1. Introduction
1.1 Motivation
We study percolative systems, and address the question: What is the value of the critical percolation threshold? A specific answer is only possible in very exceptional cases. We are pursuing a different route instead, namely an asymptotic expression of the critical threshold as a function of the dimension
$d$
of the encapsulating space in the
$d\to \infty$
limit. This has been solved for percolation on the hyper-cubic lattice
$\mathbb{Z}^d$
: for bond percolation on the hyper-cubic lattice it is known that
\begin{equation} p_c^{\text{bond}}(\mathbb Z^d)=\frac 1{2d}+\frac{1}{\left (2d\right )^2} +\frac{7}{2}\frac{1}{\left (2d\right )^3} +\mathcal{O}\!\left (\frac 1{d^4}\right )\quad \text{as $d\to \infty $,} \end{equation}
cf. [Reference Hara and SladeHS95, Reference van der Hofstad and SladeHS05], whereas for hyper-cubic site percolation on
$\mathbb{Z}^d$
it is
\begin{equation} p_c^{\text{site}}(\mathbb Z^d)= \frac 1{2d}+\frac{5}{2}\frac{1}{\left (2d\right )^2} +\frac{31}{4}\frac{1}{\left (2d\right )^3} +\mathcal{O}\!\left (\frac 1{d^4}\right )\quad \text{as $d\to \infty $,} \end{equation}
cf. [Reference Heydenreich and MatzkeHM22]. Mertens and Moore [Reference Mertens and MooreMM18] use involved numerical enumeration to identify a few more terms (without a rigorous bound on the error). In the present work, we address a corresponding question for continuum percolation. Interestingly, our analysis establishes an exponentially decaying series rather than an algebraic decay as on lattices. We shall discuss this point further in the discussion section.
1.2 The model
To this end, we are considering the random connection model (henceforth abbreviated RCM), a spatial random graph model whose points are given as a homogeneous Poison process
$\eta$
on
${\mathbb{R}}^d$
with intensity measure
$\lambda \operatorname{Leb}$
, and we refer to
$\lambda \gt 0$
as the intensity of the model. Each pair of vertices
$x,y$
in the support of
$\eta$
are connected independently with probability
$\varphi (x-y)$
, where
\begin{equation*}\varphi \colon {\mathbb{R}}^d\to [0,1]\end{equation*}
is integrable and symmetric (i.e.
$\varphi (x)=\varphi ({-}x)$
for all
$x\in{\mathbb{R}^d}$
). The classical example is the Gilbert disk model [Reference GilbertGil61] or hyper-sphere random connection model with
\begin{equation*} \varphi (x) = {\unicode {x1D7D9}}_{\left \{\lvert {x}\rvert \lt R\right \}}\end{equation*}
for some
$R\gt 0$
: two vertices are connected whenever their (Euclidean) distance is at most
$R$
.
We are interested in the percolation phase transition of the RCM, that is, the critical intensity
$\lambda _c$
given as the infimum of those values of
$\lambda$
such that the resulting random graph has an infinite connected component:
\begin{equation*} \lambda _c=\inf \{\lambda \mid \text {the RCM with intensity $\lambda $ has an infinite component}\}.\end{equation*}
See [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Section 2] for a more formal definition.
Penrose [Reference PenrosePen91] uses the ‘method of generations’ to show that for all dimensions
$d\geq 1$
the critical intensity is strictly positive. In particular he derives the lower bound
\begin{equation} q_\varphi \lambda _c \geq 1, \end{equation}
where
$q_\varphi = \int \varphi (x){\mathrm{d}} x$
. He also uses a coarse-graining argument to show that for
$d\geq 2$
the critical intensity is finite if
$q_\varphi \gt 0$
. Meester, Penrose and Sarkar [Reference Meester, Penrose and SarkarMPS97] prove the
$0^{th}$
order asymptotics of
$\lambda _c$
for radial non-increasing
$\varphi$
(with uniform bounds on the variance of the jumps taken by random walk with jump intensity proportional to
$\varphi$
). Specifically they prove that for such models
\begin{equation} q_\varphi \lambda _c \to 1 \end{equation}
as
$d\to \infty$
. In the present work we significantly expand their result by identifying several additional terms.
1.3 Results
We shall now make a couple of assumptions before formulating our main result. Throughout this paper we will denote the convolution of two non-negative functions
$f,g\colon{\mathbb{R}^d}\to{\mathbb{R}}_{\geq 0}$
to be
\begin{equation} f\star g(x) \,:\!= \int f(x-u)g(u){\mathrm{d}} u, \end{equation}
and
$f^{\star n}(x)$
to be the convolution of
$n$
copies of
$f$
. In particular,
$f^{\star 1}\equiv f$
. We will also denote the Fourier transform of an integrable function
$f\colon{\mathbb{R}^d}\to \mathbb{R} $
by
\begin{equation} \widehat{f}(k) \,:\!= \int{\text{e}}^{ik\cdot x}f(x){\mathrm{d}} x, \end{equation}
for all
$k\in{\mathbb{R}^d}$
.
Our first set of assumptions are exactly those that allow us to use the results relating to lace expansion arguments.
Assumption A.
We require
$\varphi$
to satisfy the following two properties:
-
(A.1) There exists a function
$g\colon \mathbb N \to{\mathbb{R}}_{\geq 0}$
with the following three properties. Firstly, that
$g(d)\to 0$
as
$d\to \infty$
. Secondly, that for
$m\geq 3$
, the
$m$
-fold convolution
$\varphi ^{\star m}$
of
$\varphi$
satisfies
(1.7)Thirdly, that the Lebesgue volume
\begin{equation} \frac{1}{q_\varphi ^{m-1}}\sup _{x\in{\mathbb{R}^d}}\varphi ^{\star m}(x) \leq g(d). \end{equation}
(1.8)
\begin{equation} \frac{1}{q_\varphi }\left\lvert{\left \{x\in{\mathbb{R}^d}\colon \frac{1}{q_\varphi } \varphi \star \varphi (x) \gt g\!\left (d\right )\right \}}\right\rvert \leq g(d). \end{equation}
-
(A.2) There are constants
$b,c_1,c_2\gt 0$
(independent of
$d$
) such that the Fourier transform
$\widehat \varphi$
satisfies
(1.9)
\begin{equation} \inf _{0\lt \lvert{k}\rvert \leq b}\frac{1}{\lvert{k}\rvert ^2}\left (1-\frac{1}{q_\varphi }\widehat \varphi (k)\right ) \gt c_1,\qquad \inf _{\lvert{k}\rvert \gt b}\left (1-\frac{1}{q_\varphi }\widehat \varphi (k)\right ) \gt c_2. \end{equation}
Remark 1.1. We believe that the condition (1.8) is not necessary for our results. In [Reference Dickson and HeydenreichDH22] it was required by the lace expansion argument to provide the skeleton of an argument that would work for ‘spread out’ models in dimensions
$d=7,8$
(in addition to
$d\geq 9$
). However, we are concerned here with taking
$d\to \infty$
, and so it should not be required.
$\diamond$
It will sometimes be more natural to work with a parameter
$\beta (d)$
that is related to
$g(d)$
. From [Reference Dickson and HeydenreichDH22] it is defined by
\begin{equation} \beta (d) \,:\!= \begin{cases} g(d)^{\frac{1}{4} -\frac{3}{2d}}d^{-\frac{3}{2}} &\colon \lim _{d\to \infty }g(d)\rho ^{-d}\Gamma \!\left (\frac{d}{2}+1\right )^2 = 0 \qquad \forall \rho \gt 0,\\ g(d)^{\frac{1}{4}}&\colon \text{otherwise}. \end{cases} \end{equation}
Note that the Assumption (B.1) below implies that
$\beta (d)=g(d)^{\frac{1}{4}}$
.
Our second set of assumptions allow us to keep suitable control of asymptotic properties. Let us define
$h\colon \mathbb{N}\to{\mathbb{R}}_{\geq 0}$
and
$N\colon \mathbb{N}\to \mathbb{N}$
by
\begin{align} h(d) \,:\!= & \frac{1}{q_\varphi ^5}\varphi ^{\star 6}\left (\mathbf{0}\right ) + \frac{1}{q_\varphi ^4}\int \varphi (x)\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x + \frac{1}{q_\varphi ^4}\int \!\left (\varphi ^{\star 2}(x)\right )^3{\mathrm{d}} x, \end{align}
\begin{align} N(d) \,:\!= & \left\lceil{\frac{\log h(d)}{\log \beta (d)}}\right\rceil. \end{align}
Assumption B. We require that:
-
(B.1) There exists
$\rho \gt 0$
such that
$\liminf _{d\to \infty }\rho ^{-d}q_\varphi ^{-5}\varphi ^{\star 6}\left (\mathbf{0}\right )\gt 0$
. -
(B.2)
$\limsup _{d\to \infty }N(d) \lt \infty$
.
Mind that Assumption (B.2) is in practice a lower bound on
$h(d)$
because
$\beta (d)\lt 1$
for large
$d$
.
Remark 1.2. The factor
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
appears in Assumption (B.1) only because
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
is the precision at which we stop our expansion. If we wished to proceed up to the
$\varphi ^{\star m}\left (\mathbf{0}\right )$
term, then we would need a version of (B.1) with
$\varphi ^{\star m}\left (\mathbf{0}\right )$
replacing
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
. Assumption (B.1) appears in our proof via Lemma 2.4. A close inspection of the proof would reveal that it is a slightly stronger condition than is needed there. However the version presented is more concise and sufficient for the models we consider here.
The requirement that (B.2) holds becomes apparent through Proposition 3.4. We take great care in describing the asymptotics of the first few terms in the expansion of
$\widehat \Pi _{\lambda _c}(0)$
because they dictate the behaviour of
$\lambda _c$
that we are interested in. On the other hand we can utilise pre-existing bounds for the tail of the expansion to show that it can be neglected in our calculations. If we fix a cut-off
$N\geq 1$
in this expansion then these pre-existing bounds are of order
$\beta ^N$
. Assumption (B.2) ensures that we can choose a fixed
$N$
such that this tail error is smaller than the error terms arising elsewhere in the expansion. If this was not the case, we may try to let
$N\to \infty$
as
$d\to \infty$
, but then we would be summing a diverging number of ‘small’ terms prior to the cut-off and we would not have a good control on this.
$\diamond$
Definition 1.3. In addition to using the convolution operation to combine two non-negative functions
$f,g\colon{\mathbb{R}^d}\to{\mathbb{R}}_{\geq 0}$
, we will also find it convenient to use
$f\cdot g$
to denote the pointwise multiplication of
$f$
and
$g$
:
\begin{equation} f\cdot g(x) \,:\!= f(x)g(x). \end{equation}
Furthermore, for
$n_1,n_2,n_3\geq 1$
, we will denote
\begin{equation} \varphi ^{\star n_1\star n_2 \cdot n_3}\left (\mathbf{0}\right ) \,:\!= \varphi ^{\star n_1}\star \left (\varphi ^{\star n_2}\cdot \varphi ^{\star n_3}\right )\left (\mathbf{0}\right ) = \int \varphi ^{\star n_1}(x)\varphi ^{\star n_2}(x)\varphi ^{\star n_3}(x){\mathrm{d}} x. \end{equation}
This expression shows that
$\varphi ^{\star n_1\star n_2 \cdot n_3}\left (\mathbf{0}\right )$
is invariant under the permutation of
$n_1$
,
$n_2$
, and
$n_3$
.
Theorem 1.4. Suppose Assumptions A and B are satisfied. Then as
$d\to \infty$
,
\begin{align} \lambda _c& = \frac{1}{q_\varphi } + \frac{1}{q_\varphi ^3}\varphi ^{\star 3}\left (\mathbf{0}\right ) + \frac{3}{2}\frac{1}{q_\varphi ^4}\varphi ^{\star 4}\left (\mathbf{0}\right ) + 2\frac{1}{q_\varphi ^5}\varphi ^{\star 5}\left (\mathbf{0}\right ) - \frac{5}{2}\frac{1}{q_\varphi ^4}\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) + 2\frac{1}{q_\varphi ^5}\left (\varphi ^{\star 3}\left (\mathbf{0}\right )\right )^2 \nonumber \\ &\quad + \mathcal{O}\!\left (\frac{1}{q_\varphi ^6}\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right ) + \frac{1}{q_\varphi ^7}\left (\varphi ^{\star 3}\left (\mathbf{0}\right )\right )^3 + \frac{1}{q_\varphi ^6}\varphi ^{\star 6}\left (\mathbf{0}\right ) + \frac{1}{q_\varphi ^5}\varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) + \frac{1}{q_\varphi ^5}\varphi ^{\star 1 \star 2\cdot 3}\left (\mathbf{0}\right )\right ). \end{align}
Remarks on graphical notation. It will often be convenient and clearer to represent the objects like
$\varphi ^{\star n}\left (\mathbf{0}\right )$
and
$\varphi ^{\star n_1 \star n_2 \cdot n_3}\left (\mathbf{0}\right )$
pictorially. By expanding out the convolutions in these expressions it is clear that they are integrals over some finite set of points with functions associating pairs of these points (and sometimes the origin). We are therefore able to represent these integrals pictorially as rooted graphs. In these we represent the spatial origin
$\mathbf{0}\in{\mathbb{R}^d}$
with the root vertex
$\circ$
, and an integral of some
$x\in{\mathbb{R}^d}$
with the vertex
$\bullet$
. If we can interpret a
$\varphi$
function to be ‘connecting’ two
$\mathbb{R}^d$
values, then we draw a line
$-\!\!-\!\!-\!\!-\!\!-\!\!-$
between the vertices corresponding to the two
$\mathbb{R}^d$
values. For example, this allows us to graphically represent objects such as
Observe that convolution is a commutative binary relation. This means for example that various diagrams the position of the root vertex
$\circ$
is not important. The most common example of this in our arguments will relate to
$\varphi ^{\star 1 \star 2 \cdot 2}\left (\mathbf{0}\right )$
. By first recalling that
$\varphi ^{\star n_1\star n_2 \cdot n_3}\left (\mathbf{0}\right )$
is invariant under the permutation of
$n_1$
,
$n_2$
and
$n_3$
, and then using the commutativity property of convolution, we find
We will tend to prefer 
over 
, as we find the former slightly easier to read.
This graphical notation allows us to write the expansion of Theorem1.4 in a form that is much easier to read. By a rescaling argument (see [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Section 5.1] for the details), we may assume without loss of generality that
\begin{equation} q_\varphi =\int \varphi (x){\mathrm{d}} x =1, \end{equation}
and we shall silently make this assumption in our analysis. Under this scaling choice, the expansion (1.15) is represented pictorially by
For some calculations, we will want to integrate a
$\tau _\lambda$
function instead of a
$\varphi$
(
$\tau _\lambda$
is defined below at (2.1)). We will also sometimes find it convenient to write the sum of two integrals as one integral by using
$1-\varphi$
to associate two
$\mathbb{R}^d$
values. When we can interpret a
$\tau _\lambda$
function to be ‘connecting’ two
$\mathbb{R}^d$
values, then we draw a blue line 
between the vertices corresponding to the two
$\mathbb{R}^d$
values, and similarly we draw a red line 
when a
$1-\varphi$
connects two values. As examples, we can use these to represent the following two integrals:
1.4 Applications
The result of Theorem1.4 is very general in that the Assumptions A and B apply to very many models. We now apply it to a number of examples.
1.4.1 The Gilbert disk model resp. the hyper-sphere RCM
For
$R\gt 0$
, the hyper-sphere RCM is defined by having
\begin{equation} \varphi (x) ={\unicode{x1D7D9}}_{\left \{\lvert{x}\rvert \lt R\right \}}. \end{equation}
This is the classical model for Boolean percolation studied by Gilbert in 1961 [Reference GilbertGil61]. Figure 1 shows a representation of part of such a model. Writing
$B\!\left (x;\,a,b\right )=\int ^x_0t^{a-1}\left (1-t\right )^{b-1}{\mathrm{d}} t$
for the incomplete Beta function and
$\Gamma \!\left (x\right )=\int ^\infty _0t^{x-1}{\text{e}}^{-t}{\mathrm{d}} t$
is the Gamma function, we obtain the following expansion of the critical intensity:
Corollary 1.5. For the hyper-sphere RCM with radius
$R=R(d)\gt 0$
,
\begin{equation} \frac{\pi ^{\frac{d}{2}}}{\Gamma \!\left (\frac{d}{2}+1\right )}R^{d}\lambda _c =1 + \frac{3}{2\sqrt{\pi }}\frac{\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}B\!\left (\frac{3}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right ) + \mathcal{O}\!\left (\frac{1}{\sqrt{d}}\left (\frac{16}{27}\right )^{\frac{d}{2}}\right ). \end{equation}
Remark 1.6. Here we only expand as far as the
$\varphi ^{\star 3}\left (\mathbf{0}\right )$
term, and our error is the asymptotic size of the
$\varphi ^{\star 4}\left (\mathbf{0}\right )$
term. This is because these are the only terms for which we have rigorous closed-form expressions for their asymptotic size. Conjecture A.8 gives the expected terms in the expansion based on numerical estimates of their asymptotic behaviour.
$\diamond$
Left: the hyper-sphere RCM – two Poisson points are connected whenever the circles of radius
$R/2$
overlap. Right: the hyper-cube RCM – two Poisson points are connected whenever the cubes of side length
$L/2$
overlap.
1.4.2 The hyper-cube RCM
While the hyper-sphere model is a good example showing that the numerical integration of the various convolutions of the adjacency function in (1.15) can get fairly involved, the calculations simplify significantly for the hyper-cubic RCM given by
\begin{equation} \varphi (x) = \prod ^d_{j=1}{\unicode{x1D7D9}}_{\left \{\lvert{x_j}\rvert \leq L/2\right \}}, \end{equation}
where
$x=\left (x_1,\ldots, x_d\right )\in{\mathbb{R}^d}$
and
$L\gt 0$
is a parameter. Figure 1 shows a representation of part of such a hyper-cube RCM.
Corollary 1.7. For the hyper-cubic RCM with side length
$L=L(d)\gt 0$
, as
$d\to \infty$
\begin{equation} L^d\lambda _c = 1 + \left (\frac{3}{4}\right )^d + \frac{3}{2}\left (\frac{2}{3}\right )^d + 2\!\left (\frac{115}{192}\right )^d - \frac{5}{2}\left (\frac{7}{12}\right )^d + 2\!\left (\frac{9}{16}\right )^d + \mathcal{O}\!\left (\left (\frac{11}{20}\right )^d\right ). \end{equation}
1.4.3 The Gaussian RCM
For
$\sigma ^2\gt 0$
and
$0\lt \mathcal{A}\leq \left (2\pi \sigma ^2\right )^{\frac{d}{2}}$
, the Gaussian RCM is defined by having
\begin{equation} \varphi \!\left (x\right ) = \frac{\mathcal{A}}{\left (2\pi \sigma ^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2\sigma ^2}\lvert{x}\rvert ^2\right ). \end{equation}
The parameter
$\sigma$
is a length-scale parameter while the
$\mathcal{A}$
factor ensures
$\mathcal{A} = \int \varphi (x){\mathrm{d}} x$
. The upper bound on
$\mathcal{A}$
is only there to ensure
$\varphi$
is
$\left [0,1\right ]$
-valued. Then we have the following expansion:
Corollary 1.8.
For the Gaussian RCM with
$\mathcal{A}=\mathcal{A}(d)\gt 0$
and
$\sigma =\sigma (d)\gt 0$
such that
$\liminf _{d\to \infty }\varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
, as
$d\to \infty$
\begin{equation} \mathcal{A}\lambda _c = 1 + \mathcal{A}\!\left (6\pi \sigma ^2\right )^{-\frac{d}{2}} + \frac{3}{2}\mathcal{A}\!\left (8\pi \sigma ^2\right )^{-\frac{d}{2}} + 2\mathcal{A}\!\left (10\pi \sigma ^2\right )^{-\frac{d}{2}} + \mathcal{O}\!\left (\mathcal{A}\!\left (12\pi \sigma ^2\right )^{-\frac{d}{2}}\right ). \end{equation}
In particular, if
$\varphi \!\left (\mathbf{0}\right )=\mathcal{A}\!\left (2\pi \sigma ^2\right )^{-\frac{d}{2}} = 1$
, then
\begin{equation} \mathcal{A}\lambda _c = 1 + 3^{-\frac{d}{2}} + \frac{3}{2}\times 4^{-\frac{d}{2}} +2\times 5^{-\frac{d}{2}} + \mathcal{O}\big(6^{-\frac{d}{2}}\big). \end{equation}
1.4.4 The coordinate-wise Cauchy RCM
In a similar flavour to the previous example, let
$\gamma \gt 0$
and
$0\lt \mathcal{A}\leq \left (\gamma \pi \right )^d$
, and define the Coordinate-Cauchy RCM through
\begin{equation} \varphi (x) = \frac{\mathcal{A}}{\left (\gamma \pi \right )^d}\prod ^d_{j=1}\frac{\gamma ^2}{\gamma ^2+x^2_j}, \end{equation}
where
$x=\left (x_1,\ldots, x_d\right )\in{\mathbb{R}^d}$
. Like for the Gaussian RCM we have a length-scale parameter
$\gamma$
while the
$\mathcal{A}$
factor ensures
$\mathcal{A} = \int \varphi (x){\mathrm{d}} x$
and the upper bound on
$\mathcal{A}$
is only there to ensure
$\varphi$
is
$\left [0,1\right ]$
-valued. Then the expansion of the critical intensity is as follows:
Corollary 1.9. For the Coordinate-Cauchy RCM with
$\mathcal{A}=\mathcal{A}(d)\gt 0$
and
$\gamma =\gamma (d)\gt 0$
such that
$\liminf _{d\to \infty }\varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
, as
$d\to \infty$
\begin{equation} \mathcal{A}\lambda _c = 1 + \mathcal{A}\!\left (3\gamma \pi \right )^{-d} + \frac{3}{2}\mathcal{A}\!\left (4\gamma \pi \right )^{-d} + 2\mathcal{A}\!\left (5\gamma \pi \right )^{-d} + \mathcal{O}\!\left (\mathcal{A}\!\left (6\gamma \pi \right )^{-d}\right ). \end{equation}
In particular, if
$\varphi \!\left (\mathbf{0}\right ) = \mathcal{A}\!\left (\gamma \pi \right )^{-d} = 1$
, then
\begin{equation} \mathcal{A} \lambda _c = 1 + 3^{-d} + \frac{3}{2}\times 4^{-d} + 2\times 5^{-d} + \mathcal{O}(6^{-d}). \end{equation}
Remark 1.10. The condition on
$\varphi \!\left (\mathbf{0}\right )$
appearing in Corollaries 1.8 and 1.9 is to ensure that (B.1) is satisfied. If this were not imposed, then the terms in our expansion could be so small that extra error terms arising from the volume of small balls of fixed radius could become significant and dominate.
$\diamond$
1.5 Discussion
Our results reveal a remarkable difference between continuum percolation models and lattice percolation: while the expansion in (1.1) and (1.2) decays algebraically in
$d$
, we observe that the expansions in Corollaries1.5–
1.9 decay exponentially in
$d$
. Interestingly, the expansion in (1.15) resp. (1.20) is indeed algebraic, and it is the calculation of the convolutions of
$\varphi$
that transform it to an exponentially decaying series. This is reflected in the observation that the hyper-cubic lattice is a ’sparse’ graph in high dimensional Euclidean space. Indeed, the analysis in [Reference Handa, Kamijima and SakaiHKS20] suggests that we do have exponential decay on lattices that use the space more efficiently such as the body-centred cubic lattice.
Torquato [Reference TorquatoTor12] has provided an expansion for
$\lambda _c$
using exact calculations. Interestingly, for the hyper-cubic Boolean model, we seem to get a slightly different expansion as the
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
term is absent from their expression.
It is clear that the value of
$\lambda _c$
is highly sensitive to the choice of the connectivity function
$\varphi$
. As a result, we get fairly different expansions for the four models in Section 1.4. Jonasson [Reference JonassonJon01] has shown that for Boolean models,
$\lambda _c$
is maximised for the hyper-sphere model, and minimised for a certain triangular shape.
Our analysis is based on the lace expansion for the (plain) random connection model derived in [Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22]. A key quantity in that expansion is the lace expansion coefficient
$\Pi _{\lambda _c}(x)$
(defined in Definition3.2 below), see (3.6). The main insight is that
$\int \Pi _{\lambda _c}(x){\mathrm{d}} x$
encodes
$\lambda _c$
, see (3.17), and we therefore need to investigate this integral as the dimension
$d$
increases. While the original lace expansion only needs (fairly crude) upper bounds on the different terms that constitute
$\Pi _{\lambda _c}(x)$
, in the present work we need to improve and refine these bounds to get asymptotically matching upper and lower bounds. This is the content of Section 3.
In our main expansion in (1.15), there are various terms appearing on the right-hand side. Apart from the constant term
${q_\varphi }^{-1}$
, the main contribution is given by the single loop diagram
${q_\varphi ^{-3}}\varphi ^{\star 3}\left (\mathbf{0}\right )$
. However, the order of the further terms may depend on the particular form of
$\varphi$
, e.g. compare Corollaries1.7 and 1.8.
It is an open problem to extend this analysis to the marked random connection model, for which the lace expansion has recently been derived in [Reference Dickson and HeydenreichDH22].
2. Preliminaries
Recall that
$\eta$
denotes the homogeneous Poisson point process on
$\mathbb{R}^d$
that gives the vertex set of the RCM. We then let
$\xi$
denote the vertex set and the edge set together – the whole random graph. We also want to consider the augmented configurations
$\eta ^x$
and
$\xi ^x$
. Here
$\eta ^x$
is produced by introducing an extra vertex at
$x\in{\mathbb{R}^d}$
, and
$\xi ^x$
then takes this augmented vertex set, copies the old edges, and independently forms edges between the old vertices and the new vertex. This can also be extended to get
$\eta ^{x,y}$
and
$\xi ^{x,y}$
for
$x,y\in{\mathbb{R}^d}$
, or for any finite number of augmenting vertices. For the full details of this construction see [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Section 2.2].
Recall that
$\varphi (x)$
returns the probability that a vertex at the origin and a vertex at
$x$
have a common edge, or are adjacent. Given two vertices
$x,y\in{\mathbb{R}^d}$
, we say that
$x$
and
$y$
are connected in
$\xi ^{x,y}$
, or
$x \longleftrightarrow y\:\mathrm{ in }\: \xi ^{x,y}$
, if there exists a finite sequence of distinct vertices
$x=u_0,u_1,\ldots, u_k,u_{k+1}=y\in \eta ^{x,y}$
(with
$k\in \mathbb{N}_0$
) such that
$u_i\sim u_{i+1}$
for all
$0\leq i\leq k$
. We can then define the two-point (or pair-connectedness) function
$\tau _\lambda \colon{\mathbb{R}^d} \to \left [0,1\right ]$
by
\begin{equation} \tau _\lambda (x) \,:\!= \mathbb{P}_\lambda \!\left (\mathbf{0} \longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right ). \end{equation}
Note that in general it is possible to have two different vertices at the same position, but that since
$\eta$
is distributed as a Poisson point process with a non-atomic intensity measure, the position almost surely defines the vertex for vertices in
$\eta$
. If we augment
$\eta$
with two vertices at the same position, then these are indeed distinct vertices and not just the same vertex – i.e.
$\xi ^{x,x}\ne \xi ^{x}$
. In particular, this means that while we take the convention that a vertex is always connected to itself,
$\tau _\lambda (\mathbf{0})$
is the probability that two distinct vertices at the same position are connected, and is therefore not guaranteed to be equal to
$1$
for all models.
Now we introduce two preliminary results that we will use on many occasions in this paper: Mecke’s (multivariate) equation, and the BK inequality.
Mecke’s equation. Since our vertex set
$\eta$
is a Poisson point process, we will often rely on a result called Mecke’s Equation to use integral expressions to describe the expected number of certain configurations in our RCM. For a discussion of this result see [[Reference Last and PenroseLP18], Chapter 4]. Given
$m\in \mathbb{N}$
and a measurable non-negative function
$f=f(\xi, \vec{x})$
, the Mecke equation for
$\xi$
states that
\begin{equation}{\mathbb{E}}_\lambda \!\left [ \sum _{\vec x \in \eta ^{(m)}} f(\xi, \vec{x})\right ] = \lambda ^m \int{\mathbb{E}}_\lambda \!\left [ f\!\left (\xi ^{x_1, \ldots, x_m}, \vec x\right )\right ]{\mathrm{d}} \vec{x}, \end{equation}
where
$\vec x=(x_1,\ldots, x_m)$
and
$\eta ^{(m)}=\{(x_1,\ldots, x_m)\colon x_i \in \eta, x_i \neq x_j \text{ for } i \neq j\}$
.
BK inequality. We give an overview here, but the full details can be found in [Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22]. Given two increasing events
$E_1$
and
$E_2$
, we define
$E_1\circ E_2$
to be the event that
$E_1$
and
$E_2$
both occur, but do so on disjoint subsets of the vertices
$\eta$
. Note that in the case of
$E_1=\left \{x \longleftrightarrow y\:\mathrm{ in }\: \xi ^{x,y}\right \}$
and
$E_2=\left \{u \longleftrightarrow v\:\mathrm{ in }\: \xi ^{u,v}\right \}$
,
$E_1\circ E_2$
can still occur if
$x\in \left \{u,v\right \}$
or
$y\in \left \{u,v\right \}$
– the intermediate vertices need to be disjoint. The BK inequality then gives us a simple upper bound on the probability of this disjoint occurrence.
Theorem 2.1 (BK inequality). Let
$E_1$
and
$E_2$
be two increasing events that live on some bounded measurable subset on
$\mathbb{R}^d$
. Then
\begin{equation} \mathbb{P}_\lambda \!\left (E_1\circ E_2\right ) \leq \mathbb{P}_\lambda \!\left (E_1\right )\mathbb{P}_\lambda \!\left (E_2\right ). \end{equation}
Proof. See [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Theorem2.1].
Definition 2.2. We make use of a bootstrap function also used in [Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22] (itself adapted from an argument in [Reference Heydenreich, van der Hofstad and SakaiHHS08]). Recall that we are using the scaling choice that
$\widehat \varphi (0) = q_\varphi = 1$
. For
$\lambda \geq 0$
and
$k,l\in{\mathbb{R}^d}$
, we define
\begin{align}{\mu _\lambda } &\,:\!= 1- \frac{1}{\widehat \tau _\lambda (0)} \end{align}
\begin{align} \widehat G_{{\mu _\lambda }}(k) &\,:\!= \frac{1}{1-{\mu _\lambda }\widehat \varphi (k)}. \end{align}
Note that
$\widehat G_{{\mu _\lambda }}$
can be interpreted as the Fourier transform of the Green’s function of a random walk with transition density
${\mu _\lambda }\varphi$
. We can define
$f \colon{\mathbb{R}}_{\geq 0}\to{\mathbb{R}}_{\geq 0}$
with
\begin{equation} f(\lambda )\,:\!= \sup _{k\in{\mathbb{R}^d}}\frac{\lvert{\widehat \tau _\lambda (k)}\rvert }{\widehat G_{{\mu _\lambda }}(k)}. \end{equation}
Proposition 2.3. Suppose Assumption A holds. Then for
$d$
sufficiently large,
$f(\lambda )\leq 2$
for all
$\lambda \in \left [0,\lambda _c\right )$
.
Proof. This is implied by [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Proposition 5.10].
Lemma 2.4. Suppose Assumption A holds and that there exists
$\rho \gt 0$
such that
$\liminf _{d\to \infty }\rho ^{-d}\varphi ^{\star m}\left (\mathbf{0}\right ) \gt 0$
. Let
$d$
be sufficiently large,
$m\geq 1$
be even,
$s\geq 1$
, and
$\lambda \in \left [0,\lambda _c\right ]$
. Then there exists
$K_{s}\lt \infty$
independent of
$d$
,
$m$
, and
$\lambda$
such that
\begin{equation} \sup _{x\in{\mathbb{R}^d}}\varphi ^{\star m}\star \tau _\lambda ^{\star s}\left (x\right ) \leq K_{s}\varphi ^{\star m}\left (\mathbf{0}\right ). \end{equation}
This is a key lemma in our proof as it allows us to identify leading order decay for convolutions of the adjacency function and the two-point function.
Proof. First let us consider
$\lambda \lt \lambda _c$
. We slightly adapt [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Lemma 5.4] for our purposes. From the Fourier inverse formula,
\begin{equation} \sup _{x\in{\mathbb{R}^d}}\varphi ^{\star m}\star \tau _\lambda ^{\star s}\left (x\right ) \leq \sup _{x\in{\mathbb{R}^d}}\int{\text{e}}^{-ik\cdot x}\widehat \varphi (k)^m\widehat \tau _\lambda (k)^s\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} \leq \int \widehat \varphi (k)^m\lvert{\widehat \tau _\lambda (k)}\rvert ^s\frac{{\mathrm{d}} k}{\left (2\pi \right )^d}. \end{equation}
We can omit
$\lvert{\cdot }\rvert$
from around
$\widehat \varphi (k)^m$
because
$\widehat \varphi (k)$
is real and
$m$
is even. From the definition of the bootstrap function
$f(\lambda )$
, we can bound
$\lvert{\widehat \tau _\lambda (k)}\rvert$
with
$f(\lambda )\widehat G_{{\mu _\lambda }}(k)$
and then use
${\mu _\lambda }\leq 1$
to get
\begin{equation} \sup _{x\in{\mathbb{R}^d}}\varphi ^{\star m}\star \tau _\lambda ^{\star s}\left (x\right ) \leq f(\lambda )^{s}\int \frac{\widehat \varphi (k)^m}{\left (1-{\mu _\lambda }\widehat \varphi (k)\right )^s}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} \leq f(\lambda )^{s}\int \frac{\widehat \varphi (k)^m}{\left (1-\widehat \varphi (k)\right )^s}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d}. \end{equation}
Recall the parameter
$b\gt 0$
arising from Assumption (A.2). We partition the integral on the right-hand side of (2.9) into one integral over
$\lvert{k}\rvert \leq b$
, and one integral over
$\lvert{k}\rvert \gt b$
. For
$\lvert{k}\rvert \leq b$
, (A.2) tells us that there exists
$c_1\gt 0$
such that
$\left (1-\widehat \varphi (k)\right )^{-1}\leq c_1^{-1}\lvert{k}\rvert ^{-2}$
, and therefore
\begin{equation} \int _{\lvert{k}\rvert \leq b} \frac{\widehat \varphi (k)^m}{\left (1-\widehat \varphi (k)\right )^s}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} \leq \frac{1}{c_1^s}\int _{\lvert{k}\rvert \leq b} \frac{1}{\lvert{k}\rvert ^{2s}} \frac{{\mathrm{d}} k}{\left (2\pi \right )^d} = \frac{1}{c_1^s}\frac{\mathfrak{S}_{d-1}}{d-2s}\frac{b^{d-2s}}{\left (2\pi \right )^{d}}, \end{equation}
where
$\mathfrak{S}_{d-1} = d\pi ^{\frac{d}{2}}\Gamma \!\left (1 + \frac{d}{2}\right )^{-1}$
is the surface area of a dimension
$d$
hyper-sphere with unit radius. An application of Stirling’s formula tells us that for all
$\rho \gt 0$
we have
$\frac{\mathfrak{S}_{d-1}}{d-2s}\frac{b^{d-2s}}{\left (2\pi \right )^{d}} \leq \rho ^d$
for sufficiently large
$d$
. Therefore this contribution is negligible for our purposes. For
$\lvert{k}\rvert \gt b$
, (A.2) tells us that
$\left (1-\widehat \varphi (k)\right )^{-1}\leq c_2^{-1}$
, and therefore
\begin{equation} \int _{\lvert{k}\rvert \gt b} \frac{\widehat \varphi (k)^m}{\left (1-\widehat \varphi (k)\right )^s}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} \leq \frac{1}{c_2^s}\int \widehat \varphi (k)^m \frac{{\mathrm{d}} k}{\left (2\pi \right )^d} = \frac{1}{c_2^s}\varphi ^{\star m}\left (\mathbf{0}\right ). \end{equation}
In conjunction with (2.9) and Proposition2.3, this proves the result for
$\lambda \lt \lambda _c$
.
To extend the result to
$\lambda \leq \lambda _c$
, we note that
$\tau _\lambda (x)$
is monotone increasing in
$\lambda$
for all
$x\in{\mathbb{R}^d}$
. Monotone convergence and the independence of the bound on
$\lambda$
then proves the full result.
Definition 2.5. For
$n\in \mathbb{N}$
and
$x,y\in{\mathbb{R}^d}$
,
$x$
is connected to
$y$
in
$\xi ^{x,y}$
by a path of length exactly
$n$
if there exists a sequence of vertices
$x=u_0,u_1,\ldots, u_{n-1},u_n=y$
such that
$u_i\sim u_{i+1}$
for
$0\leq i \leq n-1$
. We then define
$\left \{x \stackrel{\,\,=n\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi ^{x,y}\right \}$
as the event that
$x$
is connected to
$y$
in
$\xi ^{x,y}$
by a path of length exactly
$n$
, but no path of length
$\lt n$
. For
$\lambda \gt 0$
we denote
\begin{equation} \varphi ^{[n]}(x) \,:\!= \mathbb{P}_\lambda \!\left (\mathbf{0} \stackrel{\,\,=n\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right ). \end{equation}
In particular,
$\varphi ^{[1]}\equiv \varphi$
.
Additionally define for finite
$A\subset{\mathbb{R}^d}$
,
\begin{equation} \varphi ^{[n]}_{\langle A \rangle }\left (x,y\right ) \,:\!= \mathbb{P}_\lambda \!\left (x \stackrel{\,\,=n\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi ^{x,y}_{\langle A \rangle }\right ). \end{equation}
That is,
$\varphi ^{[n]}_{\langle A \rangle }\left (x,y\right )$
is the probability that there exists a path of length
$n$
connecting
$x$
and
$y$
in
$\xi ^{x,y}$
, but none of the interior vertices in this path (that is, not the endpoints
$x$
and
$y$
) are adjacent to any vertices in
$A$
and there is no path connecting
$x$
and
$y$
in
$\xi ^{x,y}$
that is of length
$\lt n$
. A more formal definition of
$\xi ^{x,y}_{\langle A \rangle }$
can be found below in Definition 3.1.
Lemma 2.6. Let
$x,y\in{\mathbb{R}^d}$
be distinct,
$\lambda \gt 0$
, and
$A\subset{\mathbb{R}^d}$
be a finite number of singletons. Then for
$n\geq 1$
,
\begin{align} \varphi ^{[1]}_{\langle A \rangle }(x,y) &= \varphi (x-y) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{align}
\begin{align} \varphi ^{[n+1]}_{\langle A \rangle }(x,y) &= \left (1-\varphi (x-y)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (v-y)\varphi ^{[n]}_{\langle A\cup \left \{y\right \} \rangle }(x,v)\prod _{z\in A}\left (1-\varphi (v-z)\right ){\mathrm{d}} v\right )\right ) \end{align}
\begin{align} \varphi ^{[n+1]}(x) &= \left (1-\varphi (x)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\varphi ^{[n]}_{\langle \mathbf{0} \rangle }(x,v){\mathrm{d}} v\right )\right ). \qquad\qquad\qquad\qquad\end{align}
In particular,
\begin{align} \ \; \ \varphi ^{[2]}(x) &= \left (1-\varphi (x)\right )\left (1-\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )\right ) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{align}
\begin{align} \varphi ^{[3]}(x) = \left (1-\varphi (x)\right )&\left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\left (1-\varphi (x-v)\right )\right .\right .\nonumber \\ & \qquad \left .\left .\times \left (1 - \exp \left ({-}\lambda \int \varphi (w-v)\varphi (x-w)\left (1-\varphi (w)\right ){\mathrm{d}} w\right )\right ){\mathrm{d}} v\right )\right ). \end{align}
Proof. To show (2.14), observe that if
$x \sim y\:\mathrm{ in }\: \xi ^{x,y}$
then there are no interior points on this path to be adjacent to
$A$
. Therefore
$\varphi ^{[1]}_{\langle A \rangle }(x,y) = \varphi ^{[1]}(x-y) = \varphi (x-y)$
.
For (2.15), we first note that the existence of a single edge connecting
$x$
and
$y$
is independent of everything else. Since we cannot have this edge, we have a factor of
$1-\varphi (x-y)$
outside everything else. Let us now consider the neighbours of
$y$
in
$\eta$
. The event
$\left \{x \stackrel{\,\,= n+1\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi ^{x,y}_{\langle A \rangle }\right \}$
occurs exactly when
$x\not \sim y$
and there exists a neighbour
$v$
of
$y$
that is not adjacent to any point in
$A$
and has a path of length
$n$
from
$v$
to
$x$
that does not use any vertex adjacent to
$A$
or adjacent to
$y$
(otherwise a ‘shortcut’ would exist). The existence of such a path is exactly the event
$\left \{x \stackrel{\,\,=n\,\,}{\longleftrightarrow } v\:\mathrm{ in }\: \xi ^{x,v}_{\langle A\cup \left \{y\right \} \rangle }\right \}$
. Since
$\eta$
is a Poisson point process, the number of such vertices is a Poisson distributed random variable with mean given by (via Mecke’s equation)
\begin{align} &{\mathbb{E}}_\lambda \!\left [\#\left \{v\in \eta \colon v\sim y, x \stackrel{\,\,=n\,\,}{\longleftrightarrow } v\:\mathrm{ in }\: \xi ^{x}_{\langle A\cup \left \{y\right \} \rangle }, v\not \sim z \text{ for all }z\in A\right \}\right ] \nonumber \\ &\qquad \qquad = \lambda \int \varphi (v-y)\varphi ^{[n]}_{\langle A\cup \left \{y\right \} \rangle }(x,v)\prod _{z\in A}\left (1-\varphi (v-z)\right ){\mathrm{d}} v. \end{align}
If
$X$
is a Poisson random variable with mean
$M$
, then
$\mathbb{P}\!\left (X\geq 1\right ) = 1 -{\text{e}}^{-M}$
. Since the number
$\#\left \{v\in \eta \colon v\sim y, x \stackrel{\,\,=n\,\,}{\longleftrightarrow } v\:\mathrm{ in }\: \xi ^{x}_{\langle A\cup \left \{y\right \} \rangle }, v\not \sim z \text{ for all }z\in A\right \}$
is a Poisson random variable, this returns the required second factor in (2.15).
To get (2.16), use (2.15) with
$A=\emptyset$
and
$y=\mathbf{0}$
.
To calculate
$\varphi ^{[2]}$
and
$\varphi ^{[3]}$
, we iteratively use (2.14), (2.15), and (2.16). For
$\varphi ^{[2]}$
we have
\begin{align} \varphi ^{[2]}(x) &= \left (1-\varphi (x)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\varphi ^{[1]}_{\langle \mathbf{0} \rangle }(x,v){\mathrm{d}} v\right )\right )\nonumber \\ &= \left (1-\varphi (x)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\varphi (x-v){\mathrm{d}} v\right )\right ) \nonumber \\ &= \left (1-\varphi (x)\right )\left (1-\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )\right ). \end{align}
Similarly, we find
\begin{equation} \varphi ^{[2]}_{\langle \mathbf{0} \rangle }(x,v) = \left (1-\varphi (x-v)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (w-v)\varphi (x-w)\left (1-\varphi (w)\right ){\mathrm{d}} w\right )\right ), \end{equation}
and therefore
\begin{align} \varphi ^{[3]}(x) &= \left (1-\varphi (x)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\varphi ^{[2]}_{\langle \mathbf{0} \rangle }(x,v){\mathrm{d}} v\right )\right )\nonumber \\ &= \left (1-\varphi (x)\right )\left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\left (1-\varphi (x-v)\right )\right .\right .\nonumber \\ &\qquad \left .\left .\times \left (1 - \exp \!\left ({-}\lambda \int \varphi (w-v)\varphi (x-w)\left (1-\varphi (w)\right ){\mathrm{d}} w\right )\right ){\mathrm{d}} v\right )\right ). \end{align}
Lemma 2.7. For
$n\geq 1$
,
$\lambda \gt 0$
, and
$x\in{\mathbb{R}^d}$
,
\begin{equation} \varphi ^{[n]}(x) \leq \lambda ^{n-1}\varphi ^{\star n}(x). \end{equation}
Proof. The expression
$\varphi ^{[n]}(x)$
gives the probability that there exists at least one path from
$\mathbf{0}$
to
$x$
of length
$n$
, and no shorter paths. We can bound this by the probability that there exists at least one path from
$\mathbf{0}$
to
$x$
of length
$n$
. Then by Markov’s inequality this is bounded by the expected number of paths from
$\mathbf{0}$
to
$x$
of length
$n$
. By Mecke’s equation this is given by
$\lambda ^{n-1}\varphi ^{\star n}(x)$
.
Lemma 2.8.
For
$m,n\geq 1$
,
$\lambda \gt 0$
, and
$x\in{\mathbb{R}^d}$
,
\begin{equation} \sum ^{m}_{i=1}\varphi ^{[i]}(x) \leq \tau _\lambda (x) \leq \sum ^{n}_{i=1}\varphi ^{[i]}(x) + \lambda ^n\varphi ^{\star (n+1)}(x) + \lambda ^{n+1}\varphi ^{\star (n+1)}\star \tau _\lambda (x). \end{equation}
Proof. First note that the events
$\left \{\left \{\mathbf{0} \stackrel{\,\,=i\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right \}\right \}_{i\in \mathbb{N}}$
are pairwise disjoint. They are also all contained in the event
$\left \{\mathbf{0} \longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right \}$
. Therefore
$\sum ^{m}_{i=1}\varphi ^{[i]}(x) \leq \tau _\lambda (x)$
.
For the upper bound, the above comments imply that
$\tau _\lambda (x)-\sum ^{n+1}_{i=1}\varphi ^{[i]}(x)$
is the probability that
$\mathbf{0}$
and
$x$
are connected in
$\xi ^{\mathbf{0},x}$
by some path of length
$n+2$
or longer. We can then use Markov’s inequality to bound this probability by the expected number of paths of length
$n+2$
or longer. By using Mecke’s equation, we get
\begin{align} \tau _\lambda (x)-\sum ^{n+1}_{i=1}\varphi ^{[i]}(x) &\leq{\mathbb{E}}_{\lambda }\left [\sum _{y\in \eta }{\unicode{x1D7D9}}_{\left \{\mathbf{0} \stackrel{\,\,=n+1\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi ^{\mathbf{0}}\right \}\circ \left \{y \longleftrightarrow x\:\mathrm{ in }\: \xi ^x\right \}}\right ]\nonumber \\ &= \lambda \int \mathbb{P}_\lambda \!\left (\left \{\mathbf{0} \stackrel{\,\,=n+1\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi ^{\mathbf{0},y}\right \}\circ \left \{y \longleftrightarrow x\:\mathrm{ in }\: \xi ^{y,x}\right \}\right ){\mathrm{d}} y\nonumber \\ &\leq \lambda \int \varphi ^{[n+1]}(y)\tau _\lambda (x-y){\mathrm{d}} y. \end{align}
In this last inequality we have used the BK inequality to bound the probability of the vertex-disjoint occurrence. We therefore have
\begin{equation} \tau _\lambda (x) \leq \sum ^{n}_{i=1}\varphi ^{[i]}(x) + \varphi ^{[n+1]}(x) + \lambda \varphi ^{[n+1]}\star \tau _\lambda (x). \end{equation}
Bounding
$\varphi ^{[n+1]}(x) \leq \lambda ^{n}\varphi ^{\star (n+1)}(x)$
(as shown in Lemma2.7) in these last two terms then gives the result.
Lemma 2.9.
If
$n_1,n_2,n_3\geq 2$
, then
\begin{equation} \int \varphi ^{\star n_1}(x)\varphi ^{\star n_2}(x)\varphi ^{\star n_3}(x){\mathrm{d}} x \leq \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2+n_3 - 6} \int \left (\varphi ^{\star 2}(x)\right )^3{\mathrm{d}} x. \end{equation}
If
$n_1,n_2\geq 2$
and
$n_1+n_2\geq 6$
, then
\begin{equation} \int \varphi ^{\star n_1}(x)\varphi ^{\star n_2}(x)\varphi (x){\mathrm{d}} x \leq \varphi ^{\star (n_1+n_2)}\left (\mathbf{0}\right ) \leq \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2 - 6}\varphi ^{\star 6}\left (\mathbf{0}\right ). \end{equation}
Proof. Recall that the Fourier transform of the convolution of two functions equals the pointwise product of their individual Fourier transforms, and the Fourier transform of the pointwise product of two functions equals the convolution of their individual Fourier transforms. Therefore
\begin{equation} \int \varphi ^{\star n_1}(x)\varphi ^{\star n_2}(x)\varphi ^{\star n_3}(x){\mathrm{d}} x = \int \widehat \varphi (k)^{n_1}\widehat \varphi (k-l)^{n_2}\widehat \varphi (l)^{n_3}\frac{{\mathrm{d}} k{\mathrm{d}} l}{\left (2\pi \right )^{2d}}. \end{equation}
We then note that having
$\varphi (x)\geq 0$
implies
$\sup _{k}\lvert{\widehat \varphi (k)}\rvert = \widehat \varphi (0) = \int \varphi (x){\mathrm{d}} x$
. Therefore a supremum bound implies
\begin{align} \int \widehat \varphi (k)^{n_1}\widehat \varphi (k-l)^{n_2}\widehat \varphi (l)^{n_3}\frac{{\mathrm{d}} k{\mathrm{d}} l}{\left (2\pi \right )^{2d}} &\leq \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2+n_3 - 6} \int \lvert{\widehat \varphi (k)}\rvert ^{2}\lvert{\widehat \varphi (k-l)}\rvert ^{2}\lvert{\widehat \varphi (l)}\rvert ^{2}\frac{{\mathrm{d}} k{\mathrm{d}} l}{\left (2\pi \right )^{2d}}\nonumber \\ & = \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2+n_3 - 6} \int \widehat \varphi (k)^{2}\widehat \varphi (k-l)^{2}\widehat \varphi (l)^{2}\frac{{\mathrm{d}} k{\mathrm{d}} l}{\left (2\pi \right )^{2d}}\nonumber \\ & = \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2+n_3 - 6} \int \left (\varphi ^{\star 2}(x)\right )^3{\mathrm{d}} x. \end{align}
For the second inequality, we bound
$\varphi (x)\leq 1$
to leave the convolution
$\varphi ^{\star n_1}\star \varphi ^{\star n_2}\left (\mathbf{0}\right )= \varphi ^{\star (n_1+n_2)}\left (\mathbf{0}\right )$
. Then like above we have
\begin{align} \varphi ^{\star (n_1+n_2)}\left (\mathbf{0}\right ) &= \int \widehat \varphi (k)^{n_1 + n_2}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} \nonumber \\ &\leq \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2 - 6} \int \widehat \varphi (k)^{6}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} = \left (\int \varphi (x){\mathrm{d}} x\right )^{n_1+n_2 - 6}\varphi ^{\star 6}\left (\mathbf{0}\right ). \end{align}
3. Lace expansion coefficients
The key to our proof is a decomposition of the lace expansion coefficients. In preparation for defining them, we need a few more elementary definitions. The full definitions can be found in [Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22].
Definition 3.1 (Thinnings and Pivotal Points). Let
$x,y\in{\mathbb{R}^d}$
and
$A\subset{\mathbb{R}^d}$
be a locally finite set.
-
1. Let
$\eta$
be a vertex set. We produce a vertex set
$\eta _{\langle A \rangle }$
by retaining each
$\omega \in \eta$
with probability
$\overline{\varphi }(A,\omega )\,:\!= \prod _{z\in A}\left (1-\varphi (\omega, z)\right )$
. We call
$\eta _{\langle A \rangle }$
an
$A$
thinning of
$\eta$
. A similar procedure can be followed to define
$\eta ^x_{\langle A \rangle }$
from
$\eta ^x$
. -
2. Define
$\left \{x \stackrel{\,\,A\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi \right \}$
to be the event that
$x,y\in \eta$
and
$x$
is connected to
$y$
in
$\xi$
, but that this connection does not survive an
$A$
thinning of
$\eta \setminus \left \{x\right \}$
. In particular, the connection does not survive if
$y$
is thinned out. -
3. The vertex
$u\in{\mathbb{R}^d}$
is pivotal and
$u\in \mathsf{Piv}(x,y,\xi )$
if every path on
$\xi ^{x,y}$
that connects
$x$
to
$y$
uses the vertex
$u$
. The end points
$x$
and
$y$
are never said to be pivotal. -
4. Define
(3.1)If one considers the pivotal points from
\begin{equation} E\!\left (x,y;\,A,\xi \right ) \,:\!= \left \{x \stackrel{\,\,A\,\,}{\longleftrightarrow } y\:\mathrm{ in }\: \xi \right \}\cap \left \{\not \exists w\in \mathsf{Piv}(x,y;\,\xi )\colon x \stackrel{\,\,A\,\,}{\longleftrightarrow } w\:\mathrm{ in }\: \xi \right \}. \end{equation}
$x$
to
$y$
in
$\xi$
in sequence, then this is the event that an
$A$
thinning breaks the connection after the last pivotal point and not before.
-
5. Define
(3.2)Note that this is equal to the event that
\begin{equation} \left \{x \Longleftrightarrow y\:\mathrm{ in }\: \xi ^{x,y}\right \}\,:\!= \left \{x \longleftrightarrow y\:\mathrm{ in }\: \xi ^{x,y}\right \}\circ \left \{x \longleftrightarrow y\:\mathrm{ in }\: \xi ^{x,y}\right \}. \end{equation}
$x$
and
$y$
are adjacent or there exist vertices
$u,v$
in
$\eta$
that are adjacent to
$x$
and have disjoint paths to
$y$
that both do not contain
$x$
. Alternatively, there are no pivotal points for the connection of
$x$
and
$y$
in
$\xi ^{x,y}$
.
We are now able to define the lace expansion coefficients, which will be the main objects of study in the remainder of the paper.
Definition 3.2.
For
$n\in \mathbb{N}$
,
$x\in{\mathbb{R}^d}$
, and
$\lambda \in \left [0,\lambda _c\right ]$
we define
\begin{align} \Pi _\lambda ^{(0)}(x) \,:\!= \mathbb{P}_\lambda \!\left (\mathbf{0} \Longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right ) - \varphi (x),\end{align}
\begin{align} \Pi _\lambda ^{(n)}(x) \,:\!= \lambda ^n \int \mathbb{P}_\lambda \!\left ( \{\mathbf{0} \Longleftrightarrow u_0\:\mathrm{ in }\: \xi ^{\mathbf{0}, u_0}_{0}\} \cap \bigcap _{i=1}^{n} E\!\left (u_{i-1},u_i;\,{\mathscr{C}}_{i-1}, \xi ^{u_{i-1}, u_i}_{i}\right ) \right ){\mathrm{d}} \vec u_{[0,n-1]}, \end{align}
where
$u_n=x$
, and
$\left \{\xi _i\right \}_{i\geq 0}$
are independent copies of
$\xi$
. We have also used
${\mathscr{C}}_{i} ={\mathscr{C}}\left (u_{i-1}, \xi ^{u_{i-1}}_{i}\right )$
to denote the cluster of
$u_{i-1}$
in
$\xi ^{u_{i-1}}_i$
for
$i\geq 1$
, and similarly
${\mathscr{C}}_0 ={\mathscr{C}}\left (\mathbf{0}, \xi ^{\mathbf{0}}_{0}\right )$
. The measure
${\mathrm{d}} \vec u_{[0,n-1]}$
denotes the Lebesgue product measure
$\prod ^{n-1}_{i=0}{\mathrm{d}} u_i$
on
$\big({\mathbb{R}^d}\big)^n$
. Then we further define
\begin{equation} \Pi _\lambda (x) = \sum _{n=0}^\infty \left ({-}1\right )^n\Pi _\lambda ^{(n)}(x). \end{equation}
Note that [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Corollary 6.1] proves that
$\Pi _{\lambda _c}^{(n)}(x) = \lim _{\lambda \nearrow \lambda _c}\Pi _\lambda ^{(n)}(x)$
, and (in the proof) that
$\widehat \Pi _{\lambda _c}^{(n)}(0) = \lim _{\lambda \nearrow \lambda _c}\widehat \Pi _\lambda ^{(n)}(0)$
and
$\widehat \Pi _{\lambda _c}(0) = \lim _{\lambda \nearrow \lambda _c}\widehat \Pi _\lambda (0)$
.
Proposition 3.3. Suppose Assumption A holds and
$d$
is sufficiently large. Then for all
$\lambda \leq \lambda _c$
and
$x\in{\mathbb{R}^d}$
\begin{equation} \tau _\lambda (x) = \varphi (x) + \Pi _\lambda (x) + \lambda \!\left (\varphi + \Pi _\lambda \right )\star \tau _\lambda (x). \end{equation}
Proof. This is the Ornstein-Zernike equation for the random connection model, and it is proven in [Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22]. The
$\lambda \lt \lambda _c$
result is in Corollary 5.3, and the
$\lambda =\lambda _c$
result is in Corollary 6.1.
Our main result for this section is the following proposition.
Proposition 3.4.
Suppose Assumptions A and (B.1) hold. Also let
$n_0\geq 4$
and
$N\geq 1$
be fixed. Then as
$d\to \infty$
,
\begin{align} \lambda _c\sum ^\infty _{n=N}\left ({-}1\right )^n\widehat \Pi _{\lambda _c}^{(n)}(0) &= \mathcal{O}\!\left (\beta ^N\right ). \end{align}
Note that when Assumption (B.2) holds we can choose a fixed finite
$N^*$
such that
for all
$d\in \mathbb{N}$
. If we then let
$N=N^*$
in (3.12), the bound becomes
Proof. The corollary follows from
$\widehat \Pi _{\lambda _c}(0)=\sum ^\infty _{n=0}\left ({-}1\right )^n\widehat \Pi _{\lambda _c}^{(n)}(0)$
and the bounds in Proposition3.4.
We prove Proposition3.4 in the remainder of the section: (3.7) is proved in Section 3.1, (3.8) is proved in Section 3.2, (3.9) is proved in Section 3.3, (3.10) and (3.11) are proven in Section 3.1. But first we show how it implies our main result.
Proof of Theorem 1.4. By applying the Fourier transform to both sides of (3.6), we can rearrange terms to find
\begin{equation} \widehat \tau _\lambda (k) = \frac{\widehat \varphi (k) + \widehat \Pi _\lambda (k)}{1- \lambda \!\left (\widehat \varphi (k) + \widehat \Pi _\lambda (k)\right )} \end{equation}
for all
$k\in{\mathbb{R}^d}$
and
$\lambda \leq \lambda _c$
(where we interpret the right-hand side as
$=\infty$
if the denominator vanishes). Since Mecke’s equation implies
$\chi \!\left (\lambda \right ) = 1 + \lambda \widehat \tau _\lambda (0)$
, and
$\lambda _c = \inf \left \{\lambda \gt 0 \colon \chi \left (\lambda \right )=\infty \right \}$
, this tells us that
$\lambda _c$
satisfies
\begin{equation} \lambda _c\left (1+\widehat \Pi _{\lambda _c}(0)\right ) = 1, \end{equation}
where we have used
$\widehat \varphi (0)=1$
. We now aim to use our expansion for
$\widehat \Pi _{\lambda _c}(0)$
to get an expansion for
$\lambda _c$
.
Let us denote 
, 
, 
, and 
. Using Corollary3.5, (3.17) becomes
\begin{equation} \lambda _c - a\lambda _c^2 - b\lambda _c^3 - c\lambda _c^4 + \mathcal{O}\!\left (r\right )= 1. \end{equation}
We can rearrange this to get
\begin{equation} \lambda _c = 1+ a\lambda _c^2 + b\lambda _c^3 + c\lambda _c^4 + \mathcal{O}\!\left (r\right )\!, \end{equation}
and by substituting this into itself produces
\begin{align} \lambda _c &= 1 + a\!\left (1+ a\lambda _c^2 + b\lambda _c^3 + c\lambda _c^4 + \mathcal{O}\!\left (r\right )\right )^2 + b\!\left (1+ a\lambda _c^2 + b\lambda _c^3 + c\lambda _c^4 + \mathcal{O}\!\left (r\right )\right )^3\nonumber \\ &\quad + c\left (1+ a\lambda _c^2 + b\lambda _c^3 + c\lambda _c^4 + \mathcal{O}\!\left (r\right )\right )^4 + \mathcal{O}\!\left (r\right )\nonumber \\ &= 1 + a + 2a^2\lambda _c^2 + \mathcal{O}\!\left (ab\lambda _c^3 + a^3\lambda _c^4\right ) +b + \mathcal{O}\!\left (ab\lambda ^2_c\right ) + c + \mathcal{O}\!\left (ac\lambda _c^2\right ) + \mathcal{O}\!\left (r\right )\nonumber \\ & = 1 + a + b+ c+ 2a^2 + \mathcal{O}\!\left (ab + a^3 + r\right ). \end{align}
Finally, note that 
and so the last term is exactly as stated in our result.
3.1 Bounds on the zeroth lace expansion coefficient
In this subsection we prove (3.7).
Upper bound on
$\widehat \Pi _{\lambda _c}^{(0)}(0)$
Lemma 3.6. Suppose Assumption A holds. Then as
$d\to \infty$
,
Proof. We first consider
$\mathbb{P}_\lambda \!\left (\mathbf{0} \Longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right )$
. Since the existence of an edge between
$\mathbf{0}$
and
$x$
is independent of everything else,
\begin{align} &\mathbb{P}_\lambda \!\left (\mathbf{0} \Longleftrightarrow x\:\mathrm{ in } \:\xi ^{\mathbf{0},x}\right ) \nonumber \\ &= \varphi (x) + \left (1-\varphi (x)\right )\mathbb{P}_\lambda \!\left (\exists u,v\in \eta \colon \mathbf{0}\sim u, \mathbf{0}\sim v, \left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\circ \left \{v \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\right ). \end{align}
Then note that the disjoint occurrence is a subset of the intersection of each occurrence:
$\left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\circ \left \{v \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \} \subset \left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\cap \left \{v \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}$
. Therefore
\begin{align} &\mathbb{P}_\lambda \!\left (\exists u,v\in \eta \colon u\ne v, \mathbf{0}\sim u, \mathbf{0}\sim v, \left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\circ \left \{v \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\right ) \nonumber \\ &\leq \mathbb{P}_\lambda \!\left (\#\left \{u\in \eta \colon \mathbf{0}\sim u, u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\geq 2\right ). \end{align}
Since
$\eta$
is a Poisson point process, the number of such vertices is Poisson distributed and Mecke’s equation tells us that the expected number of such vertices is given by
\begin{equation}{\mathbb{E}}_{\lambda }\left [\#\left \{u\in \eta \colon \mathbf{0}\sim u, u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\right ] = \lambda \int \varphi (v)\tau _\lambda (x-v){\mathrm{d}} v = \lambda \varphi \star \tau _\lambda (x). \end{equation}
Therefore
\begin{align} \mathbb{P}_\lambda \!\left (\exists u,v\in \eta \colon u \ne v, \mathbf{0}\sim u, \right. & \left.\mathbf{0}\sim v, \left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\circ \left \{v \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\right ) \nonumber \\[4pt] & \leq 1 - \mathbb{P}_\lambda \!\left (\#\left \{u\in \eta \colon \mathbf{0}\sim u, u \longleftrightarrow x\:\mathrm{ in }\: \xi ^{x}\right \}\leq 1\right ) \nonumber\\[4pt] &= 1 - \exp \!\left ({-}\lambda \varphi \star \tau _\lambda (x)\right ) - \lambda \varphi \star \tau _\lambda (x)\exp \!\left ({-}\lambda \varphi \star \tau _\lambda (x)\right ).\end{align}
Using this with
$1-{\text{e}}^{-x} - x{\text{e}}^{-x}\leq \frac{1}{2}x^2 + \frac{1}{6}x^3$
for all
$x\in{\mathbb{R}}$
and (3.3), we can get
\begin{align} \lambda \widehat \Pi _\lambda ^{(0)}(0) &\leq \lambda \int \!\left (1-\varphi (x)\right )\left (1 - \exp \!\left ({-}\lambda \varphi \star \tau _\lambda (x)\right ) - \lambda \varphi \star \tau _\lambda (x)\exp \!\left ({-}\lambda \varphi \star \tau _\lambda (x)\right )\right ){\mathrm{d}} x \nonumber \\ & \leq \frac{1}{2}\lambda \int \!\left (1-\varphi (x)\right )\left (\lambda \varphi \star \tau _\lambda (x)\right )^2{\mathrm{d}} x + \frac{1}{6}\lambda \int \!\left (1-\varphi (x)\right )\left (\lambda \varphi \star \tau _\lambda (x)\right )^3{\mathrm{d}} x. \end{align}
By applying
$\tau _\lambda (x) \leq \varphi (x) + \lambda \varphi \star \tau _\lambda (x)$
iteratively, we get
\begin{align} &\int \!\left (1-\varphi (x)\right )\left (\varphi \star \tau _\lambda (x)\right )^2{\mathrm{d}} x \nonumber \\ &\quad \leq \int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^2{\mathrm{d}} x + 2\lambda \int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x\nonumber\\ & \qquad + 2\lambda ^2\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 4}(x){\mathrm{d}} x\nonumber \\ &\qquad + 2\lambda ^3\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 4}\star \tau _\lambda (x){\mathrm{d}} x + \lambda ^2\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}(x)^2{\mathrm{d}} x \nonumber \\ &\qquad + 2\lambda ^3\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}(x)\varphi ^{\star 3}\star \tau _\lambda (x){\mathrm{d}} x + \lambda ^4\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}\star \tau _\lambda (x)^2{\mathrm{d}} x\nonumber \\ &\quad \leq \int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^2{\mathrm{d}} x + 2\lambda \int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x\nonumber \\ & \qquad + 3\lambda ^2\varphi ^{\star 6}\left (\mathbf{0}\right ) + 4\lambda ^3\varphi ^{\star 6}\star \tau _\lambda \!\left (\mathbf{0}\right ) + \lambda ^4\varphi ^{\star 6}\star \tau _\lambda ^{\star 2}\left (\mathbf{0}\right ). \end{align}
From Lemma2.4, we know that for
$\lambda \leq \lambda _c$
these last three terms are all
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
. By further expanding the first two terms via the
$\left (1-\varphi (x)\right )$
factors, we find
\begin{align} \int \!\left (1-\varphi (x)\right )\left (\varphi \star \tau _{\lambda _c}(x)\right )^2{\mathrm{d}} x &= \varphi ^{\star 4}\left (\mathbf{0}\right ) - \int \varphi (x)\varphi ^{\star 2}(x)^2{\mathrm{d}} x + 2\lambda _c\varphi ^{\star 5}\left (\mathbf{0}\right )\nonumber \\ & \quad + \mathcal{O}\!\left (\int \varphi (x)\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x + \varphi ^{\star 6}\left (\mathbf{0}\right )\right ). \end{align}
By the same approach, we find
\begin{align} &\int \!\left (1-\varphi (x)\right )\left (\varphi \star \tau _\lambda (x)\right )^3{\mathrm{d}} x \nonumber \\ &\quad \leq \int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^3{\mathrm{d}} x + 3\lambda \int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^2\varphi ^{\star 3}(x){\mathrm{d}} x \nonumber \\ &\qquad + 3\lambda ^2\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^2\varphi ^{\star 4}(x){\mathrm{d}} x + 3\lambda ^3\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^2\varphi ^{\star 4}\star \tau _\lambda (x){\mathrm{d}} x \nonumber \\ &\qquad + 3\lambda ^2\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 3}(x)^2{\mathrm{d}} x + 6\lambda ^3\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 3}(x)\varphi ^{\star 4}(x){\mathrm{d}} x \nonumber \\ &\qquad + 6\lambda ^4\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 3}(x)\varphi ^{\star 4}\star \tau _\lambda (x){\mathrm{d}} x + 3\lambda ^4\int \!\left (1-\varphi (x)\right )\varphi ^{\star 2}(x)\varphi ^{\star 3}\star \tau _\lambda (x)^2{\mathrm{d}} x \nonumber \\ &\qquad + \lambda ^3\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}(x)^3{\mathrm{d}} x + 3\lambda ^4\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}(x)^2\varphi ^{\star 3}\star \tau _\lambda (x){\mathrm{d}} x \nonumber \\ &\qquad + 3\lambda ^5\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}(x)\varphi ^{\star 3}\star \tau _\lambda (x)^2{\mathrm{d}} x + \lambda ^6\int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}\star \tau _\lambda (x)^3{\mathrm{d}} x\nonumber \end{align}
\begin{align} &\quad \leq \int \varphi ^{\star 2}(x)^3{\mathrm{d}} x + 3\lambda \int \varphi ^{\star 2}(x)^2\varphi ^{\star 3}(x){\mathrm{d}} x \nonumber \\ &\qquad + 3\lambda ^2\varphi ^{\star 6}\left (\mathbf{0}\right )\int \varphi (v){\mathrm{d}} v + 3\lambda ^3\varphi ^{\star 6}\star \tau _\lambda \!\left (\mathbf{0}\right )\int \varphi (v){\mathrm{d}} v \nonumber \\ &\qquad +3\lambda ^2\varphi ^{\star 6}\left (\mathbf{0}\right )\int \varphi (v){\mathrm{d}} v+6\lambda ^3\varphi ^{\star 6}\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^2 \nonumber \\ &\qquad + 6\lambda ^4\varphi ^{\star 6}\star \tau _\lambda \!\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^2 + 3\lambda ^4\varphi ^{\star 6}\star \tau _\lambda ^{\star 2}\left (\mathbf{0}\right )\int \varphi (v){\mathrm{d}} v \nonumber \\ &\qquad + \lambda ^3\varphi ^{\star 6}\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^2 + 3\lambda ^4\varphi ^{\star 6}\star \tau _\lambda \!\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^2 \nonumber \\ &\qquad + 3\lambda ^5\varphi ^{\star 6}\star \tau _\lambda ^{\star 2}\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^2 + \lambda ^6\varphi ^{\star 6}\star \tau _\lambda ^{\star 2}\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^3. \end{align}
Note that in this last inequality we first bound
$1-\varphi (x)\leq 1$
and then identify two paths that form a loop – this contributes the terms
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 6}\star \tau _\lambda \!\left (\mathbf{0}\right )$
, etc. This leaves a third path from
$\mathbf{0}$
to
$x$
. We deal with this by bounding one of the steps in the convolution by
$1$
and the remaining steps form a ‘loose’ integration. For example,
\begin{align} & \int \!\left (1-\varphi (x)\right )\varphi ^{\star 3}\star \tau _\lambda (x)^3{\mathrm{d}} x \leq \int \varphi ^{\star 3}\star \tau _\lambda (x)^2\left (\int \varphi ^{\star 3}(u)\tau _\lambda (x-u){\mathrm{d}} u\right ){\mathrm{d}} x \nonumber \\ & \quad \leq \int \varphi ^{\star 3}\star \tau _\lambda (x)^2\left (\int \varphi ^{\star 3}(u){\mathrm{d}} u\right ){\mathrm{d}} x = \varphi ^{\star 6}\star \tau _\lambda ^{\star 2}\left (\mathbf{0}\right )\left (\int \varphi (v){\mathrm{d}} v\right )^3. \end{align}
Recall that we have chosen the scaling
$\int \varphi (v){\mathrm{d}} v=1$
for our proof. Then through applications of Lemma2.4 and Lemma2.9 we find that
\begin{equation} \int \!\left (1-\varphi (x)\right )\left (\varphi \star \tau _{\lambda _c}(x)\right )^3{\mathrm{d}} x = \mathcal{O}\!\left (\int \varphi ^{\star 2}(x)^3{\mathrm{d}} x + \varphi ^{\star 6}\left (\mathbf{0}\right )\right ). \end{equation}
In summary, these bounds give us
\begin{align} \lambda _c\widehat \Pi _{\lambda _c}^{(0)}(0) & \leq \frac{1}{2}\lambda _c^3\int \varphi ^{\star 2}(x)^2{\mathrm{d}} x - \frac{1}{2}\lambda _c^3\int \varphi (x) \varphi ^{\star 2}(x)^2{\mathrm{d}} x + \lambda _c^4\int \varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x\nonumber \\ & \quad + \mathcal{O}\!\left (\int \varphi ^{\star 2}(x)^3{\mathrm{d}} x + \int \varphi (x)\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x + \varphi ^{\star 6}\left (\mathbf{0}\right )\right ) \end{align}
as required.
Lower bound on
$\widehat \Pi _{\lambda _c}^{(0)}(0)$
Lemma 3.7.
Proof. We lower bound
$\Pi _\lambda ^{(0)}(x)$
by identifying an appropriate subset of
$\left \{\mathbf{0} \Longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right \}$
. Consider
$\mathcal{F} \,:\!= \mathcal{F}_1 \cup \mathcal{F}_2 \cup \mathcal{F}_3$
, where
\begin{align} \mathcal{F}_1 \,:\!=& \left \{\mathbf{0}\sim x\right \} \end{align}
\begin{align} \mathcal{F}_2 \,:\!=& \left \{\mathbf{0}\not \sim x\right \} \cap \left \{\#\left \{u\in \eta \colon \mathbf{0}\sim u\sim x\right \}\geq 2\right \} \end{align}
\begin{align} \mathcal{F}_3 \,:\!=& \left \{\mathbf{0}\not \sim x\right \} \cap \left \{\#\left \{u\in \eta \colon \mathbf{0}\sim u\sim x\right \}= 1\right \} \cap \left \{\#\left \{v\in \eta \colon \mathbf{0} \sim v\:\mathrm{ in }\: \xi ^{\mathbf{0}}, v \stackrel{\,\,=2\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{v,x}_{\langle \mathbf{0} \rangle }\right \}\geq 1\right \}. \end{align}
In each, either
$\mathbf{0}$
is adjacent to
$x$
or there exist two vertex-disjoint paths from
$\mathbf{0}$
to
$x$
. Therefore
$\mathcal{F}\subset \left \{\mathbf{0} \Longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right \}$
. The components
$\mathcal{F}_1$
,
$\mathcal{F}_2$
, and
$\mathcal{F}_3$
are also all disjoint by construction, so
\begin{equation} \mathbb{P}_\lambda \!\left (\mathbf{0} \Longleftrightarrow x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right ) \geq \mathbb{P}_\lambda \!\left (\mathcal{F}_1\right ) + \mathbb{P}_\lambda \!\left (\mathcal{F}_2\right ) + \mathbb{P}_\lambda \!\left (\mathcal{F}_3\right ). \end{equation}
Firstly be definition we have
\begin{equation} \mathbb{P}_\lambda \!\left (\mathcal{F}_1\right ) = \varphi (x). \end{equation}
Since
$\eta$
is distributed as a Poisson point process on
$\mathbb{R}^d$
with intensity
$\lambda$
, we can use Mecke’s equation to get
\begin{equation}{\mathbb{E}}_\lambda \!\left [\#\left \{u\in \eta \colon \mathbf{0}\sim u\sim x\right \}\right ] = \lambda \int \varphi (x-u)\varphi (u){\mathrm{d}} u =\lambda \varphi ^{\star 2}(x), \end{equation}
which fully describes the distribution of the Poisson random variable
$\#\left \{u\in \eta \colon \mathbf{0}\sim u\sim x\right \}$
. The independence of this event from
$\left \{\mathbf{0}\not \sim x\right \}$
then gives
\begin{equation} \mathbb{P}_\lambda \!\left (\mathcal{F}_2\right ) = \left (1-\varphi (x)\right )\left (1-\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right ) - \lambda \varphi ^{\star 2}(x)\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )\right ). \end{equation}
For
$\mathcal{F}_3$
, we first define
$\eta _B\,:\!=\left \{u\in \eta \colon \mathbf{0}\sim u\sim x\right \}$
(blue vertices), and
$\eta _R\,:\!=\eta \setminus \eta _B$
(red vertices). Observe that
$\eta _B$
is distributed as a Poisson point process on
$\mathbb{R}^d$
with intensity measure
$\lambda \varphi (x-u)\varphi (u){\mathrm{d}} u$
. In particular,
${\mathbb{E}}_\lambda \!\left [\#\eta _B\right ]=\lambda \varphi ^{\star 2}(x) \leq \lambda \int \varphi (u){\mathrm{d}} u\lt \infty$
and therefore
$\mathbb{P}_\lambda \!\left (\#\eta _B=0\right )\gt 0$
. We similarly define
$\xi _B$
and
$\xi _R$
, which retain the edges between vertices in their respective vertex sets only. By the definition of Poisson point processes,
$\eta _B$
and
$\eta _R$
are independent and so
$\xi _B$
and
$\xi _R$
are also independent. If we let
$A$
denote a measurable set of configurations, this gives
\begin{align} \mathbb{P}_\lambda \!\left (\left \{\#\eta _B=1\right \}\cap \left \{\xi _R\in A\right \}\right ) &= \mathbb{P}_\lambda \!\left (\#\eta _B=1\right )\mathbb{P}_\lambda \!\left (\xi _R\in A\right )\nonumber \\ &= \mathbb{P}_\lambda \!\left (\#\eta _B=1\right )\frac{\mathbb{P}_\lambda \!\left (\#\eta _B=0\right )\mathbb{P}_\lambda \!\left (\xi _R\in A\right )}{\mathbb{P}_\lambda \!\left (\#\eta _B=0\right )} \nonumber \\ &\geq \mathbb{P}_\lambda \!\left (\#\eta _B=1\right ) \mathbb{P}_\lambda \!\left (\#\eta _B=0\right )\mathbb{P}_\lambda \!\left (\xi _R\in A\right )\nonumber \\ &=\mathbb{P}_\lambda \!\left (\#\eta _B=1\right ) \mathbb{P}_\lambda \!\left (\left \{\#\eta _B=0\right \}\cap \left \{\xi _R\in A\right \}\right ), \end{align}
where
$\mathbb{P}_\lambda \!\left (\#\eta _B=1\right ) = \lambda \varphi ^{\star 2}(x)\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )$
. Now we fix
\begin{equation} A=\left \{\#\left \{v\in \eta \colon \mathbf{0} \sim v\:\mathrm{ in }\: \xi ^{\mathbf{0}}, v \stackrel{\,\,=2\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{v,x}_{\langle \mathbf{0} \rangle }\right \}\geq 1\right \}, \end{equation}
or ‘
$x$
and
$\mathbf{0}$
are connected by at least one path of length
$3$
that uses only one vertex adjacent to
$\mathbf{0}$
.’ We thus see that
$\left \{\mathbf{0}\not \sim x\right \}\cap \left \{\#\eta _B=1\right \}\cap \left \{\xi _R\in A\right \}= \mathcal{F}_3$
, and
$\left \{\mathbf{0}\not \sim x\right \}\cap \left \{\#\eta _B=0\right \}\cap \left \{\xi _R\in A\right \} = \left \{\mathbf{0} \stackrel{\,\,=3\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right \}$
. The independence of everything else from
$\left \{\mathbf{0}\not \sim x\right \}$
then gives
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{F}_3\right ) \geq & \lambda \varphi ^{\star 2}(x)\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )\varphi ^{[3]}(x)\nonumber \\ =& \lambda \varphi ^{\star 2}(x)\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )\left (1-\varphi (x)\right ) \times \left (1 - \exp \!\left ({-}\lambda \int \varphi (v)\left (1-\varphi (x-v)\right )\right.\right.\nonumber\\ & \left.\left.\left (1 - \exp \!\left ({-}\lambda \int \varphi (x-w)\left (1-\varphi (w)\right )\varphi (w-v){\mathrm{d}} w\right )\right ){\mathrm{d}} v\right )\right ), \end{align}
where we have used our expression for
$\varphi ^{[3]}(x)$
from Lemma2.6.
Therefore we now have
\begin{equation} \widehat \Pi _{\lambda _c}^{(0)}(0) \geq \int \!\left (\mathbb{P}_{\lambda _c}\!\left (\mathcal{F}_2\right ) + \mathbb{P}_{\lambda _c}\!\left (\mathcal{F}_3\right )\right ){\mathrm{d}} x, \end{equation}
and we want to lower bound the integrals of
$\mathbb{P}_{\lambda _c}\!\left (\mathcal{F}_2\right )$
and
$\mathbb{P}_{\lambda _c}\!\left (\mathcal{F}_3\right )$
.
By using
$1 -{\text{e}}^{-x} - x{\text{e}}^{-x} \geq \frac{1}{2}x^2 - \frac{1}{2}x^3$
for all
$x\in{\mathbb{R}}$
,
We find our lower bound on
$\lambda \int \mathbb{P}_\lambda \!\left (\mathcal{F}_3\right ){\mathrm{d}} x$
in a few more steps. Since we have
$x{\text{e}}^{-x} \geq x - x^2$
for all
$x\in{\mathbb{R}}$
,
Since we have
$1-{\text{e}}^{-x} \geq x - \frac{1}{2}x^2$
for all
$x\in{\mathbb{R}}$
,
and
Since
$x\mapsto 1-{\text{e}}^{-x}$
is monotone increasing and
$1-{\text{e}}^{-x} \geq x - \frac{1}{2}x^2$
for all
$x\in{\mathbb{R}}$
,
When we combine (3.46) and (3.49) and integrate over
$x$
, we can simplify many of the resulting diagrams by bounding
$1-\varphi \leq 1$
. Suitably doing this for all but the ‘first’ diagram and applying Lemma2.9 allows us to show that these are 
. As an example, the diagram resulting from the first term in (3.46) and the second term in (3.49) becomes
Here in the first inequality we bounded the
$1-\varphi \leq 1$
, and in the equalities we have been able to move the origin (the hollow node) because convolution is a commutative binary relation – like we did in (1.18). The second inequality is a direct application of Lemma2.9. Therefore
where in the last equality we have expanded the
$1-\varphi$
edges in the first diagram to produce two diagrams for each such expanded edge. One such resulting diagram is the
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
term, while we can bound the others in the same way as above and put them in the error term.
We then have a lower bound on
$\widehat \Pi _\lambda ^{(0)}(0)$
for any
$\lambda \gt 0$
, and this gives the required result.
3.2 Bounds on the first lace expansion coefficient
In this subsection we prove (3.8).
Upper bound on
$\widehat \Pi _{\lambda _c}^{(1)}(0)$
Lemma 3.8. Suppose Assumption A holds. Then as
$d\to \infty$
,
We borrow from [Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22] in bounding
$\mathbb{P}_\lambda \!\left ( \{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\} \cap E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right ) \right )$
, but we need to make refinements so that our lower bound will match the upper bound at the precision we are interested in. We begin by bounding
$\{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\} \cap E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )$
by a slightly different event.
Definition 3.9. Let
$\xi _0,\xi _1$
be independent instances of the random graph with locally finite vertex sets
$\eta _0$
and
$\eta _1$
.
-
• Let
$\left \{u \leftrightsquigarrow x\:\mathrm{ in }\: (\xi _0, \xi _1) \right \}$
denote the event that
$u\in \eta _0$
and
$x\in \eta _1$
, but that
$x$
does not survive a
${\mathscr{C}}\!\left (u, \xi ^{u}_0\right )$
thinning of
$\eta _1$
. -
• Let
$m\in \mathbb{N}$
and
$\vec x, \vec y \in \big({\mathbb{R}^d}\big)^m$
. We define
$\bigcirc _m^\leftrightarrow ((x_j, y_j)_{1 \leq j \leq m};\, \xi )$
as the event that
$\{x_j \longleftrightarrow y_j\:\mathrm{ in }\: \xi \}$
occurs for every
$1 \leq j \leq m$
with the additional requirement that every point in
$\eta$
is the interior vertex of at most one of the
$m$
paths, and none of the
$m$
paths contains an interior vertex in the set
$\left \{x_j\colon j\in [m]\right \} \cup \left \{y_j: j\in [m]\right \}$
. -
• Let
$\bigcirc _m^\leftrightsquigarrow \left ( (x_j,y_j)_{1 \leq j \leq m};\, (\xi _0,\xi _1)\right )$
be the intersection of the following two events. Firstly, that
$\bigcirc _{m-1}^\leftrightarrow \left ((x_j,y_j)_{1 \leq j \lt m};\,\xi _0\right )$
occurs but no path uses
$x_{m}$
or
$y_{m}$
as an interior vertex. Secondly, that
$\{x_{m} \leftrightsquigarrow y_{m}\:\mathrm{ in }\: (\xi _0[\eta _0\setminus \{x_i, y_i\}_{1 \leq i \lt m}],\xi _1) \}$
occurs in such a way that at least one point
$z$
in
$\xi _0$
that is responsible for thinning out
$y_m$
is connected to
$x_m$
by a path
$\gamma$
so that
$z$
as well as all interior vertices of
$\gamma$
are not contained in any path of the
$\bigcirc _{m-1}^\leftrightarrow ((x_j,y_j)_{1 \leq j \lt m};\,\xi _0)$
event.
Now let
$t,u,w,x,z\in{\mathbb{R}^d}$
. Then define
\begin{align} F^{(1)}_0\left (w,u,z;\,\xi _0,\xi _1\right ) &\,:\!= \left \{\mathbf{0} \not \sim u\:\mathrm{ in }\: \xi _0\right \}\cap \bigcirc _4^\leftrightsquigarrow \left (\left (\mathbf{0},u\right ),\left (\mathbf{0},w\right ),\left (u,w\right ),\left (w,z\right );\,\left (\xi _0,\xi _1\right )\right ) \end{align}
\begin{align} F^{(2)}_0\left (w,u,z;\,\xi _0,\xi _1\right ) &\,:\!= \left \{w=\mathbf{0}\right \}\cap \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi _1\right \}\cap \left \{w \leftrightsquigarrow z\:\mathrm{ in }\: \left (\xi _0\setminus \{u\},\xi _1\right ) \right \} \end{align}
\begin{align} F_1^{(1)}\left (u,t,z,x;\,\xi _1\right ) &\,:\!= \left \{\#\left \{t,z,x\right \}=3\right \}\cap \bigcirc _4^\leftrightarrow \left (\left (u,t\right ),\left (t,z\right ),\left (t,x\right ),\left (z,x\right );\,\xi _1\right )\cap \left \{t \not \sim x\:\mathrm{ in }\: \xi _1\right \} \end{align}
\begin{align} F_1^{(2)}\left (u,t,z,x;\,\xi _1\right ) &\,:\!=\left \{t=z=x\right \}\cap \left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi _1\right \}. \end{align}
Also let
$F_0 \,:\!= F^{(1)}_0 \cup F^{(2)}_0$
and
$F_1 \,:\!= F^{(1)}_1 \cup F^{(2)}_1$
.
Lemma 3.10. Let
$x,u\in{\mathbb{R}^d}$
be distinct points. Then
\begin{equation}{\unicode{x1D7D9}}_{\left \{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\right \}}{\unicode{x1D7D9}}_{E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )} \leq \sum _{z\in \eta _1^x}\left (\sum _{w\in \eta _0^{\mathbf{0}}}{\unicode{x1D7D9}}_{F_0\!\left (w,u,z;\,\xi _0^{\mathbf{0},u},\xi _1^{u,x}\right )}\right )\left (\sum _{t\in \eta _1^{u,x}}{\unicode{x1D7D9}}_{F_1\!\left (u,t,z,x;\,\xi _1^{u,x}\right )}\right ). \end{equation}
Proof. We first prove that
\begin{equation} {\unicode{x1D7D9}}_{E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )} \leq \sum _{z\in \eta _1^x}\sum _{t\in \eta _1^{u,x}}{\unicode{x1D7D9}}_{F_1\!\left (u,t,z,x;\,\xi _1^{u,x}\right )}{\unicode{x1D7D9}}_{\left \{\mathbf{0} \leftrightsquigarrow z\:\mathrm{ in }\: \left (\xi _0^{\mathbf{0}},\xi _1^{u,x}\right ) \right \}}. \end{equation}
Note that the event
$E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )$
is contained in the event that
$u$
is connected to
$x$
and that this connection fails after a
${\mathscr{C}}_0$
thinning of
$\eta _1^x$
. There are two cases under which this can happen.
Case (a): The point
$x$
itself is thinned out. In this case
\begin{align} & E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right ) \subset \left \{u \longleftrightarrow x\:\mathrm{ in }\: \xi _1^{u,x}\right \}\cap \left \{\mathbf{0} \leftrightsquigarrow x\:\mathrm{ in }\: \left (\xi _0^{\mathbf{0}},\xi _1^{u,x}\right ) \right \}\nonumber \\ &= F^{(2)}_1\left (u,x,x,x;\,\xi _1^{u,x}\right ) \cap \left \{\mathbf{0} \leftrightsquigarrow x\:\mathrm{ in }\: \left (\xi _0^{\mathbf{0}},\xi _1^{u,x}\right ) \right \}. \end{align}
Case (b): The point
$x$
is not thinned out. This implies that there is at least one interior point on the path between
$u$
and
$x$
, and that at least one of these interior points is thinned out by
${\mathscr{C}}_0$
. Let
$t$
be the last pivotal point in
$\mathsf{Piv}(u,x;\,\xi _1^{u,x})$
, and set
$t=u$
if
$\mathsf{Piv}(u,x;\,\xi _1^{u,x})=\emptyset$
. Since
$t$
is the last pivotal point (or there are no pivotal points), we have
$\left \{t \Longleftrightarrow x\:\mathrm{ in }\: \xi _1^x\right \}$
. The event
$E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )$
implies that all the paths from
$t$
to
$x$
fail after a
${\mathscr{C}}_0$
thinning, but that
$t$
is not thinned out. We can pick any thinned out point on a path from
$t$
to
$x$
to be our
$z$
, while noting that
$t$
and
$x$
cannot be adjacent. Therefore this case corresponds to the possible occurrences of
$F^{(1)}_1$
, and we have proven (3.58).
Now it only remains to prove that
\begin{equation}{\unicode{x1D7D9}}_{\left \{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\right \}}{\unicode{x1D7D9}}_{\left \{\mathbf{0} \leftrightsquigarrow z\:\mathrm{ in }\: \left (\xi _0^{\mathbf{0}},\xi _1^{u,x}\right ) \right \}} \leq \sum _{w\in \eta _0^{\mathbf{0}}}{\unicode{x1D7D9}}_{F_0\!\left (w,u,z;\,\xi _0^{\mathbf{0},u},\xi _1^{u,x}\right )}. \end{equation}
The event
$\left \{\mathbf{0} \leftrightsquigarrow z\:\mathrm{ in }\: \left (\xi _0^{\mathbf{0}},\xi _1^{u,x}\right ) \right \}$
implies that there exists at least one point in
${\mathscr{C}}_0$
that is responsible for thinning out
$z$
. Let
$\gamma$
denote a the path from
$\mathbf{0}$
to this point in
${\mathscr{C}}_0$
. We once again now have two cases to consider.
Case (a):
$\mathbf{0} \not \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0$
. Then
$\left \{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\right \}$
implies that there exist two disjoint paths (denoted
$\gamma '$
and
$\gamma ''$
) from
$\mathbf{0}$
to
$u$
. Both of these paths are necessarily of length greater than or equal to
$2$
. Let
$w$
denote the last vertex
$\gamma$
shares with either
$\gamma '$
or
$\gamma ''$
(allowing for the possibility that
$w=\mathbf{0}$
). Requiring that
$\gamma$
,
$\gamma '$
, and
$\gamma ''$
exist results precisely in the event
$F_0^{(1)}$
.
Case (b):
$\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0$
. Now we fix
$w=\mathbf{0}$
immediately. The existence of the path from
$\mathbf{0}$
to the thinning point implies the event
$F_0^{(2)}$
.
Definition 3.11 (The
$\psi$
functions). Let
$r,s,u,w,x,y\in{\mathbb{R}^d}$
and
$n\geq 1$
. We first set
$\tau _\lambda ^\circ (x) \,:\!= \lambda ^{-1}\delta _{x,\mathbf{0}} + \tau _\lambda (x)$
and
$\tau _\lambda ^{(\geq 2)}(x) \,:\!= \mathbb{P}_\lambda \!\left (\mathbf{0} \stackrel{\,\,\geq 2\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{\mathbf{0},x}\right ) = \tau _\lambda (x) - \varphi (x)$
. Also define
\begin{align*} \psi _0^{(1)}(w,u) &\,:\!= \lambda ^2\tau _\lambda ^{(\geq 2)}(u)\tau _\lambda (u-w)\tau _\lambda (w),\\[3pt] \psi _0^{(2)}(w,u) &\,:\!=\lambda ^2 \delta _{w,\mathbf{0}} \tau _\lambda ^{(\geq 2)}(u)\int \tau _\lambda (u-t)\tau _\lambda (t){\mathrm{d}} t,\\[3pt] \psi _0^{(3)}(w,u) &\,:\!= \lambda \varphi (u) \delta _{w,\mathbf{0}},\\[3pt] \psi ^{(1)}(w,u,r,s) &\,:\!= \lambda ^4\tau _\lambda (w-u)\int \tau _\lambda ^\circ (t-s) \tau _\lambda (t-w)\tau _\lambda (u-z)\tau _\lambda (z-t)\tau _\lambda (z-r){\mathrm{d}} z{\mathrm{d}} t, \\[3pt] \psi ^{(2)}(w,u,r,s) &\,:\!= \lambda ^4\tau _\lambda ^\circ (w-s)\int \tau _\lambda (t-z)\tau _\lambda (z-u)\tau _\lambda (u-t)\tau _\lambda ^\circ (t-w)\tau _\lambda (z-r){\mathrm{d}} z{\mathrm{d}} t, \\[3pt] \psi ^{(3)}(w,u,r,s) &\,:\!= \lambda ^2\tau _\lambda (u-w)\tau _\lambda (w-s)\tau _\lambda (u-r),\\[3pt] \psi ^{(4)}(w,u,r,s) &\,:\!= \lambda \delta _{w,s}\tau _\lambda (u-w)\tau _\lambda (u-r),\\[3pt] \psi _n^{(1)} (x,r,s) &\,:\!= \lambda ^3\int \tau _\lambda ^\circ (t-s)\tau _\lambda (z-r)\tau _\lambda (t-z)\tau _\lambda (z-x)\tau _\lambda ^{(\geq 2)}(x-t){\mathrm{d}} z{\mathrm{d}} t,\\[3pt] \psi _n^{(2)}(x,r,s) &\,:\!=\lambda \tau _\lambda (x-s)\tau _\lambda (x-r), \end{align*}
and set
$\psi _0 \,:\!= \psi _0^{(1)}+\psi _0^{(2)}+\psi _0^{(3)}$
,
$\psi _n \,:\!= \psi _n^{(1)} + \psi _n^{(2)}$
, and
$\psi \,:\!= \psi ^{(1)}+\psi ^{(2)}+\psi ^{(3)} + \psi ^{(4)}$
. These functions are represented as diagrams in Figure 2.
Diagrams of the
$\psi _0$
,
$\psi$
, and
$\psi _n$
functions.
For our bounds on
$\widehat \Pi _{\lambda _c}^{(1)}(0)$
we will only require
$\psi _0$
and
$\psi _1$
. Later we will also use
$\psi$
to bound
$\widehat \Pi _{\lambda _c}^{(n)}(0)$
for
$n\geq 2$
.
Lemma 3.12.
Proof. The first inequality is proven in very nearly exactly the same way as [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Proposition 7.2], but with the extra refinement of our Lemma3.10. The first difference is that our event
$F^{(1)}_1$
has the intersection with
$\left \{t \not \sim x\:\mathrm{ in }\: \xi _1\right \}$
. Since the event
$\bigcirc _4^\leftrightarrow \left (\left (u,t\right ),\left (t,z\right ),\left (t,x\right ),\left (z,x\right );\,\xi _1\right )$
ensures that
$t$
and
$x$
are connected in
$\xi _1$
, this means that the event
$t \stackrel{\,\,\geq 2\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi _1$
occurs. This then manifests in the end result as the occurrence of a
$\tau _\lambda ^{(\geq 2)}$
function rather than a
$\tau _\lambda$
function in the integral in
$\psi ^{(1)}_1$
. Similarly, the event
$F^{(1)}_0$
implies that
$\mathbf{0} \stackrel{\,\,\geq 2\,\,}{\longleftrightarrow } u\:\mathrm{ in }\: \xi _0$
, and this results in the
$\tau _\lambda ^{(\geq 2)}(u)$
appearing rather than
$\tau _\lambda (u)$
in
$\psi ^{(1)}_0$
and
$\psi ^{(2)}_0$
.
For the equality, we first note that
\begin{equation} \int \psi ^{(3)}_0\left (w,u\right )\psi ^{(2)}_1\left (x,w,u\right ){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x = \lambda ^2\int \varphi (u)\tau _\lambda (x)\tau _\lambda (u-x){\mathrm{d}} u{\mathrm{d}} x. \end{equation}
Our task in then to show that all the other terms
$\int \psi ^{(j_0)}_0\left (w,u\right )\psi ^{(j_1)}_1\left (x,w,u\right ){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x$
are error terms. To make it clearer what we are trying to bound, we present the integral
$\int \psi _0\left (w,u\right )\psi _1\!\left (x,w,u\right ){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x$
diagrammatically:
The last of these six diagrams will be the only relevant one for our level of precision. To demonstrate how we bound these other five, we examine the second:
By using Lemma2.8 (with
$n=5$
) we can bound
$\tau _\lambda ^{(\geq 2)}(u)$
in terms of
$\varphi ^{[i]}(x)$
(for
$i=2,3,4,5$
),
$\varphi ^{\star 6}(u)$
, and
$\varphi ^{\star 6}\star \tau _\lambda (u)$
. Then we can further bound this by using Lemma2.7 to get
$\tau _\lambda ^{(\geq 2)}(u)\leq \lambda \varphi ^{\star 2}(u) + \lambda ^2\varphi ^{\star 3}(u) + \lambda ^3\varphi ^{\star 4}(u) + \lambda ^4\varphi ^{\star 5}(u) + \lambda ^5\varphi ^{\star 6}(u) + \lambda ^6\varphi ^{\star 6}\star \tau _\lambda (u)$
. For the two diagrams that result from the last two terms in this expansion, we can bound
$\tau _\lambda (w-u)\leq 1$
to get terms of the form
$\lambda ^{j+5}\varphi ^{\star 6}\star \tau _\lambda ^{\star j}\left (\mathbf{0}\right )$
for
$j\in \left \{3,4\right \}$
. From Lemma2.4, both of these are
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
when
$\lambda \leq \lambda _c$
. For the remaining diagrams we then bound
$\tau _\lambda (w) \leq \varphi (w) + \lambda \varphi ^{\star 2}(w) + \lambda ^2\varphi ^{\star 3}(w) + \lambda ^3\varphi ^{\star 4}(w) + \lambda ^4\varphi ^{\star 4}\star \tau _\lambda (w)$
and if the diagrams contain a loop of at least six
$\varphi$
functions and maybe some
$\tau _\lambda$
functions, we once again bound
$\tau _\lambda (w-u)\leq 1$
and use Lemma2.4 to show that they are
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
. For the remaining diagrams we bound
$\tau _\lambda (x-w) \leq \varphi (x-w) + \lambda \varphi ^{\star 2}(x-w) + \lambda ^2\varphi ^{\star 3}(x-w) + \lambda ^3\varphi ^{\star 3}\star \tau _\lambda (x-w)$
. Again bounding
$\tau _\lambda (w-u)\leq 1$
allows us to use Lemma2.4 to show that some of these diagrams are
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
. After bounding
$\tau _\lambda (x-u) \leq \varphi (x-u) + \lambda \varphi ^{\star 2}(x-u) + \lambda ^2\varphi ^{\star 2}\star \tau _\lambda (x-u)$
and showing that some terms are
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
, we arrive at
Then we bound
$\tau _\lambda (w-u)\leq \varphi (w-u) + \lambda \varphi ^{\star 2}(w-u) + \lambda ^2\varphi ^{\star 3}(w-u) + \lambda ^3\varphi ^{\star 3}\star \tau _\lambda (w-u)$
to get
\begin{align} &\lambda ^4\int \varphi ^{\star 2}(u)\varphi (w)\tau _\lambda (w-u)\varphi (x-w)\varphi (x-u){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x \nonumber \\[4pt] &\hspace{4cm}\leq \lambda ^4\int \varphi ^{\star 2}(u)\varphi (w)\varphi (w-u)\varphi (x-w)\varphi (x-u){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x \nonumber \\[4pt] &\hspace{5cm}+ \lambda ^5\int \varphi ^{\star 2}(u)\varphi (w)\varphi ^{\star 2}(w-u)\varphi (x-w)\varphi (x-u){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x \nonumber \\[4pt] &\hspace{5cm}+ \lambda ^6\int \varphi ^{\star 2}(u)\varphi (w)\varphi ^{\star 3}(w-u)\varphi (x-w)\varphi (x-u){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x\nonumber \\[4pt] &\hspace{5cm}+ \lambda ^7\int \varphi ^{\star 2}(u)\varphi (w)\varphi ^{\star 3}\star \tau _\lambda (w-u)\varphi (x-w)\varphi (x-u){\mathrm{d}} u{\mathrm{d}} w{\mathrm{d}} x. \end{align}
From the commutativity of convolution, observe that the first two terms are
$\mathcal{O}\!\left (\int \varphi (x)\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x\right )$
. For the last two terms we bound
$\int \varphi (x-w)\varphi (x-u){\mathrm{d}} x \leq \int \varphi (x-w){\mathrm{d}} x = 1$
for all
$u,w\in{\mathbb{R}^d}$
. Therefore we can apply Lemma2.4 to show that these diagrams are
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
when
$\lambda \leq \lambda _c$
. In summary, we have
Repeating these ideas for the other diagrams produces
Lemma 3.13. Let
$x,u\in{\mathbb{R}^d}$
. Then
\begin{align} \tau _\lambda (x)\tau _\lambda (u-x)&\leq \varphi (x)\varphi (u-x) + \varphi (x)\varphi ^{[2]}(u-x) + \varphi ^{[2]}(x)\varphi (u-x) + \varphi (x)\varphi ^{[3]}(u-x)\nonumber\\[3pt]& \quad + \varphi ^{[3]}(x)\varphi (u-x) + \varphi ^{[2]}(x)\varphi ^{[2]}(u-x) + \lambda ^3\varphi (x)\varphi ^{\star 4}(u-x)\nonumber \\[3pt]&\quad + \lambda ^4\varphi (x)\varphi ^{\star 4}\star \tau _\lambda (u-x)+ \lambda ^3\varphi ^{\star 2}(x)\varphi ^{\star 3}(u-x) + \lambda ^4\varphi ^{\star 2}(x)\varphi ^{\star 3}\star \tau _\lambda (u-x)\nonumber \\[3pt] &\quad + \lambda ^3\varphi ^{\star 3}(x)\varphi ^{\star 2}(u-x) + \lambda ^4\varphi ^{\star 3}(x)\varphi ^{\star 2}\star \tau _\lambda (u-x) + \lambda ^3\varphi ^{\star 4}(x)\varphi (u-x)\nonumber \\[3pt] &\quad + \lambda ^4\varphi ^{\star 4}(x)\varphi \star \tau _\lambda (u-x) + \lambda ^4\varphi ^{\star 4}\star \tau _\lambda (x)\varphi (u-x)+ \lambda ^5\varphi ^{\star 4}\star \tau _\lambda (x)\varphi \star \tau _\lambda (u-x). \end{align}
Therefore
Proof. For equation (3.72) we first expand
$\tau _\lambda (x)$
using the upper bound from Lemma2.8 with
$n=3$
. This produces
\begin{align} \tau _\lambda (x)\tau _\lambda (u-x) & \leq \varphi (x)\tau _\lambda (u-x) + \varphi ^{[2]}(x)\tau _\lambda (u-x) + \varphi ^{[3]}(x)\tau _\lambda (u-x) + \lambda ^3\varphi ^{\star 4}(x)\tau _\lambda (u-x)\nonumber \\[5pt] & \quad + \lambda ^4\varphi ^{\star 4}\star \tau _\lambda (x)\tau _\lambda (u-x). \end{align}
We then expand the first
$\tau _\lambda (u-x)$
using the upper bound from Lemma2.8 with
$n=3$
, the second with
$n=2$
, the third with
$n=1$
, and the last two with
$n=0$
. This produces
\begin{align} \tau _\lambda (x)\tau _\lambda (u-x) & \leq \varphi (x)\varphi (u-x) + \varphi (x)\varphi ^{[2]}(u-x) + \varphi (x)\varphi ^{[3]}(u-x) + \lambda ^3\varphi (x)\varphi ^{\star 4}(u-x)\nonumber\\ & \quad + \lambda ^4\varphi (x)\varphi ^{\star 4}\star \tau _\lambda (u-x)+ \varphi ^{[2]}(x)\varphi (u-x) + \varphi ^{[2]}(x)\varphi ^{[2]}(u-x)\nonumber \\ & \quad + \lambda ^2\varphi ^{[2]}(x)\varphi ^{\star 3}(u-x) + \lambda ^3\varphi ^{[2]}(x)\varphi ^{\star 3}\star \tau _\lambda (u-x) + \varphi ^{[3]}(x)\varphi (u-x)\nonumber\\ & \quad + \lambda \varphi ^{[3]}(x)\varphi ^{\star 2}(u-x)+ \lambda ^2\varphi ^{[3]}(x)\varphi ^{\star 2}\star \tau _\lambda (u-x) + \lambda ^3\varphi ^{\star 4}(x)\varphi (u-x)\nonumber \\ & \quad + \lambda ^4\varphi ^{\star 4}(x)\varphi \star \tau _\lambda (u-x) + \lambda ^4\varphi ^{\star 4}\star \tau _\lambda (x)\varphi (u-x) + \lambda ^5\varphi ^{\star 4}\star \tau _\lambda (x)\varphi \star \tau _\lambda (u-x). \end{align}
For the terms with
$5$
or more ‘factors of
$\varphi$
’, we then bound
$\varphi ^{[m]}(x)\leq \lambda ^{m-1}\varphi ^{\star m}(x)$
as in Lemma2.7 to get (3.72).
For the second part, we use (3.72) in conjunction with Lemma2.4 to show that many of the terms are
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
and produce the first inequality. We then immediately have 
, and simply bounding
$\varphi ^{[3]}\leq \lambda ^2\varphi ^{\star 3}$
and
$\varphi ^{[2]}\leq \lambda \varphi ^{\star 2}$
gives
To bound
$2\lambda ^2\varphi ^{\star 2}\star \varphi ^{[2]}\left (\mathbf{0}\right )$
appropriately requires a little more care though. Recall from (2.17) that
$\varphi ^{[2]}(x) = \left (1-\varphi (x)\right )\left (1-\exp \!\left ({-}\lambda \varphi ^{\star 2}(x)\right )\right )$
. Using
$1-{\text{e}}^{-x} \leq x - \frac{1}{2}x^2 + \frac{1}{6}x^3$
for all
$x\in{\mathbb{R}}$
then gives
\begin{align} \lambda ^2\varphi ^{\star 2}\star \varphi ^{[2]}\left (\mathbf{0}\right ) & \leq \lambda ^3\int \left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^2{\mathrm{d}} x \nonumber \\[5pt] & \quad -\frac{1}{2}\lambda ^4\int \left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^3{\mathrm{d}} x + \frac{1}{6}\lambda ^5\int \left (1-\varphi (x)\right )\varphi ^{\star 2}(x)^4{\mathrm{d}} x. \end{align}
The second term we can safely neglect, and for the third term we use
$1-\varphi (x)\leq 1$
and
$\varphi ^{\star 2}(x)\leq \int \varphi (x){\mathrm{d}} x=1$
for all
$x\in{\mathbb{R}^d}$
. This produces
as required.
This concludes the proof of Lemma3.8.
Lower bound on
$\widehat \Pi _{\lambda _c}^{(1)}(0)$
Lemma 3.14.
Proof. Our strategy is to identify disjoint events contained in
$\{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\} \cap E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )$
for each
$u,x\in{\mathbb{R}^d}$
, and show that the integral of their probabilities is equal to our upper bound to the required precision. Our disjoint events are the following:
\begin{align} \mathcal{G}_1 &\,:\!= \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\right \}\cap \left \{u \sim x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}\cap \left \{x\notin \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\right \} \end{align}
\begin{align} \mathcal{G}_2 &\,:\!= \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\right \}\cap \left \{u \stackrel{\,\, 2\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}\cap \left \{x\notin \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\right \} \end{align}
\begin{align} \mathcal{G}_3 &\,:\!= \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\right \}\cap \left \{u \sim x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}\cap \left \{x\in \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\right \}\cap \left \{\exists v\in \eta _0\colon \mathbf{0} \sim v\:\mathrm{ in }\: \xi ^{\mathbf{0}}_0, x\notin \eta ^{u_0,x}_{1,\langle v \rangle }\right \} \end{align}
\begin{align} \mathcal{G}_4 &\,:\!= \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\right \}\cap \left \{u \stackrel{\,\, 3\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}\cap \left \{x\notin \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\right \} \end{align}
\begin{align} \mathcal{G}_5 &\,:\!= \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\right \}\cap \left \{u \stackrel{\,\, 2\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}\cap \left \{x\in \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\right \}\cap \left \{\exists v\in \eta _0\colon \mathbf{0} \sim v\:\mathrm{ in }\: \xi ^{\mathbf{0}}_0, x\notin \eta ^{u_0,x}_{1,\langle v \rangle }\right \} \end{align}
\begin{align} \mathcal{G}_6 &\,:\!= \left \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\right \}\cap \left \{u \sim x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}\cap \left \{x\in \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\right \} \cap \left \{\not \exists v\in \eta _0\colon \mathbf{0} \sim v\:\mathrm{ in }\: \xi ^{\mathbf{0}}_0, x\in \eta ^{u_0,x}_{1,\langle v \rangle }\right \}\nonumber \\[5pt] & \quad \cap \left \{\exists w\in \eta _0\colon \mathbf{0} \stackrel{\,\,2\,\,}{\longleftrightarrow } w\:\mathrm{ in }\: \xi ^{\mathbf{0}}_0, x\notin \eta ^{u_0,x}_{1,\langle w \rangle }\right \} \end{align}
Observe that these events are indeed disjoint, and all are subsets of
$\big\{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi ^{\mathbf{0}, u}_{0}\big\} \cap E\!\left (u,x;\,{\mathscr{C}}_{0}, \xi ^{u, x}_{1}\right )$
. We now want to bound their probabilities from below. For
$\mathcal{G}_1$
, the events
$\big \{\mathbf{0} \sim u\:\mathrm{ in }\: \xi ^{\mathbf{0},u}_0\big \}$
and
$\left \{u \sim x\:\mathrm{ in }\: \xi ^{u,x}_1 \right \}$
are clearly independent. The event
$\big \{x\notin \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\big \}$
is also independent of these previous two, because it uses a thinning random variable from
$\eta _1^{u,x}$
. The probability that
$x$
is thinned out by the single vertex
$\mathbf{0}$
is also equal to the probability that an edge forms between these vertices. Therefore
$\mathbb{P}_\lambda \!\left (\mathcal{G}_1\right ) = \varphi (u)\varphi (x-u)\varphi (x)$
. The other events proceed similarly with a few points to note. All the events that are intersected to compose the
$\mathcal{G}_i$
are independent because they use different (independent) edge random variables and thinning random variables. Also, the events like
$\big \{u \stackrel{\,\,n\,\,}{\longleftrightarrow } x\:\mathrm{ in }\: \xi ^{u,x}_1 \big \}$
have probability given exactly by
$\varphi ^{[n]}(x-u)$
by definition of that function. The event
$\big \{x\in \eta ^{u,x}_{1,\langle \mathbf{0} \rangle }\big \}\cap \big \{\exists v\in \eta _0\colon \mathbf{0} \sim v\:\mathrm{ in }\: \xi ^{\mathbf{0}}_0, x\notin \eta ^{u_0,x}_{1,\langle v \rangle }\big \}$
says that
$x$
is not thinned out by
$\mathbf{0}$
, and that there exists a
$v$
that forms an edge with
$\mathbf{0}$
and thins out
$x$
. This has probability equal to that of the event that no edge forms between
$\mathbf{0}$
and
$x$
, and that they have at least one mutual neighbour. This is precisely the probability given by
$\varphi ^{[2]}(x)$
. Similar considerations allow us to find factors of
$\varphi ^{[2]}$
and
$\varphi ^{[3]}$
in the probability of the remaining events. The lower bounds we use are summarised here:
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{G}_1\right ) &= \varphi (u)\varphi (x-u)\varphi (x),\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \end{align}
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{G}_2\right ) &= \varphi (u)\varphi ^{[2]}(x-u)\varphi (x)\nonumber \\ &\geq \varphi (u)\varphi (x)\left (1-\varphi (x-u)\right )\left (\lambda \varphi ^{\star 2}(x-u) - \frac{1}{2}\lambda ^2\varphi ^{\star 2}(x-u)^2\right ),\end{align}
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{G}_3\right ) &= \varphi (u)\varphi (x-u)\varphi ^{[2]}(x)\nonumber \\ &\geq \varphi (u)\varphi (x-u)\left (1-\varphi (x)\right )\left (\lambda \varphi ^{\star 2}(x) - \frac{1}{2}\lambda ^2\varphi ^{\star 2}(x)^2\right ), \end{align}
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{G}_4\right ) &= \varphi (u)\varphi ^{[3]}(x-u)\varphi (x),\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \end{align}
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{G}_5\right ) &= \varphi (u)\varphi ^{[2]}(x-u)\varphi ^{[2]}(x)\nonumber \\[3pt] &\geq \varphi (u)\left (1-\varphi (x-u)\right )\left (1-\varphi (x)\right )\left (\lambda ^2\varphi ^{\star 2}(x-u)\varphi ^{\star 2}(x) - \frac{1}{2}\lambda ^3\varphi ^{\star 2}(x-u)^2\varphi ^{\star 2}(x)\right .\nonumber \\[3pt] & \quad \left .- \frac{1}{2}\lambda ^3\varphi ^{\star 2}(x-u)\varphi ^{\star 2}(x)^2 + \frac{1}{4}\lambda ^4\varphi ^{\star 2}(x-u)^2\varphi ^{\star 2}(x)^2\right ), \end{align}
\begin{align} \mathbb{P}_\lambda \!\left (\mathcal{G}_6\right ) &= \varphi (u)\varphi (x-u)\varphi ^{[3]}(x).\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad \end{align}
For
$\mathbb{P}_\lambda \!\left (\mathcal{G}_2\right )$
,
$\mathbb{P}_\lambda \!\left (\mathcal{G}_3\right )$
, and
$\mathbb{P}_\lambda \!\left (\mathcal{G}_5\right )$
we have used a lower bound on
$\varphi ^{[2]}$
by observing
$1-{\text{e}}^{-x}\geq x-\frac{1}{2}x^2$
for all
$x\in{\mathbb{R}}$
and using this with the expression for
$\varphi ^{[2]}$
in (2.17).
From these we can bound
To bound the integrals of
$\mathbb{P}_\lambda \!\left (\mathcal{G}_4\right )$
and
$\mathbb{P}_\lambda \!\left (\mathcal{G}_6\right )$
, recall
Then
Summing the integrals of the probabilities of the
$\mathcal{G}_i$
events gives us our desired lower bound for any
$\lambda \gt 0$
, and in particular
$\lambda =\lambda _c$
.
3.3 Bounds on the second lace expansion coefficient
We start with an upper bound, which we need both for
$n=1$
and also for
$n\ge 3$
in the next subsection.
Proposition 3.15.
Let
$n \geq 1$
,
$x\in{\mathbb{R}^d}$
, and
$\lambda \in \left [0,\lambda _c\right ]$
. Then
\begin{equation} \lambda \Pi _\lambda ^{(n)}(x) \leq \int \psi _n(x,w_{n-1},u_{n-1}) \left ( \prod _{i=1}^{n-1} \psi (w_i,u_i,w_{i-1},u_{i-1}) \right ) \psi _0(w_0,u_0){\mathrm{d}}\left ( \left (\vec w, \vec u\right )_{[0,n-1]} \right ). \end{equation}
Furthermore, for
$\lambda \in \left [0,\lambda _c\right ]$
there exists
$c\gt 0$
(independent of
$\lambda$
and
$d$
) such that
\begin{equation} \sum _{n=N}^\infty \widehat \Pi _\lambda ^{(n)}(0) \leq c\beta ^N. \end{equation}
Proof. As in Lemma3.12, the first inequality is proven in very nearly exactly the same way as [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Proposition 7.2]. The proof of [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Corollary 5.3] only needs to be slightly adjusted to get the second part of our result for
$\lambda \lt \lambda _c$
, and a dominated convergence argument like that appearing in the proof of [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Corollary 6.1] allows us to extend the result to
$\lambda =\lambda _c$
.
Upper bound on
$\widehat \Pi _{\lambda _c}^{(2)}(0)$
Lemma 3.16. Suppose Assumption A holds. Then as
$d\to \infty$
,
Proof. The first inequality is an application of (3.99). After expanding
$\psi _0$
,
$\psi$
, and
$\psi _2$
, we get
\begin{align} & \int \psi _0(w_0,u_0)\psi (w_1,u_1,w_0,u_0)\psi _2(x,w_1,u_1){\mathrm{d}} w_0{\mathrm{d}} u_0{\mathrm{d}} w_1{\mathrm{d}} u_1{\mathrm{d}} x \nonumber \\[4pt] &= \sum _{j_0=1}^3\sum _{j_1=1}^4\sum _{j_2=1}^2\int \psi ^{(j_0)}_0(w_0,u_0)\psi ^{(j_1)}(w_1,u_1,w_0,u_0)\psi ^{(j_2)}_2(x,w_1,u_1){\mathrm{d}} w_0{\mathrm{d}} u_0{\mathrm{d}} w_1{\mathrm{d}} u_1{\mathrm{d}} x. \end{align}
We can index the
$3\times 4\times 2=24$
resulting diagrams by
$\left (j_0,j_1,j_2\right )$
. For
$\left (j_0,j_1,j_2\right )\notin \left \{\left (3,2,2\right ),\left (3,3,2\right ),\left (3,4,2\right )\right \}$
, we can identify a cycle of length
$6$
or longer that visits each vertex. For each factor of
$\tau _\lambda$
that is not part of this cycle we can then bound by
$1$
. For each factor of
$\tau _\lambda$
that is part of the cycle, we bound
$\tau _\lambda \leq \varphi + \lambda \varphi \star \tau _\lambda$
. For
$\lambda \leq \lambda _c$
, Lemma2.4 then lets us bound each of these diagrams by
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
.
For the
$\left (3,2,2\right )$
diagram, we first expand out the
$\tau _\lambda ^\circ$
edges. In many of the resulting diagrams we can apply the above strategy of finding a cycle and bounding the excess edges to bound the diagram by
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
. The result is that for
$\lambda \leq \lambda _c$
we have
\begin{align} &\int \psi ^{(3)}_0(w_0,u_0)\psi ^{(2)}(w_1,u_1,w_0,u_0)\psi ^{(2)}_2(x,w_1,u_1){\mathrm{d}} w_0{\mathrm{d}} u_0{\mathrm{d}} w_1{\mathrm{d}} u_1{\mathrm{d}} x = \nonumber \\[5pt] &\lambda ^4\int \varphi (u_0)\tau _\lambda (z)\tau _\lambda (z-u_0)\tau _\lambda (u_1-z)\tau _\lambda (u_1-u_0)\tau _\lambda (x-u_1)\tau _\lambda (x-u_0){\mathrm{d}} u_0{\mathrm{d}} u_1{\mathrm{d}} z{\mathrm{d}} x +\mathcal{O}\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right ). \end{align}
In this first integral we can bound
$\tau _\lambda (u_1-u_0)\leq 1$
,
$\tau _\lambda (z-u_0)\leq \varphi (z-u_0) + \lambda \int \varphi (x){\mathrm{d}} x$
, and the other
$\tau _\lambda \leq \varphi + \lambda \varphi \star \tau _\lambda$
to find
For the
$\left (3,3,2\right )$
diagram we bound
$\tau _\lambda (u_1-z)\leq \varphi (u_1-z) + \lambda \int \varphi (x){\mathrm{d}} x$
, and the other
$\tau _\lambda \leq \varphi + \lambda \varphi \star \tau _\lambda$
to find
For the
$\left (3,4,2\right )$
diagram we bound
$\tau _\lambda (u_1-u_0)\leq \varphi (u_1-u_0) + \lambda \varphi ^{\star 2}(u_1-u_0) + \lambda ^2\left (\int \varphi (x){\mathrm{d}} x\right )^2$
, and the other
$\tau _\lambda \leq \varphi + \lambda \varphi \star \tau _\lambda$
to find
Lower bound on
$\widehat \Pi _{\lambda _c}^{(2)}(0)$
Lemma 3.17.
Proof. We begin by identifying a suitable event for each
$u_0,u_1,x\in{\mathbb{R}^d}$
that is contained in
$\left \{\mathbf{0} \Longleftrightarrow u\:\mathrm{ in }\: \xi _0^{\mathbf{0},u_0}\right \}\cap E\!\left (u_0,u_1,{\mathscr{C}}_0,\xi ^{u_0,u_1}_1\right )\cap E\!\left (u_1,x,{\mathscr{C}}_1,\xi _2^{u_1,x}\right )$
. We choose the following event
$\mathcal{H}_1$
:
\begin{equation} \mathcal{H}_1 = \left \{\mathbf{0} \sim u_0\:\mathrm{ in }\: \xi ^{\mathbf{0},u_0}_0\right \}\cap \left \{u_0 \sim u_1\:\mathrm{ in }\: \xi ^{u_0,u_1}_1\right \}\cap \left \{u_1\notin \eta ^{u_0,u_1}_{1,\langle \mathbf{0} \rangle }\right \}\cap \left \{u_1 \sim x\:\mathrm{ in }\: \xi ^{u_1,x}_2\right \}\cap \left \{x\notin \eta ^{u_1,x}_{2,\langle u_0 \rangle }\right \} \end{equation}
and note that
$\lambda \widehat \Pi _\lambda ^{(2)}(0)\geq \lambda ^3\int \mathbb{P}_{\lambda }\!\left (\mathcal{H}_1\right ){\mathrm{d}} u_0{\mathrm{d}} u_1{\mathrm{d}} x$
.
This event is constructed so that all the intersecting events are independent, and the probability of each is easily calculated – once we recall that the probability that a singleton thins out a vertex is equal to the probability that an edge forms between the singleton and the vertex. Therefore
\begin{equation} \mathbb{P}_\lambda \!\left (\mathcal{H}_1\right ) \,:\!= \varphi (u_0)\varphi (u_1-u_0)\varphi (u_1)\varphi (x-u_1)\varphi (x-u_0). \end{equation}
Integrating
$\mathbb{P}_\lambda \!\left (\mathcal{H}_1\right )$
then gives a lower bound for
$\lambda _c\widehat \Pi _{\lambda _c}^{(2)}(0)$
. This lower bound is then
This gives us our desired lower bound for any
$\lambda \gt 0$
, and in particular
$\lambda =\lambda _c$
.
3.4 Bounds on later lace expansion coefficients
We first prove 3.10 and then 3.11.
Upper bound on
$\widehat \Pi _{\lambda _c}^{(3)}(0)$
. We are first dealing with the case
$n=3$
, which required a special treatment, and subsequently with the general case
$n\ge 4$
.
Lemma 3.18. Suppose Assumption A holds. Then as
$d\to \infty$
,
Proof. From (3.99) we have
\begin{equation} \lambda _c\widehat \Pi _{\lambda _c}^{(3)}(0) \leq\! \int\!\! \psi _0(w_0,u_0)\psi (w_1,u_1,w_0,u_0)\psi (w_2,u_2,w_1,u_1)\psi _3(x,w_2,u_2){\mathrm{d}} w_0{\mathrm{d}} u_0{\mathrm{d}} w_1{\mathrm{d}} u_1{\mathrm{d}} w_2{\mathrm{d}} u_2{\mathrm{d}} x. \end{equation}
Then as in the proof of Lemma3.16 we expand out the
$\psi _0$
,
$\psi$
, and
$\psi _3$
functions. Then for each integral we aim to identify a cycle of length
$6$
or longer that visits each vertex. For each factor of
$\tau _{\lambda _c}$
that is not part of this cycle we can then bound by
$1$
. For each factor of
$\tau _{\lambda _c}$
that is part of the cycle, we bound
$\tau _{\lambda _c}\leq \varphi + \lambda _c\varphi \star \tau _{\lambda _c}$
. Lemma2.4 then lets us bound each of these diagrams by
$\mathcal{O}\!\left (\varphi ^{\star 6}\left (\mathbf{0}\right )\right )$
.
The only integral that we cannot perform this strategy for corresponds to the integral
\begin{align} & \int \psi ^{(3)}_0(w_0,u_0)\psi ^{(4)}(w_1,u_1,w_0,u_0)\psi ^{(4)}(w_2,u_2,w_1,u_1)\psi ^{(2)}_3(x,w_2,u_2){\mathrm{d}} w_0{\mathrm{d}} u_0{\mathrm{d}} w_1{\mathrm{d}} u_1{\mathrm{d}} w_2{\mathrm{d}} u_2{\mathrm{d}} x \nonumber \\ &=\lambda _c^4\int \varphi (u_0)\tau _{\lambda _c}(u_1)\tau _{\lambda _c}(u_1-u_0)\tau _{\lambda _c}(u_2-u_0)\tau _{\lambda _c}(u_2-u_1)\tau _{\lambda _c}(x-u_1)\tau _{\lambda _c}(x-u_2){\mathrm{d}} u_0{\mathrm{d}} u_1{\mathrm{d}} u_2{\mathrm{d}} x. \end{align}
If we bound
$\tau _{\lambda _c}(u_2-u_1)\leq \varphi (u_2-u_1) + \lambda _c\varphi ^{\star 2}(u_2-u_1) + \lambda _c^2\varphi ^{\star 3}(u_2-u_1) + \lambda _c^3\left (\int \varphi (x){\mathrm{d}} x\right )^3$
, and the other
$\tau _{\lambda _c}\leq \varphi + \lambda _c\varphi \star \tau _{\lambda _c}$
we find that
Upper bound on
$\widehat \Pi _{\lambda _c}^{(n)}(0)$
for
$n\geq 4$
Lemma 3.19. Suppose Assumption A holds and
$n\geq 1$
. Then as
$d\to \infty$
,
\begin{equation} \lambda _c\widehat \Pi _{\lambda _c}^{(n)}(0) = \begin{cases} \mathcal{O}\!\left (\varphi ^{\star (n+2)}\left (\mathbf{0}\right )\right )&\colon n \text{ is even,}\\[4pt] \mathcal{O}\!\left (\varphi ^{\star (n+1)}\left (\mathbf{0}\right )\right )&\colon n \text{ is odd.}\\ \end{cases} \end{equation}
Proof. We begin this proof by using Proposition3.15 to get an upper bound for
$\lambda _c\widehat \Pi _{\lambda _c}^{(n)}(0)$
in terms of a sum of integrals of
$\psi ^{(j_0)}_0$
,
$\psi ^{(j)}$
, and
$\psi ^{(j_n)}_n$
. Our strategy to bound each of these diagrams is to identify a loop of length at least
$n+2$
around each of these diagrams.
For each
$\psi$
function we provide an upper bound in a
$\overline{\psi }$
function, so that when they are applied to our integral bounds we get terms of the form
$\lambda ^{m-1}\tau _\lambda ^{\star m}\left (\mathbf{0}\right )$
for some
$m\geq n+1$
. We have
\begin{align*} \psi _0^{(1)}(w,u) &\leq \overline{\psi }_0^{(1)}(w,u) \,:\!= \lambda ^2\tau _\lambda (u)\tau _\lambda (w),\\[3pt] \psi _0^{(2)}(w,u) &\leq \overline{\psi }_0^{(2)}(w,u) \,:\!= \lambda ^2 \delta _{w,\mathbf{0}}\int \tau _\lambda (u-t)\tau _\lambda (t){\mathrm{d}} t,\\[3pt] \psi _0^{(3)}(w,u) & \leq \overline{\psi }_0^{(3)}(w,u) \,:\!= \lambda \tau _\lambda (u) \delta _{w,\mathbf{0}},\\[3pt] \psi ^{(1)}(w,u,r,s) &\leq \overline{\psi }^{(1)}(w,u,r,s) \,:\!= \lambda ^4\int \tau _\lambda ^\circ (t-s)\tau _\lambda (w-t){\mathrm{d}} t \int \tau _\lambda (u-z)\tau _\lambda (z-r){\mathrm{d}} z, \\[3pt] \psi ^{(2)}(w,u,r,s) &\leq \overline{\psi }^{(2)}(w,u,r,s) \,:\!= \lambda ^4\tau _\lambda ^\circ (w-s)\int \tau _\lambda (z-r)\tau _\lambda (t-z)\tau _\lambda (u-t){\mathrm{d}} z{\mathrm{d}} t \\[3pt] &\hspace{5cm}+ \lambda ^3\tau _\lambda ^\circ (w-s)\int \tau _\lambda (z-r)\tau _\lambda (u-z){\mathrm{d}} z, \\[3pt] \psi ^{(3)}(w,u,r,s) &\leq \overline{\psi }^{(3)}(w,u,r,s) \,:\!= \lambda ^2\tau _\lambda (w-s)\tau _\lambda (u-r),\\[3pt] \psi ^{(4)}(w,u,r,s) &\leq \overline{\psi }^{(4)}(w,u,r,s) \,:\!= \lambda \delta _{w,s}\tau _\lambda (u-r),\\[3pt] \psi _n^{(1)} (x,r,s) &\leq \overline{\psi }_n^{(1)}(w,r,s) \,:\!= \lambda ^3\int \tau _\lambda ^\circ (t-s)\tau _\lambda (z-r)\tau _\lambda (z-x)\tau _\lambda (x-t){\mathrm{d}} z{\mathrm{d}} t,\\[3pt] \psi _n^{(2)}(x,r,s) &= \overline{\psi }_n^{(2)}(x,r,s) \,:\!= \lambda \tau _\lambda (x-s)\tau _\lambda (x-r). \end{align*}
We can also define
$\overline{\psi }_0$
,
$\overline{\psi }$
, and
$\overline{\psi }_n$
analogously to how we defined
$\psi _0$
,
$\psi$
, and
$\psi _n$
. These functions are represented as diagrams in Figure3.
Diagrams of the
$\overline{\psi }_0$
,
$\overline{\psi }$
, and
$\overline{\psi }_n$
functions, which we use to bound the
$\psi _0$
,
$\psi$
, and
$\psi _n$
functions.
First off, we leave
$\psi _n^{(2)}$
alone. For
$\psi _0^{(3)}$
we bound
$\varphi \leq \tau _\lambda$
. For most of the others, the bound is achieved only by bounding
$\tau _\lambda ^{(\geq 2)}\leq \tau _\lambda$
and
$\tau _\lambda \leq 1$
in the appropriate places. The bound for
$\psi ^{(2)}$
deserves a little more explanation. Here we first expand
$\tau _\lambda ^\circ (t-w) = \tau _\lambda (t-w) + \lambda ^{-1}\delta _{t,w}$
to get two expressions. We then bound
$\tau _\lambda \leq 1$
as for the others, but in different ways for each of the two expressions.
We therefore find that
$\overline{\psi }_0$
can contribute one or two factors of
$\tau _\lambda$
,
$\overline{\psi }$
can contribute one, two, three, or four factors of
$\tau _\lambda$
, and
$\overline{\psi }_n$
can contribute two, three, or four factors of
$\tau _\lambda$
. Our bound is therefore a sum of terms of
$\tau _\lambda ^{\star m}\left (\mathbf{0}\right )$
where
$m$
is at least
$1+1\times (n-1)+2 = n+2$
and at most
$2 + 4\times (n-1)+4 = 4n+2$
. Therefore
\begin{equation} \lambda _c\widehat \Pi _{\lambda _c}^{(n)}\left (0\right ) \leq \mathcal{O}\!\left (\sum ^{4n+2}_{m=n+2}\tau _\lambda ^{\star m}\left (\mathbf{0}\right )\right ). \end{equation}
For each factor of
$\tau _\lambda$
here we now bound
$\tau _\lambda \leq \varphi + \lambda \varphi \star \tau _\lambda$
to get
\begin{equation} \lambda _c\widehat \Pi _{\lambda _c}^{(n)}(0) \leq \mathcal{O}\!\left (\sum _{m=n+2}^{4n+2}\sum _{j=0}^{4n+2}\varphi ^{\star m}\star \tau _\lambda ^{\star j}\left (\mathbf{0}\right )\right ). \end{equation}
If
$m$
is odd and
$j\geq 1$
then we bound
$\varphi ^{\star m}\star \tau _\lambda ^{\star j}\left (\mathbf{0}\right ) \leq \left (\int \varphi (x){\mathrm{d}} x\right )\varphi ^{\star (m-1)}\star \tau _\lambda ^{\star j}\left (\mathbf{0}\right )$
. Then Lemma2.4 gives us
\begin{equation} \lambda _c\widehat \Pi _{\lambda _c}^{(n)}(0) \leq \begin{cases} \mathcal{O}\!\left (\sum _{m=n+2}^{4n+2}\varphi ^{\star m}\left (\mathbf{0}\right )\right ) &\colon n\text{ is even,}\\[4pt] \mathcal{O}\!\left (\sum _{m=n+1}^{4n+2}\varphi ^{\star m}\left (\mathbf{0}\right )\right ) &\colon n\text{ is odd.} \end{cases} \end{equation}
If
$n$
is even we bound
$\varphi ^{\star m}\left (\mathbf{0}\right )\leq \left (\int \varphi (x){\mathrm{d}} x\right )^{m-n-2}\varphi ^{\star (n+2)}\left (\mathbf{0}\right )$
to get our result, and if
$n$
is odd we bound
$\varphi ^{\star m}\left (\mathbf{0}\right )\leq \left (\int \varphi (x){\mathrm{d}} x\right )^{m-n-1}\varphi ^{\star (n+1)}\left (\mathbf{0}\right )$
to get our result.
Acknowledgements
This work is supported by Deutsche Forschungsgemeinschaft (project number 443880457) through priority programme ‘Random Geometric Systems’ (SPP 2265). The authors thank Sabine Jansen and Günter Last for their inspiring discussions.
A. Calculations for Specific Models
We now provide details for the specific percolation models in Section 1.4. To this end, we need to show that each of the four models satisfies Assumptions A and B and find the specific values of the integrals of
$\varphi$
appearing in (1.15).
A.1 Hyper-sphere calculations
Recall that for radius
$R\gt 0$
, the hyper-sphere RCM is defined by having
\begin{equation} \varphi (x) ={\unicode{x1D7D9}}_{\left \{\lvert{x}\rvert \lt R\right \}}. \end{equation}
Throughout this section we choose a scaling of
$\mathbb{R}^d$
such that
$R=R(d)$
is the radius of the unit
$d$
-volume ball. Therefore
$R(d)=\pi ^{-\frac{1}{2}}\Gamma \!\left (\frac{d}{2}+1\right )^{\frac{1}{d}} = \sqrt{\frac{d}{2\pi \text{e}}}\left (1+o\!\left (1\right )\right )$
(by an application of Stirling’s formula).
Lemma A.1. The hyper-sphere RCM satisfies Assumption A .
Proof. It is proven in [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Proposition 1.1] that the hyper-sphere RCM satisfies Assumption A with
$g(d) = \varrho ^d$
for some
$\varrho \in \left (0,1\right )$
.
Lemma A.2. The hyper-sphere RCM satisfies Assumption B .
Proof. In order to prove that (B.1) holds, we need to get a lower bound on
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
. We begin using the Fourier inverse formula to get
\begin{equation} \varphi ^{\star 6}\left (\mathbf{0}\right ) = \int _{{\mathbb{R}^d}}\widehat \varphi (k)^6\frac{{\mathrm{d}} k}{\left (2\pi \right )^d}. \end{equation}
Since
$\varphi$
is symmetric,
$\widehat \varphi (k)$
is real, and therefore we can get a lower bound on
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
by getting a lower bound on
$\widehat \varphi (k)^6$
.
From [[Reference GrafakosGra08], Appendix B.5], we can find that
\begin{equation} \widehat \varphi (k) = \left (\frac{2\pi R(d)}{\lvert{k}\rvert }\right )^{\frac{d}{2}} J_{\frac{d}{2}}\left (\lvert{k}\rvert R(d)\right )\!, \end{equation}
where
$J_{\frac{d}{2}}$
is the Bessel function of the first kind of order
$\frac{d}{2}$
, and
$R(d)$
is the radius of the unit volume ball in
$d$
dimensions. In Figure 4 we highlight three important values of
$\lvert{k}\rvert$
in the shape of
$\widehat \varphi (k)$
. The Bessel function
$J_{\frac{d}{2}}$
achieves its global maximum (in absolute value) at its first non-zero stationary point,
$j'_{\frac{d}{2},1}$
. From [[Reference Abramowitz and StegunAS64], p.371], we have
$j'_{\frac{d}{2},1} = \frac{d}{2} + \gamma _1\left (\frac{d}{2}\right )^{\frac{1}{3}} + \mathcal{O}\!\left (d^{-\frac{1}{3}}\right )$
for a given
$\gamma _1 \approx 0.81$
, and
$J_{\frac{d}{2}}\left (j'_{\frac{d}{2},1}\right ) = \Gamma _1 d^{-\frac{1}{3}} + \mathcal{O}\left (d^{-1}\right )$
, where
$\Gamma _1\approx 0.54$
. Then
$J_{\frac{d}{2}}$
has its first zero at
$j_{\frac{d}{2},1}\gt j'_{\frac{d}{2},1}$
, where
$j_{\frac{d}{2},1} = \frac{d}{2} + \gamma _2\!\left (\frac{d}{2}\right )^{\frac{1}{3}} + \mathcal{O}\!\left (d^{-\frac{1}{3}}\right )$
and
$\gamma _2\approx 1.86$
(again, see [Reference Abramowitz and StegunAS64]). From differential inequalities relating Bessel functions (see [Reference GrafakosGra08]), we have
\begin{equation} \frac{{\mathrm{d}} }{{\mathrm{d}} \lvert{k}\rvert } \widehat \varphi (k) = -R(d)\left (\frac{2\pi R(d)}{\lvert{k}\rvert }\right )^{\frac{d}{2}}J_{\frac{d}{2}+1}\left (\lvert{k}\rvert R(d)\right ). \end{equation}
Therefore
$\widehat \varphi (k)$
is decreasing in
$\lvert{k}\rvert$
until
$\lvert{k}\rvert R(d) = j_{\frac{d}{2}+1,1} = \frac{d}{2} + \gamma _2\!\left (\frac{d}{2}\right )^{\frac{1}{3}} + 1 +\mathcal{O}\!\left (d^{-\frac{1}{3}}\right )$
. In particular,
$j_{\frac{d}{2}+1,1}\gt j_{\frac{d}{2},1}$
.
Sketch of
$\widehat \varphi (k)$
against
$\vert{k}\vert$
. It approaches its maximum quadratically as
$\vert{k}\vert \to 0$
. The first local maximum of
$J_{\frac{d}{2}}$
occurs at
$j'_{\frac{d}{2},1}\sim \frac{d}{2}+\gamma _1(\frac{d}{2})^{\frac{1}{3}}$
. The first zero of
$\widehat \varphi (k)$
occurs at
$\vert{k}\vert R(d) = j_{\frac{d}{2},1}\sim \frac{d}{2}+\gamma _2(\frac{d}{2})^{\frac{1}{3}}$
where
$\gamma _2\gt \gamma _1$
. Furthermore,
$\widehat \varphi (k)$
is strictly decreasing until
$\vert{k}\vert R(d) = j_{\frac{d}{2}+1,1}\sim \frac{d}{2}+\gamma _2(\frac{d}{2})^{\frac{1}{3}} + 1$
.
The significance of these points is that they allow us to bound
\begin{equation} \lvert{\widehat \varphi (k)}\rvert \geq \widehat \varphi \!\left ({j'_{\frac{d}{2},1}}/{R(d)}\right ){\unicode{x1D7D9}}_{\left \{\lvert{k}\rvert \leq{j'_{\frac{d}{2},1}}/{R(d)}\right \}}. \end{equation}
Since
$R(d)$
is the radius of the unit volume ball in
$d$
dimensions,
\begin{equation} \int _{{\mathbb{R}^d}}{\unicode{x1D7D9}}_{\left \{\lvert{k}\rvert \leq{j'_{\frac{d}{2},1}}/{R(d)}\right \}}\frac{{\mathrm{d}} k}{\left (2\pi \right )^d} = \left (\frac{j'_{\frac{d}{2},1}}{2\pi R(d)^2}\right )^d. \end{equation}
Therefore we can arrive at
\begin{equation} \varphi ^{\star 6}\left (\mathbf{0}\right ) \geq \left (\frac{2\pi R(d)^2}{j'_{\frac{d}{2},1}}\right )^{2d} J_{\frac{d}{2}}\left (j'_{\frac{d}{2},1}\right )^6 = \Gamma _1^6 \frac{1}{d^2}\left (\frac{2}{\text{e}}+o\!\left (1\right )\right )^{2d}\left (1+ o(1)\right ). \end{equation}
Here we have used the leading order asymptotics of
$R(d)$
,
$j'_{\frac{d}{2},1}$
, and
$J_{\frac{d}{2}}\left (j'_{\frac{d}{2},1}\right )$
we described above. From this lower bound, we know that
$\rho$
will satisfy the bound in (B.1) if
$\rho \lt 4{\text{e}}^{-2}$
.
From the above argument we have an exponential lower bound on
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
and therefore a linear lower bound on
$h(d)$
. It is proven in [[Reference Heydenreich, van der Hofstad, Last and MatzkeHHLM22], Proposition 1.1a] that
$g(d)=\varrho ^{d}$
for some
$\varrho \in \left (0,1\right )$
, and therefore
$\beta (d) = \varrho ^{\frac{d}{4}}$
. We can then bound
$N(d)$
to show (B.2) holds.
Lemma A.3.
For
$n\geq 3$
,
\begin{equation} \varphi ^{\star n}\left (\mathbf{0}\right ) = d2^{d\!\left (\frac{n}{2}-1\right )}\Gamma \!\left (\frac{d}{2}+1\right )^{n-2}\int ^\infty _0 x^{-1-d\!\left (\frac{n}{2}-1\right )}\left (J_{\frac{d}{2}}(x)\right )^n{\mathrm{d}} x. \end{equation}
In particular,
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right ) &= \frac{d\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\int ^1_0x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right ){\mathrm{d}} x\nonumber\\[5pt]& = \frac{3}{2}\frac{\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}B\!\left (\frac{3}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right ), \end{align}
\begin{align} \varphi ^{\star 4}\left (\mathbf{0}\right ) &= \frac{d\Gamma \!\left (\frac{d}{2}+1\right )^2}{\Gamma \!\left (\frac{1}{2}\right )^2\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^2}\int ^2_0x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^2{\mathrm{d}} x, \end{align}
\begin{align} \varphi ^{\star 5}\left (\mathbf{0}\right ) &= \frac{d2^d\Gamma \!\left (\frac{d}{2}+1\right )^3}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\int ^2_0x^{\frac{d}{2}}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )\left (\int ^\infty _0 k^{-d}\left (J_{\frac{d}{2}}(k)\right )^3J_{\frac{d}{2}-1}(kx){\mathrm{d}} k\right ){\mathrm{d}} x, \end{align}
\begin{align} \varphi ^{\star 6}\left (\mathbf{0}\right ) &= \frac{d2^{\frac{3d}{2}}\Gamma \!\left (\frac{d}{2}+1\right )^4}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\int ^2_0x^{\frac{d}{2}}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )\left (\int ^\infty _0 k^{-\frac{3d}{2}}\left (J_{\frac{d}{2}}(k)\right )^4J_{\frac{d}{2}-1}(kx){\mathrm{d}} k\right ){\mathrm{d}} x. \end{align}
Furthermore,
\begin{align} \varphi ^{\star 1\star 2 \cdot 2}\left (\mathbf{0}\right ) &= \frac{d\Gamma \!\left (\frac{d}{2}+1\right )^2}{\Gamma \!\left (\frac{1}{2}\right )^2\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^2}\int _0^1 x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^2{\mathrm{d}} x, \end{align}
\begin{align} \varphi ^{\star 2\star 2 \cdot 2}\left (\mathbf{0}\right ) &= \frac{d\Gamma \!\left (\frac{d}{2}+1\right )^3}{\Gamma \!\left (\frac{1}{2}\right )^3\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^3}\int _0^2 x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^3{\mathrm{d}} x, \end{align}
\begin{align} \varphi ^{\star 1\star 2 \cdot 3}\left (\mathbf{0}\right ) &= \frac{d2^d\Gamma \!\left (\frac{d}{2}+1\right )^3}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\int ^1_0x^{\frac{d}{2}}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )\left (\int ^\infty _0 k^{-d}\left (J_{\frac{d}{2}}(k)\right )^3J_{\frac{d}{2}-1}(kx){\mathrm{d}} k\right ){\mathrm{d}} x. \end{align}
Proof. Let
$R=R(d)$
denote the radius of the unit volume
$d$
-dimensional Euclidean ball, i.e.
$R(d)=\frac{1}{\sqrt{\pi }}\Gamma \!\left (\frac{d}{2}+1\right )^{\frac{1}{d}}$
. In particular, note the relation
\begin{equation} 1 = \mathfrak{S}_{d-1}\int ^R_0r^{d-1}{\mathrm{d}} r = \frac{\mathfrak{S}_{d-1}}{d}R^d, \end{equation}
where
$\mathfrak{S}_{d-1}=\frac{d\pi ^{\frac{d}{2}}}{\Gamma \!\left (\frac{d}{2}+1\right )}$
is the surface area of the unit radius
$d$
-dimensional Euclidean ball.
The general formula (A.8) follows from a Fourier decomposition. By the Fourier inversion formula,
\begin{equation} \varphi ^{\star n}(x) = \frac{1}{\left (2\pi \right )^d}\int \widehat \varphi (k)^n{\mathrm{d}} k. \end{equation}
Recall the expression (A.3) for the Fourier transform
$\widehat \varphi (k)$
. Then
\begin{align} \varphi ^{\star n}\left (\mathbf{0}\right ) &= \frac{1}{\left (2\pi \right )^d}\left (2\pi \right )^{\frac{d}{2}n}R^{\frac{d}{2}n}\mathfrak{S}_{d-1}\int ^\infty _0 k^{d-1-\frac{d}{2}n}\left (J_{\frac{d}{2}}\left (Rk\right )\right )^n{\mathrm{d}} k\nonumber \\[5pt] &=\left (2\pi \right )^{d\!\left (\frac{n}{2}-1\right )}R^d\mathfrak{S}_{d-1}\int ^\infty _0x^{-1-d\!\left (\frac{n}{2}-1\right )}\left (J_{\frac{d}{2}}\left (x\right )\right )^n{\mathrm{d}} x. \end{align}
Then observing that
$R^d\mathfrak{S}_{d-1}=d$
produces the result.
In the cases
$n=3,4$
, a more geometric approach may be taken. First note that
$\varphi ^{\star 2}(x)$
can be interpreted as the
$d$
-volume of the intersection of a hyper-sphere of radius
$R$
at the origin with a hyper-sphere of radius
$R$
at the position
$x$
. An expression for this volume is given by [Reference LiLi11] using incomplete Beta functions:
\begin{equation} \varphi ^{\star 2}(x) = \frac{\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}B\!\left (1-\frac{\lvert{x}\rvert ^2}{4R^2};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right ), \qquad \text{for }\lvert{x}\rvert \leq 2R. \end{equation}
Clearly
$\varphi ^{\star 2}(x)=0$
for
$\lvert{x}\rvert \gt 2R$
. It then follows that
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right ) &= \int \varphi (x)\varphi ^{\star 2}(x){\mathrm{d}} x = \frac{\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\mathfrak{S}_{d-1}\int ^R_0 r^{d-1}B\!\left (1-\frac{r^2}{4R^2};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right ){\mathrm{d}} r\nonumber \\ &= \frac{\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\mathfrak{S}_{d-1}R^d\int ^1_0 x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right ){\mathrm{d}} x. \end{align}
Again, noting that
$R^d\mathfrak{S}_{d-1} = d$
produces the required first equality in (A.9). It was noted in [Reference TorquatoTor12] that for the hyper-sphere model we have
$\varphi ^{\star 3}\left (\mathbf{0}\right )=\frac{3}{2}\varphi ^{\star 2}(\tilde{x})$
, where
$\lvert{\tilde{x}}\rvert =R$
. This can be proven by writing out the incomplete Beta function as an integral to get a double integral, partitioning the domain appropriately, and using a suitable trigonometric substitution on each part of the domain. We omit the details here. This relation allows us to get the second equality in (A.9).
For the specific form of
$\varphi ^{\star 4}\left (\mathbf{0}\right )$
, we do a similar calculation to that above:
\begin{align} \varphi ^{\star 4}\left (\mathbf{0}\right ) &= \int \varphi ^{\star 2}(x)^2{\mathrm{d}} x = \frac{\Gamma \!\left (\frac{d}{2}+1\right )^2}{\Gamma \!\left (\frac{1}{2}\right )^2\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^2}\mathfrak{S}_{d-1}\int ^{2R}_0 r^{d-1}B\!\left (1-\frac{r^2}{4R^2};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^2{\mathrm{d}} r\nonumber \\ &= \frac{\Gamma \!\left (\frac{d}{2}+1\right )^2}{\Gamma \!\left (\frac{1}{2}\right )^2\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^2}\mathfrak{S}_{d-1}R^d\int ^2_0 x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^2{\mathrm{d}} x. \end{align}
Using
$R^d\mathfrak{S}_{d-1} = d$
gives the result.
For
$\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 2\star 2\cdot 1}\left (\mathbf{0}\right )$
this approach also works. We find
\begin{align} \varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) = \int \varphi (x)\varphi ^{\star 2 }(x)^2{\mathrm{d}} x &= \frac{\Gamma \!\left (\frac{d}{2}+1\right )^2}{\Gamma \!\left (\frac{1}{2}\right )^2\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^2}\mathfrak{S}_{d-1}\int ^{R}_0 r^{d-1}B\!\left (1-\frac{r^2}{4R^2};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^2{\mathrm{d}} r\nonumber \\ &= \frac{\Gamma \!\left (\frac{d}{2}+1\right )^2}{\Gamma \!\left (\frac{1}{2}\right )^2\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^2}\mathfrak{S}_{d-1}R^d\int ^1_0 x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^2{\mathrm{d}} x, \end{align}
\begin{align} \varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) = \int \varphi ^{\star 2 }(x)^3{\mathrm{d}} x &= \frac{\Gamma \!\left (\frac{d}{2}+1\right )^3}{\Gamma \!\left (\frac{1}{2}\right )^3\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^3}\mathfrak{S}_{d-1}\int ^{2R}_0 r^{d-1}B\!\left (1-\frac{r^2}{4R^2};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^3{\mathrm{d}} r\nonumber \\ &= \frac{\Gamma \!\left (\frac{d}{2}+1\right )^3}{\Gamma \!\left (\frac{1}{2}\right )^3\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )^3}\mathfrak{S}_{d-1}R^d\int ^2_0 x^{d-1}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )^3{\mathrm{d}} x. \end{align}
As before, using
$R^d\mathfrak{S}_{d-1} = d$
gives the result.
Evaluating
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
, and
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
is more challenging than the above expressions because we don’t have such a nice expression for
$\varphi ^{\star 3}(x)$
as we did for
$\varphi ^{\star 2}(x)$
. We can nevertheless use Fourier transforms to get an expression. Using the well-known expression
\begin{equation} J_{\nu }(x) = \frac{x^\nu }{2^\nu \Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\nu +\frac{1}{2}\right )}\int ^\pi _0{\text{e}}^{ix\cos \theta }\left (\sin \theta \right )^{2\nu }{\mathrm{d}} \theta, \qquad \mathrm{Re}\,\nu \geq -\frac{1}{2} \end{equation}
from [[Reference Abramowitz and StegunAS64], p.360, Eqn.(9.1.20)], we can write
\begin{align} \varphi ^{\star 3}(x) &= \frac{\mathfrak{S}_{d-2}}{\left (2\pi \right )^d}\int ^\infty _{0} k^{d-1}\widehat \varphi (k)^3\left (\int ^\pi _0{\text{e}}^{ik\lvert{x}\rvert \cos \theta }\left (\sin \theta \right )^{d-2}{\mathrm{d}} \theta \right ){\mathrm{d}} k \nonumber \\& = \left (2\pi \right )^d R^{\frac{3}{2}d}\lvert{x}\rvert ^{1-\frac{d}{2}}\int ^\infty _0k^{-d}\left (J_{\frac{d}{2}}\left (kR\right )\right )^3 J_{\frac{d}{2}-1}\left (k\lvert{x}\rvert \right ){\mathrm{d}} k. \end{align}
Using this expression with the expression for
$\varphi ^{\star 2}(x)$
used previously then gives the result:
\begin{align} \varphi ^{\star 1 \star 2\cdot 3}\left (\mathbf{0}\right ) &= \int \varphi (x)\varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x \nonumber \\&= \frac{\Gamma \!\left (\frac{d}{2}+1\right )}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\left (2\pi \right )^d R^{\frac{3}{2}d}\mathfrak{S}_{d-1}\int ^R_0 r^{d-1}r^{1-\frac{d}{2}}B\!\left (1-\frac{r^2}{4R^2};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )\nonumber \\ & \quad \times \left (\int ^\infty _0k^{-d}\left (J_{\frac{d}{2}}\left (kR\right )\right )^3 J_{\frac{d}{2}-1}\left (kr\right ){\mathrm{d}} k\right ){\mathrm{d}} r\nonumber \\ &= d2^d\frac{\Gamma \!\left (\frac{d}{2}+1\right )^3}{\Gamma \!\left (\frac{1}{2}\right )\Gamma \!\left (\frac{d}{2}+\frac{1}{2}\right )}\!\int ^1_0\! x^{\frac{d}{2}}B\!\left (1-\frac{x^2}{4};\,\frac{d}{2}+\frac{1}{2},\frac{1}{2}\right )\!\left (\int ^\infty _0\! k^{-d}\left (J_{\frac{d}{2}}(k)\right )^3\!J_{\frac{d}{2}-1}(kx){\mathrm{d}} k\right )\!{\mathrm{d}} x, \end{align}
where we explicitly use
$R^d=\pi ^{-\frac{d}{2}}\Gamma \!\left (\frac{d}{2}+1\right )$
. Writing
$\varphi ^{\star 5}\left (\mathbf{0}\right ) = \int \varphi ^{\star 2}(x)\varphi ^{\star 3}(x){\mathrm{d}} x$
and using the same strategy gives its result.
Getting the expression for
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
requires an expression for
$\varphi ^{\star 4}\left (x\right )$
. Using the same strategy as for
$\varphi ^{\star 3}\left (x\right )$
above, we get
\begin{align} \varphi ^{\star 4}(x) &= \frac{\mathfrak{S}_{d-2}}{\left (2\pi \right )^d}\int ^\infty _{0} k^{d-1}\widehat \varphi (k)^4\left (\int ^\pi _0{\text{e}}^{ik\lvert{x}\rvert \cos \theta }\left (\sin \theta \right )^{d-2}{\mathrm{d}} \theta \right ){\mathrm{d}} k \nonumber \\[6pt] & = \left (2\pi \right )^{\frac{3}{2}d} R^{2d}\lvert{x}\rvert ^{1-\frac{d}{2}}\int ^\infty _0k^{-\frac{3}{2}d}\left (J_{\frac{d}{2}}\left (kR\right )\right )^4 J_{\frac{d}{2}-1}\left (k\lvert{x}\rvert \right ){\mathrm{d}} k. \end{align}
Using this with
$\varphi ^{\star 6}\left (\mathbf{0}\right ) = \int \varphi ^{\star 2}(x)\varphi ^{\star 4}(x){\mathrm{d}} x$
then gives the required expression.
We now turn towards asymptotic values of the terms appearing in LemmaA.3. For the terms
$\varphi ^{\star 3}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 4}\left (\mathbf{0}\right )$
, the asymptotics have already been worked out.
Lemma A.4. For the hyper-sphere RCM,
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right ) &\sim \left (\frac{27}{2\pi d}\right )^{\frac{1}{2}}\left (\frac{3}{4}\right )^{\frac{d}{2}}, \end{align}
\begin{align} \varphi ^{\star 4}\left (\mathbf{0}\right ) &\sim \left (\frac{32}{3\pi d}\right )^{\frac{1}{2}}\left (\frac{16}{27}\right )^{\frac{d}{2}}. \end{align}
Proof. These follow from the calculations in [Reference Luban and BaramLB82, Reference JoslinJos82].
Remark A.5. These asymptotics naturally also give the asymptotics of
$\varphi ^{\star 3}\left (\mathbf{0}\right )^2$
,
$\varphi ^{\star 3}\left (\mathbf{0}\right )^3$
, and
$\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )$
. For
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right )$
, and
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
we don’t have any rigorous description of their asymptotic behaviour. Nevertheless we can use our expressions from Lemma A.3 and numerical integration to calculate their values for a range of dimensions. Figure
5
presents the results of these calculations. Here we used MATLAB to plot
$\frac{1}{d}\log \left (\cdot \right )$
(where
$\log$
is the natural logarithm) for each of our diagrams against the dimension
$d$
. We chose this function of the diagrams because if a diagram was of the form
$A(d)\varrho ^d$
for some constant
$\varrho \gt 0$
and some slowly varying
$A(d)$
, then our plot should approach
$\log \varrho$
as
$d\to \infty$
. The data in Figure 5a are consistent with this behaviour (indeed we know it to be true for
$\varphi ^{\star 3}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 4}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )^2$
). We only plot the data up to
$d=50$
because the calculations of
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 1\star 2\star 3}\left (\mathbf{0}\right )$
fail for
$d\gt 54$
– we comment on this more later. The data in Figure 5b appear a little less definitive, but the authors argue these are still consistent with the hypothesised behaviour (we know it to be true for
$\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )^3$
). Note that the vertical scale is over a much narrower range than in Figure 5a, which gives the false impression that the plots are increasing with
$d$
faster than they indeed are. We are also further restricting the domain of
$d$
to
$d\leq 36$
. This is because the calculation of
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
fails for
$d\gt 36$
.
Plots of
$\frac{1}{d}\log (\cdot )$
for each of the diagrams for the hyper-sphere RCM. For comparison,
$\frac{1}{d}\log \varphi ^{\star 1\star 2\cdot 3}(\mathbf{0})$
is represented in both plots – it is the smallest of the larger diagrams and the largest of the smaller diagrams in the higher dimensions.
Remark A.6. We comment here on our choices of the range of dimensions
$d$
presented in the data in Figure 5. We found that the limiting factor in our ability to calculate the expressions in Lemma A.3 were the prefactors of powers of
$2$
and gamma functions. If we wanted to use (A.8) to calculate
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
for
$d=25$
we would have to deal with
$d2^{2d}\Gamma \left (\tfrac{d}{2}+1\right )^4 \approx 2.41\times 10^{53}$
, while
$\varphi ^{\star 6}\left (\mathbf{0}\right )\approx 5.34\times 10^{-6}$
. Fortunately MATLAB has the function betainc(
$x$
,
$a$
,
$b$
), which calculates the (normalised) incomplete beta function
\begin{equation} \frac{\Gamma \!\left (a+b\right )}{\Gamma (a)\Gamma (b)}\int ^x_0 t^{a-1}\left (1-t\right )^{b-1}{\mathrm{d}} t = \frac{\Gamma \!\left (a+b\right )}{\Gamma (a)\Gamma (b)}B\!\left (x;\,a,b\right ). \end{equation}
This betainc function is more efficient at dealing with the different sizes of the prefactor and integral than our naïve attempts, and this is why we put the extra effort in Lemma A.3 to include factors of
$B\!\left (x;\,a,b\right )$
. In particular, this makes
$\varphi ^{\star 3}\left (\mathbf{0}\right )$
very easy to calculate: MATLAB got to over
$d=5000$
before it produced an error (for
$d=5000$
,
$\varphi ^{\star 3}\left (\mathbf{0}\right )\approx 1.32\times 10^{-314}$
). We can also calculate our expressions for
$\varphi ^{\star 4}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )$
, and
$\varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right )$
over dimension
$d=1000$
. Unfortunately the use of betainc does not deal with the whole prefactor for
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
, and
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
, and this affects the dimension we can run up to. For
$\varphi ^{\star 5}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
we can run up to
$d=54$
(where they are
$\approx 9.06\times 10^{-10}$
and
$\approx 2.29\times 10^{-11}$
respectively). For
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
we can only run to
$d=36$
, where find
$\varphi ^{\star 6}\left (\mathbf{0}\right )\approx 3.58\times 10^{-8}$
.
$\diamond$
Remark A.7. Upon inspecting Figure 5a, the plots of
$\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )^2$
appear very close together. The plots of
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )$
in Figure 5b also appear to be tracking closely together. In Figure 6 we plot how the ratio of these similar terms vary with dimension.
Plots of the ratio of diagrams of similar sizes for the hyper-sphere RCM.
Since we are able to evaluate
$\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )^2$
to relatively high dimensions, we are able to plot their ratio all the way up to
$d=2500$
in Figure 6a. From this plot it is very tempting to suggest that their ratio is approaching a finite and positive limit. In fact, since
${\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )}/{\varphi ^{\star 3}\left (\mathbf{0}\right )^2}\approx 1.329$
at
$d=2500$
, it is tempting to suggest that the ratio approaches
$\tfrac{4}{3}$
as
$d\to \infty$
. Since we rigorously have the asymptotics of
$\varphi ^{\star 3}\left (\mathbf{0}\right )$
, this would imply the asymptotics of
$\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )$
.
We are not able to evaluate
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
to similarly high dimensions – we can only reach
$d=54$
. Nevertheless, the slope of the plot in Figure 6b is shallowing and it is tempting to suggest that the ratio
${\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right )}/{\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )}$
approaches a finite and positive limit. While we don’t conjecture a value for the limit here, the existence of such a limit would allow us to find the asymptotic scale of
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
.
Plots relating
$\varphi ^{\star 6}(\mathbf{0})$
,
$\varphi ^{\star 2\star 2\cdot 2}(\mathbf{0})$
, and
$\varphi ^{\star 3}(\mathbf{0})^3$
for the hyper-sphere RCM. These suggest the possibility that these three terms only differ by a polynomial factor in
$d$
.
If we look at the ratio of the other pairs of diagrams it is usually very clear that one is far larger than the other, with the ratio apparently growing at an exponential rate. The only exceptions are the trio of
$\varphi ^{\star 6}\left (\mathbf{0}\right )$
,
$\varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right )$
, and
$\varphi ^{\star 3}\left (\mathbf{0}\right )^3$
. While the ratios appear to be growing for each pair in this trio, the rate seems to be slowing. If
$\varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )^3$
were both decaying at the same exponential rate but had different polynomial corrections, then we would have
${\left (\log \varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right )-3\log \varphi ^{\star 3}\left (\mathbf{0}\right )\right )}/{\log d}$
approaching a non-zero limit as
$d\to \infty$
. In Figure 7 we plot this comparison for the two independent pairs in the trio, and it indeed seems plausible that the plots are approaching a non-zero limit. Nevertheless, these three terms look to be far smaller than the
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )$
terms, and so will both be negligible for our discussion.
$\diamond$
The observations made in RemarksA.5–A.7 and the plots in Figures 5 and 6 allow us to make the following conjecture. We use the notation
$f\gg g$
to indicate
$\frac{f(d)}{g(d)}\to \infty$
, and
$f\asymp g$
to indicate
$\frac{f(d)}{g(d)}$
and
$\frac{g(d)}{f(d)}$
are both bounded as
$d\to \infty$
.
Conjecture A.8. For the hyper-sphere RCM, as
$d\to \infty$
,
\begin{equation} \varphi ^{\star 3}\left (\mathbf{0}\right ) \gg \varphi ^{\star 4}\left (\mathbf{0}\right ) \gg \varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) \asymp \left (\varphi ^{\star 3}\left (\mathbf{0}\right )\right )^2 \gg \varphi ^{\star 5}\left (\mathbf{0}\right ) \gg \varphi ^{\star 1 \star 2\cdot 3}\left (\mathbf{0}\right ) \asymp \varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right ), \end{equation}
and
\begin{equation} \varphi ^{\star 6}\left (\mathbf{0}\right ) + \varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) + \left (\varphi ^{\star 3}\left (\mathbf{0}\right )\right )^3 = \mathcal{O}\!\left (\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )\right ). \end{equation}
Therefore
\begin{align} q_\varphi \lambda _c & = 1 + \frac{1}{q_\varphi ^2}\varphi ^{\star 3}\left (\mathbf{0}\right ) + \frac{3}{2}\frac{1}{q_\varphi ^3}\varphi ^{\star 4}\left (\mathbf{0}\right ) - \frac{5}{2}\frac{1}{q_\varphi ^3}\varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) + 2\frac{1}{q_\varphi ^4}\left (\varphi ^{\star 3}\left (\mathbf{0}\right )\right )^2 + 2\frac{1}{q_\varphi ^4}\varphi ^{\star 5}\left (\mathbf{0}\right )\nonumber\\&\quad + \mathcal{O}\!\left (\frac{1}{d}\left (\frac{2}{3}\right )^d\right ). \end{align}
Note that this would be a different order of terms than that we found for the hyper-cube RCM in Corollary1.7.
A.2 Hyper-cube calculations
Recall that for side length
$L\gt 0$
, the hyper-cubic RCM is defined by having
\begin{equation} \varphi (x) = \prod ^d_{j=1}{\unicode{x1D7D9}}_{\left \{\lvert{x_j}\rvert \leq L/2\right \}}, \end{equation}
where
$x=\left (x_1,\ldots, x_d\right )\in{\mathbb{R}^d}$
. Throughout this section we choose a scaling of
$\mathbb{R}^d$
such that
$L=1$
.
Lemma A.9. For the hyper-cube RCM with side length
$L=1$
,
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right ) &= \left (\frac{3}{4}\right )^d = \left (0.75\right )^d, \end{align}
\begin{align} \varphi ^{\star 4}\left (\mathbf{0}\right ) &= \left (\frac{2}{3}\right )^d \approx \left (0.66667\right )^d, \end{align}
\begin{align} \varphi ^{\star 5}\left (\mathbf{0}\right ) &= \left (\frac{115}{192}\right )^d \approx \left (0.59896\right )^d, \end{align}
\begin{align} \varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) &= \left (\frac{7}{12}\right )^d \approx \left (0.58333\right )^d, \end{align}
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right )^2 &= \left (\frac{9}{16}\right )^d =\left (0.5625\right )^d, \end{align}
\begin{align} \varphi ^{\star 6}\left (\mathbf{0}\right ) &= \left (\frac{11}{20}\right )^d = \left (0.55\right )^d, \end{align}
\begin{align} \varphi ^{\star 7}\left (\mathbf{0}\right ) & = \left (\frac{5887}{11520}\right )^d \approx \left (0.51102\right )^d, \end{align}
\begin{align} \varphi ^{\star 1 \star 2\cdot 3}\left (\mathbf{0}\right ) &= \left (\frac{49}{96}\right )^d \approx \left (0.51042\right )^d, \end{align}
\begin{align} \varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) &= \left (\frac{1}{2}\right )^d = \left (0.5\right )^d, \end{align}
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right ) &= \left (\frac{1}{2}\right )^d = \left (0.5\right )^d, \end{align}
\begin{align} \varphi ^{\star 8}\left (\mathbf{0}\right ) & = \left (\frac{151}{315}\right )^d \approx \left (0.47937\right )^d, \end{align}
\begin{align} \varphi ^{\star 3}\left (\mathbf{0}\right )^3 &= \left (\frac{27}{64}\right )^d \approx \left (0.42188\right )^d. \end{align}
Proof. First note that the hyper-cubic adjacency function factorises into the
$d$
dimensions:
\begin{equation} \varphi (x) = \prod ^d_{i=1}{\unicode{x1D7D9}}_{\left \{\lvert{x_i}\rvert \lt \frac{1}{2}\right \}}, \end{equation}
where
$x=\left (x_1,x_2,\ldots, x_d\right )$
. Therefore to find the desired expressions, we only need to evaluate them for dimension
$1$
, and then take the result to the power
$d$
to get the result for dimension
$d$
. Let us denote the
$1$
-dimensional adjacency function
$\varphi _1\colon \mathbb{R} \to \left [0,1\right ]$
,
\begin{equation} \varphi _1(x) = \begin{cases} 1&\colon \lvert{x}\rvert \lt \frac{1}{2}\\[5pt] 0&\colon \lvert{x}\rvert \geq \frac{1}{2}. \end{cases} \end{equation}
By direct calculation (these can be easily verified by Mathematica, for example), one finds
\begin{align} \varphi _1^{\star 2}(x) &= \begin{cases} 1-\lvert{x}\rvert &\colon \lvert{x}\rvert \lt 1\\ 0 &\colon \lvert{x}\rvert \geq 1, \end{cases} \end{align}
\begin{align} \varphi _1^{\star 3}(x) &=\begin{cases} \frac{1}{4}\left (3-4x^2\right ) &\colon \lvert{x}\rvert \lt \frac{1}{2}\\[5pt] \frac{1}{8}\left (3-2\lvert{x}\rvert \right )^2 &\colon \frac{1}{2}\leq \lvert{x}\rvert \lt \frac{3}{2}\\[5pt] 0 &\colon \lvert{x}\rvert \geq \frac{3}{2}, \end{cases} \end{align}
\begin{align} \varphi _1^{\star 4}(x) &=\begin{cases} \frac{1}{6}\left (4-6x^2+3\lvert{x}\rvert ^3\right ) &\colon \lvert{x}\rvert \lt 1\\[5pt] \frac{1}{6}\left (2-\lvert{x}\rvert \right )^3 &\colon 1\leq \lvert{x}\rvert \lt 2\\[5pt] 0 &\colon \lvert{x}\rvert \geq 2. \end{cases} \end{align}
In particular, this means
$\varphi _1^{\star 3}\left (\mathbf{0}\right ) =\frac{3}{4}$
and
$\varphi _1^{\star 4}\left (\mathbf{0}\right ) = \frac{2}{3}$
. Taking these to the power
$d$
returns the required results for
$\varphi ^{\star 3}\left (\mathbf{0}\right )$
and
$\varphi ^{\star 4}\left (\mathbf{0}\right )$
. These also give the results for
$\varphi ^{\star 3}\left (\mathbf{0}\right )^2$
,
$\varphi ^{\star 3}\left (\mathbf{0}\right )^3$
and
$\varphi ^{\star 3}\left (\mathbf{0}\right )\varphi ^{\star 4}\left (\mathbf{0}\right )$
.
Then let us observe and calculate
\begin{align} \varphi _1^{\star 5}\left (\mathbf{0}\right ) &= \int ^1_{-1}\varphi _1^{\star 2}(x)\varphi _1^{\star 3}(x){\mathrm{d}} x = \frac{115}{192} \approx 0.59896, \end{align}
\begin{align} \varphi _1^{\star 6}\left (\mathbf{0}\right )&= \int ^{\frac{3}{2}}_{-\frac{3}{2}}\varphi _1^{\star 3}(x)\varphi _1^{\star 3}(x){\mathrm{d}} x = \frac{11}{20} = 0.55, \end{align}
\begin{align} \varphi _1^{\star 7}\left (\mathbf{0}\right ) &= \int ^{\frac{3}{2}}_{-\frac{3}{2}}\varphi _1^{\star 3}(x)\varphi _1^{\star 4}(x){\mathrm{d}} x = \frac{5887}{11520}\approx 0.51102, \end{align}
\begin{align} \varphi _1^{\star 8}\left (\mathbf{0}\right ) &= \int ^2_{-2}\varphi _1^{\star 4}(x)\varphi _1^{\star 4}(x){\mathrm{d}} x = \frac{151}{315}\approx 0.47937. \end{align}
Similarly, we find
\begin{align} \varphi _1^{\star 1\star 2\cdot 2} \left (\mathbf{0}\right ) &= \int ^{\frac{1}{2}}_{-\frac{1}{2}}\varphi _1^{\star 2}(x)^2{\mathrm{d}} x = \frac{7}{12}\approx 0.58333 \end{align}
\begin{align} \varphi _1^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) &= \int ^1_{-1}\varphi _1^{\star 2}(x)^3{\mathrm{d}} x = \frac{1}{2} = 0.5 \end{align}
\begin{align} \varphi _1^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right ) &= \int ^{\frac{1}{2}}_{-\frac{1}{2}}\varphi _1^{\star 2}(x)\varphi _1^{\star 3}(x){\mathrm{d}} x = \frac{49}{96} \approx 0.51042. \end{align}
Finally taking these values to the
$d^{\mathrm{th}}$
power gives the required results.
Proof. For Assumption (A.1), recall that
\begin{equation} \varphi ^{\star 2}(x) = \prod _{i=1}^d \left (1-\lvert{x_i}\rvert \right ){\unicode{x1D7D9}}_{\left \{\lvert{x_i}\rvert \leq 1\right \}}, \end{equation}
where
$x=\left (x_1,x_2,\ldots, x_d\right )$
. In conjunction with LemmaA.9, we see that Assumption (A.1) is satisfied with
$g(d)=\left (\frac{3}{4}\right )^d$
.
For (A.2) we note that
\begin{equation} \widehat \varphi (k) = \prod ^d_{i=1}\left (\frac{2}{k_i}\sin \frac{k_i}{2}\right ), \end{equation}
where
$k=\left (k_1,k_2,\ldots, k_d\right )$
. Since
$\sin x \leq x - \frac{1}{6}x^3 + \frac{1}{120}x^5$
for all
$x\in{\mathbb{R}}$
,
\begin{equation} \widehat \varphi (k) \leq \prod ^d_{i=1}\left (1-\frac{1}{24}k_i^2 + \frac{1}{1920}k_i^4\right ). \end{equation}
Therefore for
$\max _i\lvert{k_i}\rvert \leq 3$
we have
\begin{align} \widehat \varphi (k) &\leq \prod ^d_{i=1}\left (1-\frac{71}{1920}k_i^2\right )\nonumber \\ &= 1 - \frac{71}{1920}\lvert{k}\rvert ^2 + \left (\frac{71}{1920}\right )^2\sum _{\substack{i,j=1\\ i\lt j}}^dk_i^2k_j^2 - \left (\frac{71}{1920}\right )^3\sum _{\substack{i,j,l=1\\ i\lt j\lt l}}^dk_i^2k_j^2k_l^2 + \ldots \pm \left (\frac{71}{1920}\right )^dk_1^2\ldots k_d^2\nonumber \\ &\leq 1 - \frac{71}{1920}\lvert{k}\rvert ^2 + \left (\frac{71}{1920}\right )^2\lvert{k}\rvert ^4 + 0 + \left (\frac{71}{1920}\right )^4\lvert{k}\rvert ^8 + 0 + \ldots\nonumber\\& +\begin{cases} \left (\frac{71}{1920}\right )^{d}\lvert{k}\rvert ^{2d} &\colon d \text{ is even}\\[6pt] \left (\frac{71}{1920}\right )^{d-1}\lvert{k}\rvert ^{2d-2} &\colon d \text{ is odd.} \end{cases} \end{align}
Here we have bounded the later negative terms above by
$0$
, and bounded the positive terms above by powers of
$\lvert{k}\rvert ^4$
. Therefore if
$\lvert{k}\rvert ^2 \lt \frac{1920}{71}$
then we have
\begin{align} \widehat \varphi (k) & \leq 1 - \frac{71}{1920}\lvert{k}\rvert ^2 + \sum ^\infty _{n=1}\left (\frac{71}{1920}\right )^{2n}\lvert{k}\rvert ^{4n} = 1 - \frac{71}{1920}\lvert{k}\rvert ^2\nonumber\\[5pt] & \quad + \left (\frac{71}{1920}\right )^{2}\lvert{k}\rvert ^{4}\left (1 - \left (\frac{71}{1920}\right )^{2}\lvert{k}\rvert ^{4}\right )^{-1}. \end{align}
Note that
$\lvert{k}\rvert \leq 3 \implies \max _i\lvert{k_i}\rvert \leq 3$
, and therefore
$\lvert{k}\rvert \leq 3$
also implies
\begin{equation} \widehat \varphi (k) \leq 1 - \frac{71}{1920}\lvert{k}\rvert ^2\left (1- \frac{213}{640}\left (1 - \left (\frac{213}{640}\right )^{2}\right )^{-1}\right ) \leq 1 - \frac{5}{8}\times \frac{71}{1920}\lvert{k}\rvert ^2, \end{equation}
where we have used
$\frac{213}{640}\lt \frac{1}{3}$
. Therefore we have constants
$b,c_1\gt 0$
such that
$\lvert{k}\rvert \leq b$
implies that
$\widehat \varphi (k) \leq 1 - c_1\lvert{k}\rvert ^2$
.
From (A.60) it is clear that
$\widehat \varphi (k)$
is radially decreasing and non-negative on the set
$\left \{k\in{\mathbb{R}^d}\colon \max _{i}\lvert{k_i}\rvert \leq 2\pi \right \}$
. On the other hand if there exists
$i^*\in \left \{1,2,\ldots, d\right \}$
such that
$\lvert{k_{i^*}}\rvert \gt 2\pi$
, then
$\lvert{\widehat \varphi (k)}\rvert \lt \frac{1}{\pi }$
. Therefore if
$\lvert{k}\rvert \gt 3$
we can bound
\begin{equation} \widehat \varphi (k) \leq 1 - \frac{5}{3}\times \frac{71}{1920}\times 3^2 \lt \frac{1}{2}. \end{equation}
We have therefore proven that (A.2) holds with
$b=3$
,
$c_1= \frac{5}{3}\times \frac{71}{1920}$
, and
$c_2=\frac{1}{2}$
.
LemmaA.9 and our above observation that we can have
$g(d)=\left (\frac{3}{4}\right )^d$
ensures that Assumption B holds.
A.3 Gaussian calculations
Recall that for
$\sigma ^2\gt 0$
and
$0\lt \mathcal{A}\leq \left (2\pi \sigma ^2\right )^{\frac{d}{2}}$
, the Gaussian RCM is defined by having
\begin{equation} \varphi \left (x\right ) = \frac{\mathcal{A}}{\left (2\pi \sigma ^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2\sigma ^2}\lvert{x}\rvert ^2\right ). \end{equation}
Lemma A.11. For the Gaussian RCM,
\begin{align} \varphi ^{\star n}\left (\mathbf{0}\right ) &= \mathcal{A}^n\left (2n\pi \sigma ^2\right )^{-\frac{d}{2}}\qquad \forall n\geq 1, \end{align}
\begin{align} \varphi ^{\star n_1 \star n_2 \cdot n_3}\left (\mathbf{0}\right ) &= \mathcal{A}^{n_1+n_2+n_3}\left (\left (n_1n_2 + n_1n_3 + n_2n_3\right )\left (2\pi \sigma ^2\right )^2\right )^{-\frac{d}{2}}\qquad \forall n_1,n_2,n_3\geq 1. \end{align}
In particular,
\begin{align} \varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) &= \mathcal{A}^5\left (32\pi ^2\sigma ^4\right )^{-\frac{d}{2}} = \mathcal{A}^5\left (8\times \left (2\pi \sigma ^2\right )^2\right )^{-\frac{d}{2}} \end{align}
\begin{align} \varphi ^{\star 1 \star 2\cdot 3}\left (\mathbf{0}\right ) &= \mathcal{A}^6\left (44\pi ^2\sigma ^4\right )^{-\frac{d}{2}} = \mathcal{A}^6\left (11\times \left (2\pi \sigma ^2\right )^2\right )^{-\frac{d}{2}} \end{align}
\begin{align} \varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) &= \mathcal{A}^6\left (48\pi ^2\sigma ^4\right )^{-\frac{d}{2}} = \mathcal{A}^6\left (12\times \left (2\pi \sigma ^2\right )^2\right )^{-\frac{d}{2}}. \end{align}
Proof. Without loss of generality, we scale space so that
$q_\varphi = \mathcal{A} = 1$
.
First we note that the convolution of two unit-mass Gaussian functions is itself a unit-mass Gaussian function whose ‘variance’ parameter is the sum of the variance parameters of the two initial Gaussian functions:
\begin{align} &\int _{{\mathbb{R}^d}}\frac{1}{\left (2\pi \sigma _1^2\right )^{\frac{d}{2}}}\exp \left ({-}\frac{1}{2\sigma _1^2}\lvert{x-y}\rvert ^2\right )\frac{1}{\left (2\pi \sigma _2^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2\sigma _2^2}\lvert{y}\rvert ^2\right ){\mathrm{d}} y \nonumber \\ &= \frac{1}{\left (2\pi \left (\sigma _1^2 + \sigma _2^2\right )\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2\left (\sigma _1^2+\sigma _2^2\right )}\lvert{x}\rvert ^2\right ). \end{align}
It therefore follows that
\begin{equation} \varphi ^{\star n}(x) = \frac{1}{\left (2\pi n\sigma ^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2n\sigma ^2}\lvert{x}\rvert ^2\right ), \end{equation}
and
$\varphi ^{\star n}\left (\mathbf{0}\right )=\left (2\pi n\sigma ^2\right )^{-\frac{d}{2}}$
.
For the remaining expressions we write the pointwise product of two unit-mass Gaussian functions as a constant multiple of a unit-mass Gaussian function:
\begin{align} &\frac{1}{\left (2\pi \sigma _1^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2\sigma _1^2}\lvert{x}\rvert ^2\right )\frac{1}{\left (2\pi \sigma _2^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{1}{2\sigma _2^2}\lvert{x}\rvert ^2\right )=\frac{1}{\left (4\pi ^2\sigma _1^2\sigma _2^2\right )^{\frac{d}{2}}}\exp \!\left ({-}\frac{\sigma _1^2+\sigma _2^2}{2\sigma _1^2\sigma _2^2}\lvert{x}\rvert ^2\right ) \nonumber \\& = \frac{1}{\left (2\pi \!\left (\sigma _1^2+\sigma _2^2\right )\right )^{\frac{d}{2}}}\left (\frac{\sigma _1^2+\sigma _2^2}{2\pi \sigma _1^2\sigma _2^2}\right )^{\frac{d}{2}}\exp \!\left ({-}\frac{\sigma _1^2+\sigma _2^2}{2\sigma _1^2\sigma _2^2}\lvert{x}\rvert ^2\right ). \end{align}
Using this expression, we find
\begin{align} \varphi ^{\star n_1 \star n_2 \cdot n_3}\left (\mathbf{0}\right ) &= \left (2\pi \sigma ^2\left (n_2+n_3\right )\right )^{-\frac{d}{2}}\left (2\pi \sigma ^2\left (n_1 + \frac{n_2n_3}{n_2+n_3}\right )\right )^{-\frac{d}{2}} \nonumber \\& = \left (4\pi ^2\sigma ^4\left (n_1n_2 + n_1n_3 + n_2n_3\right )\right )^{-\frac{d}{2}}. \end{align}
This produces the results.
Lemma A.12.
The Gaussian RCM with
$\liminf _{d\to \infty }\varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
satisfies Assumptions A and B
.
Proof. For this proof we make the scaling choice that the total mass of the adjacency function in each dimension is set to be equal to
$1$
. Clearly this maps
$\mathcal{A} \mapsto \widetilde{\mathcal{A}}\equiv 1$
, but since
$\varphi \!\left (\mathbf{0}\right )=\mathcal{A}\!\left (2\pi \sigma ^2\right )^{-\frac{d}{2}}$
is left invariant, we also have
$\sigma \mapsto \widetilde{\sigma }=\sigma \mathcal{A}^{-\frac{1}{d}}$
. The condition that
$\liminf \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
now means that
$\limsup \widetilde{\sigma }\lt \infty$
, and the trivial condition that
$\varphi \!\left (\mathbf{0}\right )\leq 1$
means that
$\widetilde{\sigma }^2\geq{1}/{2\pi }$
.
The results of LemmaA.11 proves that (A.1) holds with the choice
$g(d)= \left (4\pi \widetilde{\sigma }^2\right )^{-\frac{d}{2}} = 2^{-\frac{d}{2}}\varphi \!\left (\mathbf{0}\right )$
and therefore
$\beta (d) = 2^{-\frac{d}{8}}\varphi \!\left (\mathbf{0}\right )^{\frac{1}{4}}$
(here we use
$\limsup \widetilde{\sigma }\lt \infty$
to get the appropriate form of
$\beta$
from (1.10)). Now observe that the Fourier transform of
$\varphi (x)$
is given by
\begin{equation} \widehat \varphi (k) = \exp \!\left ({-}\frac{1}{2}\widetilde{\sigma }^2\lVert{k}\rVert_2^2\right ) \leq \exp \!\left ({-}\frac{1}{4\pi }\lVert{k}\rVert_2^2\right ), \end{equation}
where the inequality follows from
$\widetilde{\sigma }^2\geq{1}/{2\pi }$
. Therefore (A.2) holds.
For Assumption B, we first use LemmaA.11 to see that
$\varphi ^{\star 6}\left (\mathbf{0}\right ) = 6^{-\frac{d}{2}}q_\varphi ^5\varphi \!\left (\mathbf{0}\right )$
. Therefore (B.1) can be seen to hold with
$\rho =6^{-\frac{1}{2}}\liminf \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
. This also provides a lower bound on
$h(d)$
. After noting that
$\log \beta (d)\lt 0$
, we have
\begin{equation} \frac{\log h(d)}{\log \beta (d)} \leq \frac{-\frac{d}{2}\log 6 + \log \varphi \!\left (\mathbf{0}\right )}{-\frac{d}{8}\log 2 + \frac{1}{4}\log \varphi \!\left (\mathbf{0}\right )} \leq \frac{4\log 6 - 8\log \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}}{\log 2 - 2 \log \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}}. \end{equation}
Note that
$\log \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\leq 0$
. By taking the derivative of the map
$x\mapsto \frac{4\log 6 - 8x}{\log 2-2x}$
for
$x\leq 0$
we can find that it is maximised at
$x=0$
. Therefore
\begin{equation} \frac{\log h(d)}{\log \beta (d)} \leq 4\frac{\log 6}{\log 2} = 4\log _2 6. \end{equation}
Since this is finite, we have proven Assumption (B.2).
A.4 Coordinate-Cauchy calculations
Recall that for
$\gamma \gt 0$
and
$0\lt \mathcal{A}\leq \left (\gamma \pi \right )^d$
, the Coordinate-Cauchy RCM is defined by having
\begin{equation} \varphi (x) = \frac{\mathcal{A}}{\left (\gamma \pi \right )^d}\prod ^d_{j=1}\frac{\gamma ^2}{\gamma ^2+x^2_j}, \end{equation}
where
$x=\left (x_1,\ldots, x_d\right )\in{\mathbb{R}^d}$
.
Lemma A.13. For the Coordinate-Cauchy RCM,
\begin{align} \varphi ^{\star n}\left (\mathbf{0}\right ) &= \mathcal{A}^n\left (\frac{1}{n\gamma \pi }\right )^d, \qquad \forall n\geq 1, \end{align}
\begin{align} \varphi ^{\star n_1\star n_2 \cdot n_3}\left (\mathbf{0}\right ) &= \mathcal{A}^{n_1+n_2+n_3}\left (\frac{n_1+n_2+n_3}{\left (n_1+n_2\right )\left (n_1+n_3\right )\left (n_2+n_3\right )\gamma ^2\pi ^2}\right )^d, \qquad \forall n_1,n_2,n_3\geq 1. \end{align}
In particular,
\begin{equation} \varphi ^{\star 1\star 2\cdot 2}\left (\mathbf{0}\right ) = \mathcal{A}^5\left (\frac{5}{36\gamma ^2\pi ^2}\right )^d,\quad \varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right ) = \mathcal{A}^6\left (\frac{1}{10\gamma ^2\pi ^2}\right )^d,\quad \varphi ^{\star 2\star 2\cdot 2}\left (\mathbf{0}\right ) = \mathcal{A}^6\left (\frac{3}{32\gamma ^2\pi ^2}\right )^d. \end{equation}
Proof. We begin with the simplification that
$q_\varphi =\mathcal{A}$
is set to be equal to
$1$
(by a spatial scaling choice).
Like for the hyper-cubic model, the factorisable structure of the adjacency function means that we only need to evaluate the answers for the
$1$
-dimensional model, and then we can take the result to the power
$d$
to get the
$d$
-dimensional answer. Let the
$1$
-dimensional adjacency function be denoted
\begin{equation} \varphi _1(x) = \frac{\gamma }{\pi \!\left (\gamma ^2+x^2\right )}. \end{equation}
By well-known complex analysis techniques, the Fourier transform of this function is given by
\begin{equation} \widehat \varphi _1(k) ={\text{e}}^{-\gamma \lvert{k}\rvert } \end{equation}
for
$k\in{\mathbb{R}}$
. Then by the Fourier inversion formula, for
$n\geq 1$
,
\begin{equation} \varphi _1^{\star n}\left (\mathbf{0}\right ) = \frac{1}{2\pi }\int ^\infty _{-\infty }{\text{e}}^{-n\gamma \lvert{k}\rvert }{\mathrm{d}} k = \frac{1}{\gamma \pi }\int ^\infty _{0}{\text{e}}^{-n k}{\mathrm{d}} k = \frac{1}{n\gamma \pi }. \end{equation}
The calculation is a little more complicated for the remaining objects. For
$n_1,n_2,n_3\geq 1$
,
\begin{align} \varphi _1^{\star n_1 \star n_2 \cdot n_3}\left (\mathbf{0}\right ) & = \frac{1}{\left (2\pi \right )^2}\int ^\infty _{-\infty }\int ^\infty _{-\infty }{\text{e}}^{-n_1 \gamma \lvert k\rvert - n_2\gamma \lvert k-l\rvert - n_3\gamma \lvert l\rvert }{\mathrm{d}} k{\mathrm{d}} l \nonumber \\ &= \frac{1}{\left (2\gamma \pi \right )^2}\int ^\infty _{-\infty }\int ^\infty _{-\infty }{\text{e}}^{-n_1 \lvert k\rvert - n_2\lvert k-l\rvert - n_3\lvert l\rvert }{\mathrm{d}} k{\mathrm{d}} l. \end{align}
For
$l\geq 0$
, the
$k$
-integral can then be partitioned as
\begin{align} \int ^\infty _{-\infty }{\text{e}}^{-n_1 \lvert k\rvert - n_2\lvert k-l\rvert }{\mathrm{d}} k &= \int ^\infty _{l}{\text{e}}^{-n_1k - n_2k + n_2l}{\mathrm{d}} k + \int ^{l}_{0}{\text{e}}^{-n_1k + n_2k-n_2l}{\mathrm{d}} k + \int ^0_{-\infty }{\text{e}}^{n_1k + n_2k-n_2l}{\mathrm{d}} k\nonumber \\ &= \frac{1}{n_1+n_2}\left ({\text{e}}^{-n_1l}+{\text{e}}^{-n_2l}\right ) + \begin{cases} l{\text{e}}^{-n_1l} &\colon n_1=n_2\\ \frac{1}{n_1-n_2}\left ({\text{e}}^{-n_2l}-{\text{e}}^{-n_1l}\right ) &\colon n_1\ne n_2 \end{cases}\nonumber \\ &= \begin{cases} \left (\frac{1}{n_1} + l\right ){\text{e}}^{-n_1l} &\colon n_1=n_2\\[7pt] \frac{2n_1}{n_1^2- n_2^2}{\text{e}}^{-n_2l} - \frac{2n_2}{n_1^2-n_2^2}{\text{e}}^{-n_1l} &\colon n_1\ne n_2. \end{cases} \end{align}
The calculation is performed similarly for
$l\lt 0$
, and we get
\begin{equation} \int ^\infty _{-\infty }{\text{e}}^{-n_1 \lvert k\rvert - n_2\lvert k-l\rvert }{\mathrm{d}} k = \begin{cases} \left (\frac{1}{n_1} + \lvert{l}\rvert \right ){\text{e}}^{-n_1\lvert{l}\rvert } &\colon n_1=n_2\\[7pt] \frac{2n_1}{n_1^2- n_2^2}{\text{e}}^{-n_2\lvert{l}\rvert } - \frac{2n_2}{n_1^2-n_2^2}{\text{e}}^{-n_1\lvert{l}\rvert } &\colon n_1\ne n_2 \end{cases} \end{equation}
for all
$l\in{\mathbb{R}}$
.
For
$n_1=n_2$
we then get
\begin{align} \varphi _1^{\star n_1\star n_1\cdot n_3}\left (\mathbf{0}\right ) &= \frac{1}{4\gamma ^2\pi ^2}\int ^\infty _{-\infty }\left (\frac{1}{n_1}+\lvert{l}\rvert \right ){\text{e}}^{-\left (n_1+n_3\right )\lvert{l}\rvert }{\mathrm{d}} l\nonumber \\[4pt] &= \frac{1}{2\gamma ^2\pi ^2}\left (\frac{1}{n_1\left (n_1+n_3\right )} + \frac{1}{\left (n_1+n_3\right )^2}\right ) \nonumber \\[4pt] &=\frac{2n_1 + n_3}{2n_1\left (n_1+n_3\right )^2}\frac{1}{\gamma ^2\pi ^2}. \end{align}
Using
$n_1=n_2=2$
and
$n_3=1$
, and
$n_1=n_2=2$
and
$n_3=2$
gives us two of our desired results. We are only left with
$\varphi ^{\star 1\star 2\cdot 3}\left (\mathbf{0}\right )$
.
For
$n_1\ne n_2$
we get
\begin{align} \frac{1}{\left (2\gamma \pi \right )^2}\!\int ^\infty _{-\infty }\int ^\infty _{-\infty }{\text{e}}^{-n_1 \lvert k\rvert - n_2\lvert k-l\rvert - n_3\lvert l\rvert }{\mathrm{d}} k{\mathrm{d}} l &= \frac{1}{2\gamma ^2\pi ^2}\!\int ^\infty _{0}\!\left (\frac{2n_1}{n_1^2- n_2^2}{\text{e}}^{-n_2l} - \frac{2n_2}{n_1^2-n_2^2}{\text{e}}^{-n_1l}\right ){\text{e}}^{-n_3l}{\mathrm{d}} l\nonumber \\ &= \frac{1}{\gamma ^2\pi ^2}\!\left (\frac{n_1}{\left (n_1^2-n_2^2\right )\left (n_2+n_3\right )} - \frac{n_2}{\left (n_1^2-n_2^2\right )\left (n_1+n_3\right )}\right )\nonumber \\ &= \frac{n_1+n_2+n_3}{\left (n_1+n_2\right )\left (n_1+n_3\right )\left (n_2+n_3\right )}\frac{1}{\gamma ^2\pi ^2}. \end{align}
Note that this expression reduces to the case (A.89) if
$n_1=n_2$
.
Lemma A.14. The Coordinate-Cauchy RCM with
$\liminf _{d\to \infty }\varphi \left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
satisfies Assumptions A and B.
Proof. For simplicity we scale space so that
$q_\varphi =\mathcal{A}=1$
. As argued analogously for the Gaussian RCM in LemmaA.12, the condition
$\liminf \varphi \left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
then becomes
$\limsup \gamma \lt \infty$
, and
$\varphi \!\left (\mathbf{0}\right )\leq 1$
becomes
$\gamma \geq{1}/{\pi }$
.
Since
$\widehat \varphi (k) ={\text{e}}^{-\gamma \lVert{k}\rVert_1}\geq 0$
, we know that
$\mathrm{ess\,sup}_{x\in{\mathbb{R}^d}}\varphi ^{\star m}(x) = \varphi ^{\star m}\left (\mathbf{0}\right )$
for all
$m\geq 1$
. Therefore
\begin{equation} \mathrm{ess\,sup}_{x\in{\mathbb{R}^d}}\varphi ^{\star m}(x) = \left (m\gamma \pi \right )^{-d}. \end{equation}
Since
$\gamma \pi \geq 1$
, this approaches zero for all
$m\geq 2$
. Therefore (A.1) holds with the choice
$g(d) = \left (2\gamma \pi \right )^{-d}= 2^{-d}\varphi \!\left (\mathbf{0}\right )$
and
$\beta (d)=2^{-\frac{d}{4}}\varphi \!\left (\mathbf{0}\right )^{\frac{1}{4}}$
(here we use
$\limsup \gamma \lt \infty$
to get the appropriate form of
$\beta$
from (1.10)). Furthermore,
$\gamma$
cannot approach
$0$
and therefore our expression for
$\widehat \varphi (k)$
implies (A.2) holds too.
From our prior calculations we have
$\varphi ^{\star 6}\left (\mathbf{0}\right ) =6^{-d}q_\varphi ^5\varphi \!\left (\mathbf{0}\right )$
and therefore (B.1) can be seen to hold with
$\rho =6^{-1}\liminf \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\gt 0$
. It also allows us to lower bound
$h(d)$
. Noting that
$\log \beta (d)\lt 0$
, this implies that
\begin{equation} \frac{\log h(d)}{\log \beta (d)} \leq \frac{-d\log 6 + \log \varphi \!\left (\mathbf{0}\right )}{-\frac{d}{4}\log 2 + \frac{1}{4}\log \varphi \!\left (\mathbf{0}\right )} \leq \frac{4\log 6 - 4\log \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}}{\log 2 - \log \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}}. \end{equation}
Note that
$\log \varphi \!\left (\mathbf{0}\right )^{\frac{1}{d}}\leq 0$
. By taking the derivative of the map
$x\mapsto \frac{4\log 6 - 4x}{\log 2-x}$
for
$x\leq 0$
we can find that it is maximised at
$x=0$
. Therefore
\begin{equation} \frac{\log h(d)}{\log \beta (d)} \leq 4\frac{\log 6}{\log 2} = 4\log _2 6. \end{equation}
Since this is finite, we have proven Assumption (B.2).
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