_{i}=\begin{cases}\inf\left\\{n>\kappa_{i-1}\,:\,\mathrm{d}_{\mathscr{B}}(Y(n))>\delta\sqrt{F(Y(n))}\right\\}&\text{if
$i$ is odd,}\\\ \kappa_{i-1}+n_{1}&\text{if $i$ is even.}\end{cases}$

By Corollary 3, a.s. $\kappa_{i}<\infty$ for all $i$. By Lemma 19, for each
$m\geq 1$,

$\mathbb{P}\left(\left.\mathrm{d}_{\mathscr{B}}(Y(\kappa_{2m}))>\Delta\sqrt{F(Y(\kappa_{2m}))}\,\right|\,\mathscr{F}_{\textstyle{\kappa_{2m-2}}}\right)>\epsilon\,,$

so

$\mathbb{P}\left(\mathrm{d}_{\mathscr{B}}(Y(\kappa_{2m}))\leq\Delta\sqrt{F(Y(\kappa_{2m}))}\text{
for all }i<2m\right)<(1-\epsilon)^{m}\,,$

giving the conclusion by letting $m\to\infty$. ∎

We can now give the proof of Theorem 2.

###### Proof.

In Lemma 9, we take $\epsilon=\frac{1}{2}$. Define

$\Delta=\frac{2}{M\sqrt{M-1}}\left(n_{0}(1/2)+\frac{1}{1-\gamma^{1/4}}\right)+\frac{M+1}{M\sqrt{M-1}}\,.$

On the event that $\mathrm{d}_{\mathscr{B}}(Y(n))>\Delta\sqrt{F(Y(n))}$, we
have

$\mathbb{P}\left(\left.\|\mu(Y(n+j))-\mu(Y(n))\|+\frac{M+1}{M\sqrt{M-1}}\sqrt{F(Y(n))}\leq\Delta\sqrt{F(Y(n))}\text{
for all }j\geq 0\,\right|\,\mathscr{F}_{n}\right)\geq\frac{1}{2}$

and hence, using (6) to compare $A(Y(n+j))$ with $\sqrt{F(Y(n+j))}$,

$\mathbb{P}\left(\left.\|\mu(Y(n+j))-\mu(Y(n))\|+\frac{M+1}{M-1}A(Y(n))\leq\Delta\sqrt{F(Y(n))}\text{
for all }j\geq 0\,\right|\,\mathscr{F}_{n}\right)\geq\frac{1}{2}\,.$

Now by Lemma 3,

$\mathbb{P}\left(\mathrm{Keep}\left(Y(n+j);\mathbb{R}^{d}\right)=\mathrm{Keep}(Y(n+j);\mathscr{B})\text{
for all }j\geq 0\,|\,\mathscr{F}_{n}\right)\geq\frac{1}{2}\,.$

Define an increasing sequence of stopping times
$\left(\theta_{i}\right)_{i\geq 0}$ as follows. $\theta_{0}=0$ and then
inductively for $i=1,2,3,\dots$

$\theta_{i}=\begin{cases}\inf\left\\{n>\theta_{i-1}\,:\,\mathrm{d}_{\mathscr{B}}(Y(n))>\Delta\sqrt{F(Y(n))}\right\\}&\text{if
$i$ is odd,}\\\
\inf\left\\{n>\theta_{i-1}\,:\,\mathrm{Keep}\left(Y(n);\mathbb{R}^{d}\right)\neq\mathrm{Keep}\left(Y(n);\mathscr{B}\right)\right\\}&\text{if
$i$ is even.}\end{cases}$

(If for some $i$ we have $\theta_{i}=\infty$ then for all $j>i$ we also have
$\theta_{j}=\infty$.) For each $m\geq 1$, on the event that
$\theta_{2m-2}<\infty$, firstly $\theta_{2m-1}<\infty$ a.s., by Corollary 4,
and then
$\mathbb{P}\left(\theta_{2m}=\infty\,\left|\,\mathscr{F}_{\textstyle{\theta_{2m-1}}}\right.\right)\geq\frac{1}{2}$.
Hence a.s. there is a finite $m$ such that $\theta_{2m-1}<\infty$ but
$\theta_{2m}=\infty$.

We have shown that the distribution of $\xi$ is a mixture of distributions,
each of which is (for some $m$, $n$, and $X$) the conditional distribution of
$\xi$ given $\theta_{2m-1}=n$, $Y(n)=X$ and $\theta_{2m}=\infty$. Each such
distribution is equal to the conditional distribution of $z_{\infty}$ for the
$\mathbb{R}^{d}$-valued Jante’s law process $Z(\cdot)$ started at $Z(0)=X$,
conditioned on the positive probability event that for all $j\geq 0$,
$\mathrm{Keep}\left(Z(j);\mathbb{R}^{d}\right)=\mathrm{Keep}\left(Z(j);\mathscr{B}\right)$.
By Theorem 1, this distribution is continuous. Thus the distribution of $\xi$
is a mixture of continuous distributions, and is therefore continuous. ∎

## Acknowledgment

The research of S.V. is partially supported by the Swedish Science Foundation
grant VR 2019-04173 and the Crafoord foundation grant no. 20190667. S.V. would
like to acknowledge the hospitality of the University of Bristol during his
visits to Bristol. The research of E.C. is supported by the Heilbronn
Institute for Mathematical Research. We would like to thank John Mackay for
pointing us to the notion of uniform geometry in [5] and Rami Atar for
pointing out the relevant Brownian bees model.

## References

  * [1] Grinfeld, Michael; Volkov, Stanislav; Wade, Andrew R. _Convergence in a multidimensional randomized Keynesian beauty contest_ , Adv. in Appl. Probab. 47 (2015), no. 1, 57–82.
  * [2] Kennerberg, Philip; Volkov, Stanislav. _Jante’s law process_ , Adv. in Appl. Probab. 50 (2018), no. 2, 414–439.
  * [3] Kennerberg, Philip; Volkov, Stanislav. _Convergence in the p-contest_ , Journal of Statistical Physics 178 (2020), 1096–1125.
  * [4] Kennerberg, Philip; Volkov, Stanislav. _A Local Barycentric Version of the Bak-Sneppen Model_ , Journal of Statistical Physics 182:42 (2021), 17 pp.
  * [5] Leonardi, Gian Paolo; Ritoré, Manuel; Vernadakis, Efstratios. _Isoperimetric Inequalities in Unbounded Convex Bodies_. Memoirs of the American Math. Soc., no. 1354, Vol. 276 (2022).
  * [6] Siboni, Maor; Sasorov, Pavel and Meerson, Baruch. _Fluctuations of a swarm of Brownian bees_ , Phys. Rev. E 104 (2021), 7 pp.

## Appendix 1: Continuity of probability distributions

Here are the basic facts which we use about continuous $\mathbb{R}^{d}$-valued
random variables.

  1. 1.

Definitions. Let $\mathscr{B}(\mathbb{R}^{d})$ be the Borel $\sigma$-algebra
of Euclidean space $\mathbb{R}^{d}$. Let $W:M\to\mathbb{R}^{d}$ be a Borel-
measurable random variable on a probability space
$(M,\mathfrak{S},\mathbb{P})$, meaning that for every
$A\in\mathscr{B}(\mathbb{R}^{d})$, $W^{-1}(A)\in\mathfrak{S}$. The
distribution of $W$ is the unique Borel probability measure $\nu$ on
$\mathscr{B}(\mathbb{R}^{d})$ such that for every
$A\in\mathscr{B}(\mathbb{R}^{d})$ we have $\nu(A)=\mathbb{P}(W\in A)$. (It
suffices to check that this holds for every open set $A\subset\mathbb{R}^{d}$,
or for every closed set.) $W$ and $\nu$ are both said to be _continuous_ when
$\nu$ is absolutely continuous with respect to the Lebesgue measure $\lambda$.
This means that for every Borel set $A\subset\mathbb{R}^{d}$,

$(\lambda(A)=0)\implies(\nu(A)=\mathbb{P}(W\in A)=0)\,.$

Equivalently, there exists an element $f\in L^{1}(\mathbb{R}^{d},\lambda)$ of
norm $1$ such that for every Borel set $A\subseteq\mathbb{R}^{d}$ we have

$\mathbb{P}(W\in A)=\nu(A)=\int_{A}f(x)\,d\lambda(x).$

This element $f$ is called the density of $\nu$, or the density of $W$. Every
element of $L^{1}(\mathbb{R}^{d},\lambda)$ has a Borel-measurable
representative, so we may assume that $f$ is Borel-measurable. The density of
a continuous random variable is not necessarily representable by a continuous
function, and if $M$ happens to be equipped with a topology which generates
$\mathfrak{S}$, the continuity of $W$ as a random variable is unrelated to the
continuity of $W$ as a function.

  2. 2.

Mixtures of continuous random variables are continuous. Suppose that $\pi$ is
a Borel probability measure on $L^{1}(\mathbb{R}^{d})$ and $\nu$ is a random
continuous probability measure defined on $(M,\mathfrak{S},\mathbb{P})$ whose
density $d\nu/d\lambda$ with respect to Lebesgue measure $\lambda$ is
distributed according to $\pi$. Then for any Lebesgue measurable set
$A\subseteq\mathbb{R}^{d}$ we have (by Tonelli’s theorem)

$\displaystyle\mathbb{E}^{\pi}(\nu)(A):=\int\nu(A)\,d\pi(\nu)$
$\displaystyle=$
$\displaystyle\int\int\frac{d\nu}{d\lambda}(x)\mathbf{1}_{A}(x)\,d\lambda(x)\,d\pi(\nu)$
$\displaystyle=$
$\displaystyle\int\int\frac{d\nu}{d\lambda}(x)\,d\pi(\nu)\mathbf{1}_{A}(x)\,d\lambda(x)\,,$

so the mixture $\mathbb{E}^{\pi}(\nu)$ is a continuous probability measure
with density $\int\frac{d\nu}{d\lambda}\,d\pi(\nu)$. To see that the integral
of this density with respect to $\lambda$ is $1$, take $A=\mathbb{R}^{d}$
above. Even if $\nu$ a.s. has a $C^{\infty}$ density, it need not be true that
$\mathbb{E}^{\pi}(\nu)$ has a continuous density.

  3. 3.

Conditioning on positive probability events. Let $\nu$ be a Borel probability
measure on $\mathbb{R}^{d}$ and $W$ an $\mathbb{R}^{d}$-valued random variable
on $(M,\mathfrak{S},\mathbb{P})$ with distribution $\nu$. If
$E\in\mathfrak{S}$ is an event such that $\mathbb{P}(E)>0$, then the
conditional distribution of $W$ given $E$ is also continuous, with density
$\frac{\mathbb{P}(E\,|\,W=x)}{\mathbb{P}(E)}\frac{d\nu}{d\lambda}(x)$, since
for any Lebesgue-measurable set $A\subseteq\mathbb{R}^{d}$ we have

$\mathbb{P}(W\in A\,|\,E)=\frac{\mathbb{P}(E\cap(W\in
A))}{\mathbb{P}(E)}=\int\mathbf{1}_{A}(x)\frac{\mathbb{P}(E\,|W=x)}{\mathbb{P}(E)}\,\frac{d\nu}{d\lambda}(x)\,d\lambda(x)\,.$

Again, taking $A=\mathbb{R}^{d}$ shows that the integral of this density is
$1$.

## Appendix 2: Proof of Lemma 17

###### Proof.

By replacing $\mathscr{B}$ with $\mathscr{B}\cap B(x,R+r+1)$, which does not
alter the set whose volume is to be estimated, we may assume that
$\mathscr{B}$ is bounded.

Next, we reduce to the case where $\mathscr{B}$ is a polyhedron. Let
$\epsilon>0$. We claim that there is a convex polyhedron $P$ (with finitely
many facets) such that $\mathscr{B}\subseteq P\subset
N_{\epsilon}(\mathscr{B})$. To see this, let $y$ be any point in the interior
of $\mathscr{B}$ and let $\pi_{y}$ denote the radial projection about $y$ from
$\mathbb{R}^{d}\setminus\\{y\\}$ onto the unit sphere $\mathbb{S}_{y}$ that is
centered on $y$. For each point $z\in\partial\mathscr{B}$ and each supporting
hyperplane $H$ of $\mathscr{B}$ at $z$, the radial projection
$\pi_{y}\left(H\cap N_{\epsilon}(\mathscr{B})\right)$ is a neighbourhood of
$\pi_{y}(z)$ in the sphere $\mathbb{S}_{y}$. Since $\mathbb{S}_{y}$ is
compact, some finite collection of such projections covers it. The
corresponding finite collection of support hyperplanes defines a suitable
polyhedron $P$ (by taking the intersection of the half-spaces that they bound
which contain $\mathscr{B}$).

For each $z\in\mathscr{B}$, there exists a point $b\in\partial\mathscr{B}$
such that $\|z-b\|=\mathrm{dist}(z,\mathscr{B}^{c})$, and a supporting
hyperplane $H$ of $\mathscr{B}$ such that $H$ passes through $b$ and if $z\neq
B$ then $H$ is orthogonal to $b-z$. Then the translate of $H$ by distance
$\epsilon$ in the direction $b-z$ is a supporting hyperplane of
$N_{\epsilon}(\mathscr{B})$ and it follows that

$\mathrm{dist}\left(z,N_{\epsilon}\left(\mathscr{B})^{c}\right)\right)=\mathrm{dist}\left(z,\mathscr{B}^{c}\right)+\epsilon\,.$

Since $P\subseteq N_{\epsilon}(\mathscr{B})$, we have

$\mathrm{dist}\left(z,P^{c}\right)\leq\mathrm{dist}\left(z,\mathscr{B}^{c}\right)\,.$

It follows that

$\mathscr{B}\cap N_{r}\left(\mathscr{B}^{c}\right)\subset P\cap
N_{r+\epsilon}\left(P^{c}\right)\,.$

Hence if inequality (27) always holds for a bounded convex polyhedron $P$ with
finitely many faces in place of $\mathscr{B}$, (and $r+\epsilon$ in place of
$r$), then by taking the limit as $\epsilon\searrow 0$, we obtain (27) for a
general bounded convex body $\mathscr{B}$ and hence also for any convex body
$\mathscr{B}$.

Finally, we prove (27) in the case where $\mathscr{B}$ is a bounded convex
polyhedron $P$ which has finitely many faces. Suppose the facets of $P$ are
$F_{1},\dots,F_{n}$. (A facet is a face of co-dimension one.) For each
$i=1,\dots,n$, let $u_{i}$ be the outward-pointing normal vector of $F_{i}$,
and let the facet $F_{i}$ be contained in the hyperplane
$H_{i}=\\{z\in\mathbb{R}^{d}\,:\,z.u_{i}=h_{i}\\}$, so that

$P=\\{z\in\mathbb{R}^{d}\,:\,z.u_{i}<h_{i}\text{ for $i=1,\dots,n$}\\}\,.$

Let $\mathrm{Cut}(P)$ be the _cut-set_ of $P$, which is the set of points in
$P$ that are equidistant from at least two points in $\partial P$. Note that
$\lambda(\mathrm{Cut}(P))=0$ since $\mathrm{Cut}(P)$ is contained in the union
of the finite set of hyperplanes which bisect the dihedral angles of pairs of
hyperplanes $H_{i},H_{j}$. Also, $\lambda(\partial{P})=0$. So to prove (27),
if suffices to define a volume-preserving injection $\varphi$ from
$B(x,R)\cap(P^{\circ}\setminus\mathrm{Cut}(P))\cap N_{r}(P^{c})$ to
$B(x,R)\setminus B(x,R-r)$.

Partition the open set $(P^{\circ}\setminus\mathrm{Cut}(P))\cap N_{r}(P^{c})$
into finitely many pieces $Q_{1},\dots,Q_{n}$, where each piece $Q_{i}$
consists of the points whose closest facet of $P$ is a given facet $F_{i}$.
$Q_{i}$ is an open convex polyhedron contained in the $r$-neighbourhood of the
facet $F_{i}$.

The map $\varphi$ will translate each point of $Q_{i}\cap B(x,R)$ by a non-
negative multiple of $u_{i}$, where the multiple may vary from point to point.
Specifically, for any $z\in Q_{i}\cap B(x,R)$, define

$a(z)=\sup(\\{t\in\mathbb{R}\,:\,z+tu_{i}\in
Q_{i}\\})=h_{i}-z.u_{i}=\mathrm{dist}(z,P^{c})\,,$

and

$b(z)=\sup(\\{t\in\mathbb{R}\,;\,z+tu_{i}\in B(x,R)\\})\,.$

Then define

$\varphi(z):=z+\max(0,b(z)-a(z))\,.$

Note that if $z^{\prime}=z+tu_{i}$ is also in $Q_{i}\cap B(x,R)$ then
$b(z^{\prime})=b(z)-t$ and $a(z^{\prime})=a(z)-t$, since both $B(x,R)$ and
$Q_{i}$ are convex. Hence $\varphi(z^{\prime})=\varphi(z)+tu_{i}$. Informally,
the map $\varphi$ slides each line segment of the form
$\\{z+tu_{i}:t\in\mathbb{R},z+tu_{i}\in B(x,R)\cap Q_{i}\\}$ as far as
possible in the direction $u_{i}$ (parallel to the line segment) such that it
remains a subset of $B(x,R)$. Each of these line segments has length at most
$r$. It follows that every point in the image of $\varphi$ is within distance
$r$ of the boundary of $B(x,R)$. The restriction of $\varphi$ to $Q_{i}\cap
B(x,R)$ is a smooth volume-preserving map, since both $a(z)$ and $b(z)$ depend
smoothly on $z$, and in an appropriate coordinate system its Jacobian matrix
is upper unitriangular.

Finally, we must check that $\varphi$ is an injection. From the description in
terms of sliding line segments, we see that the restriction of $\varphi$ to
each piece $Q_{i}\cap B(x,R)$ is an injection. Secondly, if $z\in Q_{i}\cap
B(x,R)$, then the unique closest facet of $P$ to $\varphi(z)$ is $F_{i}$.
Therefore the images of distinct pieces under $\varphi$ are disjoint and we
are done.

∎