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Fluctuations of the connectivity threshold and largest nearest-neighbour link

Mathew D. Penrose1 and Xiaochuan Yang1
University of Bath and Brunel University London
(April 2, 2025)
Abstract

Consider a random uniform sample of nn points in a compact region AA of Euclidean dd-space, d2d\geq 2, with a smooth or (when d=2d=2) polygonal boundary. Fix kk\in\mathbb{N}. Let Tn,kT_{n,k} be the threshold rr at which the geometric graph on these nn vertices with distance parameter rr becomes kk-connected. We show that if d=2d=2 then n(π/|A|)Tn,12lognn(\pi/|A|)T_{n,1}^{2}-\log n is asymptotically standard Gumbel. For (d,k)(2,1)(d,k)\neq(2,1), it is n(θd/|A|)Tn,kd(22/d)logn(42k2/d)loglognn(\theta_{d}/|A|)T_{n,k}^{d}-(2-2/d)\log n-(4-2k-2/d)\log\log n that converges in distribution to a nondegenerate limit, where θd\theta_{d} is the volume of the unit ball. The limit is Gumbel with scale parameter 2 except when (d,k)=(2,2)(d,k)=(2,2) where the limit is two component extreme value distributed. The different cases reflect the fact that boundary effects are more more important in some cases than others. We also give similar results for the largest kk-nearest neighbour link Un,kU_{n,k} in the sample, and show Tn,k=Un,kT_{n,k}=U_{n,k} with high probability. We provide estimates on rates of convergence and give similar results for Poisson samples in AA. Finally, we give similar results even for non-uniform samples, with a less explicit sequence of centring constants.
Keywords: Connectivity threshold; weak limit; Poisson process; Gumbel distribution.

11footnotetext: Supported by EPSRC grant EP/T028653/1

1 Introduction

1.1 Overview and motivation

This paper is concerned with the threshold at which the random geometric graph becomes connected. This graph is defined as follows. Let dd\in\mathbb{N}. Given a finite set 𝒳d\mathcal{X}\subset\mathbb{R}^{d}, and r>0r>0, the geometric graph G(𝒳,r)G(\mathcal{X},r) has vertex set 𝒳\mathcal{X}, with an edge drawn between any two vertices at Euclidean distance at most rr from each other. We say G(𝒳,r)G(\mathcal{X},r) is kk-connected if it is not possible to disconnect the graph by removing k1k-1 or fewer vertices. (in particular, 1-connectivity is the same as connectivity). The kk-connectivity threshold of 𝒳\mathcal{X} is the number

Mk(𝒳):=inf{r>0:G(𝒳,r) is k-connected}.\displaystyle M_{k}(\mathcal{X}):=\inf\{r>0:G(\mathcal{X},r)\mbox{ is $k$-connected}\}.

An alternative characterisation of M1(𝒳)M_{1}(\mathcal{X}) is in terms of minimal spanning tree (MST). A MST on 𝒳\mathcal{X} is a tree spanning 𝒳\mathcal{X} that minimises the total (Euclidean) length of the edges. It is not hard to see that M1(𝒳)M_{1}(\mathcal{X}) equals the longest edge length of a MST on 𝒳\mathcal{X}.

For the random geometric graph, the vertex set 𝒳\mathcal{X} is given by the set of points of a Poisson point process 𝒫n\mathcal{P}_{n} in d\mathbb{R}^{d} with intensity measure nνn\nu, where ν\nu is a probability measure on d\mathbb{R}^{d} with probability density function f:d[0,)f:\mathbb{R}^{d}\to[0,\infty), and n(0,)n\in(0,\infty) is the mean number of vertices. Alternatively, for nn\in\mathbb{N} we can take 𝒳=𝒳n\mathcal{X}=\mathcal{X}_{n}, where 𝒳n\mathcal{X}_{n} denotes a binomial point process whose points are nn independent random dd-vectors with common density ff. Since the vertices are placed randomly in d\mathbb{R}^{d}, the thresholds Mk(𝒳n)M_{k}(\mathcal{X}_{n}) and

Mn,k:=Mk(𝒫n)=inf{r>0:G(𝒫n,r) is k-connected}\displaystyle M_{n,k}:=M_{k}(\mathcal{P}_{n})=\inf\{r>0:G(\mathcal{P}_{n},r)\mbox{ is $k$-connected}\} (1.1)

are random variables.

In this paper we investigate the limiting behaviour of the connectivity threshold Mn,kM_{n,k} and Mk(𝒳n)M_{k}(\mathcal{X}_{n}) for large nn and fixed kk\in\mathbb{N}. We assume throughout that d2d\geq 2. We consider a broad class of measures ν\nu, subject to the working assumption that ν\nu has compact support AdA\subset\mathbb{R}^{d}, and its density ff is continuous and bounded away from zero on AA. As nn grows, the spacing between vertices becomes smaller, so one expects to have Mn,k0M_{n,k}\to 0 as nn\to\infty. A simple consideration by computing typical spacing of vertices leads to the belief that Mn,kM_{n,k} should decay more slowly than n1/dn^{-1/d}, in the sense that nMn,kdnM_{n,k}^{d} should tend to infinity in probability. In the special case where ν\nu is the uniform distribution on [0,1]d[0,1]^{d}, it is known [9, 11, 8], that there is an explicit sequence of centring constants (an)n1(a_{n})_{n\geq 1} satisfying ana_{n}\to\infty as nn\to\infty such that

nMn,kdan𝑑X;nMk(𝒳n)dan𝑑X,\displaystyle nM_{n,k}^{d}-a_{n}\overset{d}{\longrightarrow}X;\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ nM_{k}(\mathcal{X}_{n})^{d}-a_{n}\overset{d}{\longrightarrow}X, (1.2)

where XX is an explicit non-degenerate random variable. Clearly (1.2) is what is needed to determine limn[Mn,krn]\lim_{n\to\infty}\mathbb{P}[M_{n,k}\leq r_{n}] for any sequence (rn)n1(r_{n})_{n\geq 1} such that the limit exists.

In the present paper, we show that (1.2) holds for suitable ana_{n} and XX, for a broad class of measures ν\nu. While one might perhaps anticipate that the limiting behaviour of the form (1.2) would carry over from the uniform distribution on [0,1]d[0,1]^{d} to more general ν\nu satisfying our working assumption, proving this seems to be considerably harder that one might naively expect, and in the last 20 years or so there has been limited progress in proving such results. It turns out that even in the uniform case where ff is constant on AA, boundary effects are important because the ‘most isolated’ vertex is likely to lie near the boundary when d3d\geq 3. Thus the formula for ana_{n}, even when ν\nu is uniform on [0,1]d[0,1]^{d}, is quite complicated (see (1.9) below) due to having to consider all of the different kinds of faces making up the boundary, and does not necessarily provide much insight into the appropriate choice of centring constants for other AA. In the non-uniform case, determining appropriate constants ana_{n} is even harder because they depend in a delicate way on how ff approaches its minimum, both in the interior and on the boundary of AA.

In this work we chiefly consider the case where A\partial A is smooth. In the uniform case we determine an explicit sequence of centring constants ana_{n} such that (1.2) holds for suitable XX. In the non-uniform case we are still (in most cases) able to derive (1.2) on taking ana_{n} to be the median of the distribution of nMn,kdnM_{n,k}^{d}. Part of our proof involves approximating AA with a polyhedral set AnA_{n} with the spacing between vertices of AnA_{n} decreasing slowly as nn becomes large. This technique was developed recently for certain random coverage problems in [15, 4], and its availability is one reason why this problem is more tractable now than it was before.

We also address the case where d=2d=2 and AA is polygonal; this case could be of importance for some applications, and it turns out that the effects of corners are asymptotically negligible for d=2d=2. We hope to deal with the case of polytopal AA in dimension d3d\geq 3 in future work.

Understanding the connectivity threshold is important for a variety of applications. In telecommunications, the vertices could represent mobile transceivers and one might be interested in whether the network of transceivers is connected; see e.g. [3]. In topological data analysis (TDA), detecting connectivity is a fundamental step for inspecting all other higher dimensional topological features (here the dimension of the ambient space may be very high). See also applications in machine learning (clustering), statistical tests (e.g. for outliers), spatial epidemics or forest fires (see the description in [9])

Note that (1.2) implies the weaker statement that the sequence (nMn,kdan)n1(nM_{n,k}^{d}-a_{n})_{n\geq 1} is tight as nn\to\infty, and even in the (rather exceptional) cases where we cannot prove (1.2), we shall prove this weaker statement. Tightness in turn implies nMn,kd/an1nM_{n,k}^{d}/a^{\prime}_{n}\to 1 in probability as nn\to\infty, for any sequence (an)n1(a^{\prime}_{n})_{n\geq 1} satisfying an/an1a^{\prime}_{n}/a_{n}\to 1 as nn\to\infty. Another direction of research (not followed in the present paper) is to improve this to almost sure convergence, i.e. a strong law of large numbers (SLLN), under the natural coupling of (𝒳n,n1)(\mathcal{X}_{n},n\geq 1):

nMk(𝒳n)dana.s.1,\displaystyle\frac{nM_{k}(\mathcal{X}_{n})^{d}}{a^{\prime}_{n}}\overset{a.s.}{\longrightarrow}1, (1.3)

or to establish (1.3) for some ana^{\prime}_{n} even in cases when (1.2) is not known; see (1.7), (1.8) below.

1.2 The largest kk-nearest-neighbour link

Given xdx\in\mathbb{R}^{d} and r>0r>0, we denote the closed Euclidean ball of radius rr, centred at xx, by either Br(x)B_{r}(x) or B(x,r)B(x,r). Given finite 𝒳d\mathcal{X}\subset\mathbb{R}^{d} we define the largest kk-nearest-neighbour link Lk(𝒳)L_{k}(\mathcal{X}) of 𝒳\mathcal{X}, by

Lk(𝒳):={maxx𝒳(inf{r>0:𝒳(Br(x))>k})if|𝒳|k+1,0otherwise,\displaystyle L_{k}(\mathcal{X}):=\begin{cases}\max_{x\in\mathcal{X}}\left(\inf\{r>0:\mathcal{X}(B_{r}(x))>k\}\right)&{\rm if}\leavevmode\nobreak\ |\mathcal{X}|\geq k+1,\\ 0&{\rm otherwise,}\end{cases} (1.4)

where |𝒳||\mathcal{X}| denotes the number of elements of 𝒳\mathcal{X} and 𝒳():=|𝒳|\mathcal{X}(\cdot):=|\mathcal{X}\cap\cdot| denotes the counting measure associated with 𝒳\mathcal{X}. Note that if |𝒳|k+1|\mathcal{X}|\geq k+1 and x𝒳x\in\mathcal{X}, then inf{r>0:𝒳(Br(x))>k}\inf\{r>0:\mathcal{X}(B_{r}(x))>k\} is the distance from xx to its kk-nearest neighbour in 𝒳\mathcal{X}. Note also that Lk(𝒳)Mk(𝒳)L_{k}(\mathcal{X})\leq M_{k}(\mathcal{X}) since if r<Lk(𝒳)r<L_{k}(\mathcal{X}) and |𝒳|k+1|\mathcal{X}|\geq k+1, then G(𝒳,r)G(\mathcal{X},r) has at least one vertex with degree less than kk and therefore is not kk-connected, so rMk(𝒳)r\leq M_{k}(\mathcal{X}).

Our analysis of Mn,kM_{n,k} and Mk(𝒳n)M_{k}(\mathcal{X}_{n}) will involve first investigating Ln,k:=Lk(𝒫n)L_{n,k}:=L_{k}(\mathcal{P}_{n}) and Lk(𝒳n)L_{k}(\mathcal{X}_{n}). A priori, it is not obvious that Lk(𝒳n)L_{k}(\mathcal{X}_{n}) is a sharp lower bound for Mk(𝒳n)M_{k}(\mathcal{X}_{n}); nevertheless, it is known in some cases (see Section 1.3) that Mk(𝒳n)M_{k}(\mathcal{X}_{n}) enjoys the same limiting behaviour as Lk(𝒳n)L_{k}(\mathcal{X}_{n}), and even sometimes that

limn[Lk(𝒳n)=Mk(𝒳n)]=1.\displaystyle\lim_{n\to\infty}\mathbb{P}[L_{k}(\mathcal{X}_{n})=M_{k}(\mathcal{X}_{n})]=1. (1.5)

Equation (1.5), when true, says that with probability tending to 1 as nn\to\infty the point set 𝒳n\mathcal{X}_{n} has the following property: If we start with no edges between the vertices of 𝒳n\mathcal{X}_{n}, and then add edges one by one in order of increasing Euclidean length until we arrive at a kk-connected graph, then just before the addition of the last edge we still have a vertex of degree less than kk; if k=1k=1 then just before the addition of the last edge we have exactly two components, one of which is a singleton.

The largest kk-nearest neighbour link Lk(𝒳n)L_{k}(\mathcal{X}_{n}) (or Lk(𝒫n)L_{k}(\mathcal{P}_{n})) is of interest in its own right. To quote [1], it ‘comes up in almost all discussions of computational complexity involving nearest neighbours’. See e.g. [9] for further motivation. As with Mn,kM_{n,k}, its limiting behaviour has previously been studied on the torus, and the unit cube, and only at the level of a SLLN for regions with smooth or polytopal boundary. By providing a limiting distribution for Lk(𝒳n)L_{k}(\mathcal{X}_{n}) for regions with smooth or polygonal boundary, we here add significantly to this body of work, too.

1.3 Literature review

Before stating our main results, let us give a literature review on this topic. Under our working assumption (WA), we use throughout the notation

f0:=infxAf(x);f1:=infxAf(x);fmax:=supxAf(x).\displaystyle f_{0}:=\inf_{x\in A}f(x);\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ f_{1}:=\inf_{x\in\partial A}f(x);\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ f_{\rm max}:=\sup_{x\in A}f(x). (1.6)

Note that 0<f0f1fmax<0<f_{0}\leq f_{1}\leq f_{\rm max}<\infty under our WA. Let θd\theta_{d} denote the volume of a dd-dimensional unit ball. i.e. θd:=πd/2/Γ(1+d/2)\theta_{d}:=\pi^{d/2}/\Gamma(1+d/2),

In the case where AA is the flat torus 𝕋d\mathbb{T}^{d} of any dimension, it is known [8, Theorem 13.6] that under the natural coupling of (𝒳n,n1)(\mathcal{X}_{n},n\geq 1) we have

limn(θdn(Mk(𝒳n))dlogn)=1f0a.s.\displaystyle\lim_{n\to\infty}\left(\frac{\theta_{d}n(M_{k}(\mathcal{X}_{n}))^{d}}{\log n}\right)=\frac{1}{f_{0}}\quad a.s.

The dimensionality and the density play a crucial role when one considers compact sets with boundaries. More precisely, if AA has a smooth boundary, it is proved for k=1k=1 in [12, 13], and for general kk in [8], that

limn(θdn(Mk(𝒳n))dlogn)=limn(θdn(Lk(𝒳n))dlogn)=max(1f0,22/df1)a.s.\displaystyle\lim_{n\to\infty}\left(\frac{\theta_{d}n(M_{k}(\mathcal{X}_{n}))^{d}}{\log n}\right)=\lim_{n\to\infty}\left(\frac{\theta_{d}n(L_{k}(\mathcal{X}_{n}))^{d}}{\log n}\right)=\max\Big{(}\frac{1}{f_{0}},\frac{2-2/d}{f_{1}}\Big{)}\quad a.s. (1.7)

When AA is a convex polytope, it is proved in [17] that

limn(n(Mk(𝒳n))dlogn)=limn(n(Lk(𝒳n))dlogn)=maxφΦ(A)(D(φ)fφρφd)a.s.\displaystyle\lim_{n\to\infty}\left(\frac{n(M_{k}(\mathcal{X}_{n}))^{d}}{\log n}\right)=\lim_{n\to\infty}\left(\frac{n(L_{k}(\mathcal{X}_{n}))^{d}}{\log n}\right)=\max_{\varphi\in\Phi^{*}(A)}\Big{(}\frac{D(\varphi)}{f_{\varphi}\rho_{\varphi}d}\Big{)}\quad a.s. (1.8)

where Φ\Phi^{*} denotes the collection of all faces of φ\varphi of all dimensions (including AA itself, considered as a face of dimension dd), and where D(φ)D(\varphi) is the dimension of face φ\varphi, and where fφf_{\varphi} denotes the infimum of ff over face φ\varphi and ρφ\rho_{\varphi} is the angular volume of face φ\varphi.

Less is known about the fluctuations of nMk(𝒳n)dannM_{k}(\mathcal{X}_{n})^{d}-a_{n}. Weak limit results of the type (1.2) are proved for two cases in [9, 11]. The first case is when ff is uniform on 𝕋d\mathbb{T}^{d} for any d2d\geq 2. In this case, by [8, Corollary 13.20], one has

θdn(Mk(𝒳n))dlogn(k1)loglogn+log((k1)!)𝑑𝖦𝗎,\displaystyle\theta_{d}n(M_{k}(\mathcal{X}_{n}))^{d}-\log n-(k-1)\log\log n+\log((k-1)!)\overset{d}{\longrightarrow}{\mathsf{Gu}},

where 𝖦𝗎{\mathsf{Gu}} denotes a standard Gumbel random variable, i.e. one with cumulative distribution function F(x)=exp(ex),xF(x)=\exp(-e^{-x}),x\in\mathbb{R}. (For a,b>0a\in\mathbb{R},b>0 the random variable b𝖦𝗎+ab{\mathsf{Gu}}+a is said to be Gumbel distributed with scale parameter bb and location parameter aa.)

The second case is when ff is uniform on [0,1]d[0,1]^{d}. For this case, we describe only the results for k=1k=1 from [8] but the case of general kk is also treated there. When ff is uniform on [0,1]d[0,1]^{d}, one has by [8, Corollary 13.21] that

22dθdn(M1(𝒳n))d(2/d)logn+(d3+2/d)loglogn\displaystyle 2^{2-d}\theta_{d}n(M_{1}(\mathcal{X}_{n}))^{d}-(2/d)\log n+(d-3+2/d)\log\log n
+log((222/dd(d1))(θdd)3d2/dθd1d2)𝑑𝖦𝗎.\displaystyle+\log\left(\Big{(}\frac{2^{2-2/d}}{d(d-1)}\Big{)}(\theta_{d}d)^{3-d-2/d}\theta_{d-1}^{d-2}\right)\overset{d}{\longrightarrow}{\mathsf{Gu}}. (1.9)

Similar results hold for Lk(𝒳n)L_{k}(\mathcal{X}_{n}); see [8, Theorems 8.3 and 8.4]. Moreover it is known that (1.5) holds. Also these results hold for 𝒫n\mathcal{P}_{n} instead of 𝒳n\mathcal{X}_{n}.

So far as we know, there is no weak limit result for other shaped boundaries or for non-uniform distributions (until now). In the special case where d=2d=2 and ff is uniformly distributed in a disk, [3] gives a partial result in the direction of a weak limit.

The main results of this paper considerably generalise previous findings and deepen the understanding of the connectivity threshold in terms of the geometry of AA. In the uniform case, we also provide a bound on the rate of convergence that is new for all shapes under the WA.

We end this section by mentioning some related results. It is natural to ask what happens if we drop the working assumption and take the support of ff to be unbounded. Penrose [10] found that the scaling is completely different in the case of standard Gaussian density in d\mathbb{R}^{d}; see also Hsing and Rootzén [5] for a significant extension in dimension two, where a class of elliptically contoured distributions are included, e.g. Gaussian densities with correlated coordinates. Gupta and Iyer [2] give a limiting distribution and SLLN for Ln,1L_{n,1} for a class of radially symmetric densities with unbounded support, including cases where Ln,1L_{n,1} (and hence also Mn,1M_{n,1}) does not even tend to zero.

1.4 Main results

Throughout this paper, cc and cc^{\prime} denote positive finite constants whose values may vary from line to line and do not depend on nn. Also if n0(0,)n_{0}\in(0,\infty) and f(n),g(n)f(n),g(n) are two functions, defined for all nn0n\geq n_{0} with g(n)>0g(n)>0 for all nn0n\geq n_{0}, the notation f(n)=O(g(n))f(n)=O(g(n)) as nn\to\infty means that lim supn(|f(n)|/g(n))<\limsup_{n\to\infty}(|f(n)|/g(n))<\infty. If also f(n)>0f(n)>0 for all nn0n\geq n_{0}, we use notation f(n)=Θ(g(n))f(n)=\Theta(g(n)) to mean that both f(n)=O(g(n))f(n)=O(g(n)) and g(n)=O(f(n))g(n)=O(f(n)).

Given d,kd,k\in\mathbb{N}, define the constant

cd,k:=θd11θd11/d(22/d)k2+1/d21k/(k1)!\displaystyle c_{d,k}:=\theta_{d-1}^{-1}\theta_{d}^{1-1/d}(2-2/d)^{k-2+1/d}2^{1-k}/(k-1)! (1.10)

In this paper, given AdA\subset\mathbb{R}^{d}, we say that AA has C2C^{2} boundary (or for short: AC2\partial A\in C^{2}) if for each xAx\in\partial A, the topological boundary of AA, there exists a rigid motion ρx\rho_{x} of d\mathbb{R}^{d}, an open set Uxd1U_{x}\subset\mathbb{R}^{d-1} and a C2C^{2} function fx:d1f_{x}:\mathbb{R}^{d-1}\to\mathbb{R} such that ρx(AU)=ρx(U)epi(fx)\rho_{x}(A\cap U)=\rho_{x}(U)\cap\mathrm{epi}(f_{x}), where epi(fx):={(u,z)d1×:zfx(u)}\mathrm{epi}(f_{x}):=\{(u,z)\in\mathbb{R}^{d-1}\times\mathbb{R}:z\geq f_{x}(u)\}, the closed epigraph of fxf_{x}.

For compact AdA\subset\mathbb{R}^{d} with C2C^{2} or polytopal boundary, let |A||A| denote the volume (Lebesgue measure) of AA, and |A||\partial A| the perimeter of AA, i.e. the (d1)(d-1)-dimensional Hausdorff measure of A\partial A. Also define

σA:=|A||A|11/d.\displaystyle\sigma_{A}:=\frac{|\partial A|}{|A|^{1-1/d}}. (1.11)

Note that σAd\sigma_{A}^{d} is sometimes called the isoperimetric ratio of AA, and is at least ddθdd^{d}\theta_{d} by the isoperimetric inequality.

Theorem 1.1 (Weak limits in the uniform case).

Suppose either that d2d\geq 2 and AA a compact subset of d\mathbb{R}^{d} with C2C^{2} boundary, or that d=2d=2 and AA is a convex polygon. Let ff0𝟏Af\equiv f_{0}{\bf 1}_{A} with f0=|A|1f_{0}=|A|^{-1}. Let β\beta\in\mathbb{R}. Then, if d=2d=2, we have as nn\to\infty that

[nf0πM1(𝒳n)2lognβ]=exp(σAπ1/2eβ/22(logn)1/2)eeβ+O((logn)1);\displaystyle\mathbb{P}[nf_{0}\pi M_{1}(\mathcal{X}_{n})^{2}-\log n\leq\beta]=\exp\Big{(}-\frac{\sigma_{A}\pi^{1/2}e^{-\beta/2}}{2(\log n)^{1/2}}\Big{)}e^{-e^{-\beta}}+O((\log n)^{-1}); (1.12)
[nf0πMn,12lognβ]=exp(σAπ1/2eβ/22(logn)1/2)eeβ+O((logn)1).\displaystyle\mathbb{P}[nf_{0}\pi M_{n,1}^{2}-\log n\leq\beta]=\exp\Big{(}-\frac{\sigma_{A}\pi^{1/2}e^{-\beta/2}}{2(\log n)^{1/2}}\Big{)}e^{-e^{-\beta}}+O((\log n)^{-1}). (1.13)

Also, given kk\in\mathbb{N}, set

un,k\displaystyle u_{n,k} :=[nθdf0Mn,kd(22/d)logn+(42k2/d)loglognβ];\displaystyle:=\mathbb{P}[n\theta_{d}f_{0}M_{n,k}^{d}-(2-2/d)\log n+(4-2k-2/d)\log\log n\leq\beta];
un,k\displaystyle u^{\prime}_{n,k} :=[nθdf0Mk(𝒳n)d(22/d)logn+(42k2/d)loglognβ].\displaystyle:=\mathbb{P}[n\theta_{d}f_{0}M_{k}(\mathcal{X}_{n})^{d}-(2-2/d)\log n+(4-2k-2/d)\log\log n\leq\beta].

If d=2d=2 then as nn\to\infty

un,2=exp(π1/2σAeβ/2loglogn8logneβloglognlogn)exp(eβπ1/2σAeβ/24)\displaystyle u_{n,2}=\exp\Big{(}-\frac{\pi^{1/2}\sigma_{A}e^{-\beta/2}\log\log n}{8\log n}-\frac{e^{-\beta}\log\log n}{\log n}\Big{)}\exp\Big{(}-e^{-\beta}-\frac{\pi^{1/2}\sigma_{A}e^{-\beta/2}}{4}\Big{)}
+O(1logn),\displaystyle+O\big{(}\frac{1}{\log n}\big{)},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ (1.14)

and likewise for un,2u^{\prime}_{n,2}. If d3d\geq 3, or if d=2,k3d=2,k\geq 3 we have as nn\to\infty that

un,k=exp(cd,kσAeβ/2(k2+1/d)2loglogn(11/d)logn)exp(cd,kσAeβ/2)+O(1logn),\displaystyle u_{n,k}=\exp\Big{(}-\frac{c_{d,k}\sigma_{A}e^{-\beta/2}(k-2+1/d)^{2}\log\log n}{(1-1/d)\log n}\Big{)}\exp(-c_{d,k}\sigma_{A}e^{-\beta/2})+O\big{(}\frac{1}{\log n}\big{)}, (1.15)

and likewise for un,ku^{\prime}_{n,k}. Also (1.5) holds, and all of the above results hold with Ln,kL_{n,k} (resp. Lk(𝒳n)L_{k}(\mathcal{X}_{n})) instead of Mn,kM_{n,k} (resp. Mk(𝒳n)M_{k}(\mathcal{X}_{n})).

Remark 1.2.
  1. i)

    The statements about Ln,kL_{n,k} and Lk(𝒳n)L_{k}(\mathcal{X}_{n}) in this theorem are spelt out in Corollaries 6.3 and 6.4.

  2. ii)

    These results imply certain convergence in distribution results. Namely, nθdf0Mn,kdn\theta_{d}f_{0}M_{n,k}^{d}, suitably centred, is asymptotically Gumbel distributed with scale parameter 11 when d=2,k=1d=2,k=1 but with scale parameter 22 when d3d\geq 3 or d=2,k3d=2,k\geq 3. When d=2,k=2d=2,k=2 the limiting distribution is a so-called two-component extreme value distribution (TCEV), that is, a probability distribution with cumulative distribution function (cdf) given by the product of two Gumbel cdfs with different scale parameters, in this case 1 and 2. The terminology TCEV was introduced in the hydrology literature [18].

  3. iii)

    We have included a multiplicative correction factor in each of (1.12)–(1.15), namely the first factor in the right hand side in each case, because this factor tends to 1 very slowly, especially in (1.12) and (1.13) where the correction factor is 1+O((logn)1/2)1+O((\log n)^{-1/2}) so for moderately large values of nn the limiting Gumbel distribution with cdf eeβe^{-e^{-\beta}} is not very close to the centred distribution of nf0θdMn,1dnf_{0}\theta_{d}M_{n,1}^{d}. If d3d\geq 3 it is possible to give some extra terms in the correction and improve the error bound to O((logn)ε2)O((\log n)^{\varepsilon-2}).

  4. iv)

    The error bounds above are for fixed β\beta but we would need to make them uniform in β\beta for an error bound in the Kolmogorov distance between probability distributions. We do not address this in this paper.

  5. v)

    It seems likely that our results carry over to the case where d=2d=2 and AA is a non-convex polygon. Our main reason for restricting attention to polygons that are convex its that some of our arguments in Section 4 are based on results from [17] that are stated there only for convex polytopes.

We now give a result for the general non-uniform case; that is, we still use our WA on ff, but drop the stronger assumption that ff is constant on AA. Recall f0,f1f_{0},f_{1} defined at (1.6). In this more general case, subject to the condition f1f0(22/d)f_{1}\neq f_{0}(2-2/d), we still provide a result along the lines of (1.2), but now, instead of using the explicit centring constants an=(22/d)logn(42k2/d)𝟏{d3ork2}loglogna_{n}=(2-2/d)\log n-(4-2k-2/d){\bf 1}\{d\geq 3\leavevmode\nobreak\ {\rm or}\leavevmode\nobreak\ k\geq 2\}\log\log n as in Theorem 1.1, we take ana_{n} to be the median of the distribution nMndnM_{n}^{d}. In the case f1=f0(22/d)f_{1}=f_{0}(2-2/d) we prove only the weaker result that our sequence of centred random variables is tight.

Given a random variable XX, let μ(X):=inf{x:[Xx]1/2}\mu(X):=\inf\{x\in\mathbb{R}:\mathbb{P}[X\leq x]\geq 1/2\}, the median of the distribution of XX. Note that μ(𝖦𝗎)=log(log2)\mu({\mathsf{Gu}})=-\log(\log 2), so for α>0\alpha>0, the random variable α(𝖦𝗎+log(log2))\alpha({\mathsf{Gu}}+\log(\log 2)) has a Gumbel distribution with median 0 and with scale parameter α\alpha.

Theorem 1.3 (Weak limit in the non-uniform case).

Suppose our working assumption applies, either with d2d\geq 2 and AA a compact subset of d\mathbb{R}^{d} with C2C^{2} boundary, or with d=2d=2 and AA a convex polygon. Let kk\in\mathbb{N}.

(i) If f1>f0(22/d)f_{1}>f_{0}(2-2/d), then as nn\to\infty,

nMk(𝒳n)dnμ(Mk(𝒳n))d𝑑(θdf0)1(𝖦𝗎+loglog2);\displaystyle nM_{k}(\mathcal{X}_{n})^{d}-n\mu(M_{k}(\mathcal{X}_{n}))^{d}\overset{d}{\longrightarrow}(\theta_{d}f_{0})^{-1}({\mathsf{Gu}}+\log\log 2); (1.16)
nLk(𝒳n)dnμ(Lk(𝒳n))d𝑑(θdf0)1(𝖦𝗎+loglog2);\displaystyle nL_{k}(\mathcal{X}_{n})^{d}-n\mu(L_{k}(\mathcal{X}_{n}))^{d}\overset{d}{\longrightarrow}(\theta_{d}f_{0})^{-1}({\mathsf{Gu}}+\log\log 2); (1.17)
nMn,kdnμ(Mn,k)d𝑑(θdf0)1(𝖦𝗎+loglog2);\displaystyle nM_{n,k}^{d}-n\mu(M_{n,k})^{d}\overset{d}{\longrightarrow}(\theta_{d}f_{0})^{-1}({\mathsf{Gu}}+\log\log 2); (1.18)
nLn,kdnμ(Ln,k)d𝑑(θdf0)1(𝖦𝗎+loglog2).\displaystyle nL_{n,k}^{d}-n\mu(L_{n,k})^{d}\overset{d}{\longrightarrow}(\theta_{d}f_{0})^{-1}({\mathsf{Gu}}+\log\log 2). (1.19)

(ii) If f1<f0(22/d)f_{1}<f_{0}(2-2/d), then as nn\to\infty,

nMk(𝒳n)dnμ(Mk(𝒳n))d𝑑(2/(θdf1))(𝖦𝗎+loglog2);\displaystyle nM_{k}(\mathcal{X}_{n})^{d}-n\mu(M_{k}(\mathcal{X}_{n}))^{d}\overset{d}{\longrightarrow}(2/(\theta_{d}f_{1}))({\mathsf{Gu}}+\log\log 2); (1.20)
nLk(𝒳n)dnμ(Lk(𝒳n))d𝑑(2/(θdf1))(𝖦𝗎+loglog2).\displaystyle nL_{k}(\mathcal{X}_{n})^{d}-n\mu(L_{k}(\mathcal{X}_{n}))^{d}\overset{d}{\longrightarrow}(2/(\theta_{d}f_{1}))({\mathsf{Gu}}+\log\log 2). (1.21)
nMn,kdnμ(Mn,kd)𝑑(2/(θdf1))(𝖦𝗎+loglog2);\displaystyle nM_{n,k}^{d}-n\mu(M_{n,k}^{d})\overset{d}{\longrightarrow}(2/(\theta_{d}f_{1}))({\mathsf{Gu}}+\log\log 2); (1.22)
nLn,kdnμ(Ln,k)d𝑑(2/(θdf1))(𝖦𝗎+loglog2);\displaystyle nL_{n,k}^{d}-n\mu(L_{n,k})^{d}\overset{d}{\longrightarrow}(2/(\theta_{d}f_{1}))({\mathsf{Gu}}+\log\log 2); (1.23)

(iii) In all cases, including when f1=f0(22/d)f_{1}=f_{0}(2-2/d), (1.5) holds, and also the family of random variables (n(Mn,kdμ(Mn,k)d))n1(n(M_{n,k}^{d}-\mu(M_{n,k})^{d}))_{n\geq 1} is tight. Likewise the collection of random variables (n(Ln,kdμ(Ln,k)d))n1(n(L_{n,k}^{d}-\mu(L_{n,k})^{d}))_{n\geq 1} is tight, as are the sequences (n(Mk(𝒳n)dμ(Mk(𝒳n))d))n1(n(M_{k}(\mathcal{X}_{n})^{d}-\mu(M_{k}(\mathcal{X}_{n}))^{d}))_{n\geq 1} and (n(Lk(𝒳n)dμ(Lk(𝒳n))d))n1(n(L_{k}(\mathcal{X}_{n})^{d}-\mu(L_{k}(\mathcal{X}_{n}))^{d}))_{n\geq 1}.

Before proceeding to proofs, we give a rough calculation indicating why, in the uniform case with ff0𝟏Af\equiv f_{0}{\bf 1}_{A}, we might expect to see qualitative differences between the cases with d=2d=2 and k=1k=1 or k=2k=2, and other cases, as seen in Theorem 1.1. Suppose we take a sequence of distance parameters rnr_{n} with nf0θdrnd=logn+(k1)loglogn+cnf_{0}\theta_{d}r_{n}^{d}=\log n+(k-1)\log\log n+c for some constant cc. Given rnr_{n}, let FnoF^{o}_{n}, FnF_{n}^{\partial} be the number of vertices of degree less than kk in the interior of AA, respectively near the boundary of AA. We give a rough calculation suggesting that for d3d\geq 3 we have 𝔼[Fn]𝔼[Fno]\mathbb{E}[F_{n}^{\partial}]\gg\mathbb{E}[F_{n}^{o}] so the boundary region dominates, while for d=2d=2, it depends on the value of kk whether the interior or boundary region dominates. Firstly,

𝔼[Fno]n((nf0θdrnd)k1/(k1)!)enf0θdrnd(ec)/(k1)!.\mathbb{E}[F_{n}^{o}]\approx n((nf_{0}\theta_{d}r_{n}^{d})^{k-1}/(k-1)!)e^{-nf_{0}\theta_{d}r_{n}^{d}}\sim(e^{-c})/(k-1)!.

For FnF_{n}^{\partial}, note that for small positive ss the volume of the intersection of AA with a ball of radius rnr_{n} centred at distance srnsr_{n} from A\partial A is about (θd/2)rnd+θd1srnd(\theta_{d}/2)r_{n}^{d}+\theta_{d-1}sr_{n}^{d}, suggesting

𝔼[Fn]\displaystyle\mathbb{E}[F_{n}^{\partial}] nrn(f0θdnrnd/2)k1(|A|/(k1)!)enf0θdrnd/20enf0θd1srnd𝑑s\displaystyle\approx nr_{n}(f_{0}\theta_{d}nr_{n}^{d}/2)^{k-1}(|\partial A|/(k-1)!)e^{-nf_{0}\theta_{d}r_{n}^{d}/2}\int_{0}^{\infty}e^{-nf_{0}\theta_{d-1}sr_{n}^{d}}ds
const.×n(1/2)(1/d)(nrnd)(1/d)+k2(logn)(1k)/2\displaystyle\approx{\rm const.}\times n^{(1/2)-(1/d)}(nr_{n}^{d})^{(1/d)+k-2}(\log n)^{(1-k)/2}
const.×n(1/2)(1/d)(logn)(1/d)+((k3)/2).\displaystyle\approx{\rm const.}\times n^{(1/2)-(1/d)}(\log n)^{(1/d)+((k-3)/2)}.

If d3d\geq 3 this tends to infinity (regardless of kk). Thus the boundary effects dominate in this case; we should choose a slightly bigger rnr_{n} to make 𝔼[Fn]\mathbb{E}[F_{n}^{\partial}] tend to a constant, and then 𝔼[Fno]\mathbb{E}[F_{n}^{o}] will tend to zero.

If d=2d=2, the last expression for 𝔼[Fn]\mathbb{E}[F_{n}^{\partial}] tends to zero if k=1k=1 and to infinity if k3k\geq 3, so the interior contribution dominates when k=1k=1 but the boundary contribution dominates when k3k\geq 3. When k=2k=2 the interior and boundary effects are of comparable size.

Having chosen rnr_{n} so that 𝔼[Fno+Fn]\mathbb{E}[F_{n}^{o}+F_{n}^{\partial}] tends to a constant, we shall use Poisson approximation to show that [Lnorn]\mathbb{P}[L_{n}^{o}\leq r_{n}] tends to a non-trivial constant, and then some percolation arguments to show the same limit holds for [Mn,1rn]\mathbb{P}[M_{n,1}\leq r_{n}].

The rest of the paper is organised as follows. After the preparation of geometrical ingredients in Section 2, we prove Poisson approximation for the number of kk-isolated vertices in Section 3, asymptotic equivalence of Ln,kL_{n,k} and Mn,kM_{n,k} in Section 4, the weak law in the nonuniform case (Theorem 1.3) in Section 5 and finally the weak law in the uniform case (Theorem 1.1) in Section 6.

2 Geometrical preliminaries

In this section, we prepare some geometrical ingredients for later use. Let AA be a compact subset of d\mathbb{R}^{d} with d2d\geq 2.

Given B,CdB,C\subset\mathbb{R}^{d}, set BC:={x+y:xB,yC}B\oplus C:=\{x+y:x\in B,y\in C\}. Let oo denote the origin in d\mathbb{R}^{d}. Given xdx\in\mathbb{R}^{d}, we write B+xB+x for B{x}B\oplus\{x\}.

Given s>0s>0, and ΓA\Gamma\subset A, we write Γ(s)\Gamma^{(s)} for (ΓBs(o))A(\Gamma\oplus B_{s}(o))\cap A, the set of points in AA distant at most ss from Γ\Gamma. Also we set diam(Γ):=supx,yΓyx\operatorname{diam}(\Gamma):=\sup_{x,y\in\Gamma}\|y-x\|, or zero if Γ=\Gamma=\varnothing.

We write A(s)A^{(-s)} for A(A)(s)A\setminus(\partial A)^{(s)}, the set of points in AA distant more than ss from the boundary A\partial A of AA.

When AA is polygonal, we denote by 𝖢𝗈𝗋\mathsf{Cor} the set of corners of AA.

Lemma 2.1.

Suppose AA has a C2C^{2} boundary. Let ε(0,1]\varepsilon\in(0,1]. Then:

(i) For all small enough r>0r>0 we have

|Br(x)A|((θd/2)+(θdε/4))rd,xA(εr).\displaystyle|B_{r}(x)\cap A|\geq((\theta_{d}/2)+(\theta_{d}\varepsilon/4))r^{d},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall\leavevmode\nobreak\ x\in A^{(-\varepsilon r)}. (2.1)

(ii) There exists δ>0\delta>0 and r0>0r_{0}>0 such that if 0<r<s<2r<r00<r<s<2r<r_{0}, then

|ABs(x)Br(x)|((θd/2)+ε)(sdrd),x(A)(δs).\displaystyle|A\cap B_{s}(x)\setminus B_{r}(x)|\leq((\theta_{d}/2)+\varepsilon)(s^{d}-r^{d}),\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall x\in(\partial A)^{(\delta s)}. (2.2)
Proof.

Clearly (2.1) holds for xA(r)x\in A^{(-r)}.

Let x(A)(r)A(K0r2)x\in(\partial A)^{(r)}\cap A^{(-K_{0}r^{2})} with K0K_{0} to be chosen later. Without loss of generality (after a translation and rotation), we can assume that the closest point of A\partial A to xx lies at the origin, and x=hedx=he_{d} for some h>0h>0 (where ede_{d} is the ddth coordinate vector), and for some convex open VdV\subset\mathbb{R}^{d} with oVo\in V and some open convex neighbourhood UU of the origin in d1\mathbb{R}^{d-1}, and some C2C^{2} function ϕ:d1\phi:\mathbb{R}^{d-1}\to\mathbb{R} we have that AV=Vepi(ϕ)A\cap V=V\cap\mathrm{epi}(\phi).

Since z=oz=o is the closest point in A\partial A to xx, we must have ϕ(o)=o\nabla\phi(o)=o. By a compactness argument, we can also assume i=1dj=1d|ij2ϕ|K/(9d2)\sum_{i=1}^{d}\sum_{j=1}^{d}|\partial^{2}_{ij}\phi|\leq K/(9d^{2}) on UU for some constant KK (depending on AA).

Now suppose ud1u\in\mathbb{R}^{d-1} with u3r\|u\|\leq 3r (assume rr is small enough that all such uu lie in UU). By the Mean Value theorem ϕ(u)=uϕ(w)\phi(u)=u\cdot\nabla\phi(w) for some w[o,u]w\in[o,u], and for 1id1\leq i\leq d, iϕ(w)=wiϕ(v)\partial_{i}\phi(w)=w\cdot\nabla\partial_{i}\phi(v) for some v[o,w]v\in[o,w]. Hence

|ϕ(u)|(K/9)u2Kr2,ud1withu3r.\displaystyle|\phi(u)|\leq(K/9)\|u\|^{2}\leq Kr^{2},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall u\in\mathbb{R}^{d-1}\leavevmode\nobreak\ {\rm with}\leavevmode\nobreak\ \|u\|\leq 3r. (2.3)

For a>0a>0 set g(a):=|B1(o)(d1×[0,a])|g(a):=|B_{1}(o)\cap(\mathbb{R}^{d-1}\times[0,a])|. Then g(a)/ag(a)/a is decreasing in aa, so for 0a10\leq a\leq 1 we have g(a)/ag(1)=θd/2g(a)/a\geq g(1)=\theta_{d}/2.

Let π:dd1\pi:\mathbb{R}^{d}\to\mathbb{R}^{d-1} denote projection onto the first d1d-1 coordinates and let h:dh:\mathbb{R}^{d}\to\mathbb{R} denote projection onto the last coordinate (hh stands for ‘height’). Take K0=2KK_{0}=2K. Then h(x)=d(x,A)K0r2h(x)=d(x,\partial A)\geq K_{0}r^{2}. Also h(x)rh(x)\leq r. For zBr(x)(d1×[Kr2,))z\in B_{r}(x)\cap(\mathbb{R}^{d-1}\times[Kr^{2},\infty)) we have π(z)r\|\pi(z)\|\leq r so that by (2.3) we have |ϕ(π(z))|Kr2h(z)|\phi(\pi(z))|\leq Kr^{2}\leq h(z), and hence zAz\in A. Therefore

|Br(x)A|\displaystyle|B_{r}(x)\cap A| |Br(x)(d1×[Kr2,))|\displaystyle\geq|B_{r}(x)\cap(\mathbb{R}^{d-1}\times[Kr^{2},\infty))|
(θd/2)rd+rdg((h(x)Kr2)/r)\displaystyle\geq(\theta_{d}/2)r^{d}+r^{d}g((h(x)-Kr^{2})/r)
(θd/2)rd+(θd/2)rd1(h(x)Kr2)\displaystyle\geq(\theta_{d}/2)r^{d}+(\theta_{d}/2)r^{d-1}(h(x)-Kr^{2})
(θd/2)rd+(θdh(x)/4)rd1,\displaystyle\geq(\theta_{d}/2)r^{d}+(\theta_{d}h(x)/4)r^{d-1}, (2.4)

where the last line is because Kr2=K0r2/2h(x)/2Kr^{2}=K_{0}r^{2}/2\leq h(x)/2.

Let ε>0\varepsilon>0. Provided rr is small enough we have εrK0r2\varepsilon r\geq K_{0}r^{2}, so that A(εr)A(K0r2)A^{(-\varepsilon r)}\subset A^{(-K_{0}r^{2})}. Hence for x(A)(r)A(εr)x\in(\partial A)^{(r)}\cap A^{(-\varepsilon r)} we have (2.4), and therefore since h(x)εrh(x)\geq\varepsilon r,

|Br(x)A|((θd/2)+(θdε/4))rd.|B_{r}(x)\cap A|\geq((\theta_{d}/2)+(\theta_{d}\varepsilon/4))r^{d}.

Thus we have (i).

Refer to caption
Figure 1: The horizontal lines are at height 0, 2δs-2\delta s and 4δs-4\delta s. The circles are of radius r,sr,s centred on the origin.

For part (ii), let δ>0\delta>0, to be chosen later. Suppose that s>0s>0 and x(A)(δs)x\in(\partial A)^{(\delta s)}. Let r(s/2,s)r\in(s/2,s). Provided ss is small enough, by (2.3) we have ϕ(u)δs\phi(u)\geq-\delta s for all ud1u\in\mathbb{R}^{d-1} with u3s\|u\|\leq 3s. Therefore

ABs(x)Br(x)(Bs(x)Br(x))(d1×[h(x)2δs,)).A\cap B_{s}(x)\setminus B_{r}(x)\subset(B_{s}(x)\setminus B_{r}(x))\cap(\mathbb{R}^{d-1}\times[h(x)-2\delta s,\infty)).

Define the set Sδ:={xB1(o):πd(x1x)4δ}.S_{\delta}:=\{x\in B_{1}(o):\pi_{d}(\|x\|^{-1}x)\geq-4\delta\}. Then since s2rs\leq 2r, as shown in Figure 1,

|ABs(x)Br(x)|(sdrd)|Sδ|.\displaystyle|A\cap B_{s}(x)\setminus B_{r}(x)|\leq(s^{d}-r^{d})|S_{\delta}|.

On taking δ\delta small enough so that |Sδ|θd((1/2)+ε)|S_{\delta}|\leq\theta_{d}((1/2)+\varepsilon), we obtain (2.2), completing the proof of part (ii). ∎

Given xAx\in A, set dist(x,A):=infzA{zx}{\rm dist}(x,\partial A):=\inf_{z\in\partial A}\{\|z-x\|\}.

Lemma 2.2.

Suppose AdA\subset\mathbb{R}^{d} is compact with C2C^{2} boundary.

(i) Given ε>0\varepsilon>0, there exists r0=r0(d,A,ε)>0r_{0}=r_{0}(d,A,\varepsilon)>0 such that

|Br(x)A|((θd/2)ε)rd,xA,r(0,r0).\displaystyle|B_{r}(x)\cap A|\geq((\theta_{d}/2)-\varepsilon)r^{d},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall x\in A,r\in(0,r_{0}). (2.5)

(ii) There is a constant r1=r1(d,A)r_{1}=r_{1}(d,A), such that if r(0,r1)r\in(0,r_{1}) and x,yAx,y\in A with yx2r\|y-x\|\leq 2r and dist(x,A)dist(y,A){\rm dist}(x,\partial A)\leq{\rm dist}(y,\partial A), then

|ABr(y)Br(x)|8dθd1rd1yx.\displaystyle|A\cap B_{r}(y)\setminus B_{r}(x)|\geq 8^{-d}\theta_{d-1}r^{d-1}\|y-x\|. (2.6)

(iii) Given ε>0\varepsilon>0, there exists r2=r2(d,A,ε)>0r_{2}=r_{2}(d,A,\varepsilon)>0 such that

|ABs(x)Br(x)|((θd/2)ε)(sdrd),xA,s(0,r2),r(0,s).\displaystyle|A\cap B_{s}(x)\setminus B_{r}(x)|\geq((\theta_{d}/2)-\varepsilon)(s^{d}-r^{d}),\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall x\in A,s\in(0,r_{2}),r\in(0,s).
Proof.

For (i) and (ii), see [4, Lemma 3.2]. For (iii), as in the proof of Lemma 2.1, it suffices to consider xAx\in A such that the closest point of A\partial A to xx is at oo with x=hedx=he_{d} for some h(0,s]h\in(0,s], and moreover A\partial A near oo is the graph of a C2C^{2} function ϕ:U\phi:U\to\mathbb{R} with UU an open neighbourhood of the origin in d1\mathbb{R}^{d-1} and with i=1dj=1d|ij2ϕ|K/(9d2)\sum_{i=1}^{d}\sum_{j=1}^{d}|\partial^{2}_{ij}\phi|\leq K/(9d^{2}) on UU for some constant KK depending only on AA.

Then, as at (2.3), provided r1r_{1} is small enough we have |ϕ(u)|(K/9)u2|\phi(u)|\leq(K/9)\|u\|^{2} for u<r1\|u\|<r_{1}. Therefore given δ>0\delta>0, if also u<δ/K\|u\|<\delta/K we have |ϕ(u)|δu|\phi(u)|\leq\delta\|u\|.

Choose δ\delta as follows. Given a>0a>0, define the set Λad\Lambda_{a}\subset\mathbb{R}^{d} by

Λa:={(u,z):ud1,z,zau,u2+z21}.\Lambda_{a}:=\{(u,z):u\in\mathbb{R}^{d-1},z\in\mathbb{R},z\geq a\|u\|,\|u\|^{2}+z^{2}\leq 1\}.

Then |Λa|θd/2|\Lambda_{a}|\uparrow\theta_{d}/2 as a0a\downarrow 0 so we can and do choose δ>0\delta>0 such that |Λδ|(θd/2)ε|\Lambda_{\delta}|\geq(\theta_{d}/2)-\varepsilon.

For 0r<s0\leq r<s, given xx as above define the set S:=((sΛδ)(rΛδ))+x.S:=((s\Lambda_{\delta})\setminus(r\Lambda_{\delta}))+x. Then

|S|=|(sΛδ)(rΛδ)|=(sdrd)|Λδ|((θd/2)ε)(sdrd).|S|=|(s\Lambda_{\delta})\setminus(r\Lambda_{\delta})|=(s^{d}-r^{d})|\Lambda_{\delta}|\geq((\theta_{d}/2)-\varepsilon)(s^{d}-r^{d}).

Now suppose also that s<δ/Ks<\delta/K. For y=(v,h)Sy=(v,h^{\prime})\in S, we have vs\|v\|\leq s and

h(hh)δvϕ(v),h^{\prime}\geq(h^{\prime}-h)\geq\delta\|v\|\geq\phi(v),

so yAy\in A and SAS\subset A. Also SBs(x)Br(x)S\subset B_{s}(x)\setminus B_{r}(x). This gives the result, with r2=δ/Kr_{2}=\delta/K. ∎

Recall that 𝖢𝗈𝗋\mathsf{Cor} denotes the set of corners of AA when AA is polygonal.

Lemma 2.3.

Assume d=2d=2 and AA is polygonal, then there exist K>0K>0 and r1>0r_{1}>0 depending on AA such that for all r(0,r1)r\in(0,r_{1}), x,yA𝖢𝗈𝗋(Kr)x,y\in A\setminus\mathsf{Cor}^{(Kr)} with dist(x,A)dist(y,A){\rm dist}(x,\partial A)\leq{\rm dist}(y,\partial A) and xy3r\|x-y\|\leq 3r, the lower bound (2.6) holds.

Proof.

Let r1r_{1} be small enough such that non-overlapping edges of AA are distant at least 8r18r_{1} from each other. Consider xA𝖢𝗈𝗋(Kr)x\in A\setminus\mathsf{Cor}^{(Kr)} with r<r1r<r_{1} where KK is made explicit later. We can assume that the corner of AA closest to xx is formed by edges e,ee,e^{\prime} meeting at the origin with angle α(0,2π){π}\alpha\in(0,2\pi)\setminus\{\pi\}. We claim that, provided K>4+8/|sinα|K>4+8/|\sin\alpha|, the disk B(x,4r)B(x,4r) intersects at most one of the two edges. Indeed, if it intersects both edges, then taking wB(x,4r)e,wB(x,4r)ew\in B(x,4r)\cap e,w^{\prime}\in B(x,4r)\cap e^{\prime} we have ww8r\|w-w^{\prime}\|\leq 8r; hence dist(w,e)8r{\rm dist}(w,e^{\prime})\leq 8r. Then, wdist(w,e)/|sinα|8r/|sinα|\|w\|\leq{\rm dist}(w,e^{\prime})/|\sin\alpha|\leq 8r/|\sin\alpha|. However, w(K4)r\|w\|\geq(K-4)r by the triangle inequality, so we arrive at a contradiction. We have thus shown that any ball of radius 4r4r with centre distant at least KrKr from the corners of AA cannot intersect two edges at the same time, where K=5+(8/mini|sinαi|)K=5+(8/\min_{i}|\sin\alpha_{i}|) and {αi}\{\alpha_{i}\} are the angles of the corners of AA.

We have B(x,r)B(y,r)B(x,4r)B(x,r)\cup B(y,r)\subset B(x,4r); hence, the argument leading to Lemma 2.2-(ii), namely [4, Lemma 3.2], gives the estimate (2.6) in this case too. ∎

Lemma 2.4.

Let ε(0,1]\varepsilon\in(0,1]. Then for all r>0r>0 and all compact BdB\subset\mathbb{R}^{d} with diamBεr\operatorname{diam}B\geq\varepsilon r we have |BBr(o)||B|+θd(1+2d1ddεd)rd|B\oplus B_{r}(o)|\geq|B|+\theta_{d}(1+2^{-d-1}d^{-d}\varepsilon^{d})r^{d}.

Proof.

By scaling, it suffices to show that for all compact BdB\subset\mathbb{R}^{d} with diamBε\operatorname{diam}B\geq\varepsilon, we have |BB1(o)||B|+θd(1+2d1ddεd)|B\oplus B_{1}(o)|\geq|B|+\theta_{d}(1+2^{-d-1}d^{-d}\varepsilon^{d}).

Let BdB\subset\mathbb{R}^{d} with εdiamB<\varepsilon\leq\operatorname{diam}B<\infty. Without loss of generality we can assume diam(π1(B))ε/d\operatorname{diam}(\pi_{1}(B))\geq\varepsilon/d, where π1\pi_{1} denotes projection onto the first coordinate.

Let xx be a left-most point of BB, yy a right-most point of BB and uu a top-most point of BB. Here ‘left’ and ‘right’ refer to ordering using the first coordinate and ‘top’ refers to ordering using the last coordinate. Let H+H^{+} be the right half of B1(y)B_{1}(y) and HH^{-} the left-half of B1(x)B_{1}(x). Let D:=Bε/(2d)(u+(0,,0,ε/(2d)))D:=B_{\varepsilon/(2d)}(u+(0,\ldots,0,\varepsilon/(2d))), and let D+D^{+} and DD^{-} be the left half and right half of DD, respectively. Then the interiors of H+H^{+} and of HH^{-} are disjoint from BB and from each other, and the interior of either D+D^{+} or DD^{-} (or both) is disjoint from all of B,H+B,H^{+} and HH^{-}. Therefore since H+H^{+}, HH^{-} and DD are all contained in BB1(o)B\oplus B_{1}(o), we obtain that

|BB1(o)||B|+θd+(θd2d1dd)εd,|B\oplus B_{1}(o)|\geq|B|+\theta_{d}+(\theta_{d}2^{-d-1}d^{-d})\varepsilon^{d},

as required. ∎

Lemma 2.5.

Suppose AC2\partial A\in C^{2}. Let ρ,ε\rho,\varepsilon\in\mathbb{R} with 0<ε<ρ0<\varepsilon<\rho. Then there exist δ=δ(d,ρ,ε)>0\delta=\delta(d,\rho,\varepsilon)>0, and r0=r0(d,ρ,ε,A)r_{0}=r_{0}(d,\rho,\varepsilon,A), such that for all r(0,r0)r\in(0,r_{0}) and all compact BAB\subset A with εrdiamBρr\varepsilon r\leq\operatorname{diam}B\leq\rho r we have

|(BBr(o))A||B|+((θd/2)+δ)rd,\displaystyle|(B\oplus B_{r}(o))\cap A|\geq|B|+((\theta_{d}/2)+\delta)r^{d}, (2.7)

and also, letting x0x_{0} denote a closest point of BB to A\partial A, we have

|(BBr(o))A||B|+|Br(x0)A|+2δrd.\displaystyle|(B\oplus B_{r}(o))\cap A|\geq|B|+|B_{r}(x_{0})\cap A|+2\delta r^{d}. (2.8)
Proof.

It suffices to prove (2.8). Indeed, if (2.8) holds for some δ\delta and r0r_{0}, then using (2.8) and Lemma 2.2-(i) readily yields (2.7) for some new, possibly smaller, choice of r0r_{0}.

Without loss of generality we may assume ε<1<ρ\varepsilon<1<\rho. Let r>0r>0, and let BAB\subset A be compact with εrdiam(B)ρr\varepsilon r\leq\operatorname{diam}(B)\leq\rho r. If BA(r)B\subset A^{(-r)} we can use Lemma 2.4 so it suffices to consider the case where B(A)(r)B\cap(\partial A)^{(r)}\neq\varnothing.

Let x0x_{0} be a closest point of BB to A\partial A. Without loss of generality (after a rotation and translation), we can assume that the closest point of A\partial A to x0x_{0} lies at the origin, and x0=hedx_{0}=he_{d} for some h[0,r]h\in[0,r], and that within some neighbourhood of the origin, AA coincides with the closed epigraph of a function ϕ:U\phi:U\to\mathbb{R} with UU an open convex neighbourhood of the origin in d1\mathbb{R}^{d-1}. As in the proof of Lemma 2.1, we can find K=K(d,A)(1,)K=K(d,A)\in(1,\infty) such that

|ϕ(u)|(K/9)u2Kρ2r2,ud1withu2ρr.\displaystyle|\phi(u)|\leq(K/9)\|u\|^{2}\leq K\rho^{2}r^{2},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall u\in\mathbb{R}^{d-1}\leavevmode\nobreak\ {\rm with}\leavevmode\nobreak\ \|u\|\leq 2\rho r. (2.9)

Assume rε/(144dKρ2)r\leq\varepsilon/(144dK\rho^{2}). Let π:dd1\pi:\mathbb{R}^{d}\to\mathbb{R}^{d-1} denote projection onto the first d1d-1 coordinates, and for 1id1\leq i\leq d, let πi:d\pi_{i}:\mathbb{R}^{d}\to\mathbb{R} denote projection onto the iith coordinate. Define the set HH^{-} (slightly less than half a ball of radius rr: see Figure 2) by

H:={zBr(x0):πd(z)<πd(x0)Kρ2r2}.\displaystyle H^{-}:=\{z\in B_{r}(x_{0}):\pi_{d}(z)<\pi_{d}(x_{0})-K\rho^{2}r^{2}\}. (2.10)

For all wBw\in B we have π(w)wx0ρr\|\pi(w)\|\leq\|w-x_{0}\|\leq\rho r, so by (2.9),

πd(x0)=dist(x0,A)dist(w,A)πd(w)+|ϕ(π(w))|πd(w)+Kρ2r2.\displaystyle\pi_{d}(x_{0})={\rm dist}(x_{0},\partial A)\leq{\rm dist}(w,\partial A)\leq\pi_{d}(w)+|\phi(\pi(w))|\leq\pi_{d}(w)+K\rho^{2}r^{2}. (2.11)

Therefore any zHz\in H^{-}, wBw\in B satisfy πd(z)<πd(w)\pi_{d}(z)<\pi_{d}(w), so that H(BBr(o))BH^{-}\subset(B\oplus B_{r}(o))\setminus B.

We can bound above the volume difference of a half-ball and HH^{-} by the volume of a cylinder of thickness Kρ2r2K\rho^{2}r^{2} with base of radius rr. Using this and the union bound, we obtain that

|Br(x0)A|(θd/2)rd+|HA|+θd1rd1Kρ2r2.\displaystyle|B_{r}(x_{0})\cap A|\leq(\theta_{d}/2)r^{d}+|H^{-}\cap A|+\theta_{d-1}r^{d-1}K\rho^{2}r^{2}. (2.12)

For at least one i{1,,d}i\in\{1,\ldots,d\}, we must have diam(πi(B))εr/d\operatorname{diam}(\pi_{i}(B))\geq\varepsilon r/d. We distinguish the cases where this holds for i=di=d, and where it holds for some id1i\leq d-1.

Refer to caption
Refer to caption
Figure 2: Illustration of the proof of Lemma 2.5.
Left: when diam(πd(B))εr/d\operatorname{diam}(\pi_{d}(B))\geq\varepsilon r/d, the sets B,H+,H,SB,H^{+},H^{-},S are disjoint. The point xx^{-} (not indicated) could be the same as x0x_{0}; if not, it is only slightly lower than x0x_{0}.
Right: when diam(π1(B))εr/d\operatorname{diam}(\pi_{1}(B))\geq\varepsilon r/d, the sets B,H,Q+,Q,SB,H^{-},Q^{+},Q^{-},S are disjoint.

First suppose diam(πd(B))εr/d\operatorname{diam}(\pi_{d}(B))\geq\varepsilon r/d. Choose x+Bx^{+}\in B of maximal height (i.e., maximal dd-coordinate), xBx^{-}\in B of minimal height, and yBy\in B of maximal 11-coordinate (see Figure 2 (Left)).

For all zBBr(o)z\in B\oplus B_{r}(o) we have π(z)zx02ρr\|\pi(z)\|\leq\|z-x_{0}\|\leq 2\rho r so by (2.9) we have |ϕ(π(z))|Kρ2r2|\phi(\pi(z))|\leq K\rho^{2}r^{2}. Applying this in the case z=xz=x^{-}, we deduce that

πd(x+)πd(x)+εr/d(εr/d)Kρ2r2(8/9)εr/d,\displaystyle\pi_{d}(x^{+})\geq\pi_{d}(x^{-})+\varepsilon r/d\geq(\varepsilon r/d)-K\rho^{2}r^{2}\geq(8/9)\varepsilon r/d,

and thus

πd(x+)ϕ(π(z))(8/9)εr/dKρ2r2(7/9)εr/d,zBBr(o).\displaystyle\pi_{d}(x^{+})-\phi(\pi(z))\geq(8/9)\varepsilon r/d-K\rho^{2}r^{2}\geq(7/9)\varepsilon r/d,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall z\in B\oplus B_{r}(o).

Let H+:={zBr(x+):πd(z)>πd(x+)}H^{+}:=\{z\in B_{r}(x^{+}):\pi_{d}(z)>\pi_{d}(x^{+})\} (see Figure 2). Then H+B=H^{+}\cap B=\varnothing, since x+x^{+} is a point of maximal height in BB. Also H+BBr(o)H^{+}\subset B\oplus B_{r}(o), so for all zH+z\in H^{+} we have

ϕ(π(z))Kρ2r2(ε/(9d))rπd(x+)<πd(z),\phi(\pi(z))\leq K\rho^{2}r^{2}\leq(\varepsilon/(9d))r\leq\pi_{d}(x^{+})<\pi_{d}(z),

so zAz\in A. Also, since πd(z)>πd(x+)πd(x0)\pi_{d}(z)>\pi_{d}(x^{+})\geq\pi_{d}(x_{0}) we have from (2.10) that zHz\notin H^{-}. Hence H+AHH^{+}\subset A\setminus H^{-}.

Now consider yy. For 1id1\leq i\leq d let eie_{i} denote the iith unit coordinate vector. Define a point y~\tilde{y} slightly to the right of yy by

y~:={y+(εr/(8d))e1(εr/(8d))edifπd(y)(πd(x+)+πd(x))/2y+(εr/(8d))e1+(εr/(8d))edifπd(y)<(πd(x+)+πd(x))/2,\displaystyle\tilde{y}:=\begin{cases}y+(\varepsilon r/(8d))e_{1}-(\varepsilon r/(8d))e_{d}&{\rm if}\leavevmode\nobreak\ \pi_{d}(y)\geq(\pi_{d}(x^{+})+\pi_{d}(x^{-}))/2\\ y+(\varepsilon r/(8d))e_{1}+(\varepsilon r/(8d))e_{d}&{\rm if}\leavevmode\nobreak\ \pi_{d}(y)<(\pi_{d}(x^{+})+\pi_{d}(x^{-}))/2,\end{cases}

and define the small ball S:=Bεr/(9d)(y~).S:=B_{\varepsilon r/(9d)}(\tilde{y}). Then |S|=δ1rd|S|=\delta_{1}r^{d}, where we set δ1:=θd(ε/(9d))d\delta_{1}:=\theta_{d}(\varepsilon/(9d))^{d}.

Suppose πd(y)(πd(x+)+πd(x))/2\pi_{d}(y)\geq(\pi_{d}(x^{+})+\pi_{d}(x^{-}))/2 (as well as diam(πd(B))εr/d\operatorname{diam}(\pi_{d}(B))\geq\varepsilon r/d).

Then for all zSz\in S we have πd(z)πd(y)πd(x+)\pi_{d}(z)\leq\pi_{d}(y)\leq\pi_{d}(x^{+}) so zH+z\notin H^{+}. Moreover

πd(z)πd(y)εr/(4d)πd(x)+εr/(4d)πd(x0)+εr/(8d)\pi_{d}(z)\geq\pi_{d}(y)-\varepsilon r/(4d)\geq\pi_{d}(x^{-})+\varepsilon r/(4d)\geq\pi_{d}(x_{0})+\varepsilon r/(8d)

by (2.11), applied to w=xw=x^{-}. Therefore zHz\notin H^{-} by (2.10), and also (by (2.9)) πd(z)ϕ(z)\pi_{d}(z)\geq\phi(z) so zAz\in A. Thus SA(H+H)S\subset A\setminus(H^{+}\cup H^{-}) in this case.

Now suppose πd(y)<(πd(x+)+πd(x))/2\pi_{d}(y)<(\pi_{d}(x^{+})+\pi_{d}(x^{-}))/2 (as well as diam(πd(B))εr/d\operatorname{diam}(\pi_{d}(B))\geq\varepsilon r/d).

Then for all zSz\in S we have πd(z)πd(y)+εr/(4d)πd(x+)\pi_{d}(z)\leq\pi_{d}(y)+\varepsilon r/(4d)\leq\pi_{d}(x^{+}), so zH+z\notin H^{+}. Also, since πd(y)ϕ(π(y))Kρ2r2\pi_{d}(y)\geq\phi(\pi(y))\geq-K\rho^{2}r^{2}, we have

πd(z)πd(y)+εr/(72d)Kρ2r2ϕ(π(z)),\pi_{d}(z)\geq\pi_{d}(y)+\varepsilon r/(72d)\geq K\rho^{2}r^{2}\geq\phi(\pi(z)),

so zAz\in A, and also by (2.11) applied to w=yw=y, we have πd(z)πd(x0)\pi_{d}(z)\geq\pi_{d}(x_{0}), so zHz\notin H^{-}. Therefore SA(H+H)S\subset A\setminus(H^{+}\cup H^{-}) in this case too. Thus, whenever diam(πd(B))εr/d\operatorname{diam}(\pi_{d}(B))\geq\varepsilon r/d, we have

|(BBr(o))A||B|+|H+|+|S|+|HA|.\displaystyle|(B\oplus B_{r}(o))\cap A|\geq|B|+|H^{+}|+|S|+|H^{-}\cap A|. (2.13)

Combining (2.13) and (2.12), provided rδ1/(2Kρ2θd1)r\leq\delta_{1}/(2K\rho^{2}\theta_{d-1}) we have

|(BBr(o))A||B|\displaystyle|(B\oplus B_{r}(o))\cap A|\geq|B| +|Br(x0)A|+(δ1/2)rd,\displaystyle+|B_{r}(x_{0})\cap A|+(\delta_{1}/2)r^{d}, ifdiam(πd(B))εr/d.\displaystyle{\rm if}\leavevmode\nobreak\ \operatorname{diam}(\pi_{d}(B))\geq\varepsilon r/d. (2.14)

Now suppose diamπi(B)εr/d\operatorname{diam}\pi_{i}(B)\geq\varepsilon r/d for some i{1,,d1}i\in\{1,\ldots,d-1\}. We shall consider here the case where this holds for i=1i=1; the other cases may be treated similarly.

Let x,x+,yx^{-},x^{+},y be points in BB of minimal 11-coordinate, maximal 11-coordinate, and maximum height respectively. Let δ2:=δ1/(2θd1)\delta_{2}:=\delta_{1}/(2\theta_{d-1}). Define the sets QQ^{-} and Q+Q^{+} (slightly less than quarter-balls of radius rr: see Figure 2 (Right)) by

Q:={zBr(x):πd(z)πd(x)+δ2r,π1(z)<π1(x)};\displaystyle Q^{-}:=\{z\in B_{r}(x^{-}):\pi_{d}(z)\geq\pi_{d}(x^{-})+\delta_{2}r,\pi_{1}(z)<\pi_{1}(x^{-})\};
Q+:={zBr(x+):πd(z)πd(x+)+δ2r,π1(z)>π1(x+)}.\displaystyle Q^{+}:=\{z\in B_{r}(x^{+}):\pi_{d}(z)\geq\pi_{d}(x^{+})+\delta_{2}r,\pi_{1}(z)>\pi_{1}(x^{+})\}.

By (2.9), for zQz\in Q^{-} we have |ϕ(π(z))|Kρ2r2|\phi(\pi(z))|\leq K\rho^{2}r^{2}, so provided r<δ2/(2Kρ2)r<\delta_{2}/(2K\rho^{2}), for all zQz\in Q^{-} we have

πd(z)πd(x)+δ2rδ2rKρ2r2Kρ2r2ϕ(π(z)),\pi_{d}(z)\geq\pi_{d}(x^{-})+\delta_{2}r\geq\delta_{2}r-K\rho^{2}r^{2}\geq K\rho^{2}r^{2}\geq\phi(\pi(z)),

so that zAz\in A. Also by (2.11) applied to w=xw=x^{-} we have πd(z)πd(x)πd(x0)Kρ2r2\pi_{d}(z)\geq\pi_{d}(x^{-})\geq\pi_{d}(x_{0})-K\rho^{2}r^{2}, so zHz\notin H^{-} by (2.10). Thus QAHQ^{-}\subset A\setminus H^{-}, and similarly Q+AHQ^{+}\subset A\setminus H^{-}. Also (QQ+)B=(Q^{-}\cup Q^{+})\cap B=\varnothing, and |QQ+|(θd2δ2θd1)rd/2=(θdδ1)rd/2|Q^{-}\cup Q^{+}|\geq(\theta_{d}-2\delta_{2}\theta_{d-1})r^{d}/2=(\theta_{d}-\delta_{1})r^{d}/2.

We define a point y~\tilde{y} slightly above yy by

y~:={y+(εr/(8d))ed+(εr/(8d))e1ifπ1(y)(π1(x)+π1(x+))/2y+(εr/(8d))ed(εr/(8d))e1ifπ1(y)>(π1(x)+π1(x+))/2,\displaystyle\tilde{y}:=\begin{cases}y+(\varepsilon r/(8d))e_{d}+(\varepsilon r/(8d))e_{1}&{\rm if\leavevmode\nobreak\ }\pi_{1}(y)\leq(\pi_{1}(x^{-})+\pi_{1}(x^{+}))/2\\ y+(\varepsilon r/(8d))e_{d}-(\varepsilon r/(8d))e_{1}&{\rm if\leavevmode\nobreak\ }\pi_{1}(y)>(\pi_{1}(x^{-})+\pi_{1}(x^{+}))/2,\end{cases}

and set S:=Bεr/(9d)(y~)S:=B_{\varepsilon r/(9d)}(\tilde{y}): see Figure 2 (Right). Then |S|=δ1rd|S|=\delta_{1}r^{d} as before.

Then for all zSz\in S we have πd(z)>πd(y)πd(x0)\pi_{d}(z)>\pi_{d}(y)\geq\pi_{d}(x_{0}), so that zBz\notin B and zHz\notin H^{-}. Also ϕ(π(z))Kρ2r2<εr/(72d)πd(z)\phi(\pi(z))\leq K\rho^{2}r^{2}<\varepsilon r/(72d)\leq\pi_{d}(z), so zAz\in A. Moreover, if π1(y)>(π1(x)+π1(x+))/2\pi_{1}(y)>(\pi_{1}(x^{-})+\pi_{1}(x^{+}))/2 then

π1(z)π1(x)=π1(y)π1(x)+(π1(z)π1(y))(εr/(2d))(εr/(4d))>0,\pi_{1}(z)-\pi_{1}(x^{-})=\pi_{1}(y)-\pi_{1}(x^{-})+(\pi_{1}(z)-\pi_{1}(y))\geq(\varepsilon r/(2d))-(\varepsilon r/(4d))>0,

while if π1(y)(π1(x)+π1(x+))/2\pi_{1}(y)\leq(\pi_{1}(x^{-})+\pi_{1}(x^{+}))/2 then π1(z)π1(x)>π1(y)π1(x)0\pi_{1}(z)-\pi_{1}(x^{-})>\pi_{1}(y)-\pi_{1}(x^{-})\geq 0 so in both cases zQz\notin Q^{-}. Similarly zQ+z\notin Q^{+}. Thus SA(Q+QBH)S\subset A\setminus(Q^{+}\cup Q^{-}\cup B\cup H^{-}). Combining all of this and using (2.12) in the third line below yields

|(BBr(o))A|\displaystyle|(B\oplus B_{r}(o))\cap A| |B|+|QQ+|+|S|+|HA|\displaystyle\geq|B|+|Q^{-}\cup Q^{+}|+|S|+|H^{-}\cap A|
|B|+((θd+δ1)/2)rd+|HA|\displaystyle\geq|B|+((\theta_{d}+\delta_{1})/2)r^{d}+|H^{-}\cap A|
|B|+|Br(x0)A|θd1rd1Kρ2r2+(δ1/2)rd\displaystyle\geq|B|+|B_{r}(x_{0})\cap A|-\theta_{d-1}r^{d-1}K\rho^{2}r^{2}+(\delta_{1}/2)r^{d}
|B|+|Br(x0)A|+(δ1/4)rdifdiam(π1(B))εr/d,\displaystyle\geq|B|+|B_{r}(x_{0})\cap A|+(\delta_{1}/4)r^{d}\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ {\rm if}\leavevmode\nobreak\ \operatorname{diam}(\pi_{1}(B))\geq\varepsilon r/d,

provided rδ1/(4Kρ2θd1)r\leq\delta_{1}/(4K\rho^{2}\theta_{d-1}). Combined with (2.14), this shows that (2.8) holds for rr small if we take δ=δ1/8\delta=\delta_{1}/8. ∎

3 Poisson approximation for the kk-isolated vertices

Fix kk\in\mathbb{N}. We say a vertex is kk-isolated if its degree is at most k1k-1. Given n,r>0n,r>0 let ξn,r\xi_{n,r} denote the number of kk-isolated vertices in G(𝒫n,r)G(\mathcal{P}_{n},r):

ξn,r:=x𝒫n𝟏{𝒫n(B(x,r))k}.\displaystyle\xi_{n,r}:=\sum_{x\in\mathcal{P}_{n}}{\bf 1}\{\mathcal{P}_{n}(B(x,r))\leq k\}. (3.1)

The goal of this section is to prove (in Proposition 3.1 below) Poisson approximation for ξn,r\xi_{n,r} when nn is large and rr is small.

Throughout this section we adopt our working assumption on ν\nu. Moreover we assume either that d2d\geq 2 and the support AA of ν\nu is compact with C2C^{2} boundary, or that d=2d=2 and AA is a polygon. We do not assume in this section that ν\nu is necessarily uniform on AA. Recall that 𝒫n\mathcal{P}_{n} is the Poisson process in d\mathbb{R}^{d} with intensity measure nνn\nu.

A fundamental identity used throughout this paper is the Mecke equation which basically says that the law of a Poisson process 𝒫\mathcal{P} conditioned on having a point mass at xx is that of 𝒫{x}\mathcal{P}\cup\{x\}. More precisely, let 𝒫\mathcal{P} be a Poisson process on d\mathbb{R}^{d} with diffuse intensity measure λ\lambda (that is, λ\lambda does not charge atoms). The Mecke equation says that

𝔼[x𝒫f(x,𝒫)]=𝔼[f(x,𝒫{x})]λ(dx),\displaystyle\mathbb{E}\Big{[}\sum_{x\in\mathcal{P}}f(x,\mathcal{P})\Big{]}=\int\mathbb{E}\left[f(x,\mathcal{P}\cup\{x\})\right]\lambda(dx), (3.2)

for all f:d×𝐍(d)f:\mathbb{R}^{d}\times\mathbf{N}(\mathbb{R}^{d})\to\mathbb{R} such that both sides of the identity are finite, where 𝐍(d)\mathbf{N}(\mathbb{R}^{d}) denotes the space of all locally finite subsets of d\mathbb{R}^{d} - see [7, Chapter 4] for a more general statement.

By the Mecke equation, given n,r>0n,r>0 we have

𝔼[ξn,r]=nApn,r(x)ν(dx),\displaystyle\mathbb{E}[\xi_{n,r}]=n\int_{A}p_{n,r}(x)\nu(dx), (3.3)

where for each xAx\in A we set

pn,r(x):=[𝒫n(B(x,r))k1]=j=0k1(n(ν(Br(x)))j/j!)exp(nν(Br(x))).\displaystyle p_{n,r}(x):=\mathbb{P}[\mathcal{P}_{n}(B(x,r))\leq k-1]=\sum_{j=0}^{k-1}(n(\nu(B_{r}(x)))^{j}/j!)\exp(-n\nu(B_{r}(x))). (3.4)

Given random variables X,ZX,Z taking values in 0:={0}\mathbb{N}_{0}:=\mathbb{N}\cup\{0\}, define the total variation distance

dTV(X,Z):=supB0|[XB][ZB]|.\displaystyle d_{\mathrm{TV}}(X,Z):=\sup_{B\subset\mathbb{N}_{0}}|\mathbb{P}[X\in B]-\mathbb{P}[Z\in B]|.

Given α>0\alpha>0, let 𝖯𝗈α{\mathsf{Po}}_{\alpha} be Poisson distributed with mean α\alpha.

Proposition 3.1 (Poisson approximation).

Let β>0\beta^{\prime}>0. Let (rn)n1(r_{n})_{n\geq 1} be chosen so that rn0r_{n}\geq 0 for all nn and

limn𝔼[ξn,rn]=β.\displaystyle\lim_{n\to\infty}\mathbb{E}[\xi_{n,r_{n}}]=\beta^{\prime}. (3.5)

Then we have

dTV(ξn,rn,𝖯𝗈𝔼[ξn,rn])=O((logn)1d)asn.\displaystyle d_{\mathrm{TV}}(\xi_{n,r_{n}},{\mathsf{Po}}_{\mathbb{E}[\xi_{n,r_{n}}]})=O((\log n)^{1-d})\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ {\rm as}\leavevmode\nobreak\ n\to\infty. (3.6)

In particular, with Ln,k=Lk(𝒫n)L_{n,k}=L_{k}(\mathcal{P}_{n}) defined at (1.4),

[Ln,krn]exp(𝔼[ξn,rn])=O((logn)1d)asn.\displaystyle\mathbb{P}[L_{n,k}\leq r_{n}]-\exp(-\mathbb{E}[\xi_{n,r_{n}}])=O((\log n)^{1-d})\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ {\rm as}\leavevmode\nobreak\ n\to\infty. (3.7)

We prepare for proving this with three lemmas, the first of which is used repeatedly later on.

Lemma 3.2.

Under the WA, assuming either that AC2\partial A\in C^{2} or d=2d=2 and AA is polygonal, there exists a constant δ0>0\delta_{0}>0 (depending on AA and ff) such that

2δ0rdν(Br(x))θdrdfmax,xA,r(0,1].\displaystyle 2\delta_{0}r^{d}\leq\nu(B_{r}(x))\leq\theta_{d}r^{d}f_{\rm max},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall\leavevmode\nobreak\ x\in A,r\in(0,1]. (3.8)
Proof.

The second inequality is clear. The first inequality follows from Lemma 2.2-(i) in the case where AC2\partial A\in C^{2}, and can be seen directly when AA is polygonal. ∎

Lemma 3.3.

Let β(0,)\beta^{\prime}\in(0,\infty) and suppose that (rn)n1(r_{n})_{n\geq 1} satisfies (3.5). Then we have that lim infn(nrnd/logn)1/(fmaxθd)\liminf_{n\to\infty}(nr_{n}^{d}/\log n)\geq 1/(f_{\rm max}\theta_{d}), and lim supn(nrnd/logn)<\limsup_{n\to\infty}(nr_{n}^{d}/\log n)<\infty.

Proof.

Let α(0,1/(fmaxθd))\alpha\in(0,1/(f_{\rm max}\theta_{d})). If nrnd<αlognnr_{n}^{d}<\alpha\log n, then for all xAx\in A we have nν(Brn(x))nθdfmaxrndαθdfmaxlognn\nu(B_{r_{n}}(x))\leq n\theta_{d}f_{\rm max}r_{n}^{d}\leq\alpha\theta_{d}f_{\rm max}\log n. Therefore by (3.3), 𝔼[ξn,r]nenν(Brn(x))ν(dx)n1αθdfmax,\mathbb{E}[\xi_{n,r}]\geq n\int e^{-n\nu(B_{r_{n}}(x))}\nu(dx)\geq n^{1-\alpha\theta_{d}f_{\rm max}}, so the condition (3.5) implies that nrndαlognnr_{n}^{d}\geq\alpha\log n for all large enough nn. The first claim follows.

For the second claim, let δ0>0\delta_{0}>0 be as in (3.8). Take sn>0s_{n}>0 so that nsnd=δ01lognns_{n}^{d}=\delta_{0}^{-1}\log n. Using (3.8), for some constant cc we have

nApn,sn(x)ν(dx)cn(logn)k1exp(2nδ0snd)=c(logn)k1n1,\displaystyle n\int_{A}p_{n,s_{n}}(x)\nu(dx)\leq cn(\log n)^{k-1}\exp(-2n\delta_{0}s_{n}^{d})=c(\log n)^{k-1}n^{-1},

which tends to zero. Hence by (3.5) we have rnsnr_{n}\leq s_{n} for nn large, and hence the second claim. ∎

Given x,ydx,y\in\mathbb{R}^{d} and n,r>0n,r>0, setting 𝒫nx:=𝒫n{x}\mathcal{P}_{n}^{x}:=\mathcal{P}_{n}\cup\{x\}, we define the quantity

qn,r(x,y):=[𝒫ny(B(x,r))k1,𝒫nx(B(y,r))k1].\displaystyle q_{n,r}(x,y):=\mathbb{P}[\mathcal{P}^{y}_{n}(B(x,r))\leq k-1,\mathcal{P}^{x}_{n}(B(y,r))\leq k-1].

Our proof of Proposition 3.1 is based on the following estimate which was proved in [8] by the local dependence approach of Stein’s method.

Lemma 3.4 ([8, Theorem 6.7]).

Let n,r>0n,r>0. Then

dTV(ξn,r,𝖯𝗈𝔼[ξn,r])3(I1(n,r)+I2(n,r))\displaystyle d_{\mathrm{TV}}(\xi_{n,r},{\mathsf{Po}}_{\mathbb{E}[\xi_{n,r}]})\leq 3(I_{1}(n,r)+I_{2}(n,r))

where

I1(n,r)\displaystyle I_{1}(n,r) =n2𝟏{xy3r}pn,r(x)pn,r(y)ν2(d(x,y))\displaystyle=n^{2}\int{\bf 1}\{\|x-y\|\leq 3r\}p_{n,r}(x)p_{n,r}(y)\nu^{2}(d(x,y))
I2(n,r)\displaystyle I_{2}(n,r) =n2𝟏{xy3r}qn,r(x,y)ν2(d(x,y)).\displaystyle=n^{2}\int{\bf 1}\{\|x-y\|\leq 3r\}q_{n,r}(x,y)\nu^{2}(d(x,y)).
Proof of Proposition 3.1.

Observe first that whenever |𝒫n|k+1|\mathcal{P}_{n}|\geq k+1, the statement Ln,krnL_{n,k}\leq r_{n} is equivalent to ξn,rn=0\xi_{n,r_{n}}=0, so that |[Ln,krn][ξn,rn=0]|[|𝒫n|k]=O(nken).|\mathbb{P}[L_{n,k}\leq r_{n}]-\mathbb{P}[\xi_{n,r_{n}}=0]|\leq\mathbb{P}[|\mathcal{P}_{n}|\leq k]=O(n^{k}e^{-n}). Therefore (3.6) will imply (3.7), so it suffices to prove (3.6).

By (3.8), provided nn is large enough, for all yAy\in A we have

pn,rn(y)k(nfmaxθdrnd)k1exp(nν(B(y,rn)))exp(δ0nrnd).\displaystyle p_{n,r_{n}}(y)\leq k(nf_{\mathrm{max}}\theta_{d}r_{n}^{d})^{k-1}\exp(-n\nu(B(y,r_{n})))\leq\exp(-\delta_{0}nr_{n}^{d}).

Therefore, using (3.3) in the second line below we have

I1(n,rn)\displaystyle I_{1}(n,r_{n}) n(3dfmaxθdrnd)exp(δ0nrnd)npn,rn(x)ν(dx)\displaystyle\leq n(3^{d}f_{\rm max}\theta_{d}r_{n}^{d})\exp(-\delta_{0}nr_{n}^{d})n\int p_{n,r_{n}}(x)\nu(dx)
exp((δ0/2)nrnd)𝔼[ξn,rn].\displaystyle\leq\exp(-(\delta_{0}/2)nr_{n}^{d})\mathbb{E}[\xi_{n,r_{n}}]. (3.9)

Now we estimate I2:=I2(n,rn)I_{2}:=I_{2}(n,r_{n}). Since the integrand of I2I_{2} is symmetric in xx and yy,

I22n2𝟏{xy3rn,dist(x,A)dist(y,A)}qn,rn(x,y)ν2(d(x,y)).\displaystyle I_{2}\leq 2n^{2}\int{\bf 1}\{\|x-y\|\leq 3r_{n},{\rm dist}(x,\partial A)\leq{\rm dist}(y,\partial A)\}q_{n,r_{n}}(x,y)\nu^{2}(d(x,y)).

To further simplify the integral, writing Bx=B(x,rn)B_{x}=B(x,r_{n}) and likewise for B(y,rn)B(y,r_{n}), we have

qn,rn(x,y)\displaystyle q_{n,r_{n}}(x,y) [𝒫n(Bx)k1,𝒫n(ByBx)k1]\displaystyle\leq\mathbb{P}[\mathcal{P}_{n}(B_{x})\leq k-1,\mathcal{P}_{n}(B_{y}\setminus B_{x})\leq k-1]
=pn,rn(x)[𝒫n(ByBx)k1].\displaystyle=p_{n,r_{n}}(x)\mathbb{P}[\mathcal{P}_{n}(B_{y}\setminus B_{x})\leq k-1].

Consider first the case where AA has a C2C^{2} boundary. If dist(x,A)dist(y,A){\rm dist}(x,\partial A)\leq{\rm dist}(y,\partial A), setting κd:=23d1θd1\kappa_{d}:=2^{-3d-1}\theta_{d-1} and using Lemma 2.2-(ii) for the lower bound and Fubini’s theorem for the upper bound below, we have

f0κdyxrnd1ν(ByBx)fmaxθd1rnd1yx,\displaystyle f_{0}\kappa_{d}\|y-x\|r_{n}^{d-1}\leq\nu(B_{y}\setminus B_{x})\leq f_{\rm max}\theta_{d-1}r_{n}^{d-1}\|y-x\|,

and hence

qn,rn(x,y)\displaystyle q_{n,r_{n}}(x,y) pn,rn(x)j=0k1(nfmaxθd1rnd1yx)jexp(κdf0yxnrnd1).\displaystyle\leq p_{n,r_{n}}(x)\sum_{j=0}^{k-1}(nf_{\rm max}\theta_{d-1}r_{n}^{d-1}\|y-x\|)^{j}\exp(-\kappa_{d}f_{0}\|y-x\|nr_{n}^{d-1}).

Therefore, we have

I2\displaystyle I_{2} 2max(fmaxθd1,1)k1n2\displaystyle\leq 2\max(f_{\rm max}\theta_{d-1},1)^{k-1}n^{2}
×A(B(x,3rn)j=0k1(nrnd1yx)jexp(κdf0yxnrnd1)ν(dy))pn,rn(x)ν(dx).\displaystyle\times\int_{A}\left(\int_{B(x,3r_{n})}\sum_{j=0}^{k-1}(nr_{n}^{d-1}\|y-x\|)^{j}\exp(-\kappa_{d}f_{0}\|y-x\|nr_{n}^{d-1})\nu(dy)\right)p_{n,r_{n}}(x)\nu(dx).

A change of variables z=nrnd1(yx)z=nr_{n}^{d-1}(y-x) shows that the inner integral is bounded by crnd(nrnd)dc^{\prime}r_{n}^{d}(nr_{n}^{d})^{-d} for some finite constant cc^{\prime}. Together with (3.3), this yields for some further constant c′′c^{\prime\prime} that

I22c′′(nrnd)1d𝔼[ξn,rn].\displaystyle I_{2}\leq 2c^{\prime\prime}(nr_{n}^{d})^{1-d}\mathbb{E}[\xi_{n,r_{n}}]. (3.10)

This, together with (3) and (3.5), shows that I1+I2=O((nrnd)1d)I_{1}+I_{2}=O((nr_{n}^{d})^{1-d}); applying Lemmas 3.4 and 3.3 proves (3.6) as required for this case.

Now consider the other case, where d=2d=2 and AA is polygonal. Let x,yAx,y\in A with yx3rn\|y-x\|\leq 3r_{n} and dist(x,A)dist(y,A){\rm dist}(x,\partial A)\leq{\rm dist}(y,\partial A). By Lemma 2.3, there exists δ1>0\delta_{1}>0 such that ν(B(y,rn)B(x,rn))δ1xyrn\nu(B(y,r_{n})\setminus B(x,r_{n}))\geq\delta_{1}\|x-y\|r_{n}. Using this, we can estimate the contribution to I2I_{2} from x,yx,y not too close to the corners similarly to how we estimated I2I_{2} at (3.10) in the previous case.

Suppose instead that xx is close to a corner of AA and xy3rn\|x-y\|\leq 3r_{n}. By (3.8) the contribution to I2I_{2} from such pairs (x,y)(x,y) is at most c′′′n2rn4exp(δ2nrn2)c^{\prime\prime\prime}n^{2}r_{n}^{4}\exp(-\delta_{2}nr_{n}^{2}) for suitable constants c′′′<,δ2>0c^{\prime\prime\prime}<\infty,\delta_{2}>0. Hence by Lemma 3.3, this contribution is O(nδ3)O(n^{-\delta_{3}}) for some δ3>0\delta_{3}>0. The proof is now complete. ∎

4 Relating Ln,kL_{n,k} to Mn,kM_{n,k}

Throughout this section we assume that AC2\partial A\in C^{2} or that AA is a convex polygon. We adopt our WA but do not assume ff is necessarily constant on AA.

Fix kk\in\mathbb{N}. Recall that Ln,kL_{n,k} and Mn,kM_{n,k} were defined at (1.4) and (1.1). While Proposition 3.1 provides an understanding of [Ln,krn]\mathbb{P}[L_{n,k}\leq r_{n}] for suitable rnr_{n}, for Theorems 1.1 and 1.3 we also need to understand the limiting behaviour of [Mn,krn]\mathbb{P}[M_{n,k}\leq r_{n}] and [Mk(𝒳n)rn]\mathbb{P}[M_{k}(\mathcal{X}_{n})\leq r_{n}]. In this section, we work towards this by showing (in Proposition 4.6) that [Ln,kr<Mn,k]\mathbb{P}[L_{n,k}\leq r<M_{n,k}] and [Lk(𝒳n)r<Mk(𝒳n)]\mathbb{P}[L_{k}(\mathcal{X}_{n})\leq r<M_{k}(\mathcal{X}_{n})] are small for nn large and rr small.

Suppose 𝒳d\mathcal{X}\subset\mathbb{R}^{d} is finite, and r>0r>0. We adapt terminology from [8, p. 282]. For j0={0}j\in\mathbb{N}_{0}=\mathbb{N}\cup\{0\} a jj-separating pair for the geometric graph G(𝒳,r)G(\mathcal{X},r) means a pair of disjoint non-empty subsets 𝒴,𝒴\mathcal{Y},\mathcal{Y}^{\prime} of 𝒳\mathcal{X} such that G(𝒴,r)G(\mathcal{Y},r) and G(𝒴,r)G(\mathcal{Y}^{\prime},r) are both connected, G(𝒴𝒴,r)G(\mathcal{Y}\cup\mathcal{Y}^{\prime},r) is not, and 𝒳(𝒴𝒴)\mathcal{X}\setminus(\mathcal{Y}\cup\mathcal{Y}^{\prime}) contains at most jj points within distance rr of 𝒴𝒴\mathcal{Y}\cup\mathcal{Y}^{\prime}.

When we need to refer to an individual set in a separating pair, we use the terminology separating set. That is, for j0j\in\mathbb{N}_{0} a jj-separating set for the graph G(𝒳,r)G(\mathcal{X},r) is a set 𝒴𝒳\mathcal{Y}\subset\mathcal{X} such that G(𝒴,r)G(\mathcal{Y},r) is connected, and with Δ𝒴\Delta\mathcal{Y} denoting the set of sites in 𝒳𝒴\mathcal{X}\setminus\mathcal{Y} adjacent to 𝒴\mathcal{Y}, we have |Δ𝒴|j|\Delta\mathcal{Y}|\leq j and 𝒳(𝒴Δ𝒴)\mathcal{X}\setminus(\mathcal{Y}\cup\Delta\mathcal{Y})\neq\varnothing.

Lemma 4.1.

Suppose 𝒳d\mathcal{X}\subset\mathbb{R}^{d} is finite with |𝒳|k+2|\mathcal{X}|\geq k+2. Let r>0r>0, and suppose Lk(𝒳)r<Mk(𝒳)L_{k}(\mathcal{X})\leq r<M_{k}(\mathcal{X}). Then there exists a (k1)(k-1)-separating pair (𝒴,𝒴)(\mathcal{Y},\mathcal{Y}^{\prime}) for G(𝒳,r)G(\mathcal{X},r) such that neither 𝒴\mathcal{Y} nor 𝒴\mathcal{Y}^{\prime} is a singleton.

Proof.

Since Mk(𝒳)>rM_{k}(\mathcal{X})>r, the graph G(𝒳,r)G(\mathcal{X},r) is not kk-connected. Therefore by [8, Lemma 13.1], it has a (k1)(k-1)-separating pair 𝒴,𝒴𝒳\mathcal{Y},\mathcal{Y}^{\prime}\subset\mathcal{X}. Since also Lk(𝒳)rL_{k}(\mathcal{X})\leq r, every vertex x𝒳x\in\mathcal{X} has degree at least kk, which implies that neither 𝒴\mathcal{Y} nor 𝒴\mathcal{Y}^{\prime} can be a singleton. ∎

Our strategy in this section is to estimate the probability that there exists a pair of non-singleton separating sets for G(𝒫n,r)G({\cal P}_{n},r) or G(𝒳n,r)G(\mathcal{X}_{n},r). We do this in stages, according to the size of the separating sets.

For x,yAx,y\in A, we write x<yx<y if xx precedes yy in the lexicographic ordering. We define the following ordering \prec on AA, that we shall use repeatedly:

xy(dist(x,A)<dist(y,A))or(dist(x,A)=dist(y,A)andx<y).\displaystyle x\prec y\Leftrightarrow({\rm dist}(x,\partial A)<{\rm dist}(y,\partial A)){\rm\leavevmode\nobreak\ or\leavevmode\nobreak\ }({\rm dist}(x,\partial A)={\rm dist}(y,\partial A){\rm\leavevmode\nobreak\ and\leavevmode\nobreak\ }x<y). (4.1)

4.1 Small separating sets

The goal of this section is to prove that for any fixed vertex xAx\in A, the probability that xx belongs to a non-singleton (k1)(k-1)-separating set of ‘small’ diameter in G(𝒫n{x},r)G(\mathcal{P}_{n}\cup\{x\},r) is negligible compared to the probability that it has degree at most k1k-1, provided that xyx\prec y for all other yy in the separating set containing xx, where the ordering \prec was defined at (4.1).

We introduce further notation. With kk fixed, for r>0r>0 and finite 𝒳d\mathcal{X}\subset\mathbb{R}^{d}, xdx\in\mathbb{R}^{d}, let 𝒞r(x,𝒳)\mathcal{C}_{r}(x,\mathcal{X}) denote the collection of (k1)(k-1)-separating sets 𝒴\mathcal{Y} for G(𝒳{x},r)G(\mathcal{X}\cup\{x\},r) containing xx such that moreover xyx\prec y for all y𝒴{x}y\in\mathcal{Y}\setminus\{x\}. Given also ρ>0\rho>0, we are interested in the event

Ex,ρ,r(𝒳):={𝒴𝒞r(x,𝒳),0<diam(𝒴)ρr}.\displaystyle E_{x,\rho,r}(\mathcal{X}):=\{\exists\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{X}),0<\operatorname{diam}(\mathcal{Y})\leq\rho r\}. (4.2)
Lemma 4.2.

(i) Suppose d2d\geq 2 and AA has C2C^{2} boundary. Then there exist δ,r0(0,1)\delta,r_{0}\in(0,1) and c<c<\infty such that for all nk+2n\geq k+2, any xAx\in A and any r(0,r0)r\in(0,r_{0}) we have

[Ex,δ,r(𝒫n)]cpn,r(x)(nrd)1d;\displaystyle\mathbb{P}[E_{x,\delta,r}(\mathcal{P}_{n})]\leq cp_{n,r}(x)(nr^{d})^{1-d}; (4.3)
[Ex,δ,r(𝒳n1)]cpn,r(x)(nrd)1d,\displaystyle\mathbb{P}[E_{x,\delta,r}(\mathcal{X}_{n-1})]\leq cp_{n,r}(x)(nr^{d})^{1-d}, (4.4)

where pn,r(x)p_{n,r}(x) was defined at (3.4)

(ii) Suppose d=2d=2 and AA is polygonal. Then there exist K(0,)K\in(0,\infty), and δ,r0(0,1)\delta,r_{0}\in(0,1) and c<c<\infty such that for all n3n\geq 3, xA𝖢𝗈𝗋(Kr)x\in A\setminus\mathsf{Cor}^{(Kr)}, and r(0,r0)r\in(0,r_{0}), (4.3) and (4.4) hold.

Proof.

(i) Let δ(0,1)\delta\in(0,1). Suppose that Ex,δ,r(𝒫n)E_{x,\delta,r}(\mathcal{P}_{n}) occurs with some 𝒴\mathcal{Y}. Then by considering the vertex furthest from xx in 𝒴\mathcal{Y}, we see that there exists y𝒫ny\in\mathcal{P}_{n} such that 𝒴Byx(x)\mathcal{Y}\subset B_{\|y-x\|}(x) and yxδr\|y-x\|\leq\delta r. Moreover, setting Dx,y:=(Br(x)Br(y))Byx(x)D_{x,y}:=(B_{r}(x)\cup B_{r}(y))\setminus B_{\|y-x\|}(x) we have that 𝒫n(Dx,y)k1\mathcal{P}_{n}(D_{x,y})\leq k-1. By Markov’s inequality, [Ex,δ,r(𝒫n)]\mathbb{P}[E_{x,\delta,r}(\mathcal{P}_{n})] is bounded above by the expected number of y𝒫nBδr(x)y\in\mathcal{P}_{n}\cap B_{\delta r}(x) satisfying 𝒫n(Dx,y)k1{\cal P}_{n}(D_{x,y})\leq k-1, and hence by the Mecke formula

[Ex,δ,r(𝒫n)]nB(x,δr)[𝒫n(Dx,y)k1]ν(dy).\displaystyle\mathbb{P}[E_{x,\delta,r}({\cal P}_{n})]\leq n\int_{B(x,\delta r)}\mathbb{P}[{\cal P}_{n}(D_{x,y})\leq k-1]\nu(dy). (4.5)

To proceed, we need to bound the volume of Dx,yD_{x,y} from below. By Lemma 2.2-(ii), there exists r0>0r_{0}>0 such that for all r(0,r0)r\in(0,r_{0}) and x,yAx,y\in A with xyr\|x-y\|\leq r and xyx\prec y, setting κd:=23d1θd1\kappa_{d}:=2^{-3d-1}\theta_{d-1} we have

ν(Br(x)Br(y))ν(Br(x))+2κdf0rd1yx.\nu(B_{r}(x)\cup B_{r}(y))\geq\nu(B_{r}(x))+2\kappa_{d}f_{0}r^{d-1}\|y-x\|.

Hence, for r<r0r<r_{0}, for x,yAx,y\in A with yxδr\|y-x\|\leq\delta r and xyx\prec y,

ν(Dx,y)ν(Br(x))+2κdf0rd1yxfmaxθdyxd.\displaystyle\nu(D_{x,y})\geq\nu(B_{r}(x))+2\kappa_{d}f_{0}r^{d-1}\|y-x\|-f_{\rm max}\theta_{d}\|y-x\|^{d}.

Now provided δ(κdf0/(fmaxθd))1/(d1)\delta\leq(\kappa_{d}f_{0}/(f_{\rm max}\theta_{d}))^{1/(d-1)}, we have fmaxθdyxdκdf0rd1yxf_{\rm max}\theta_{d}\|y-x\|^{d}\leq\kappa_{d}f_{0}r^{d-1}\|y-x\|, yielding

ν(Dx,y)ν(Br(x))+κdf0rd1yx.\nu(D_{x,y})\geq\nu(B_{r}(x))+\kappa_{d}f_{0}r^{d-1}\|y-x\|. (4.6)

By (3.8), there is also a bound the other way, namely ν(Dx,y)ν(Br(x)Br(y))c0ν(Br(x))\nu(D_{x,y})\leq\nu(B_{r}(x)\cup B_{r}(y))\leq c_{0}\nu(B_{r}(x)) for some constant c0[1,)c_{0}\in[1,\infty). Using (4.5) and the preceding upper and lower bounds on ν(Dx,y)\nu(D_{x,y}), we have

[Ex,δ,r(𝒫n)]c0k1nB(x,δr)j=0k1((nν(Br(x)))j/j!)enν(Br(x))nrd1f0κdyxν(dy).\displaystyle\mathbb{P}[E_{x,\delta,r}(\mathcal{P}_{n})]\leq c_{0}^{k-1}n\int_{B(x,\delta r)}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))-nr^{d-1}f_{0}\kappa_{d}\|y-x\|}\nu(dy). (4.7)

Recall that pn,r(x)=j=0k1((nν(Br(x)))j/j!)enν(Br(x))p_{n,r}(x)=\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}. Changing variable to y=yxy^{\prime}=y-x, and then to z=nrd1yz=nr^{d-1}y^{\prime} leads to

[Ex,δ,r(𝒫n)]\displaystyle\mathbb{P}[E_{x,\delta,r}(\mathcal{P}_{n})] c0k1fmaxnpn,r(x)yδrenf0κdrd1y𝑑y\displaystyle\leq c_{0}^{k-1}f_{\max}np_{n,r}(x)\int_{\|y^{\prime}\|\leq\delta r}e^{-nf_{0}\kappa_{d}r^{d-1}\|y^{\prime}\|}dy^{\prime}
cpn,r(x)n(nrd1)ddef0κdz𝑑z,\displaystyle\leq c^{\prime}p_{n,r}(x)n(nr^{d-1})^{-d}\int_{\mathbb{R}^{d}}e^{-f_{0}\kappa_{d}\|z\|}dz,

for a suitable positive constant cc^{\prime}, not depending on rr or nn. This proves (4.3).

To prove (4.4), we use similar reasoning to before, now using the union bound (instead of the Mecke formula) and the binomial distribution, to obtain that

[Ex,δ,r(𝒳n1)](n1)B(x,δr)j=0k1(n2j)ν(Dx,y)j(1ν(Dx,y))n2jν(dy).\displaystyle\mathbb{P}[E_{x,\delta,r}(\mathcal{X}_{n-1})]\leq(n-1)\int_{B(x,\delta r)}\sum_{j=0}^{k-1}\binom{n-2}{j}\nu(D_{x,y})^{j}(1-\nu(D_{x,y}))^{n-2-j}\nu(dy).

As before we bound ν(Dx,y)j\nu(D_{x,y})^{j} from above by c0k1ν(Br(x))j1c_{0}^{k-1}\nu(B_{r}(x))^{j-1}. Provided rr is sufficiently small, we have for all x,yx,y and all jk1j\leq k-1 that (1ν(Dx,y))2j2(1-\nu(D_{x,y}))^{-2-j}\leq 2. Also we can bound the binomial coefficient from above by nj/j!n^{j}/j!. Combining all of these and also using the bound (1t)et(1-t)\leq e^{-t} we obtain that

[Ex,δ,r(𝒳n1)]2c0k1nB(x,δr)j=0k1(nν(Br(x)))jj!exp(nν(Dx,y))ν(dy);\displaystyle\mathbb{P}[E_{x,\delta,r}(\mathcal{X}_{n-1})]\leq 2c_{0}^{k-1}n\int_{B(x,\delta r)}\sum_{j=0}^{k-1}\frac{(n\nu(B_{r}(x)))^{j}}{j!}\exp(-n\nu(D_{x,y}))\nu(dy);

then using (4.6) and arguing similarly to the Poisson case, we obtain (4.4).

(ii) Suppose d=2d=2 and AA is polygonal. We use Lemma 2.3 in place of Lemma 2.2 to get lower bound (4.6). This together with the simple upper bound ν(Dx,y)c0ν(Br(x))\nu(D_{x,y})\leq c_{0}\nu(B_{r}(x)) and the same reasoning as in part (i) lead to (4.3) and (4.4) in this case too. ∎

4.2 Medium sized separating sets

Recall the definition of 𝒞r(x,𝒳)\mathcal{C}_{r}(x,\mathcal{X}) before the previous lemma, and pn,r(x)p_{n,r}(x) at (3.4). Given ε,ρ\varepsilon,\rho with 0<ε<ρ<0<\varepsilon<\rho<\infty, define the event

Fx,ε,ρ,r(𝒫n):={𝒴𝒞r(x,𝒫n),εr<diam(𝒴)<ρr}.\displaystyle F_{x,\varepsilon,\rho,r}({\cal P}_{n}):=\{\exists\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{P}_{n}),\leavevmode\nobreak\ \varepsilon r<\operatorname{diam}(\mathcal{Y})<\rho r\}. (4.8)

The next lemma helps us bound the probability of having a medium-sized separating set.

Lemma 4.3.

(i) Suppose AC2\partial A\in C^{2}. Given ρ,ε\rho,\varepsilon\in\mathbb{R} with 0<ε<ρ0<\varepsilon<\rho, there exist δ,r0,c>0\delta,r_{0},c>0 such that for all n2+kn\geq 2+k, r(0,r0)r\in(0,r_{0}) and all xAx\in A, we have

[Fx,ε,ρ,r(𝒫n)]cpn,r(x)eδnrd;\displaystyle\mathbb{P}[F_{x,\varepsilon,\rho,r}({\cal P}_{n})]\leq cp_{n,r}(x)e^{-\delta nr^{d}}; (4.9)
[Fx,ε,ρ,r(𝒳n1)]cpn,r(x)eδnrd.\displaystyle\mathbb{P}[F_{x,\varepsilon,\rho,r}(\mathcal{X}_{n-1})]\leq cp_{n,r}(x)e^{-\delta nr^{d}}. (4.10)

(ii) Suppose d=2d=2 and AA is polygonal. Given 0<ε<ρ<0<\varepsilon<\rho<\infty, there exists K(0,)K\in(0,\infty) and δ,r0>0\delta,r_{0}>0 such that for all n3n\geq 3, r(0,r0)r\in(0,r_{0}) and all xA𝖢𝗈𝗋(Kr)x\in A\setminus\mathsf{Cor}^{(Kr)}, we have (4.9) and (4.10).

Proof.

Later in the proof we shall use the fact that since we assume AA is compact and ff is continuous on AA with f0>0f_{0}>0,

lims0(sup{f(y)/f(x):x,yA,yxs})=1.\displaystyle\lim_{s\downarrow 0}\big{(}\sup\{f(y)/f(x):x,y\in A,\|y-x\|\leq s\}\big{)}=1. (4.11)

(i) Suppose AC2\partial A\in C^{2}. Without loss of generality, we can and do assume ε<1\varepsilon<1. Let δ1:=δ(d,ρ,ε)\delta_{1}:=\delta(d,\rho,\varepsilon) be as in Lemma 2.5. With e1e_{1} denoting an arbitrary unit vector in d\mathbb{R}^{d}, choose δ2(0,1/(99d))\delta_{2}\in(0,1/(99\sqrt{d})) such that

|B1(o)B1dδ2(o)|δ1.\displaystyle|B_{1}(o)\setminus B_{1-\sqrt{d}\delta_{2}}(o)|\leq\delta_{1}. (4.12)

Partition d\mathbb{R}^{d} into cubes of side length δ2r\delta_{2}r. Given 𝒴𝒞r(x,𝒫n)\mathcal{Y}\in\mathcal{C}_{r}(x,{\cal P}_{n}), denote by 𝖠δ2(𝒴)\mathsf{A}_{\delta_{2}}(\mathcal{Y}) the closure of the union of all the cubes in the partition that intersect 𝒴\mathcal{Y}. Here 𝖠\mathsf{A} stands for “animal” and is unrelated to our underlying domain AA. If diam𝒴(εr,ρr]\operatorname{diam}\mathcal{Y}\in(\varepsilon r,\rho r], then 𝖠δ2(𝒴)B(x,ρr+δ2d1/2r)\mathsf{A}_{\delta_{2}}(\mathcal{Y})\subset B(x,\rho r+\delta_{2}d^{1/2}r) and 𝖠δ2(𝒴)\mathsf{A}_{\delta_{2}}(\mathcal{Y}) can take at most c:=2(2(ρ/δ2)+d)dc:=2^{{(2\lceil(\rho/{\delta_{2}})+\sqrt{d}\rceil)^{d}}} different possible shapes.

If the event Fx,ε,ρ,r(𝒫n)F_{x,\varepsilon,\rho,r}({\cal P}_{n}) occurs there is at least one set 𝒴𝒞r(x,𝒫n)\mathcal{Y}\in\mathcal{C}_{r}(x,{\cal P}_{n}) with εr<diam𝒴ρr\varepsilon r<\operatorname{diam}\mathcal{Y}\leq\rho r. If there are several such sets 𝒴\mathcal{Y}, choose one of these according to some deterministic rule, and denote it by 𝒴(𝒫n)\mathcal{Y}^{*}({\cal P}_{n}).

Fix a possible shape σ\sigma that might arise as 𝖠δ2(𝒴)\mathsf{A}_{\delta_{2}}(\mathcal{Y}) for some 𝒴𝒞r(x,𝒫n)\mathcal{Y}\in\mathcal{C}_{r}(x,{\cal P}_{n}) with diam𝒴(εr,ρr]\operatorname{diam}\mathcal{Y}\in(\varepsilon r,\rho r], and suppose the event Fx,ε,ρ,r(𝒫n){𝖠δ2(𝒴(𝒫n))=σ}F_{x,\varepsilon,\rho,r}({\cal P}_{n})\cap\{\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}({\cal P}_{n}))=\sigma\} occurs. Let σ:={zσ:xz}{x}\sigma^{*}:=\{z\in\sigma:x\prec z\}\cup\{x\}. Set H:=H(σ)=(σB(1dδ2)r(o))σH:=H(\sigma)=(\sigma^{*}\oplus B_{(1-\sqrt{d}{\delta_{2}})r}(o))\setminus\sigma^{*}. By the triangle inequality, H𝒴(𝒫n)Br(o)H\subset\mathcal{Y}^{*}({\cal P}_{n})\oplus B_{r}(o). We claim that 𝒫n(H)k1\mathcal{P}_{n}(H)\leq k-1. Indeed, if there are kk or more points in 𝒫nH\mathcal{P}_{n}\cap H, then since 𝒴(𝒫n)\mathcal{Y}^{*}({\cal P}_{n}) is (k1)(k-1)-separating, necessarily one of these points, denoted by yy, belongs to 𝒴(𝒫n)\mathcal{Y}^{*}({\cal P}_{n}). Hence y𝒫nH𝒴(𝒫n)y\in\mathcal{P}_{n}\cap H\cap\mathcal{Y}^{*}({\cal P}_{n}), implying yσy\in\sigma and therefore yσσy\in\sigma\setminus\sigma^{*} (since yHy\in H), but this would contradict the assumption that xyx\prec y for all y𝒴(𝒫n){x}y\in\mathcal{Y}^{*}({\cal P}_{n})\setminus\{x\}.

Now we estimate from below the volume of HAH\cap A. Recall that δ1=δ(d,ρ,ε)\delta_{1}=\delta(d,\rho,\varepsilon) is as in Lemma 2.5. Applying (2.8) from there leads to

|HA||Br(1dδ2)(x)A|+2δ1rd.\displaystyle|H\cap A|\geq|B_{r(1-\sqrt{d}{\delta_{2}})}(x)\cap A|+2\delta_{1}r^{d}.

By (4.12), |(Br(x)Br(1dδ2)(x))A|δ1rd|(B_{r}(x)\setminus B_{r(1-\sqrt{d}\delta_{2})}(x))\cap A|\leq\delta_{1}r^{d} and hence

|Br(1dδ2)(x)A||Br(x)A|δ1rd.|B_{r(1-\sqrt{d}{\delta_{2}})}(x)\cap A|\geq|B_{r}(x)\cap A|-\delta_{1}r^{d}.

Let δ3(0,1/2)\delta_{3}\in(0,1/2) be such that δ4:=(12δ3)(1+δ1/(fmaxθd))1>0\delta_{4}:=(1-2\delta_{3})(1+\delta_{1}/(f_{\rm max}\theta_{d}))-1>0. By the preceding estimates, and (4.11), provided rr is small we have that

ν(H)\displaystyle\nu(H) (1δ3)f(x)(|Br(x)A|+δ1rd)\displaystyle\geq(1-\delta_{3})f(x)\left(|B_{r}(x)\cap A|+\delta_{1}r^{d}\right)
(12δ3)ν(Br(x))(1+δ1rdfmaxθdrd)=(1+δ4)ν(Br(x)).\displaystyle\geq(1-2\delta_{3})\nu(B_{r}(x))\left(1+\frac{\delta_{1}r^{d}}{f_{\rm max}\theta_{d}r^{d}}\right)=(1+\delta_{4})\nu(B_{r}(x)).

Let δ5=δ0δ4\delta_{5}=\delta_{0}\delta_{4}, with δ0\delta_{0} given at (3.8). Then

ν(H)ν(Br(x))+δ5rd.\displaystyle\nu(H)\geq\nu(B_{r}(x))+\delta_{5}r^{d}. (4.13)

Also, because of the upper bound on diameters and (3.8), there is a constant c1[1,)c_{1}\in[1,\infty) such that ν(H)c1ν(Br(x))\nu(H)\leq c_{1}\nu(B_{r}(x)) uniformly over all possible xx, all small rr, and all possible σ\sigma.

Using these upper and lower bounds on ν(H)\nu(H), we can deduce that

[Fx,ε,ρ,r(𝒫n){𝖠δ2(𝒴(𝒫n))=σ}]\displaystyle\mathbb{P}[F_{x,\varepsilon,\rho,r}({\cal P}_{n})\cap\{\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}({\cal P}_{n}))=\sigma\}] [𝒫n(H)k1]\displaystyle\leq\mathbb{P}[{\cal P}_{n}(H)\leq k-1]
c1k1j=0k1((nν(Br(x)))j/j!)enν(Br(x))nδ5rd\displaystyle\leq c_{1}^{k-1}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))-n\delta_{5}r^{d}}
=c1k1pn,r(x)enδ5rd.\displaystyle=c_{1}^{k-1}p_{n,r}(x)e^{-n\delta_{5}r^{d}}.

This, together with the union bound over the choice of possible shapes σ\sigma, gives us (4.9).

To prove the result (4.10) for the binomial case we use the volume estimates (4.13) and ν(H)c1ν(Br(x))\nu(H)\leq c_{1}\nu(B_{r}(x)) once more. For nn large, we have

[Fx,ε,ρ,r(𝒳n1){𝖠δ2(𝒴(𝒳n1))=σ}]\displaystyle\mathbb{P}[F_{x,\varepsilon,\rho,r}(\mathcal{X}_{n-1})\cap\{\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}(\mathcal{X}_{n-1}))=\sigma\}] [𝒳n1(H)k1]\displaystyle\leq\mathbb{P}[\mathcal{X}_{n-1}(H)\leq k-1]
=j=0k1(n1j)ν(H)j(1ν(H))n1j\displaystyle=\sum_{j=0}^{k-1}\binom{n-1}{j}\nu(H)^{j}(1-\nu(H))^{n-1-j}
2c1k1j=0k1(nj/j!)ν(Br(x))jexp(nν(H))\displaystyle\leq 2c_{1}^{k-1}\sum_{j=0}^{k-1}(n^{j}/j!)\nu(B_{r}(x))^{j}\exp(-n\nu(H))
2c1k1pn,r(x)enδ5rd,\displaystyle\leq 2c_{1}^{k-1}p_{n,r}(x)e^{-n\delta_{5}r^{d}},

and hence (4.10).

(ii) Suppose d=2d=2 and AA is polygonal. Let 0<ε<ρ<0<\varepsilon<\rho<\infty. Choose KK such that for all rr and all xA𝖢𝗈𝗋(Kr)x\in A\setminus\mathsf{Cor}^{(Kr)}, the ball B(ρ+9)r(x)B_{(\rho+9)r}(x) intersects at most one edge of AA. We can choose such a KK by a similar argument to the proof of Lemma 2.3. Then for xA𝖢𝗈𝗋(Kr)x\in A\setminus\mathsf{Cor}^{(Kr)} we can deduce (4.9) and (4.10) in the same manner as in the proof of part (i). ∎

Next we consider the probability of having a small or medium-sized separating set near the corner of a polygon in dimension 2. We do not attempt to compare this probability with pn,k1(x)p_{n,k-1}(x) because the corners have negligible area and we can get by with a less precise estimate.

Lemma 4.4.

Suppose that d=2d=2 and AA is a convex polygon. Given ρ,K(0,)\rho,K\in(0,\infty), there exist constants c,δ,r0(0,)c,\delta,r_{0}\in(0,\infty) depending only on AA, f0f_{0}, ρ\rho and KK, such that if nk+2n\geq k+2 then

supx𝖢𝗈𝗋(Kr),r(0,r0)[𝒴𝒞r(x,𝒫n),diam(𝒴)ρr]cexp(δnr2);\displaystyle\sup_{x\in\mathsf{Cor}^{(Kr)},r\in(0,r_{0})}\mathbb{P}[\exists\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{P}_{n}),\operatorname{diam}(\mathcal{Y})\leq\rho r]\leq c\exp(-\delta nr^{2}); (4.14)
supx𝖢𝗈𝗋(Kr),r(0,r0)[𝒴𝒞r(x,𝒳n1),diam(𝒴)ρr]cexp(δnr2).\displaystyle\sup_{x\in\mathsf{Cor}^{(Kr)},r\in(0,r_{0})}\mathbb{P}[\exists\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{X}_{n-1}),\operatorname{diam}(\mathcal{Y})\leq\rho r]\leq c\exp(-\delta nr^{2}). (4.15)
Proof.

Fix ρ,K(0,)\rho,K\in(0,\infty). Let αmin\alpha_{\rm min} be the smallest angle of the corners of AA. Assume without loss of generality that one of the corners of AA lies at the origin, and moreover one of the edges of AA incident to the origin is in the direction of the positive xx-axis, while the other edge is the in the anti-clockwise direction from the positive xx-axis, and therefore lies in the upper half-plane since AA is assumed convex.

Let δ2:=1/4\delta_{2}:=1/4. Define 𝖠δ2(𝒴(𝒫n))\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}({\cal P}_{n})) as in the proof of Lemma 4.3. As argued there, if there exists 𝒴𝒞r(x,𝒫n)\mathcal{Y}\in\mathcal{C}_{r}(x,{\cal P}_{n}) with diam(𝒴)ρr\operatorname{diam}(\mathcal{Y})\leq\rho r, then 𝖠δ2(𝒴(𝒫n))\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}({\cal P}_{n})) can take at most cc different possible shapes for some finite cc not depending on rr.

Suppose xBKr(o)x\in B_{Kr}(o). Fix a possible shape σ\sigma that might arise when there exists 𝒴𝒞r(x,𝒫n)\mathcal{Y}\in\mathcal{C}_{r}(x,{\cal P}_{n}) with diam(𝒴)ρr\operatorname{diam}(\mathcal{Y})\leq\rho r, and suppose the event {𝖠δ2(𝒴(𝒫n))=σ}\{\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}({\cal P}_{n}))=\sigma\} occurs. Let ymaxy_{\rm max} be the largest point of σ\sigma in the lexicographic ordering (i.e., the highest rightmost point). Let SS be a sector centred on ymaxy_{\rm max} of radius r/2r/2 and with one straight edge from yy in the direction of the positive xx-axis, while the other edge is in the anti-clockwise direction with angle min(αmin,π/2)\min(\alpha_{\rm min},\pi/2) from the first edge.

Then provided rr is small enough, SAS\subset A and the interior of SS is disjoint from σ\sigma. Also the squares making up σ\sigma have diameter less than r/2r/2, so SS is contained in 𝒫nBr(o){\cal P}_{n}\oplus B_{r}(o); hence 𝒫n(S)k1{\cal P}_{n}(S)\leq k-1. Also f0min(αmin,π/2)r2/2ν(S)(π/4)fmaxr2.f_{0}\min(\alpha_{\min},\pi/2)r^{2}/2\leq\nu(S)\leq(\pi/4)f_{\rm max}r^{2}. Therefore

[{𝒴𝒞r(x,𝒫n),diam(𝒴)ρr}{𝖠δ2(𝒴(𝒫n))=σ}]k(nfmax(π/4)r2)k1\displaystyle\mathbb{P}[\{\exists\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{P}_{n}),\operatorname{diam}(\mathcal{Y})\leq\rho r\}\cap\{\mathsf{A}_{\delta_{2}}(\mathcal{Y}^{*}({\cal P}_{n}))=\sigma\}]\leq k(nf_{\rm max}(\pi/4)r^{2})^{k-1}
×exp(nf0min(αmin,π/2)r2/2).\displaystyle\times\exp(-nf_{0}\min(\alpha_{\min},\pi/2)r^{2}/2).

Summing over all possible σ\sigma and treating other corners similarly, we obtain (4.14). Also

[{𝒴𝒞r(x,𝒳n1),diam(𝒴)ρr}\displaystyle\mathbb{P}[\{\exists\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{X}_{n-1}),\operatorname{diam}(\mathcal{Y})\leq\rho r\} {𝖠δ(𝒴(𝒳n1))=σ}]\displaystyle\cap\{\mathsf{A}_{\delta}(\mathcal{Y}^{*}(\mathcal{X}_{n-1}))=\sigma\}]
j=0k1(n1j)ν(S)j(1ν(S))n1j\displaystyle\leq\sum_{j=0}^{k-1}\binom{n-1}{j}\nu(S)^{j}(1-\nu(S))^{n-1-j}
j=0n2((nν(S))j/j!)exp(nν(S)),\displaystyle\leq\sum_{j=0}^{n}2((n\nu(S))^{j}/j!)\exp(-n\nu(S)),

and using the same upper and lower bounds on ν(S)\nu(S) as before gives us (4.15). ∎

4.3 Large separating pairs

Given r>0r>0, ρ>0\rho>0, recall that if Ln,kr<Mn,kL_{n,k}\leq r<M_{n,k} then there is a (k1)(k-1)-separating pair for G(𝒫n,r)G({\cal P}_{n},r), and each individual set in the pair is non-singleton. Then, either there exists a non-singleton (k1)(k-1)-separating set with diameter at most ρr\rho r, or both sets in the pair have diameter greater than ρr\rho r. Our next lemma deals with the latter possibility. Given ρ>0,r0\rho>0,r\geq 0, define the event

Hr,ρ(𝒳)={ a (k1)-separating pair 𝒴,𝒴 for G(𝒳,r),min(diam(𝒴),diam(𝒴))>ρr}.H_{r,\rho}(\mathcal{X})=\{\exists\text{ a }(k-1)\text{-separating pair }\mathcal{Y},\mathcal{Y}^{\prime}\text{ for }G(\mathcal{X},r),\min(\operatorname{diam}(\mathcal{Y}),\operatorname{diam}(\mathcal{Y}^{\prime}))>\rho r\}.
Lemma 4.5.

Suppose (rn)n>0(r_{n})_{n>0} satisfies (3.5) for some β(0,)\beta^{\prime}\in(0,\infty). Then there exists ρ(0,)\rho\in(0,\infty) such that [Hrn,ρ(𝒫n)]=O(n2)\mathbb{P}[H_{r_{n},\rho}(\mathcal{P}_{n})]=O(n^{-2}) and [Hrn,ρ(𝒳n)]=O(n2)\mathbb{P}[H_{r_{n},\rho}(\mathcal{X}_{n})]=O(n^{-2}) as nn\to\infty.

Proof.

Case 1: AC2\partial A\in C^{2}. See [13, Equation (3.14)]. That result is formulated only for 𝒳n\mathcal{X}_{n}, not for 𝒫n{\cal P}_{n}, and also only for the case k=0k=0. However, it uses only the probability bound that if XX is binomial with mean μ\mu then [X=0]eμ\mathbb{P}[X=0]\leq e^{-\mu}. Using a standard Chernoff bound, e.g. [8, Lemmas 1.1 and 1.2], we have that if XX is either binomial or Poisson distributed with mean μ\mu for μ\mu sufficiently large, we have [X<k]eμ/2\mathbb{P}[X<k]\leq e^{-\mu/2}, and using this we can readily adapt the argument in [13] to the generality required here.

Case 2: d=2d=2 and AA is polygonal. In this case we use the proof of [17, Lemma 3.12]. Our rnr_{n} is not quite the same as there, but the argument works for our rnr_{n} too; the properties of rnr_{n} given in Lemma 3.3 are sufficient. Again, the proof in [17] is only for 𝒳n\mathcal{X}_{n}, but it relies only on Chernoff probability bounds for a binomial random variable, which also apply for a Poisson random variable with the same mean and therefore the result holds for 𝒫n{\cal P}_{n} as well as for 𝒳n\mathcal{X}_{n}. ∎

We can now bound [Ln,krn<Mn,k]\mathbb{P}[L_{n,k}\leq r_{n}<M_{n,k}], which is the result we have been leading up to in this whole section.

Proposition 4.6.

Let β(0,)\beta^{\prime}\in(0,\infty) and suppose (rn)n1(r_{n})_{n\geq 1} satisfies (3.5). Then as nn\to\infty,

[Ln,krn<Mn,k]=O((logn)1d);\displaystyle\mathbb{P}[L_{n,k}\leq r_{n}<M_{n,k}]=O((\log n)^{1-d}); (4.16)
[Lk(𝒳n)rn<Mk(𝒳n)]=O((logn)1d).\displaystyle\mathbb{P}[L_{k}(\mathcal{X}_{n})\leq r_{n}<M_{k}(\mathcal{X}_{n})]=O((\log n)^{1-d}). (4.17)
Proof.

Given r,ρ(0,)r,\rho\in(0,\infty), and finite 𝒳d\mathcal{X}\subset\mathbb{R}^{d}, define the event

Jr,ρ(𝒳):={x𝒳,𝒴𝒞r(x,𝒳),0<diam(𝒴)ρr}.\displaystyle J_{r,\rho}(\mathcal{X}):=\{\exists x\in\mathcal{X},\mathcal{Y}\in\mathcal{C}_{r}(x,\mathcal{X}),0<\operatorname{diam}(\mathcal{Y})\leq\rho r\}.

By Lemma 4.1, if Lk(𝒳)rn<Mk(𝒳)L_{k}(\mathcal{X})\leq r_{n}<M_{k}(\mathcal{X}), then either Jrn,ρ(𝒳)J_{r_{n},\rho}(\mathcal{X}) or Hrn,ρ(𝒳)H_{r_{n},\rho}(\mathcal{X}) occurs. Hence by Lemma 4.5, it suffices to prove that for any ρ(0,)\rho\in(0,\infty), the events Jrn,ρ(𝒫n)J_{r_{n},\rho}(\mathcal{P}_{n}) and Jrn,ρ(𝒳n)J_{r_{n},\rho}(\mathcal{X}_{n}) occur with probability O((logn)1d)O((\log n)^{1-d}) as nn\to\infty.

Case 1: AC2\partial A\in C^{2}. Fix ρ(0,)\rho\in(0,\infty). Let NnN_{n} denote the (random) number of x𝒫nx\in{\cal P}_{n} such that there exists a 𝒴𝒞rn(x,𝒫n)\mathcal{Y}\in\mathcal{C}_{r_{n}}(x,{\cal P}_{n}) with 0<diam(𝒴)ρrn0<\operatorname{diam}(\mathcal{Y})\leq\rho r_{n}. By Markov’s inequality [Jρ,rn(𝒫n)]=[Nn1]𝔼[Nn]\mathbb{P}[J_{\rho,r_{n}}({\cal P}_{n})]=\mathbb{P}[N_{n}\geq 1]\leq\mathbb{E}[N_{n}].

Let δ\delta be as in Lemma 4.2 (i) and assume without loss of generality that 0<δ<ρ0<\delta<\rho. Then by the Mecke equation and the definitions of Ex,ρ,rE_{x,\rho,r} and Fx,ε,ρ,rF_{x,\varepsilon,\rho,r} at (4.2) and (4.8), and the union bound,

𝔼[Nn]\displaystyle\mathbb{E}[N_{n}] nA[Ex,δ,rn(x,𝒫n)]ν(dx)+nA[Fx,δ,ρ,rn(x,𝒫n)]ν(dx).\displaystyle\leq n\int_{A}\mathbb{P}[E_{x,\delta,r_{n}}(x,\mathcal{P}_{n})]\nu(dx)+n\int_{A}\mathbb{P}[F_{x,\delta,\rho,r_{n}}(x,\mathcal{P}_{n})]\nu(dx). (4.18)

Using Lemma 4.2 for the first integral and Lemma 4.3 for the second integral, and (3.3), we can find c,δ2(0,)c,\delta_{2}\in(0,\infty) such that for large enough nn we have that

𝔼[Nn]c((nrnd)1d+eδ2nrnd)𝔼[ξn,rn],\displaystyle\mathbb{E}[N_{n}]\leq c((nr_{n}^{d})^{1-d}+e^{-\delta_{2}nr_{n}^{d}})\mathbb{E}[\xi_{n,r_{n}}],

where ξn,r\xi_{n,r} was defined at (3.1). From this, we obtain the claimed estimate (4.16) by (3.5) and Lemma 3.3.

Case 2: d=2d=2 and AA is polygonal. Fix ρ(0,)\rho\in(0,\infty). Let δ\delta be as in Lemma 4.2 (ii); assume without loss of generality that 0<δ<ρ0<\delta<\rho. By a similar argument to (4.18) we have

[Jρ,rn(𝒫n)]\displaystyle\mathbb{P}[J_{\rho,r_{n}}({\cal P}_{n})] A𝖢𝗈𝗋(Krn)[Ex,δ,rn(x,𝒫n)]nν(dx).\displaystyle\leq\int_{A\setminus\mathsf{Cor}^{(Kr_{n})}}\mathbb{P}[E_{x,\delta,r_{n}}(x,\mathcal{P}_{n})]n\nu(dx).
+A𝖢𝗈𝗋(Krn)[Fx,δ,ρ,rn(x,𝒫n)]nν(dx).\displaystyle+\int_{A\setminus\mathsf{Cor}^{(Kr_{n})}}\mathbb{P}[F_{x,\delta,\rho,r_{n}}(x,\mathcal{P}_{n})]n\nu(dx).
+𝖢𝗈𝗋(Krn)[{𝒴𝒞rn(x,𝒫n),diam(𝒴)ρrn}]nν(dx).\displaystyle+\int_{\mathsf{Cor}^{(Kr_{n})}}\mathbb{P}[\{\exists\mathcal{Y}\in\mathcal{C}_{r_{n}}(x,\mathcal{P}_{n}),\operatorname{diam}(\mathcal{Y})\leq\rho r_{n}\}]n\nu(dx).

We can deal with the first two integrals just as we did in Case 1. By Lemma 4.4, there is a constant δ\delta^{\prime} such that the third integral is O(nrn2exp(δnrn2))O(nr_{n}^{2}\exp(-\delta^{\prime}nr_{n}^{2})), which completes the proof of (4.16) for this case.

The proof of (4.17) is identical, now relying on the binomial parts of Lemmas 4.24.5. We omit the details. ∎

5 Proof of Theorem 1.3

Throughout this section we assume that AC2\partial A\in C^{2} or that AA is a convex polygon. We adopt our WA but do not assume ff is necessarily constant on AA.

5.1 Proof of parts (i) and (ii)

Given β\beta\in\mathbb{R}, choose n0(β)n_{0}(\beta) such that n0(β)>eβn_{0}(\beta)>e^{-\beta} and enj=0k1(nj+1)/j!<eβe^{-n}\sum_{j=0}^{k-1}(n^{j+1})/j!<e^{-\beta} for all n[n0(β),)n\in[n_{0}(\beta),\infty). Recall the definition of pn,r(x)p_{n,r}(x) at (3.4). Given nn0(β)n\geq n_{0}(\beta), define rn(β)(0,)r_{n}(\beta)\in(0,\infty) by

r=rn(β)nApn,r(x)ν(dx)=eβ.\displaystyle r=r_{n}(\beta)\Longleftrightarrow n\int_{A}p_{n,r}(x)\nu(dx)=e^{-\beta}. (5.1)

By the Intermediate Value theorem, such an rn(β)r_{n}(\beta) exists and is unique (note that the integrand is nonincreasing in rr because the Poisson(λ)(\lambda) distribution is stochastically monotone in λ\lambda). Moreover, for <β<γ<-\infty<\beta<\gamma<\infty and nmax(n0(β),n0(γ))n\geq\max(n_{0}(\beta),n_{0}(\gamma)), we have rn(β)<rn(γ)r_{n}(\beta)<r_{n}(\gamma).

We first determine the first-order limiting behaviour of rn(β)r_{n}(\beta).

Lemma 5.1.

Let β\beta\in\mathbb{R} and let rn(β)r_{n}(\beta) satisfy (5.1) for all n>n0(β)n>n_{0}(\beta). Then

limn(nθdrn(β)d/logn)=max(1/f0,(22/d)/f1).\displaystyle\lim_{n\to\infty}(n\theta_{d}r_{n}(\beta)^{d}/\log n)=\max(1/f_{0},(2-2/d)/f_{1}). (5.2)

If also γ\gamma\in\mathbb{R} with β<γ\beta<\gamma, then

limnsupxA(ν(B(x,rn(γ)))/ν(B(x,rn(β))))=1.\displaystyle\lim_{n\to\infty}\sup_{x\in A}\big{(}\nu(B(x,r_{n}(\gamma)))/\nu(B(x,r_{n}(\beta)))\big{)}=1. (5.3)
Proof.

By Proposition 3.1, as tt\to\infty (through \mathbb{R}),

[Lt,krt(β)]exp(eβ).\displaystyle\mathbb{P}[L_{t,k}\leq r_{t}(\beta)]\to\exp(-e^{-\beta}). (5.4)

On the other hand, we claim that tθdLt,kd/logtmax(1/f0,(22/d)/f1)t\theta_{d}L_{t,k}^{d}/\log t\overset{\mathbb{P}}{\longrightarrow}\max(1/f_{0},(2-2/d)/f_{1}) as tt\to\infty. Indeed, writing NtN_{t} for the number of points of the Poisson process 𝒫t\mathcal{P}_{t}, we have

tθdLt,kd/logt=(NtθdLt,kd/logNt)×(t/Nt)×(logNt/logt).\displaystyle t\theta_{d}L_{t,k}^{d}/\log t=(N_{t}\theta_{d}L_{t,k}^{d}/\log N_{t})\times(t/N_{t})\times(\log N_{t}/\log t). (5.5)

Since the conditional distribution of Lt,kL_{t,k}, given Nt=nN_{t}=n, is that of Lk(𝒳n)L_{k}(\mathcal{X}_{n}), we have from (1.7) when AA has a C2C^{2} boundary, and from (1.8) when AA is a convex polygon, that the first factor in the right hand side of (5.5) tends to max(1/f0,(22/d)/f1)\max(1/f_{0},(2-2/d)/f_{1}) in probability, and by Chebyshev’s inequality the second factor also tends to 11 in probability, from which we can deduce the third factor also tends to 1 in probability. Combining these gives us the claim.

Let α<max(1/f0,(22/d)/f1)\alpha<\max(1/f_{0},(2-2/d)/f_{1}). By the preceding claim we have [tθdLt,kd/logt<α]0\mathbb{P}[t\theta_{d}L_{t,k}^{d}/\log t<\alpha]\to 0, and hence by (5.4), tθdrtd/logtαt\theta_{d}r_{t}^{d}/\log t\geq\alpha for tt large. Similarly, if α>max(1/f0,(22/d)/f1)\alpha^{\prime}>\max(1/f_{0},(2-2/d)/f_{1}) then [tθLt,kd/logtα]1\mathbb{P}[t\theta L_{t,k}^{d}/\log t\leq\alpha]\to 1, so tθdrtd/logtαt\theta_{d}r_{t}^{d}/\log t\leq\alpha^{\prime} for tt large. Combining these assertions gives us (5.2).

For (5.3), note that ν(Bs(x)Br(x))fmaxθd(sdrd)\nu(B_{s}(x)\setminus B_{r}(x))\leq f_{\rm max}\theta_{d}(s^{d}-r^{d}) for xAx\in A and 0<r<s0<r<s. Therefore using (3.8), for all xAx\in A and all large enough nn we have

ν(Brn(γ)(x)Brn(β)(x))ν(Brn(β)(x))fmaxθd(rn(γ)drn(β)d)δ0rn(β)d,\displaystyle\frac{\nu(B_{r_{n}(\gamma)}(x)\setminus B_{r_{n}(\beta)}(x))}{\nu(B_{r_{n}(\beta)}(x))}\leq\frac{f_{\rm max}\theta_{d}(r_{n}(\gamma)^{d}-r_{n}(\beta)^{d})}{\delta_{0}r_{n}(\beta)^{d},}

which tends to zero by (5.2), and (5.3) follows. ∎

Recall that μ(X)\mu(X) denotes the median of a random variable XX. For non-uniform ν\nu, it seems to be hard in general to find a formula for rn(β)r_{n}(\beta) satisfying (5.1) (even if the equality is replaced by convergence). However, if we can determine a limit for nrn(γ)dnrn(β)dnr_{n}(\gamma)^{d}-nr_{n}(\beta)^{d} for all β<γ\beta<\gamma, then we can still obtain a weak limiting distribution for nMn,kdnμ(Mn,k)dnM_{n,k}^{d}-n\mu(M_{n,k})^{d} without giving an explicit sequence for μ(Mn,k)\mu(M_{n,k}). The next lemma spells out this argument.

Lemma 5.2.

Suppose there exists α(0,)\alpha\in(0,\infty) such that for all β,γ\beta,\gamma\in\mathbb{R} with β<γ\beta<\gamma, we have as nn\to\infty that

limnn(rn(γ)drn(β)d)=α(γβ),\displaystyle\lim_{n\to\infty}n(r_{n}(\gamma)^{d}-r_{n}(\beta)^{d})=\alpha(\gamma-\beta), (5.6)

where rn(β)r_{n}(\beta) is defined by (5.1). Suppose that (Xn)n>0(X_{n})_{n>0} are random variables satisfying

limn[Xnrn(β)]=exp(eβ),β.\displaystyle\lim_{n\to\infty}\mathbb{P}[X_{n}\leq r_{n}(\beta)]=\exp(-e^{-\beta}),\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall\leavevmode\nobreak\ \beta\in\mathbb{R}. (5.7)

Then

nXndnμ(Xn)d𝑑α(𝖦𝗎+loglog2)asn.\displaystyle nX_{n}^{d}-n\mu(X_{n})^{d}\overset{d}{\longrightarrow}\alpha({\mathsf{Gu}}+\log\log 2)\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ {\rm as}\leavevmode\nobreak\ n\to\infty. (5.8)
Proof.

Set β0=loglog2\beta_{0}=-\log\log 2. Let rn=rn(β0)r_{n}=r_{n}(\beta_{0}) and let <y<x<y<-\infty<y<x<y^{\prime}<\infty. Set sn:=rn(β0+y/α)s_{n}:=r_{n}(\beta_{0}+y/\alpha) and sn:=rn(β0+y/α)s^{\prime}_{n}:=r_{n}(\beta_{0}+y^{\prime}/\alpha). Then by (5.6), n(sndrnd)yn(s_{n}^{d}-r_{n}^{d})\to y and n((sn)drnd)yn((s^{\prime}_{n})^{d}-r_{n}^{d})\to y^{\prime}. Hence for nn large we have nsnd<x+nrndns_{n}^{d}<x+nr_{n}^{d} and n(sn)d>x+nrndn(s^{\prime}_{n})^{d}>x+nr_{n}^{d}, so that by (5.7), setting F(x):=exp(ex)F(x):=\exp(-e^{-x}) we have

[nXndnrndx][nXndnsnd]F(β0+y/α),\mathbb{P}[nX_{n}^{d}-nr_{n}^{d}\leq x]\geq\mathbb{P}[nX_{n}^{d}\leq ns_{n}^{d}]\to F(\beta_{0}+y/\alpha),

and similarly [nXndnrndx][nXndn(sn)d]F(β0+y/α).\mathbb{P}[nX_{n}^{d}-nr_{n}^{d}\leq x]\leq\mathbb{P}[nX_{n}^{d}\leq n(s^{\prime}_{n})^{d}]\to F(\beta_{0}+y^{\prime}/\alpha). Since we can take yy and yy^{\prime} arbitrarily close to xx and F()F(\cdot) is continuous, we can deduce that

[nXndnrndx]F(β0+x/α),x.\mathbb{P}[nX_{n}^{d}-nr_{n}^{d}\leq x]\to F(\beta_{0}+x/\alpha),\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ x\in\mathbb{R}.

Since F(β0+z/α)=[α(𝖦𝗎+loglog2)z]F(\beta_{0}+z/\alpha)=\mathbb{P}[\alpha({\mathsf{Gu}}+\log\log 2)\leq z], we thus have

nXndnrnd𝑑α(𝖦𝗎+loglog2).\displaystyle nX_{n}^{d}-nr_{n}^{d}\overset{d}{\longrightarrow}\alpha({\mathsf{Gu}}+\log\log 2). (5.9)

Finally we need to check that nμ(Xn)dnrnd0n\mu(X_{n})^{d}-nr_{n}^{d}\to 0. Let ε>0\varepsilon>0. By (5.9), as nn\to\infty we have

[nXndnrndε]\displaystyle\mathbb{P}[nX_{n}^{d}-nr_{n}^{d}\leq\varepsilon] [α(𝖦𝗎+loglog2)ε]>1/2;\displaystyle\to\mathbb{P}[\alpha({\mathsf{Gu}}+\log\log 2)\leq\varepsilon]>1/2;
[nXndnrndε]\displaystyle\mathbb{P}[nX_{n}^{d}-nr_{n}^{d}\leq-\varepsilon] [α(𝖦𝗎+loglog2)ε]<1/2,\displaystyle\to\mathbb{P}[\alpha({\mathsf{Gu}}+\log\log 2)\leq-\varepsilon]<1/2,

so for large nn we have

μ(Xnd)nrnd=μ(nXndnrnd)[ε,ε],\mu(X_{n}^{d})-nr_{n}^{d}=\mu(nX_{n}^{d}-nr_{n}^{d})\in[-\varepsilon,\varepsilon],

so μ(Xnd)nrnd0\mu(X_{n}^{d})-nr_{n}^{d}\to 0 as nn\to\infty, and then (5.8) follows from (5.9) and the continuity of the Gumbel cdf. ∎

To use Lemma 5.2, we need to show that (5.6) holds for some α\alpha. We do this first for the case where f1>f0(22/d)f_{1}>f_{0}(2-2/d).

Lemma 5.3.

Let β,γ\beta,\gamma\in\mathbb{R} with β<γ\beta<\gamma. Define rn(β)r_{n}(\beta) by (5.1). If f1>f0(22/d)f_{1}>f_{0}(2-2/d), then

n(rn(γ)drn(β)d)(γβ)/(θdf0)asn.\displaystyle n(r_{n}(\gamma)^{d}-r_{n}(\beta)^{d})\to(\gamma-\beta)/(\theta_{d}f_{0})\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ {\rm as}\leavevmode\nobreak\ n\to\infty. (5.10)
Proof.

Given nn, set r=rn(β),s=rn(γ)r=r_{n}(\beta),s=r_{n}(\gamma). For all xA(s)x\in A^{(-s)}, ν(Bs(x)Br(x))θdf0(sdrd),\nu(B_{s}(x)\setminus B_{r}(x))\geq\theta_{d}f_{0}(s^{d}-r^{d}), and hence using (5.1), we have

eβ\displaystyle e^{-\beta} A(s)j=0k1((nν(Br(x)))j/j!)enν(Bs(x))enν(Bs(x)Br(x))nν(dx)\displaystyle\geq\int_{A^{(-s)}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}e^{n\nu(B_{s}(x)\setminus B_{r}(x))}n\nu(dx)
enθdf0(sdrd)A(s)j=1k1((nν(Br(x)))j/j!)enν(Bs(x))nν(dx).\displaystyle\geq e^{n\theta_{d}f_{0}(s^{d}-r^{d})}\int_{A^{(-s)}}\sum_{j=1}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx). (5.11)

Suppose AC2\partial A\in C^{2}. Using our assumption f1>f0(22/d)f_{1}>f_{0}(2-2/d), choose δ>0\delta>0 such that (22/d)(f12δ)1<f01(2-2/d)(f_{1}-2\delta)^{-1}<f_{0}^{-1}. Using Lemma 2.2-(i) and the continuity of f|Af|_{A} we find for all large enough nn and all x(A)(s)x\in(\partial A)^{(s)} that ν(Br(x))θd(f1δ)rd/2\nu(B_{r}(x))\geq\theta_{d}(f_{1}-\delta)r^{d}/2. Then using Lemma 5.1 and our assumption f1>f0(22/d)f_{1}>f_{0}(2-2/d), we have that

limn(nθdrd/logn)=f01>(22/d)(f12δ)1.\displaystyle\lim_{n\to\infty}\big{(}n\theta_{d}r^{d}/\log n\big{)}=f_{0}^{-1}>(2-2/d)(f_{1}-2\delta)^{-1}. (5.12)

Hence there are constants c,cc,c^{\prime} such that for nn large

(A)(s)j=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)c(logn)k1nsenθd(f1δ)rd/2\displaystyle\int_{(\partial A)^{(s)}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx)\leq c(\log n)^{k-1}nse^{-n\theta_{d}(f_{1}-\delta)r^{d}/2}
c(logn)k1+1/dn11/dexp((f1δ)(11/d)(f12δ)1logn),\displaystyle\leq c^{\prime}(\log n)^{k-1+1/d}n^{1-1/d}\exp(-(f_{1}-\delta)(1-1/d)(f_{1}-2\delta)^{-1}\log n),

which tends to zero.

Suppose instead that d=2d=2 and AA is polygonal. The preceding estimate shows (A)(s)𝖢𝗈𝗋(s)j=0k1((nν(Br(x)))j/j!)exp(nν(Br(x)))nν(dx)\int_{(\partial A)^{(s)}\setminus\mathsf{Cor}^{(s)}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)\exp(-n\nu(B_{r}(x)))n\nu(dx) tends to zero. Moreover, by (3.8) there exist c,δ0(0,)c,\delta_{0}\in(0,\infty) such that

𝖢𝗈𝗋(s)j=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)c(nr2)keδ0nr2\int_{\mathsf{Cor}^{(s)}}\sum_{j=0}^{k-1}\left((n\nu(B_{r}(x)))^{j}/j!\right)e^{-n\nu(B_{r}(x))}n\nu(dx)\leq c(nr^{2})^{k}e^{-\delta_{0}nr^{2}}

which tends to zero since nr2nr^{2}\to\infty by (5.2). Thus in both cases we have that

(A)(s)j=0k1((nν(Bs(x)))j/j!)exp(nν(Br(x)))nν(dx)0.\displaystyle\int_{(\partial A)^{(s)}}\sum_{j=0}^{k-1}((n\nu(B_{s}(x)))^{j}/j!)\exp(-n\nu(B_{r}(x)))n\nu(dx)\to 0. (5.13)

Therefore using (5.1) and (5.3) we obtain that

eγ\displaystyle e^{-\gamma} =limnA(s)j=0k1((nν(Bs(x)))j/j!)enν(Bs(x))nν(dx),\displaystyle=\lim_{n\to\infty}\int_{A^{(-s)}}\sum_{j=0}^{k-1}((n\nu(B_{s}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx),
=limnA(s)j=0k1((nν(Br(x)))j/j!)enν(Bs(x))nν(dx),\displaystyle=\lim_{n\to\infty}\int_{A^{(-s)}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx),

and then taking nn\to\infty in (5.11) we obtain that eβeγlim supn(enθdf0(sdrd)),e^{-\beta}\geq e^{-\gamma}\limsup_{n\to\infty}\left(e^{n\theta_{d}f_{0}(s^{d}-r^{d})}\right), so that

lim sup(n(sdrd))(γβ)/(θdf0).\displaystyle\limsup(n(s^{d}-r^{d}))\leq(\gamma-\beta)/(\theta_{d}f_{0}). (5.14)

For an inequality the other way, let ε>0\varepsilon>0 and let Aε:={xA:f(x)f0+4ε}A_{\varepsilon}:=\{x\in A:f(x)\leq f_{0}+4\varepsilon\}. By the assumed continuity of ff on AA, for all nn large enough, and all xAεx\in A_{\varepsilon}, we have ν(Bs(x)Br(x))θd(f0+5ε)(sdrd).\nu(B_{s}(x)\setminus B_{r}(x))\leq\theta_{d}(f_{0}+5\varepsilon)(s^{d}-r^{d}). Therefore by (5.1),

eβ\displaystyle e^{-\beta} Aεj=0k1((nν(Br(x)))j/j!)enθd(f0+5ε)(sdrd)enν(Bs(x))nν(dx)\displaystyle\leq\int_{A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{n\theta_{d}(f_{0}+5\varepsilon)(s^{d}-r^{d})}e^{-n\nu(B_{s}(x))}n\nu(dx)
+A(r)Aεj=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)\displaystyle+\int_{A^{(-r)}\setminus A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx)
+(A)(r)Aεj=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx).\displaystyle+\int_{(\partial A)^{(r)}\setminus A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx). (5.15)

The third integral on the right tends to zero by (5.13). For nn large enough, and all xA(r)Aεx\in A^{(-r)}\setminus A_{\varepsilon}, using (5.12) we have nν(Br(x))n(f0+3ε)θdrd(f0+3ε)(logn)/(f0+ε)n\nu(B_{r}(x))\geq n(f_{0}+3\varepsilon)\theta_{d}r^{d}\geq(f_{0}+3\varepsilon)(\log n)/(f_{0}+\varepsilon), and hence the second integral in (5.15) tends to zero. Therefore

lim inf(Aεj=0k1((nν(Br(x)))j/j!)enθd(f0+5ε)(sdrd)enν(Bs(x))nν(dx))eβ.\displaystyle\liminf\left(\int_{A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{n\theta_{d}(f_{0}+5\varepsilon)(s^{d}-r^{d})}e^{-n\nu(B_{s}(x))}n\nu(dx)\right)\geq e^{-\beta}.

Also, by (5.3) the second and third integrals in (5.15) still tend to zero if we change Br(x)B_{r}(x) to Bs(x)B_{s}(x), so

eγ=limAεj=0k1((nν(Bs(x)))j/j!)enν(Bs(x))nν(dx).\displaystyle e^{-\gamma}=\lim\int_{A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{s}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx).

Hence, using (5.3) again we obtain that

eγ=limAεj=0k1((nν(Br(x)))j/j!)enν(Bs(x))nν(dx).\displaystyle e^{-\gamma}=\lim\int_{A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx).

Hence lim inf(enθd(f0+5ε)(sdrd))×eγeβ\liminf(e^{n\theta_{d}(f_{0}+5\varepsilon)(s^{d}-r^{d})})\times e^{-\gamma}\geq e^{-\beta}, so that

lim inf(n(sdrd))(γβ)/(θd(f0+5ε)).\liminf(n(s^{d}-r^{d}))\geq(\gamma-\beta)/(\theta_{d}(f_{0}+5\varepsilon)).

Since ε>0\varepsilon>0 is arbitrary, combining this with (5.14) yields (5.10). ∎

Next we show that (5.6) holds for a different α\alpha in the case where f1<f0(22/d)f_{1}<f_{0}(2-2/d).

Lemma 5.4.

Let β,γ\beta,\gamma\in\mathbb{R} with β<γ\beta<\gamma. If f1<f0(22/d)f_{1}<f_{0}(2-2/d), then

limn(n(rn(γ)drn(β)d))=2(γβ)/(θdf1).\displaystyle\lim_{n\to\infty}(n(r_{n}(\gamma)^{d}-r_{n}(\beta)^{d}))=2(\gamma-\beta)/(\theta_{d}f_{1}). (5.16)

Also, if f1=f0(22/d)f_{1}=f_{0}(2-2/d), then

lim supn(n(rn(γ)drn(β)d))2(γβ)/(θdf1).\displaystyle\limsup_{n\to\infty}(n(r_{n}(\gamma)^{d}-r_{n}(\beta)^{d}))\leq 2(\gamma-\beta)/(\theta_{d}f_{1}). (5.17)
Proof.

Assume f1f0(22/d)f_{1}\leq f_{0}(2-2/d). For each nn set r:=rn(β)r:=r_{n}(\beta), s:=rn(γ)s:=r_{n}(\gamma). Suppose AC2\partial A\in C^{2}. By Lemma 2.2-(iii), given ε>0\varepsilon>0, for nn large enough and all x(A)(s)x\in(\partial A)^{(-s)}, we have ν(Bs(x)Br(x))(f1ε)θd(sdrd)/2\nu(B_{s}(x)\setminus B_{r}(x))\geq(f_{1}-\varepsilon)\theta_{d}(s^{d}-r^{d})/2. Also ν(Bs(x)Br(x))nf0θd(sdrd)\nu(B_{s}(x)\setminus B_{r}(x))\geq nf_{0}\theta_{d}(s^{d}-r^{d}) for xA(s)x\in A^{(-s)}. Hence for nn large enough

eβ\displaystyle e^{-\beta} en(f1ε)θd(sdrd)/2(A)(s)j=0k1((ν(Br(x)))j/j!)enν(Bs(x))nν(dx)\displaystyle\geq e^{n(f_{1}-\varepsilon)\theta_{d}(s^{d}-r^{d})/2}\int_{(\partial A)^{(s)}}\sum_{j=0}^{k-1}((\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)
+enf0θd(sdrd)A(s)j=0k1((ν(Br(x)))j/j!)enν(Bs(x))nν(dx)\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +e^{nf_{0}\theta_{d}(s^{d}-r^{d})}\int_{A^{(-s)}}\sum_{j=0}^{k-1}((\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)
en(f1ε)θd(sdrd)/2((A)(s)j=0k1((ν(Br(x)))j/j!)enν(Bs(x))nν(dx)\displaystyle\geq e^{n(f_{1}-\varepsilon)\theta_{d}(s^{d}-r^{d})/2}\Big{(}\int_{(\partial A)^{(s)}}\sum_{j=0}^{k-1}((\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)
+A(s)j=0k1((ν(Br(x)))j/j!)enν(Bs(x))nν(dx)),\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +\int_{A^{(-s)}}\sum_{j=0}^{k-1}((\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)\Big{)}, (5.18)

since the assumption f1f0(22/d)f_{1}\leq f_{0}(2-2/d) implies f0f1/2f_{0}\geq f_{1}/2. Hence, for nn large enough, setting ψn:=infxA(ν(Br(x))/ν(Bs(x)))\psi_{n}:=\inf_{x\in A}\big{(}\nu(B_{r}(x))/\nu(B_{s}(x))\big{)} we have

eβen(f1ε)θd(sdrd)/2eγψnk1.e^{-\beta}\geq e^{n(f_{1}-\varepsilon)\theta_{d}(s^{d}-r^{d})/2}e^{-\gamma}\psi_{n}^{k-1}.

By (5.3) we have ψn1\psi_{n}\to 1 as nn\to\infty, and thus

lim sup(n(sdrd))2(γβ)/(θd(f1ε)).\displaystyle\limsup(n(s^{d}-r^{d}))\leq 2(\gamma-\beta)/(\theta_{d}(f_{1}-\varepsilon)). (5.19)

In the other case with d=2d=2 and AA polygonal, on choosing a suitable large KK (dependent on the smallest angle of AA) we can obtain, similarly to (5.18), that

eβ\displaystyle e^{-\beta} en(f1ε)π(s2r2)/2ψnk1A𝖢𝗈𝗋(Kr)j=0k1((nν(Bs(x)))j/j!)enν(Bs(x))nν(dx)\displaystyle\geq e^{n(f_{1}-\varepsilon)\pi(s^{2}-r^{2})/2}\psi_{n}^{k-1}\int_{A\setminus\mathsf{Cor}^{(Kr)}}\sum_{j=0}^{k-1}((n\nu(B_{s}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)
=en(f1ε)π(s2r2)/2ψnk1(eγ𝖢𝗈𝗋(Kr)j=0k1((nν(Bs(x)))j/j!)enν(Bs(x))nν(dx)).\displaystyle=e^{n(f_{1}-\varepsilon)\pi(s^{2}-r^{2})/2}\psi_{n}^{k-1}\left(e^{-\gamma}-\int_{\mathsf{Cor}^{(Kr)}}\sum_{j=0}^{k-1}((n\nu(B_{s}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)\right). (5.20)

Since the integral over 𝖢𝗈𝗋(Kr)\mathsf{Cor}^{(Kr)} tends to zero by (3.8), we therefore obtain from (5.20) that (5.19) holds in this case too. Taking ε0\varepsilon\downarrow 0, we deduce in both cases that (5.17) holds whenever f1f0(22/d)f_{1}\leq f_{0}(2-2/d).

Now assume the strict inequality f1<f0(22/d)f_{1}<f_{0}(2-2/d). Again set r=rn(β)r=r_{n}(\beta), s=rn(γ)s=r_{n}(\gamma). Using (5.2) from Lemma 5.1, choose δ2>0\delta_{2}>0 such that

limnnθdrdlogn=22/df1>1+2δ2f0.\displaystyle\lim_{n\to\infty}\frac{n\theta_{d}r^{d}}{\log n}=\frac{2-2/d}{f_{1}}>\frac{1+2\delta_{2}}{f_{0}}. (5.21)

Then there exists a constant cc such that for nn large,

A(r)j=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)\displaystyle\int_{A^{(-r)}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx) cn(logn)k1enθdf0rd\displaystyle\leq cn(\log n)^{k-1}e^{-n\theta_{d}f_{0}r^{d}}
nc(logn)k1e(1+δ2)logn,\displaystyle\leq nc(\log n)^{k-1}e^{-(1+\delta_{2})\log n}, (5.22)

which tends to zero.

Take a new ε>0\varepsilon>0. Let Aε:={xA:f(x)f1+3ε}A_{\varepsilon}:=\{x\in A:f(x)\leq f_{1}+3\varepsilon\}. Using Lemma 2.1-(ii) in the case AC2\partial A\in C^{2}, take δ>0\delta>0 such that for all large enough nn and all x(A)(δr)x\in(\partial A)^{(\delta r)}, we have

|ABs(x)Br(x)|θd(sdrd)(f1+5ε)/(2(f1+4ε)).|A\cap B_{s}(x)\setminus B_{r}(x)|\leq\theta_{d}(s^{d}-r^{d})(f_{1}+5\varepsilon)/(2(f_{1}+4\varepsilon)).

Such δ\delta can also be found in the other case where AA is a convex polygon.

Then for nn large and x(A)(δr)Aεx\in(\partial A)^{(\delta r)}\cap A_{\varepsilon} we have supBs(x)Aff1+4ε\sup_{B_{s}(x)\cap A}f\leq f_{1}+4\varepsilon, and hence

ν(Bs(x)Br(x))θd(sdrd)(f1+5ε)/2,x(A)(δr)Aε.\displaystyle\nu(B_{s}(x)\setminus B_{r}(x))\leq\theta_{d}(s^{d}-r^{d})(f_{1}+5\varepsilon)/2,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \forall\leavevmode\nobreak\ x\in(\partial A)^{(\delta r)}\cap A_{\varepsilon}. (5.23)

Using Lemma 2.2-(i), in the case where AC2\partial A\in C^{2} we have for large nn, and all xAAεx\in A\setminus A_{\varepsilon}, that ν(Br(x))(f1+2ε)θdrd/2\nu(B_{r}(x))\geq(f_{1}+2\varepsilon)\theta_{d}r^{d}/2. Hence for nn large

(A)(r)Aεj=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)c(logn)knrexp(n(f1+2ε)θdrd/2)\displaystyle\int_{(\partial A)^{(r)}\setminus A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx)\leq c(\log n)^{k}nr\exp(-n(f_{1}+2\varepsilon)\theta_{d}r^{d}/2)
cn11/d(logn)k+1/dexp((f1+ε)(11/d)(logn)/f1)0,\displaystyle\leq cn^{1-1/d}(\log n)^{k+1/d}\exp(-(f_{1}+\varepsilon)(1-1/d)(\log n)/f_{1})\to 0,

where for the second inequality we have used the equality in (5.21).

By Lemma 2.1-(i), there is a constant δ1>0\delta_{1}>0 such that for x(A)(r)(A)(δr)x\in(\partial A)^{(r)}\setminus(\partial A)^{(\delta r)} (in the case where AC2\partial A\in C^{2}) or for x(A)(r)(A)(δr)𝖢𝗈𝗋(Kr)x\in(\partial A)^{(r)}\setminus(\partial A)^{(\delta r)}\setminus\mathsf{Cor}^{(Kr)} (in the case where AA is polygonal) we have ν(Br(x))(f1+2δ1)θdrd/2\nu(B_{r}(x))\geq(f_{1}+2\delta_{1})\theta_{d}r^{d}/2. Thus if AC2\partial A\in C^{2} then

(A)(r)(A)(δr)pn,r(x)nν(dx)c(logn)knrexp(n(f1+2δ1)θdrd/2),\displaystyle\int_{(\partial A)^{(r)}\setminus(\partial A)^{(\delta r)}}p_{n,r}(x)n\nu(dx)\leq c(\log n)^{k}nr\exp(-n(f_{1}+2\delta_{1})\theta_{d}r^{d}/2),

which tends to zero by (5.21). In the polygonal case we get the same conclusion using also the fact that the integral over 𝖢𝗈𝗋(Kr)\mathsf{Cor}^{(Kr)} tends to zero. Combining the last two estimates with (5.22) shows that

A((A)(δr)Aε)j=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)0,\displaystyle\int_{A\setminus((\partial A)^{(\delta r)}\cap A_{\varepsilon})}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx)\to 0, (5.24)

and therefore using (5.1) followed by (5.23) we have

eβ\displaystyle e^{-\beta} =limn(A)(δr)Aεj=0k1((nν(Br(x)))j/j!)enν(Br(x))nν(dx)\displaystyle=\lim_{n\to\infty}\int_{(\partial A)^{(\delta r)}\cap A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{r}(x))}n\nu(dx)
lim infn(enθd(sdrd)(f1+5ε)/2(A)(δr)Aεj=0k1((nν(Br(x)))j/j!)enν(Bs(x))nν(dx)).\displaystyle\leq\liminf_{n\to\infty}\left(e^{n\theta_{d}(s^{d}-r^{d})(f_{1}+5\varepsilon)/2}\int_{(\partial A)^{(\delta r)}\cap A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)\right). (5.25)

By (5.24) we have A((A)(δr)Aε)j=0k1((nν(Br(x)))j/j!)enν(Bs(x))nν(dx)0\int_{A\setminus((\partial A)^{(\delta r)}\cap A_{\varepsilon})}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)\to 0, so using (5.3) from Lemma 5.1 we have

(A)(δr)Aεj=0k1((nν(Br(x)))j/j!)enν(Bs(x))nν(dx)eγ,\int_{(\partial A)^{(\delta r)}\cap A_{\varepsilon}}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)e^{-n\nu(B_{s}(x))}n\nu(dx)\to e^{-\gamma},

and hence by (5.25),

eβlim infn(enθd(sdrd)(f1+5ε)/2)×eγ,e^{-\beta}\leq\liminf_{n\to\infty}\left(e^{n\theta_{d}(s^{d}-r^{d})(f_{1}+5\varepsilon)/2}\right)\times e^{-\gamma},

so that

lim infn(n(sdrd))2(γβ)/(θd(f1+5ε)).\liminf_{n\to\infty}(n(s^{d}-r^{d}))\geq 2(\gamma-\beta)/(\theta_{d}(f_{1}+5\varepsilon)).

Taking ε0\varepsilon\downarrow 0 and combining with (5.17) shows that (5.16) holds. ∎

Lemma 5.5 (De-Poissonization).

Let β\beta\in\mathbb{R} and suppose rn=rn(β)r_{n}=r_{n}(\beta) is given by (5.1) for nn sufficiently large. Then

limn[Lk(𝒳n)rn(β)]=limn[Ln,krn(β)]=exp(eβ).\displaystyle\lim_{n\to\infty}\mathbb{P}[L_{k}(\mathcal{X}_{n})\leq r_{n}(\beta)]=\lim_{n\to\infty}\mathbb{P}[L_{n,k}\leq r_{n}(\beta)]=\exp(-e^{-\beta}). (5.26)
Proof.

The statement about Ln,kL_{n,k} in (5.26) follows from Proposition 3.1. It remains to prove the statement about Lk(𝒳n)L_{k}(\mathcal{X}_{n}).

Given n>0,r>0n>0,r>0 define ϕn,r:=𝔼[ξn,r]\phi_{n,r}:=\mathbb{E}[\xi_{n,r}], i.e. by (3.3),

ϕn,r:=Apn,r(x)nν(dx)=Aj=0k1((nν(Br(x)))j/j!)exp(nν(Br(x)))nν(dx),\displaystyle\phi_{n,r}:=\int_{A}p_{n,r}(x)n\nu(dx)=\int_{A}\sum_{j=0}^{k-1}((n\nu(B_{r}(x)))^{j}/j!)\exp(-n\nu(B_{r}(x)))n\nu(dx), (5.27)

which is decreasing in rr. Set n:=nn3/4n^{-}:=n-n^{3/4}, n+:=n+n3/4n^{+}:=n+n^{3/4}, and let β\beta\in\mathbb{R}. Then

ϕn,rn(β)ϕn,rn(β)(nn)k1=1+O(n1/4),\frac{\phi_{n^{-},r_{n}(\beta)}}{\phi_{n,r_{n}(\beta)}}\geq\big{(}\frac{n^{-}}{n}\big{)}^{k-1}=1+O(n^{-1/4}),

and

ϕn,rn(β)ϕn,rn(β)exp(n3/4fmaxθdrn(β)d)=1+O((logn)n1/4),\frac{\phi_{n^{-},r_{n}(\beta)}}{\phi_{n,r_{n}(\beta)}}\leq\exp(n^{3/4}f_{\rm max}\theta_{d}r_{n}(\beta)^{d})=1+O((\log n)n^{-1/4}),

so that ϕn,rn(β)/ϕn,rn(β)1\phi_{n^{-},r_{n}(\beta)}/\phi_{n,r_{n}(\beta)}\to 1 as nn\to\infty, and thus ϕn,rn(β)eβ\phi_{n^{-},r_{n}(\beta)}\to e^{-\beta} as nn\to\infty. Therefore using Proposition 3.1, we have

[Ln,krn(β)]exp(eβ).\displaystyle\mathbb{P}[L_{n^{-},k}\leq r_{n}(\beta)]\to\exp(-e^{-\beta}). (5.28)

Now, following the proof of [8, Theorem 8.1], we note that with 𝒫n\mathcal{P}_{n^{-}}, 𝒫n+\mathcal{P}_{n^{+}} and 𝒳n\mathcal{X}_{n} coupled in the usual way (as described in [8]), we have

{Ln,krn(β)}{Lk(𝒳n)rn(β)}EnFnGn\{L_{n^{-},k}\leq r_{n}(\beta)\}\triangle\{L_{k}(\mathcal{X}_{n})\leq r_{n}(\beta)\}\subset E_{n}\cup F_{n}\cup G_{n}

where, setting Nt=𝒫t(A)N_{t}=\mathcal{P}_{t}(A) for all tt, we set

En\displaystyle E_{n} :={x𝒫n+𝒫n:𝒫n(Brn(β)(x))k1};\displaystyle:=\{\exists x\in\mathcal{P}_{n^{+}}\setminus\mathcal{P}_{n^{-}}:\mathcal{P}_{n^{-}}(B_{r_{n}(\beta)}(x))\leq k-1\};
Fn\displaystyle F_{n} :={x𝒫n,y𝒫n+𝒫n:𝒫n(Brn(β)(x))k,yxrn(β)};\displaystyle:=\{\exists x\in\mathcal{P}_{n^{-}},y\in\mathcal{P}_{n^{+}}\setminus\mathcal{P}_{n^{-}}:\mathcal{P}_{n^{-}}(B_{r_{n}(\beta)}(x))\leq k,\|y-x\|\leq r_{n}(\beta)\};
Gn\displaystyle G_{n} :={NnnNn+}c.\displaystyle:=\{N_{n^{-}}\leq n\leq N_{n^{+}}\}^{c}.

By Chebyshev’s inequality [Gn]=O(n1/2)\mathbb{P}[G_{n}]=O(n^{-1/2}). By Markov’s inequality,

[En]\displaystyle\mathbb{P}[E_{n}] 2n3/4A[𝒫n(Brn(β)(x))k1]ν(dx)\displaystyle\leq 2n^{3/4}\int_{A}\mathbb{P}[{\cal P}_{n^{-}}(B_{r_{n}(\beta)}(x))\leq k-1]\nu(dx)
=(2n3/4/n)ϕn,rn(β),\displaystyle=(2n^{3/4}/n^{-})\phi_{n^{-},r_{n}(\beta)},

which tends to zero. Finally, by the Mecke formula,

[Fn]\displaystyle\mathbb{P}[F_{n}] 2n3/4fmaxθdrn(β)dA[𝒫n(Brn(β)(x))k1]nν(dx)\displaystyle\leq 2n^{3/4}f_{\rm max}\theta_{d}r_{n}(\beta)^{d}\int_{A}\mathbb{P}[{\cal P}_{n^{-}}(B_{r_{n}(\beta)}(x))\leq k-1]n^{-}\nu(dx)
=2fmaxθdn3/4rn(β)dϕn,rn(β)=O((logn)n1/4).\displaystyle=2f_{\rm max}\theta_{d}n^{3/4}r_{n}(\beta)^{d}\phi_{n^{-},r_{n}(\beta)}=O((\log n)n^{-1/4}).

Therefore using (5.28) we obtain (5.26). ∎

Proof of Theorem 1.3 parts (i) and (ii).

For part (i) we assume f1>f0(22/d)f_{1}>f_{0}(2-2/d); in this case set α=1/(θdf0)\alpha=1/(\theta_{d}f_{0}). For part (ii) we assume f1<f0(22/d)f_{1}<f_{0}(2-2/d); in this case set α=2/(θdf1)\alpha=2/(\theta_{d}f_{1}).

By Lemma 5.3 in the first case, or by Lemma 5.4 in the second case, for all β,γ\beta,\gamma\in\mathbb{R} with β<γ\beta<\gamma the condition (5.6) holds.

Let β\beta\in\mathbb{R} and suppose rn=rn(β)r_{n}=r_{n}(\beta) is given by (5.1) for nn sufficiently large. By Lemma 5.5, [Ln,krn(β)]F(β)\mathbb{P}[L_{n,k}\leq r_{n}(\beta)]\to F(\beta) and [Lk(𝒳n)rn(β)]F(β)\mathbb{P}[L_{k}(\mathcal{X}_{n})\leq r_{n}(\beta)]\to F(\beta), where we set F(x):=exp(ex)F(x):=\exp(-e^{-x}). By Proposition 4.6, [Mn,krn(β)]F(β)\mathbb{P}[M_{n,k}\leq r_{n}(\beta)]\to F(\beta) as nn\to\infty.

Then by Lemma 5.2 (taking Xn=Mn,kX_{n}=M_{n,k}), we obtain that nMn,kdnμ(Mn,k)d𝑑α(𝖦𝗎+loglog2),nM_{n,k}^{d}-n\mu(M_{n,k})^{d}\overset{d}{\longrightarrow}\alpha({\mathsf{Gu}}+\log\log 2), i.e. (1.18) holds if f1>f0(22/d)f_{1}>f_{0}(2-2/d), and (1.22) holds if f1<f0(22/d)f_{1}<f_{0}(2-2/d). Also by taking Xn=Ln,kX_{n}=L_{n,k} in Lemma 5.2 we obtain (1.19) if f1>f0(22/d)f_{1}>f_{0}(2-2/d), and (1.23) if f1<f0(22/d)f_{1}<f_{0}(2-2/d).

It remains to demonstrate the results for the binomial model, i.e. (1.16), (1.17), (1.20) and (1.21). By Lemma 5.5, we can use Lemma 5.2 (now taking Xn=Lk(𝒳n)X_{n}=L_{k}(\mathcal{X}_{n})) to deduce that (1.17) holds if f1>f0(22/d)f_{1}>f_{0}(2-2/d), and (1.21) holds if f1<f0(22/d)f_{1}<f_{0}(2-2/d).

Using (5.26), and (4.17) from Proposition 4.6, we obtain that

[Mk(𝒳n)rn(β)]exp(eβ).\mathbb{P}[M_{k}(\mathcal{X}_{n})\leq r_{n}(\beta)]\to\exp(-e^{-\beta}).

Then using Lemma 5.2 (now taking Xn=Mk(𝒳n)X_{n}=M_{k}(\mathcal{X}_{n})) we can deduce that (1.16) holds if f1>f0(22/d)f_{1}>f_{0}(2-2/d), and (1.20) holds if f1<f0(22/d)f_{1}<f_{0}(2-2/d). ∎

5.2 Proof of Theorem 1.3: conclusion

It remains to prove part (iii) of Theorem 1.3. We deal first with the assertions there concerning tightness. Again in the next proof, set F(x):=exp(ex)F(x):=\exp(-e^{-x}), xx\in\mathbb{R}.

Lemma 5.6.

The collection of random variables {nMn,kdnμ(Mn,k)d}n1\{nM_{n,k}^{d}-n\mu(M_{n,k})^{d}\}_{n\geq 1} is tight. So is the collection of random variables {nLn,kdnμ(Ln,k)d}n1\{nL_{n,k}^{d}-n\mu(L_{n,k})^{d}\}_{n\geq 1}, and also the sequence (nMk(𝒳n)dnμ(Mk(𝒳n))d)n(nM_{k}(\mathcal{X}_{n})^{d}-n\mu(M_{k}(\mathcal{X}_{n}))^{d})_{n\in\mathbb{N}}, and the sequence (nLk(𝒳n)dnμ(Lk(𝒳n))d)n(nL_{k}(\mathcal{X}_{n})^{d}-n\mu(L_{k}(\mathcal{X}_{n}))^{d})_{n\in\mathbb{N}}.

Proof.

Let ε(0,1/6)\varepsilon\in(0,1/6). Choose β<β\beta<\beta^{\prime} with F(β)<ε/3F(\beta)<\varepsilon/3 and F(β)>1ε/3F(\beta^{\prime})>1-\varepsilon/3. Set rn=rn(β)r_{n}=r_{n}(\beta), sn=rn(β)s_{n}=r_{n}(\beta^{\prime}) as given by (5.1). By Lemma 5.4 there exists a constant KK such that n(sndrnd)Kn(s_{n}^{d}-r_{n}^{d})\leq K for all large enough nn. By Proposition 3.1, [Ln,krn(β)]F(β)\mathbb{P}[L_{n,k}\leq r_{n}(\beta)]\to F(\beta). By Proposition 4.6, [Mn,krn]F(β)\mathbb{P}[M_{n,k}\leq r_{n}]\to F(\beta) as nn\to\infty. Similarly, [Ln,ksn]F(β)\mathbb{P}[L_{n,k}\leq s_{n}]\to F(\beta^{\prime}) and [Mn,ksn]F(β)\mathbb{P}[M_{n,k}\leq s_{n}]\to F(\beta^{\prime}) as nn\to\infty. Therefore since F(β)<1/2<F(β)F(\beta)<1/2<F(\beta^{\prime}), we have rnμ(Ln,k)<snr_{n}\leq\mu(L_{n,k})<s_{n} and rnμ(Mn,k)<snr_{n}\leq\mu(M_{n,k})<s_{n} for nn large. Then for nn large

[nMn,kdnμ(Mn,k)dK]\displaystyle\mathbb{P}[nM_{n,k}^{d}\leq n\mu(M_{n,k})^{d}-K] [nMn,kdnsndK]\displaystyle\leq\mathbb{P}[nM_{n,k}^{d}\leq ns_{n}^{d}-K]
[Mn,krn]<ε/2,\displaystyle\leq\mathbb{P}[M_{n,k}\leq r_{n}]<\varepsilon/2,

and likewise for Ln,kL_{n,k}. Similarly for nn large

[nMn,kd>nμ(Mn,k)d+K]\displaystyle\mathbb{P}[nM_{n,k}^{d}>n\mu(M_{n,k})^{d}+K] [nMn,kd>nrnd+K]\displaystyle\leq\mathbb{P}[nM_{n,k}^{d}>nr_{n}^{d}+K]
[Mn,k>sn]<ε/2,\displaystyle\leq\mathbb{P}[M_{n,k}>s_{n}]<\varepsilon/2,

and likewise for Ln,kL_{n,k}. Thus [|n(Mn,kdμ(Mn,k)d)|>K]ε\mathbb{P}[|n(M_{n,k}^{d}-\mu(M_{n,k})^{d})|>K]\leq\varepsilon and [|n(Ln,kdμ(Ln,k)d)|>K]ε\mathbb{P}[|n(L_{n,k}^{d}-\mu(L_{n,k})^{d})|>K]\leq\varepsilon for all large enough nn, Also {n(Mn,kdμ(Mn,kd))}1nn0\{n(M_{n,k}^{d}-\mu(M_{n,k}^{d}))\}_{1\leq n\leq n_{0}} and {n(Ln,kdμ(Ln,kd))}1nn0\{n(L_{n,k}^{d}-\mu(L_{n,k}^{d}))\}_{1\leq n\leq n_{0}} are uniformly bounded for any fixed n0(0,)n_{0}\in(0,\infty). This yields the asserted tightness of (Mn,k)n1(M_{n,k})_{n\geq 1} and of (Ln,k)n1(L_{n,k})_{n\geq 1}.

The proof of tightness for Lk(𝒳n)L_{k}(\mathcal{X}_{n}) and of Mk(𝒳n)M_{k}(\mathcal{X}_{n}) is similar, except that instead of Proposition 3.1 we use (5.26). Proposition 4.6 still applies in the binomial setting. ∎

To prove (1.5) we shall adapt the ‘squeezing argument’ from [9]. For <β<γ<-\infty<\beta<\gamma<\infty we define the random variable

Un(β,γ):=x𝒳n𝟏{𝒳n(Brn(β)(x))k,𝒳n(Brn(γ)(x)Brn(β)(x))2}.\displaystyle U_{n}(\beta,\gamma):=\sum_{x\in\mathcal{X}_{n}}{\bf 1}\{\mathcal{X}_{n}(B_{r_{n}(\beta)}(x))\leq k,\mathcal{X}_{n}(B_{r_{n}(\gamma)}(x)\setminus B_{r_{n}(\beta)}(x))\geq 2\}. (5.29)
Lemma 5.7.

Let K>0K>0. Then there is a constant c(0,)c\in(0,\infty) such that for all β,γ\beta,\gamma\in\mathbb{R} with Kβ<γK-K\leq\beta<\gamma\leq K,

lim supn[Un(β,γ)1]c(γβ)2.\limsup_{n\to\infty}\mathbb{P}[U_{n}(\beta,\gamma)\geq 1]\leq c(\gamma-\beta)^{2}.
Proof.

Set rn:=rn(β)r_{n}:=r_{n}(\beta), and sn:=rn(γ)s_{n}:=r_{n}(\gamma). By the union bound,

[Un(β,γ)1]nA[𝒳n1(Brn(x))<k,𝒳n1(Bsn(x)Brn(x))2]ν(dx).\displaystyle\mathbb{P}[U_{n}(\beta,\gamma)\geq 1]\leq n\int_{A}\mathbb{P}[\mathcal{X}_{n-1}(B_{r_{n}}(x))<k,\mathcal{X}_{n-1}(B_{s_{n}}(x)\setminus B_{r_{n}}(x))\geq 2]\nu(dx). (5.30)

Let xAx\in A and set Y:=𝒳n1(Brn(x))Y:=\mathcal{X}_{n-1}(B_{r_{n}}(x)), Z:=𝒳n1(Bsn(x)Brn(x))Z:=\mathcal{X}_{n-1}(B_{s_{n}}(x)\setminus B_{r_{n}}(x)). Also set vn(x):=ν(Brn(x))v_{n}(x):=\nu(B_{r_{n}}(x)) and wn(x):=ν(Bsn(x))w_{n}(x):=\nu(B_{s_{n}}(x)). Then

[Y<k]\displaystyle\mathbb{P}[Y<k] =j=0k1(n1j)vn(x)j(1vn(x))n1j\displaystyle=\sum_{j=0}^{k-1}\binom{n-1}{j}v_{n}(x)^{j}(1-v_{n}(x))^{n-1-j}
(1vn(x))kj=0k1((nvn(x))j/j!)(1vn(x))n\displaystyle\leq(1-v_{n}(x))^{-k}\sum_{j=0}^{k-1}((nv_{n}(x))^{j}/j!)(1-v_{n}(x))^{n}
(1fmaxθdrnd)kj=0k1((nvn(x))j/j!)envn(x).\displaystyle\leq(1-f_{\rm max}\theta_{d}r_{n}^{d})^{-k}\sum_{j=0}^{k-1}((nv_{n}(x))^{j}/j!)e^{-nv_{n}(x)}.

Also, using the fact that [Z2|Y=j]\mathbb{P}[Z\geq 2|Y=j] is nonincreasing in jj, and the fact that for any binomial random variable WW with mean α\alpha we have 𝔼[W(W1)]α2\mathbb{E}[W(W-1)]\leq\alpha^{2}, we have

[Z2|Y<k]\displaystyle\mathbb{P}[Z\geq 2|Y<k] [Z2|Y=0]\displaystyle\leq\mathbb{P}[Z\geq 2|Y=0]
(1/2)𝔼[Z(Z1)|Y=0]\displaystyle\leq(1/2)\mathbb{E}[Z(Z-1)|Y=0]
(1/2)n2((wn(x)vn(x))/(1vn(x)))2\displaystyle\leq(1/2)n^{2}((w_{n}(x)-v_{n}(x))/(1-v_{n}(x)))^{2}
(1/2)n2(1fmaxθdrn(β)d)2(wn(x)vn(x))2.\displaystyle\leq(1/2)n^{2}(1-f_{\rm max}\theta_{d}r_{n}(\beta)^{d})^{-2}(w_{n}(x)-v_{n}(x))^{2}.

If nn is taken to be large enough we have (1fmaxθdrn(β)d)k22(1-f_{\rm max}\theta_{d}r_{n}(\beta)^{d})^{-k-2}\leq 2, and hence using (5.30) followed by (5.1), we have

[Un(β,γ)1]\displaystyle\mathbb{P}[U_{n}(\beta,\gamma)\geq 1] n3supyA(wn(y)vn(y))2Aj=0k1((nvn(x))j/j!)envn(x)ν(dx)\displaystyle\leq n^{3}\sup_{y\in A}(w_{n}(y)-v_{n}(y))^{2}\int_{A}\sum_{j=0}^{k-1}((nv_{n}(x))^{j}/j!)e^{-nv_{n}(x)}\nu(dx)
=(nsupyA(wn(y)vn(y)))2eβ.\displaystyle=\big{(}n\sup_{y\in A}(w_{n}(y)-v_{n}(y))\big{)}^{2}e^{-\beta}. (5.31)

By Lemmas 5.3 and 5.4,

lim supnnθd(sndrnd)(1f02f1)(γβ),\displaystyle\limsup_{n\to\infty}n\theta_{d}(s_{n}^{d}-r_{n}^{d})\leq\left(\frac{1}{f_{0}}\vee\frac{2}{f_{1}}\right)(\gamma-\beta), (5.32)

where \vee denotes maximum, and hence

lim supnsupxAn(wn(x)vn(x))lim supnnfmaxθd(sndrnd)fmax(1f02f1)(γβ),\displaystyle\limsup_{n\to\infty}\sup_{x\in A}n(w_{n}(x)-v_{n}(x))\leq\limsup_{n\to\infty}nf_{\rm max}\theta_{d}(s_{n}^{d}-r_{n}^{d})\leq f_{\rm max}\left(\frac{1}{f_{0}}\vee\frac{2}{f_{1}}\right)(\gamma-\beta),

so by (5.31) we have lim supn[Un(β,γ)1]eK(fmax(1f02f1))2(γβ)2\limsup_{n\to\infty}\mathbb{P}[U_{n}(\beta,\gamma)\geq 1]\leq e^{K}(f_{\rm max}(\frac{1}{f_{0}}\vee\frac{2}{f_{1}}))^{2}(\gamma-\beta)^{2}. ∎

Proof of Theorem 1.3 (conclusion).

It remains to prove part (iii), and by Lemma 5.6 it remains only to prove (1.5). Let ε>0\varepsilon>0. Choose K>0K>0 such that exp(eK)>1ε\exp(-e^{-K})>1-\varepsilon, and also exp(eK)<ε\exp(-e^{K})<\varepsilon. Then let cc be as in Lemma 5.7. Choose β0<<βm\beta_{0}<\ldots<\beta_{m} with β0=K\beta_{0}=-K and βm=K\beta_{m}=K such that ci=1m(βiβi1)2<εc\sum_{i=1}^{m}(\beta_{i}-\beta_{i-1})^{2}<\varepsilon. Write Ln,kL^{\prime}_{n,k} for Lk(𝒳n)L_{k}(\mathcal{X}_{n}) and Mn,kM^{\prime}_{n,k} for Mk(𝒳n)M_{k}(\mathcal{X}_{n}). Since Ln,kMn,kL^{\prime}_{n,k}\leq M^{\prime}_{n,k}, by the union bound

[Ln,kMn,k]\displaystyle\mathbb{P}[L^{\prime}_{n,k}\neq M^{\prime}_{n,k}] =[Ln,k<Mn,k]\displaystyle=\mathbb{P}[L^{\prime}_{n,k}<M^{\prime}_{n,k}]
[Ln,krn(β0)]+[Ln,k>rn(βm)]+i=1m[Ln,krn(βi)<Mn,k]\displaystyle\leq\mathbb{P}[L^{\prime}_{n,k}\leq r_{n}(\beta_{0})]+\mathbb{P}[L^{\prime}_{n,k}>r_{n}(\beta_{m})]+\sum_{i=1}^{m}\mathbb{P}[L^{\prime}_{n,k}\leq r_{n}(\beta_{i})<M^{\prime}_{n,k}]
+i=1m[rn(βi1)<Ln,k<Mn,krn(βi)].\displaystyle+\sum_{i=1}^{m}\mathbb{P}[r_{n}(\beta_{i-1})<L^{\prime}_{n,k}<M^{\prime}_{n,k}\leq r_{n}(\beta_{i})].

Using Lemma 5.5 and Proposition 4.6, we obtain that

lim supn[Ln,kMn,k]2ε+i=1mlim supn[rn(βi1)<Ln,k<Mn,krn(βi)].\displaystyle\limsup_{n\to\infty}\mathbb{P}[L^{\prime}_{n,k}\neq M^{\prime}_{n,k}]\leq 2\varepsilon+\sum_{i=1}^{m}\limsup_{n\to\infty}\mathbb{P}[r_{n}(\beta_{i-1})<L^{\prime}_{n,k}<M^{\prime}_{n,k}\leq r_{n}(\beta_{i})].

Suppose β<γ\beta<\gamma, and suppose rn(β)<Ln,k<Mn,krn(γ)r_{n}(\beta)<L^{\prime}_{n,k}<M^{\prime}_{n,k}\leq r_{n}(\gamma) and all inter-point distances in 𝒳n\mathcal{X}_{n} are distinct (the latter condition holds almost surely).

Then there exist x,y𝒳nx,y\in\mathcal{X}_{n} with xy=Mn,k\|x-y\|=M^{\prime}_{n,k}, and it is possible to remove kk vertices from G(𝒳n,Mn,k)G(\mathcal{X}_{n},M^{\prime}_{n,k}) leaving the resulting graph connected, but disconnected if the edge {x,y}\{x,y\} is also removed. Removing the same set of vertices from G(𝒳n,rn(β))G(\mathcal{X}_{n},r_{n}(\beta)) leaves xx and yy in distinct components, and if also for some fixed ρ>0\rho>0, events Hrn(β),ρ(𝒳n)H_{r_{n}(\beta),\rho}(\mathcal{X}_{n}) (defined in Lemma 4.5) and Jrn(β),ρ(𝒳n)J_{r_{n}(\beta),\rho}(\mathcal{X}_{n}) (defined in the proof of Proposition 4.6), fail to occur, then xx or yy must have at most k1k-1 other points of 𝒳n\mathcal{X}_{n} within distance rn(β)r_{n}(\beta). But 𝒳n(Brn(γ)(x){x})k+1\mathcal{X}_{n}(B_{r_{n}(\gamma)}(x)\setminus\{x\})\geq k+1 since Ln,k<yxrn(γ)L^{\prime}_{n,k}<\|y-x\|\leq r_{n}(\gamma), and similarly 𝒳n(Brn(β)(y){y})k+1\mathcal{X}_{n}(B_{r_{n}(\beta)}(y)\setminus\{y\})\geq k+1. Thus we have the event inclusion

{rn(β)<Ln,k<Mn,krn(γ)}Hrn(β),ρ(𝒳n)Jrn(β),ρ(𝒳n)Un(β,γ),\displaystyle\{r_{n}(\beta)<L^{\prime}_{n,k}<M^{\prime}_{n,k}\leq r_{n}(\gamma)\}\subset H_{r_{n}(\beta),\rho}(\mathcal{X}_{n})\cup J_{r_{n}(\beta),\rho}(\mathcal{X}_{n})\cup U_{n}(\beta,\gamma),

where Un(β,γ)U_{n}(\beta,\gamma) was defined at (5.29).

By Lemma 4.5, we can choose ρ\rho so that [Hrn(β),ρ(𝒳n)]0\mathbb{P}[H_{r_{n}(\beta),\rho}(\mathcal{X}_{n})]\to 0, and by the proof of Proposition 4.6 [Jrn(β),ρ(𝒳n)]0\mathbb{P}[J_{r_{n}(\beta),\rho}(\mathcal{X}_{n})]\to 0. Therefore using Lemma 5.7 we obtain that

lim supn[rn(β)<Ln,k<Mn,krn(γ)]lim supn[Un(β,γ)1]c(γβ)2.\displaystyle\limsup_{n\to\infty}\mathbb{P}[r_{n}(\beta)<L^{\prime}_{n,k}<M^{\prime}_{n,k}\leq r_{n}(\gamma)]\leq\limsup_{n\to\infty}\mathbb{P}[U_{n}(\beta,\gamma)\geq 1]\leq c(\gamma-\beta)^{2}.

Thus

lim supn[Ln,kMn,k]2ε+i=1mc(βiβi1)2<3ε,\displaystyle\limsup_{n\to\infty}\mathbb{P}[L^{\prime}_{n,k}\neq M^{\prime}_{n,k}]\leq 2\varepsilon+\sum_{i=1}^{m}c(\beta_{i}-\beta_{i-1})^{2}<3\varepsilon,

and since ε>0\varepsilon>0 is arbitrary this gives us (1.5). ∎

6 Proof of Theorem 1.1

Now we specialise to the uniform case. We make the same assumptions on AA as in the previous section, but now we take ff0𝟏Af\equiv f_{0}{\bf 1}_{A} with f0=|A|1f_{0}=|A|^{-1}. Recall from (1.10) the definition

cd,k:=θd11θd11/d(22/d)k2+1/d21k/(k1)!\displaystyle c_{d,k}:=\theta_{d-1}^{-1}\theta_{d}^{1-1/d}(2-2/d)^{k-2+1/d}2^{1-k}/(k-1)! (6.1)

Given kk\in\mathbb{N} and β\beta\in\mathbb{R}, let rn=rn(β)0r_{n}=r_{n}(\beta)\geq 0 be defined for all n>0n>0 by

f0nθdrnd=max((22/d)logn+(2k4+2/d)𝟏{d3 or k2}loglogn+β,0).f_{0}n\theta_{d}r_{n}^{d}=\max\big{(}(2-2/d)\log n+(2k-4+2/d){\bf 1}_{\{d\geq 3\text{ or }k\geq 2\}}\log\log n+\beta,0\big{)}. (6.2)

We now show the convergence of 𝔼[ξn,rn]\mathbb{E}[\xi_{n,r_{n}}] (with ξn,r\xi_{n,r} defined at (3.1)). That is, we show that this choice of rnr_{n} satisfies (3.5) for appropriate β\beta^{\prime}. Recall the definition of the isoperimetric ratio σA\sigma_{A} at (1.11).

Proposition 6.1 (convergence of the expectation in the uniform case with d=2d=2).

Suppose ff0𝟏Af\equiv f_{0}{\bf 1}_{A}, with d=2d=2 and either AA compact with C2C^{2} boundary, or AA polygonal. Fix kk\in\mathbb{N}, β\beta\in\mathbb{R}, and let rn,ξn,rr_{n},\xi_{n,r} be as given in (6.2) and (3.1). Then as nn\to\infty,

𝔼[ξn,rn]={eβ+σAeβ/2π2(logn)1/2+O((logn)3/2) if k=1eβ+σAeβ/2π4(1+loglogn2logn)+eβloglognlogn+O((logn)1) if k=2σAeβ/2π(k1)!2k(1+(2k3)2loglogn2logn)+O((logn)1) if k3.\displaystyle\mathbb{E}[\xi_{n,r_{n}}]=\begin{cases}e^{-\beta}+\sigma_{A}e^{-\beta/2}\frac{\sqrt{\pi}}{2}(\log n)^{-1/2}+O((\log n)^{-3/2})&\mbox{ if }k=1\\ e^{-\beta}+\sigma_{A}e^{-\beta/2}\frac{\sqrt{\pi}}{4}\Big{(}1+\frac{\log\log n}{2\log n}\Big{)}+\frac{e^{-\beta}\log\log n}{\log n}+O((\log n)^{-1})&\mbox{ if }k=2\\ \sigma_{A}e^{-\beta/2}\frac{\sqrt{\pi}}{(k-1)!2^{k}}\left(1+\frac{(2k-3)^{2}\log\log n}{2\log n}\right)+O\big{(}(\log n)^{-1}\big{)}&\mbox{ if }k\geq 3.\end{cases} (6.3)
Proof.

Define the ‘kk-vacant region’ Vn,k:={xA:𝒫n(Brn(x))<k}V_{n,k}:=\{x\in A:{\cal P}_{n}(B_{r_{n}}(x))<k\}. Recall the definition of pn,r(x)p_{n,r}(x) at (3.4). By (3.3), we have

𝔼[ξn,rn]=nf0Apn,rn(x)𝑑x=n|A|1𝔼[|Vn,k|].\displaystyle\mathbb{E}[\xi_{n,r_{n}}]=nf_{0}\int_{A}p_{n,r_{n}}(x)dx=n|A|^{-1}\mathbb{E}[|V_{n,k}|]. (6.4)

Therefore the result follows from [4, Proposition 5.1]. ∎

Proposition 6.2 (convergence of the expectation in the uniform case with d3d\geq 3).

Suppose ff0𝟏Af\equiv f_{0}{\bf 1}_{A}, with d3d\geq 3 and AA compact with AC2\partial A\in C^{2}. Fix β\beta\in\mathbb{R} and let rn(β),ξn,r,cd,k,σAr_{n}(\beta),\xi_{n,r},c_{d,k},\sigma_{A} be as given in (6.2), (3.1) and (6.1). Let ε>0\varepsilon>0. Then as nn\to\infty,

𝔼[ξn,rn]\displaystyle\mathbb{E}[\xi_{n,r_{n}}] =eβ/2cd,kσA(1+(k2+1/d)2loglogn(11/d)logn\displaystyle=e^{-\beta/2}c_{d,k}\sigma_{A}\Big{(}1+\frac{(k-2+1/d)^{2}\log\log n}{(1-1/d)\log n}
+(k2+1/d)β+4k4(22/d)logn)+O((logn)ε2).\displaystyle+\frac{(k-2+1/d)\beta+4k-4}{(2-2/d)\log n}\Big{)}+O\big{(}(\log n)^{\varepsilon-2}\big{)}. (6.5)
Proof.

Again using (6.4), we obtain this result from [4, Proposition 5.2]. ∎

Corollary 6.3.

Let d=2d=2, β\beta\in\mathbb{R}. Then (1.13) holds, and also

[nf0πLn,12lognβ]=exp(σAπ1/2eβ/22(logn)1/2)exp(eβ)+O((logn)1).\displaystyle\mathbb{P}[nf_{0}\pi L_{n,1}^{2}-\log n\leq\beta]=\exp\big{(}-\frac{\sigma_{A}\pi^{1/2}e^{-\beta/2}}{2(\log n)^{1/2}}\big{)}\exp(-e^{-\beta})+O((\log n)^{-1}). (6.6)

Moreover (1.14) holds, and

[nf0πLn,22lognloglognβ]=\displaystyle\mathbb{P}[nf_{0}\pi L_{n,2}^{2}-\log n-\log\log n\leq\beta]= exp(σAπ1/2eβ/2loglogn8logneβloglognlogn)\displaystyle\exp\Big{(}-\frac{\sigma_{A}\pi^{1/2}e^{-\beta/2}\log\log n}{8\log n}-\frac{e^{-\beta}\log\log n}{\log n}\Big{)}
×exp(eβπ1/2σAeβ/24)+O(1logn).\displaystyle\times\exp\Big{(}-e^{-\beta}-\frac{\pi^{1/2}\sigma_{A}e^{-\beta/2}}{4}\Big{)}+O\Big{(}\frac{1}{\log n}\Big{)}. (6.7)
Proof.

Let rn=rn(β)r_{n}=r_{n}(\beta) be given by (6.2) with d=2,k=1d=2,k=1; then nf0πrn2logn=βnf_{0}\pi r_{n}^{2}-\log n=\beta for all large enough nn.

Let ξn,r\xi_{n,r} be the number of isolated vertices of G(𝒫n,r)G({\cal P}_{n},r) as defined at (3.1), taking k=1k=1. By Proposition 6.1, (3.5) holds on taking β=eβ\beta^{\prime}=e^{-\beta}. Hence by Proposition 3.1,

[nf0πLn,12lognβ]=[Ln,1rn]=exp(𝔼[ξn,rn])+O(1/(logn)).\displaystyle\mathbb{P}[nf_{0}\pi L_{n,1}^{2}-\log n\leq\beta]=\mathbb{P}[L_{n,1}\leq r_{n}]=\exp(-\mathbb{E}[\xi_{n,r_{n}}])+O(1/(\log n)).

Then using Proposition 6.1, and the fact that |eλeλ||λλ||e^{-\lambda}-e^{-\lambda^{\prime}}|\leq|\lambda-\lambda^{\prime}| for any λ,λ>0\lambda,\lambda^{\prime}>0, we obtain (6.6). We can then deduce (1.13) using Proposition 4.6.

Next, let rn=rn(β)r_{n}=r_{n}(\beta) be given by (6.2) again, but now with d=2,k=2d=2,k=2. Then nf0πrn2lognloglogn=βnf_{0}\pi r_{n}^{2}-\log n-\log\log n=\beta for nn large. Repeating the previous argument gives us (6.7) and then (1.14). ∎

Corollary 6.4.

Suppose either d3d\geq 3, or d=2,k3d=2,k\geq 3. Let β\beta\in\mathbb{R}. Then (1.15) holds, and

[nf0θdLn,kd(22/d)logn+(42k2/d)loglognβ]\displaystyle\mathbb{P}[nf_{0}\theta_{d}L_{n,k}^{d}-(2-2/d)\log n+(4-2k-2/d)\log\log n\leq\beta]
=exp(cd,kσAeβ/2(k2+1/d)2loglogn(11/d)logn)exp(cd,kσAeβ/2)+O(1logn),\displaystyle=\exp\Big{(}-\frac{c_{d,k}\sigma_{A}e^{-\beta/2}(k-2+1/d)^{2}\log\log n}{(1-1/d)\log n}\Big{)}\exp(-c_{d,k}\sigma_{A}e^{-\beta/2})+O\big{(}\frac{1}{\log n}\big{)},
Proof.

The proof is the same as for Corollary 6.3, using Proposition 6.2 in place of Proposition 6.1 when d3d\geq 3. ∎

We are now ready to finish the proof of Theorem 1.1.

Proof of Theorem 1.1.

We already showed (1.13), (1.14), (1.15) and the corresponding results for Ln,kL_{n,k}, in Corollaries 6.3 and 6.4. We already proved (1.5) under weaker assumptions in Theorem 1.3. Therefore it remains only to prove (1.12) and the binomial versions of (1.14) and (1.15), along with the corresponding results for Lk(𝒳n)L_{k}(\mathcal{X}_{n}).

Let ϕn,r\phi_{n,r} be as defined at (5.27). Set n:=nn3/4n^{-}:=n-n^{3/4}. As shown in the proof of Lemma 5.5, given β\beta\in\mathbb{R} we have that

ϕn,rn(β)=(1+O(lognn1/4))ϕn,rn(β).\phi_{n^{-},r_{n}(\beta)}=\Big{(}1+O\big{(}\frac{\log n}{n^{1/4}}\big{)}\Big{)}\phi_{n,r_{n}(\beta)}.

Then by Proposition 3.1,

[Ln,krn(β)]=exp(ϕn,rn(β))+O((logn)1)\displaystyle\mathbb{P}[L_{n^{-},k}\leq r_{n}(\beta)]=\exp(-\phi_{n^{-},r_{n}(\beta)})+O((\log n)^{-1})
=exp(ϕn,rn(β))+O((logn)1).\displaystyle=\exp(-\phi_{n,r_{n}(\beta)})+O((\log n)^{-1}).

By the proof of Lemma 5.5,

[Lk(𝒳n)rn(β)]\displaystyle\mathbb{P}[L_{k}(\mathcal{X}_{n})\leq r_{n}(\beta)] =[Ln,krn(β)]+O((logn)/n1/4).\displaystyle=\mathbb{P}[L_{n^{-},k}\leq r_{n}(\beta)]+O((\log n)/n^{1/4}).
=exp(ϕn,rn(β))+O((logn)1).\displaystyle=\exp(-\phi_{n,r_{n}(\beta)})+O((\log n)^{-1}).

Plugging in the expressions for ϕn,rn(β)=𝔼[ξn,rn(β)]\phi_{n,r_{n}(\beta)}=\mathbb{E}[\xi_{n,r_{n}(\beta)}] in Lemmas 6.1 and 6.2 gives us the result (1.12) for Lk(𝒳n)L_{k}(\mathcal{X}_{n}) and the binomial versions of (1.14) and (1.15) for Lk(𝒳n)L_{k}(\mathcal{X}_{n}). Finally, applying Proposition 4.6 gives the same results for Mk(𝒳n)M_{k}(\mathcal{X}_{n}). ∎

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