2.1. Hyperbolic geometry

Before introducing the RHG model, let us review some definitions and notation concerning hyperbolic geometry. We refer to the book of Stillwell [Reference Stillwell21] for a broader introduction to hyperbolic geometry. The Poincaré disc, denoted by $\mathbb{H}$ , is the open unit disc of $\mathbb{C}$ equipped with the Riemannian metric $\textbf{g}_\mathbb{H}$ , defined at $w \in \mathbb{H}$ by

\begin{align*}\textbf{g}_\mathbb{H} \;:\!=\; \frac{4\textbf{g}_{\mathbb{C}}}{(1-|w|^2)^2}, \quad \mbox{where}\;\textbf{g}_{\mathbb{C}}\; \mbox{is the Euclidean metric on}\;\mathbb{C}.\end{align*}

We denote by $d_{\mathbb{H}}$ the distance induced by $\textbf{g}_{\mathbb{H}}$ on $\mathbb{H}$ . Throughout this paper, we make extensive use of the polar coordinates to describe points in the Poincaré disc. The polar coordinates of a point w in $\mathbb{H}$ are given by the pair $(r(w),\theta(w))$ , where r(w) denotes its hyperbolic distance to the origin and $\theta(w)$ denotes its angle in the complex plane. The quantity r(w) is also referred to as the radius of w. If x and y are two points of $\mathbb{H}$ with respective polar coordinates $(r,\theta)$ and $(s,\beta)$ , the hyperbolic distance between x and y is given by the celebrated hyperbolic law of cosines:

(2.1) \begin{align}\cosh(d_{\mathbb{H}}(x,y)) = \cosh(r)\cosh(s) - \sinh(r)\sinh(s)\cos(\theta - \beta).\end{align}

For visualization purposes, we use the native representation to draw pictures of the RHG, as done in [Reference Krioukov, Papadopoulos, Kitsak, Vahdat and Boguñá17]. This means that instead of representing the RHG directly in $\mathbb{H}$ , we represent its image under the mapping $\omega \mapsto r(w)\mathrm{e}^{i\theta(w)}$ , defined from $\mathbb{H}$ to $\mathbb{C}$ . This transformation dilates all distances to the origin, ensuring that every point $(r,\theta)$ is represented with a Euclidean distance to the origin equal to its radial coordinate r.

For a point x with polar coordinates $(r,\theta)$ in $\mathbb{H}$ and a radius $s > 0$ , we denote by $\mathcal{B}_{x}(s)$ or $\mathcal{B}_{(r,\theta)}(s)$ the open hyperbolic ball of radius s centred at x. For $0 < r_1 < r_2$ , we define $\mathcal{C}(r_1,r_2)$ as the annulus with inner radius $r_1$ and outer radius $r_2$ , i.e. $\mathcal{C}(r_1,r_2)\;:\!=\; \mathcal{B}_{0}(r_2) \setminus \mathcal{B}_{0}(r_1)$ .

2.2. RHG

We now proceed with the formal definition of the RHG $\mathcal{G}_{\alpha,\nu}(n)$ , as defined in [Reference Krioukov, Papadopoulos, Kitsak, Vahdat and Boguñá17]. Fix two parameters $\alpha > 0$ and $\nu > 0$ and, for $n \in \mathbb{N}^*$ , set

(2.2) \begin{align}R_n \;:\!=\; 2\log\bigg(\frac{n}{\nu}\bigg).\end{align}

Define a probability measure $\mu_n$ on $\mathbb{H}$ such that, if a point $(r,\theta)$ (in polar coordinates) is chosen according to $\mu_n$ , then r and $\theta$ are independent, $\theta$ is uniformly distributed in $(-\pi,\pi]$ , and the probability distribution of r has a density function on $\mathbb{R}_+$ given by

(2.3) \begin{align}\rho_n(r) \;:\!=\; \frac{\alpha \sinh(\alpha r)}{\cosh(\alpha R_n) - 1} \textbf{1}_{\{r < R_n\}}.\end{align}

Let $X_1^n,X_2^n,\dots,X_n^n$ be a sequence of n independent points sampled from the Poincaré disc according to the distribution $\mu_n$ (for brevity, the superscript n in $X_i^n$ will often be omitted). We denote by $\mathcal{G}_{\alpha,\nu}(n)$ the RHG with n nodes and parameters $\alpha$ and $\nu$ . It is defined as the undirected graph with nodes at the points $X_1^n,X_2^n,\dots,X_n^n$ , where an edge exists between two nodes if and only if their hyperbolic distance is at most $R_n$ . The degree of a node $X_i$ in the graph $\mathcal{G}_{\alpha,\nu}(n)$ is defined as the number of its direct neighbours in $\mathcal{G}_{\alpha,\nu}(n)$ and is denoted by $\deg(X_i)$ . Since we focus on the behaviour of large graphs, the value of n will always be considered large, while the parameters $\alpha$ and $\nu$ are fixed.

Figure 1.

Simulations of RHGs (native representation) with $n = 500$ , $\nu = 1$ , $\alpha = 0.45$ (left), $\alpha = 0.50$ (middle), and $\alpha = 0.55$ (right). The boundary of $\mathcal{B}_{0}(R_n)$ is represented by a black circle and its centre by a larger dot.

Observe that the n nodes of the graph $\mathcal{G}_{\alpha,\nu}(n)$ are located within $\mathcal{B}_{0}(R_n)$ . Moreover, due to the choice of the measure, the points tend to concentrate near the boundary of $\mathcal{B}_{0}(R_n)$ . Also note that the measure $\mu_n(\mathcal{B}_{(r,\theta)}(s))$ of a ball centred at $(r,\theta)$ is independent of $\theta$ . Therefore, we omit the angle $\theta$ to shorten notation and instead write $\mu_n(\mathcal{B}_{r}(s))$ . Likewise, we write $\mu_n(\mathcal{B}_{r}(s) \setminus \mathcal{B}_{0}(s'))$ instead of $\mu_n(\mathcal{B}_{(r,\theta)}(s) \setminus \mathcal{B}_{0}(s'))$ .

In the special case $\alpha = 1$ , $\mu_n$ is the uniform measure on $\mathcal{B}_{0}(R_n)$ associated with the Riemannian metric $\textbf{g}_\mathbb{H}$ . In the general case $\alpha > 0$ , $\mu_n$ corresponds to a uniform measure on the hyperbolic plane $\mathbb{H}_\alpha$ of curvature $-\alpha^2$ . More precisely, for fixed $\alpha > 0$ , multiplying the differential form in the Poincaré disc model by a factor ${1}/{\alpha^2}$ , we obtain a hyperbolic plane of curvature $-\alpha^2$ . Choosing a point according to the measure $\mu_n$ amounts to choosing a point uniformly in the ball of radius $R_n$ of $\mathbb{H}_\alpha$ and projecting it on $\mathbb{H}$ , by keeping the same polar coordinates. The measure of a ball of radius r in $\mathbb{H}_\alpha$ is

\begin{align*}\frac{2\pi}{\alpha^2}(\!\cosh(\alpha r) - 1);\end{align*}

thus, the larger $\alpha$ is, the faster it increases with r. Therefore, the larger $\alpha$ is, the more the points of the graph $\mathcal{G}_{\alpha,\nu}(n)$ concentrate near the boundary of $\mathcal{B}_{0}(R_n)$ . Thus, the maximum degree is expected to decrease with $\alpha$ (see Figure 1). The degree distribution is also influenced by the parameter $\nu$ . Increasing $\nu$ makes the domain smaller, which limits the concentration of the nodes near the boundary of the domain, resulting in a higher expected degree. In the sparse regime $\alpha > \tfrac{1}{2}$ , the expected degree evolves linearly with $\nu$ [Reference Gugelmann, Panagiotou and Peter14].

The Poissonised RHG $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ is obtained by choosing the nodes according to a Poisson point process with intensity measure $n\mu_n$ , instead of choosing n points according to the measure $\mu_n$ . All the results proved in this paper also hold for the Poissonised model. In the proof of the non-ordering result (3.2), the use of a classical Poissonisation/de-Poissonisation procedure allows us to avoid some technicalities by using the properties of Poisson processes.

8.1. Ordering up to rank $k_n$ (proof of (3.1))

The proof of (3.1) follows the same general idea as the proof of Theorem 3.1, but in a more precise form. The argument is divided into three steps. The first two steps consist in localising the $k_n$ first nodes and showing that there are large gaps between their radii. Actually, this is done by proving that two specific events, $L_n$ and $G_n$ , occur with high probability. In the third step, we show that, under the event $L_n \cap G_n$ , there are large differences between the expected degrees of the $k_n$ first nodes. The conclusion then follows from a Chernoff bound.

Step 1: Localisation of the $k_n$ first nodes. Let $w_n \;:\!=\; \log(\log(n))$ and define the localisation event $L_n$ by

\begin{align*}L_n \;:\!=\; \{t_n \leq r(X_{(1)}) \; \mbox{and} \; r(X_{(k_n)}) \leq r_n\},\end{align*}

where

\begin{align*}t_n \;:\!=\; \bigg(1-\frac{1}{2\alpha}\bigg)R_n - w_n \quad \mbox{and} \quad r_n \;:\!=\; t_n + \frac{\beta}{\alpha}\log(n).\end{align*}

Let us prove that, with this choice of $t_n$ and $r_n$ , the event $L_n$ is realised with high probability. We already know from Proposition 4.1 that $t_n \leq r(X_{(1)})$ holds with high probability. On the other hand, we note that $r(X_{(k_n)}) \leq r_n$ holds if and only if the number of nodes in the ball $\mathcal{B}_{0}(r_n)$ is larger than $k_n$ . The number of points falling in this ball has a binomial distribution with expected value

\begin{align*}n \mu_n(\mathcal{B}_{0}(r_n)) \underset{n \to \infty}{\sim} \nu n^{\beta}\log(n)^{-\alpha} \gg k_n.\end{align*}

Using a Chernoff bound it follows that $r(X_{(k_n)}) \leq r_n$ holds with high probability, so

\begin{align*}\mathbb{P}[L_n] \xrightarrow[n\to\infty]{} 1.\end{align*}

Step 2: Proving the existence of large gaps between the $k_n$ first nodes. We define the gap event $G_n$ by

\begin{align*}G_n \;:\!=\; \{\text{for all }i \leq k_n,\, r(X_{(i+1)}) - r(X_{(i)}) \geq \lambda_n(r(X_{(i)})) \},\end{align*}

where, for all n, the function $\lambda_n$ is defined by

\begin{align*}\lambda_n(s) \;:\!=\; \mathrm{e}^{\alpha(R_n - s)} n^{-(\beta+1)}\quad\text{for all }s \geq 0.\end{align*}

Let us prove that with this choice of $\lambda_n$ , the event $G_n$ is realised with high probability, which will prove that the radius gaps between the $n^{\beta}$ first nodes are relatively large.

The first $k_n+1$ nodes (i.e. $X_{(1)},X_{(2)},\dots,X_{(k_n+1)}$ ) can be sampled in the following way. First, sample n nodes in $\mathcal{B}_{0}(R_n)$ according to the distribution $\mu_n$ , select the closest to the centre as $X_{(1)}$ and erase the $(n-1)$ other points. Next, sample $(n-1)$ points according to the restriction of $\mu_n$ to the annulus $\mathcal{C}(r(X_{(1)}),R_n)$ , choose the closest to the centre as $X_{(2)}$ and erase the other ones. Repeat this process until the $(k_n + 1)$ first nodes have been sampled (at step $(i+1)$ sample $(n-i)$ points in the annulus $\mathcal{C}(r(X_{(i)}),R_n)$ , choose the closest to the centre as $X_{(i+1)}$ and erase the other ones). Considering this process, we get

\begin{align*}\mathbb{P}[G_n^c,\,L_n] &\leq \sum\limits_{i = 1}^{k_n} \mathbb{P}[Z_i^n \neq 0,\,L_n],\end{align*}

where the random variable $Z_i^n$ counts the number of points falling in the annulus $\mathcal{C}(r(X_{(i)}), r(X_{(i)}) + \lambda_n(r(X_{(i)})))$ at step $(i+1)$ . Conditionally on the position of $X_{(i)}$ , the variable $Z_i^n$ follows a binomial distribution with $(n-i)$ trials and probability $p_i^n$ given by

\begin{align*}p_i^n&\;:\!=\; \frac{\mu_n(\mathcal{C}(r(X_{(i)}),r(X_{(i)})+\lambda_n(r(X_{(i)}))))}{\mu_n(\mathcal{C}(r(X_{(i)}),R_n))}\\&= \frac{\cosh(\alpha (r(X_{(i)})+\lambda_n(r(X_{(i)})))) - \cosh(\alpha r(X_{(i)}))}{\cosh(\alpha R_n) - \cosh(\alpha r(X_{(i)}))}.\end{align*}

For $r(X_{(i)}) \leq r_n$ and as $n \to \infty$ , the denominator is asymptotically equivalent to $\mathrm{e}^{\alpha R_n}/2$ . Moreover, for $r(X_{(i)}) \geq t_n$ , we have $\lambda_n(r(X_{(i)})) = o(1)$ . Applying the mean value theorem to the numerator, it follows that, for large enough n and $t_n \leq r(X_{(i)}) \leq r_n$ ,

\begin{align*}p_i^n &\leq K \lambda_n(r(X_{(i)})) \exp(\alpha (r(X_{(i)}) - R_n)) = K n^{-(\beta+1)}.\end{align*}

Writing $\mathbb{P}_{\scriptscriptstyle \scriptscriptstyle X_{(i)}}[\cdot]$ for conditional probability with respect to the variable $X_{(i)}$ , we get

\begin{align*}\mathbb{P}[G_n^c \cap L_n] &\leq \sum_{i = 1}^{k_n} \mathbb{E}[\mathbb{P}_{\scriptscriptstyle \scriptscriptstyle X_{(i)}}[Z_i^n \neq 0] \textbf{1}_{\{t_n \leq r(X_{(i)}) \leq r_n\}}]\\&\leq k_n (1 - (1-K n^{-(\beta+1)})^{n})\\&\underset{n \to \infty}{\sim} K \log(n)^{-2\alpha}\\&\xrightarrow[n\to\infty]{} 0.\end{align*}

Since $L_n$ occurs with high probability, we conclude that

\begin{align*}\mathbb{P}[G_n] \xrightarrow[n\to\infty]{} 1.\end{align*}

Step 3: Comparing the successive expected values of the degrees. Let us denote by $O_n$ the ordering event described in (3.1). By the previous two steps, it remains to prove that

(8.1) \begin{align}\mathbb{P}[O_n^c \cap L_n \cap G_n] \xrightarrow[n\to\infty]{} 0.\end{align}

Proceeding as we did for proving (6.2) in the proof of Theorem 3.1, we get the upper bound

(8.2) \begin{align}\mathbb{P}[O_n^c \cap L_n \cap G_n]&\leq \max\limits_{(s_1,s_2)} 2n^2 \exp\bigg(- \frac{(n-2)(\mu_n(\mathcal{B}_{s_1}(R_n)) - \mu_n(\mathcal{B}_{s_2}(R_n)))^2}{8\mu_n(\mathcal{B}_{s_1}(R_n))}\bigg),\end{align}

where the maximum is taken over the couples $(s_1,s_2)$ belonging to the set

\begin{align*}E_n \;:\!=\; \{(s_1,s_2) \in [0,R_n)^2\colon t_n \leq s_1 \leq r_n,\, s_2 - s_1 \geq \lambda_n(s_1)\}.\end{align*}

We are now interested in bounding the fraction appearing in the exponential term. For $(s_1,s_2) \in E_n$ and n large, we have

\begin{align*}\mu_n(\mathcal{B}_{s_1}(R_n)) - \mu_n(\mathcal{B}_{s_2}(R_n))&= - \int_{s_1}^{s_2} \frac{\partial}{\partial r} \mu_n(\mathcal{B}_{r}(R_n)) \,\textrm{d}r\\&\geq K (1 + O(\mathrm{e}^{-(\alpha-1/2)t_n} + t_n \mathrm{e}^{-t_n})) \int_{s_1}^{s_2} \mathrm{e}^{-r/2}\,\textrm{d}r\\&= K \mathrm{e}^{-s_1/2}(1 - \mathrm{e}^{-(s_2-s_1)/2}),\end{align*}

where the second line follows from Lemma 8.1. Using (5.5) to estimate the denominator in the expression below, it follows that, for $(s_1,s_2) \in E_n$ and large n,

(8.3) \begin{align}\frac{(n-2)(\mu_n(\mathcal{B}_{s_1}(R_n)) - \mu_n(\mathcal{B}_{s_2}(R_n)))^2}{8\mu_n(\mathcal{B}_{s_1}(R_n))}&\geq K n\mathrm{e}^{-s_1/2}(1-\mathrm{e}^{-(s_2-s_1)/2})^2 \notag\\&\geq K n\mathrm{e}^{-r_n/2}\lambda_n(r_n)^2, \end{align}

Moreover, straightforward computations give

\begin{align*}n\mathrm{e}^{-r_n/2} = K n^{{(1-\beta)}/{2\alpha}} \log(n)^{1/2} \quad \mbox{and} \quad \lambda_n(r_n)= K n^{-2\beta} \log(n)^\alpha.\end{align*}

By choice of $\beta$ , we have ${(1-\beta)}/{2\alpha} = 4\beta$ . Thus, substituting the above in (8.3) and combining with (8.2) gives

\begin{align*}\mathbb{P}[O_n^c \cap L_n \cap G_n]&\leq 2n^2\exp(\!- K \log(n)^{1/2+2\alpha}) \xrightarrow[n\to\infty]{} 0.\end{align*}

This proves (8.1) and concludes the proof of (3.1).

8.2. No ordering beyond rank $n^\beta$ (proof of (3.2))

The proof is divided into four main steps. Before proceeding with these, we first introduce some notation and a Poissonised version of (3.2), which will serve as an intermediate result. Fix an arbitrary sequence $(a_n)$ diverging to $+\infty$ . Observe that if (3.2) holds for a given sequence $(a_n)$ then it also holds for any sequence larger than $(a_n)$ . Thus, without loss of generality, we may assume that $a_n \leq \log(n)$ . This assumption on $(a_n)$ will be used implicitly in the proof when establishing certain bounds. Let $w_n' \;:\!=\; \log(a_n)$ and define

\begin{align*}t_n' &\;:\!=\; \bigg(1-\frac{1}{2\alpha}\bigg)R_n + \frac{\beta}{\alpha}\log(n) + w_n' \quad \mbox{and} \quad r_n' \;:\!=\; t_n' + w_n'.\end{align*}

Recall that we aim at finding an index $i \in [n^\beta,n^\beta a_n]$ such that $\deg(X_{(i)}^n) < \deg(X_{(i+1)}^n)$ . In order to gain some independence between the degrees of the nodes, we first prove an analogue of (3.2) for the Poissonised model $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ . It turns out that, for $i \in [n^\beta,n^\beta a_n]$ , the nodes $X_{(i)}$ and $X_{(i+1)}$ have radii that are approximately located in the interval $[t_n',r_n']$ . Thus, the counterpart of the non-ordering event (3.2) for the Poissonised model $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ can be written as follows.

• • With high probability, there exist two nodes v and v ^′ in the graph $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ , such that

  (8.4) \begin{align}t_n' \leq r(v) \leq r(v') \leq r_n' \quad \mbox{and} \quad \deg(v) + \delta_n \leq \deg(v'),\end{align}
  where the sequence $(\delta_n)$ is defined by $\delta_n \;:\!=\; \sqrt{n\mathrm{e}^{-r_n'/2}} = K n^{2 \beta} \mathrm{e}^{-w_n'/2}$ .

In the first three steps of the proof, we work within the Poissonised model $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ to establish the Poissonised statement. The final step of the proof is a de-Poissonisation procedure that gives the result for $\mathcal{G}_{\alpha,\nu}(n)$ from the result for $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ . The extra gap of size $\delta_n$ that appears in the Poissonised statement is crucial for this de-Poissonisation step.

Let us fix a small $\varepsilon \in (0,1)$ and set $R_n^{\varepsilon} \;:\!=\; (1-\varepsilon)R_n$ . For $x \in \mathcal{B}_{0}(R_n)$ , we denote by $\mathcal{B}^{\varepsilon}_{x}(R_n)$ the part of the ball $\mathcal{B}_{x}(R_n)$ that lies beyond the circle of radius $R_n^{\varepsilon}$ , i.e.

\begin{align*}\mathcal{B}^{\varepsilon}_{x}(R_n) \;:\!=\; \mathcal{B}_{x}(R_n) \cap \mathcal{C}(R_n^{\varepsilon},R_n).\end{align*}

The $\varepsilon$ degree of a node X is the number of nodes (excluding X itself) contained in $\mathcal{B}^{\varepsilon}_{X}(R_n)$ . It is denoted by $\deg_{\varepsilon}(X)$ . Since the nodes of $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ are concentrated near the boundary of $\mathcal{B}_{0}(R_n)$ , the $\varepsilon$ degrees provide good estimates of the actual degrees of the nodes. Therefore, we first focus on the $\varepsilon$ degree rather than degrees directly, as it is easier to get independence results concerning the $\varepsilon$ degrees.

We can now proceed with the four steps of the proof. The first step consists in finding a number $c_n \to \infty$ of pairs of points (v, v ^′) in the Poissonised graph $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ that have a positive probability of satisfying (8.4). In the second step, we use the independence property of the Poissonised model $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ to show that, with high probability, at least one of these candidate pairs satisfies a version of (8.4) in which the degree is replaced by the $\varepsilon$ degree. In the third step, we show that the error induced by replacing the degree with the $\varepsilon$ degree is sufficiently small. It follows that (8.4) holds with high probability. The fourth and final step is the de-Poissonisation procedure.

Figure 4.

Depiction of the closeness event $C_n$ . By condition (8.5), for all i, the radius gap $r(v_i') - r(v_i)$ is small, ensuring that $\deg_{\varepsilon}(v_i) \leq \deg_{\varepsilon}(v_i')$ holds with a probability bounded away from 0. By condition (8.6), the portions of the balls $\mathcal{B}_{v_i}(R_n)$ and $\mathcal{B}_{v_i'}(R_n)$ that lie beyond the circle of radius $R_n^{\varepsilon}$ are all disjoints, ensuring the independence of the corresponding $\varepsilon$ degrees.

Step 1: Finding good pairs of candidates for the couple (v,v’). For all n, set $\lambda_n' \;:\!=\; n^{-2\beta} \mathrm{e}^{- \alpha w_n'}$ and denote by $C_n$ the following event (see Figure 4).

• • There exist two sequences $(v_i)_{\scriptscriptstyle 1 \leq i \leq c_n}$ and $(v_i')_{\scriptscriptstyle 1 \leq i \leq c_n}$ each containing $c_n$ nodes of $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ , such that $v_1,\dots,v_{c_n},v_1',\dots,v_{c_n}'$ are pairwise distinct and, for all i,

  (8.5) \begin{align}t_n' \leq r(v_i) \leq r(v_i') \leq r(v_i)+\lambda_n' \leq r_n'.\end{align}
  We also require that, for all x and y distinct in $\{v_1,\dots,v_{c_n},v_1',\dots,v_{c_n}'\}$ ,
  (8.6) \begin{align}\mathcal{B}^{\varepsilon}_{x}(R_n) \cap \mathcal{B}^{\varepsilon}_{y}(R_n) = \emptyset .\end{align}

Let us prove that this event is realised with high probability. We prove in the following two steps of the proof that, when this event occurs, there is a high probability of finding two nodes satisfying (8.4). We define $m_n \;:\!=\; w_n' n^{2\beta}$ and take $c_n = o(w_n')$ such that $c_n \to \infty$ . For all $1 \leq j \leq m_n$ , let $I_j^n \;:\!=\; [t_n' + j \lambda_n', t_n' + (j+1)\lambda_n')$ . Note that all of these intervals are contained in $[t_n',r_n']$ . We denote by $A_j^n$ the event that there exist at least two nodes of $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ with radial coordinates in the interval $I_j^n$ . The number of nodes having a radial coordinate in the interval $I_j^n$ follows a Poisson distribution with parameter $n p_j^n$ , with $p_j^n$ given by

\begin{align*}p_j^n = \frac{\cosh(\alpha (t_n' + (j+1) \lambda_n')) - \cosh(\alpha (t_n' + j \lambda_n'))}{\cosh(\alpha R_n) - 1} \geq K \lambda_n' \mathrm{e}^{\alpha (t_n' - R_n)} = K n^{-(\beta + 1)}.\end{align*}

It follows that

\begin{align*}\mathbb{P}[A_j^n] \geq Kn^{-2\beta}.\end{align*}

Moreover, the events $A_j^n$ are independent, so the number of indices $1 \leq j \leq m_n$ for which the event $A_j^n$ occurs dominates a binomial distribution with $m_n$ trials and an expected value equivalent to $Kw_n'$ . Since $c_n = o(w_n')$ , it follows that, with high probability, we can find $c_n$ indices j for which the event $A_j^n$ is realised. We conclude that, with high probability, there exist two sequences made of distinct nodes $(v_i)_{1 \leq i \leq c_n}$ and $(v_i')_{1 \leq i \leq c_n}$ satisfying condition (8.5) of the event $G_n$ .

Let us prove that these nodes also have a high probability of satisfying condition (8.6). First remark that, by the estimate of $\theta_r(y)$ (given in Lemma 5.2), if $\varepsilon$ is small enough then

\begin{align*}\theta_{t_n'}(R_n^{\varepsilon})= 2\exp\bigg(\frac{R_n-t_n' - R_n^{\varepsilon}}{2}\bigg) (1 + O(\mathrm{e}^{R_n - t_n' - R_n^{\varepsilon}})) \xrightarrow[n\to\infty]{} 0.\end{align*}

Thus, we may assume that $c_n$ was chosen such that $c_n^2 \theta_{t_n'}(R_n^{\varepsilon}) \to 0$ .

Sample all the nodes appearing in $\mathcal{C}(t_n',r_n')$ and suppose that there exist two sequences made of distinct nodes $(v_i)_{1 \leq i \leq c_n}$ and $(v_i')_{1 \leq i \leq c_n}$ in $\mathcal{C}(t_n',r_n')$ that satisfy condition (8.5) of the event $C_n$ . Suppose that the angular coordinates of the nodes of these sequences have not been sampled yet. Note that it is valid to suppose this as condition (8.5) only concerns the radial coordinates of the nodes. We now sample the angular coordinates of these nodes one by one. Each time we sample a new angular coordinate, the probability that it differs by less than $2\theta_{t_n'}(R_n^{\varepsilon})$ with an already sampled angular coordinate is upper bounded by $({2c_n}/{\pi}) \theta_{t_n'}(R_n^{\varepsilon})$ . Thus, the probability of getting a pair of angular coordinates that differ by less than $2\theta_{t_n'}(R_n^{\varepsilon})$ is upper bounded by $({4c_n^2}/{\pi})\theta_{t_n'}(R_n^{\varepsilon})$ , which tends to 0 by our choice of $c_n$ . Thus, with high probability, all the angular coordinates of the nodes $v_1,\dots,v_{c_n},v_1',\dots,v_{c_n}'$ differ by more than $2\theta_{t_n'}(R_n^{\varepsilon})$ .

For $1 \leq i \leq c_n$ , the angle of the smallest cone (with apex at 0) containing $\mathcal{B}^{\varepsilon}_{v_i}(R_n)$ is $2\theta_{r(v_i)}(R_n^{\varepsilon})$ . Since $t_n' \leq r(v_i)$ , Lemma 5.1 implies that this angle is at most $2\theta_{t_n'}(R_n^{\varepsilon})$ . The same holds for the nodes $v_i'$ . It follows that the sequences $(v_i)$ and $(v_i')$ also satisfy condition (8.6), with high probability. Thus,

(8.7) \begin{align}\mathbb{P}[C_n] \xrightarrow[n\to\infty]{} 1.\end{align}

Step 2: Finding a pair of nodes (V,V’) with $r(V) \leq r(V')$ such that $\deg_{\varepsilon}(V)$ is small and $\deg_{\varepsilon}(V')$ is large. Let us sample all the nodes that fall in the annulus $\mathcal{C}(t_n',r_n')$ . Since the event $C_n$ depends only on the point process of the nodes restricted to this region, we can suppose that the points in this annulus are such that the event $C_n$ occurs. We denote by $\mathbb{P}_{\scriptscriptstyle C_n}$ the probability measure conditioned on the event $C_n$ .

For all $K > 0$ , we define $\eta(K)$ by

\begin{align*}\eta(K) = \liminf\limits_{n \in \mathbb{N}^*} \inf\limits_{a \, \geq \, n \mu_n(\mathcal{B}^{\varepsilon}_{r_n'}(R_n))} \mathbb{P}[\mbox{Poi}(a) \geq a + K \sqrt{a}],\end{align*}

where $\mbox{Poi}(a)$ denotes a Poisson variable with parameter a. Using (5.5) and (5.6), one checks that, if $\varepsilon$ is chosen sufficiently small, then $n \mu_n(\mathcal{B}^{\varepsilon}_{r_n'}(R_n)) \to \infty$ as $n \to \infty$ . By the central limit theorem, $\eta(K)$ is then equal to the probability that a standard normal random variable exceeds K. In particular, $\eta(K) > 0$ .

Select two sequences $(v_i)_{1 \leq i \leq c_n}$ and $(v_i')_{1 \leq i \leq c_n}$ as provided by the event $C_n$ (choose them according to a fixed rule to ensure measurability). For $1 \leq i \leq c_n$ , since $r(v_i) \leq r_n'$ , it follows from Lemma 5.1 that

(8.8) \begin{align}n\mu_n(\mathcal{B}^{\varepsilon}_{v_i}(R_n))\geq n\mu_n(\mathcal{B}^{\varepsilon}_{r_n'}(R_n)) = n (\mu_n(\mathcal{B}_{r_n'}(R_n)) - \mu_n(\mathcal{B}_{r_n'}(R_n) \cap \mathcal{B}_{0}(R_n^{\varepsilon}))).\end{align}

If $\varepsilon$ is chosen small enough then $r_n' \geq \varepsilon R_n$ . Thus, we can use the approximations given by (5.5) and (5.6) to get, uniformly in $1 \leq i \leq c_n$ ,

\begin{align*}n\mu_n(\mathcal{B}^{\varepsilon}_{v_i'}(R_n)) \geq K n \mathrm{e}^{-r_n'/2} = K \delta_n^2\end{align*}

(the sequence $(\delta_n)$ is defined below (8.4)). It follows that, for a possibly different constant $K > 0$ , we have, for all $1 \leq i \leq c_n$ ,

(8.9) \begin{align}&\mathbb{P}_{\scriptscriptstyle \scriptscriptstyle C_n}[\deg_{\varepsilon}(v_i') \geq n\mu_n(\mathcal{B}^{\varepsilon}_{v_i'}(R_n)) + 2\delta_n]\nonumber\\&\qquad\geq \mathbb{P}_{\scriptscriptstyle \scriptscriptstyle C_n}\big[\deg_{\varepsilon}(v_i') \geq n\mu_n(\mathcal{B}^{\varepsilon}_{v_i'}(R_n)) + K \sqrt{n\mu_n(\mathcal{B}^{\varepsilon}_{v_i'}(R_n))}\big].\end{align}

Moreover, for all $1 \leq i \leq c_n$ , conditionally on the event $C_n$ and on the position of $v_i'$ , the variable $\deg_{\varepsilon}(v_i')$ follows a Poisson distribution with parameter $n\mu_n(\mathcal{B}^{\varepsilon}_{v_i'}(R_n))$ . Since $r(v_i') \leq r_n'$ and $\mu_n(\mathcal{B}_{r}(R_n))$ is decreasing in r, this parameter is at least $n\mu_n(\mathcal{B}^{\varepsilon}_{r_n'}(R_n))$ . Thus, combining (8.8) with (8.9) provides a constant $K > 0$ such that, for large enough n and for all $1 \leq i \leq c_n$ ,

(8.10) \begin{align}\mathbb{P}_{\scriptscriptstyle \scriptscriptstyle C_n}[\deg_{\varepsilon}(v_i') \geq n\mu_n(\mathcal{B}^{\varepsilon}_{v_i'}(R_n)) + 2\delta_n] \geq \frac{\eta(K)}{2}.\end{align}

With a similar argument, we get $\hat{\eta} > 0$ such that, for large enough n and $1 \leq i \leq c_n$ ,

(8.11) \begin{align}\mathbb{P}_{\scriptscriptstyle C_n}[\deg_{\varepsilon}(v_i) \leq n\mu_n(\mathcal{B}^{\varepsilon}_{v_i}(R_n))] \geq \hat{\eta} .\end{align}

Conditionally on the event $C_n$ and on the positions of the nodes $(v_i)_{1\leq i \leq c_n}$ and $(v_i')_{1 \leq i \leq c_n}$ , the variables $\deg_{\varepsilon}(v_1),\dots,\deg_{\varepsilon}(v_{c_n}),\deg_{\varepsilon}(v_i'),\dots,\deg_{\varepsilon}(v_{c_n}')$ are independent. Indeed, once these nodes are fixed, their $\varepsilon$ degrees correspond to the number of points of a Poisson point process in $\mathcal{B}_{0}(R_n) \setminus \mathcal{B}_{0}(R_n^\varepsilon)$ falling into certain predetermined regions: by condition (8.6) of the event $C_n$ , these regions are disjoint (see Figure 4). Since $c_n \to \infty$ and both $\eta(K)$ and $\hat{\eta}$ are positive, it follows that, with high probability, there exists an index $1 \leq i \leq c_n$ for which the events in (8.10) and (8.11) occur simultaneously. Define $i_0$ as the smallest such index and set $(V,V') = (v_{i_0},v_{i_0}')$ . We verify that this defines a random vector (V, V ^′) on an event of high probability. The definition of (V, V ^′) out of this event is not relevant.

Step 3: From $\deg_{\varepsilon}$ to $\deg$ . The previous step would allow us to conclude immediately if we were comparing the $\varepsilon$ degrees $\deg_{\varepsilon}$ of the nodes instead of their actual degrees $\deg$ . Thus, it remains to show that $\deg_{\varepsilon}$ is very close to $\deg$ . To achieve this, we introduce the following subsets of the ball $\mathcal{B}_{x}(R_n)$ :

(8.12) \begin{align}\mathcal{B}'_{x}(R_n) &\;:\!=\; \mathcal{B}_{x}(R_n) \setminus (\mathcal{C}(t_n',r_n') \cup \mathcal{C}(R_n^{\varepsilon},R_n)),\end{align}

(8.13) \begin{align}\mathcal{B}''_{x}(R_n) &\;:\!=\; \mathcal{B}_{x}(R_n) \cap \mathcal{C}(t_n',r_n'). \end{align}

For a node X, we write $\deg'(X)$ (respectively $\deg''(X)$ ) for the number of nodes (excluding X) that are contained in $\mathcal{B}'_{X}(R_n)$ (respectively $\mathcal{B}''_{X}(R_n)$ ). Note that, for any point x, the ball $\mathcal{B}_{x}(R_n)$ is the union of the balls $\mathcal{B}^{\varepsilon}_{x}(R_n)$ , $\mathcal{B}'_{x}(R_n)$ , and $\mathcal{B}''_{x}(R_n)$ . Thus, the degree of X can be decomposed as follows:

\begin{align*}\deg(X) = \deg_{\varepsilon}(X) + \deg'(X) + \deg''(X).\end{align*}

Let us first estimate the error induced by the term $\deg'$ . If $\varepsilon$ is small enough, we can use (5.6) and get $\varepsilon' > 0$ such that, for large enough n,

\begin{align*}n\mu_n(\mathcal{B}'_{V}(R_n)) \leq n\mu_n(\mathcal{B}_{V}(R_n) \cap \mathcal{B}_{0}(R_n^{\varepsilon})) \leq n^{-5\varepsilon'} n\mathrm{e}^{-t_n'/2} \leq n^{-4\varepsilon'} \delta_n^2\end{align*}

(the sequence $(\delta_n)$ is defined below (8.4)). Since $\mathcal{B}'_{V}(R_n)$ does not intersect the annulus $\mathcal{C}(t_n',r_n')$ , we find that, conditionally on $C_n$ and on the position of the node V, the variable $\deg'(V)$ follows a Poisson distribution with parameter $n\mu_n(\mathcal{B}'_{V}(R_n)) $ . Thus, by Chebyshev’s inequality,

(8.14) \begin{align}\mathbb{P}_{\scriptscriptstyle C_n}[\!\deg'(V) \leq n\mu_n(\mathcal{B}'_{V}(R_n)) + n^{-\varepsilon'}\delta_n] \xrightarrow[n\to\infty]{} 1.\end{align}

With a similar argument, we also prove that

(8.15) \begin{align}\mathbb{P}_{\scriptscriptstyle C_n}[\deg'(V') \geq n\mu_n(\mathcal{B}'_{V'}(R_n)) - n^{-\varepsilon'}\delta_n] \xrightarrow[n\to\infty]{} 1.\end{align}

Now, concerning $\deg''$ , a direct computation gives, for $r \in [0,R_n)$ ,

(8.16) \begin{align}n\mu_n(\mathcal{C}(t_n',r_n')) = o(n^{\beta} \log(n)^{3\alpha}).\end{align}

Since $\deg''(V)$ is smaller than the number of nodes that fall in the annulus $\mathcal{C}(t_n',r_n')$ , it follows from Chebyshev’s inequality that

(8.17) \begin{align}\mathbb{P}_{\scriptscriptstyle C_n}[\deg''(V) \leq n^\beta \log(n)^{3\alpha}] \xrightarrow[n\to\infty]{} 1.\end{align}

Combining the definition of V with (8.14) and (8.17), it follows that, with high probability,

\begin{align*}\deg(V)&= \deg_{\varepsilon}(V) + \deg'(V) + \deg''(V) \\&\leq n \mu_n(\mathcal{B}^{\varepsilon}_{V}(R_n)) + n \mu_n(\mathcal{B}'_{V}(R_n)) + n^{-\varepsilon'}\delta_n + n^{\beta}\log(n)^{3 \alpha}.\end{align*}

Since $\mathcal{B}^{\varepsilon}_{V}(R_n)$ and $\mathcal{B}'_{V}(R_n)$ are disjoint subsets of $\mathcal{B}_{V}(R_n)$ and since $\delta_n = n^{2 \beta + o(1)}$ , we finally obtain, with high probability,

(8.18) \begin{align}\deg(V) &\leq n \mu_n(\mathcal{B}_{V}(R_n)) + o(\delta_n).\end{align}

Likewise, combining the definition of V ^′ with (8.15) and (8.16) and using the fact that $\mathcal{B}_{V'}(R_n)$ is the union of the balls $\mathcal{B}^{\varepsilon}_{V'}(R_n)$ , $\mathcal{B}'_{V'}(R_n)$ , and $\mathcal{B}''_{V'}(R_n)$ , it follows that, with high probability,

\begin{align*}\deg(V')&\geq \deg_{\varepsilon}(V') + \deg'(V') \\&\geq n \mu_n(\mathcal{B}^{\varepsilon}_{V'}(R_n)) + 2\delta_n + n\mu_n(\mathcal{B}'_{V'}(R_n)) - n^{-\varepsilon'}\delta_n \\&\geq n \mu_n(\mathcal{B}_{V'}(R_n)) - n \mu_n(\mathcal{B}''_{V'}(R_n)) + 2\delta_n - n^{-\varepsilon'}\delta_n .\end{align*}

It follows from (8.13) and (8.16) that $n \mu_n(\mathcal{B}''_{V'}(R_n)) = o(\delta_n)$ , so we finally obtain, with high probability,

(8.19) \begin{align}\deg(V')&\geq n \mu_n(\mathcal{B}_{V'}(R_n)) + 2\delta_n + o(\delta_n).\end{align}

On the other hand, using the fact that $r(V') - r(V) \leq \lambda_n'$ , Lemma 8.1 yields

\begin{align*}n\mu_n(\mathcal{B}_{V}(R_n)) - n\mu_n(\mathcal{B}_{V'}(R_n)) \leq Kn \mathrm{e}^{-t_n'/2} \lambda_n' = o(\delta_n).\end{align*}

Combining this with (8.18) and (8.19) proves that, with high probability,

(8.20) \begin{align}\deg(V) + \delta_n \leq \deg(V').\end{align}

Since $t_n' \leq r(V) \leq r(V') \leq r_n'$ , this concludes the proof of (8.4).

Step 4: de-Poissonisation. Let us now explain how to transfer the result from the Poissonised model $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ to the original model $\mathcal{G}_{\alpha,\nu}(n)$ . We use a standard procedure that consists in coupling the original model with the Poissonised model by sampling the points of $\mathcal{G}_{\alpha,\nu}(n)$ in the following manner: sample the graph $\mathcal{G}^{\scriptscriptstyle{\text{Poi}}}_{\alpha,\nu}(n)$ and call $N_n$ the number of nodes in this graph. If $N_n > n$ , randomly remove $(N_n - n)$ nodes from the graph. If $N_n < n$ , add $(n - N_n)$ nodes, independently sampled from $\mathcal{B}_{0}(R_n)$ according to $\mu_n$ and connect all pairs of nodes that are within distance $R_n$ . The resulting graph follows the same distribution as $\mathcal{G}_{\alpha,\nu}(n)$ and is referred to as the de-Poissonised graph.

For a node x that appears in both the Poissonised and de-Poissonised graph, we denote its degree in each graph by $\deg(x)$ and $\widetilde{\deg}(x)$ , respectively. Fix $\varepsilon > 0$ small enough. The random variable $N_n$ follows a Poisson distribution with parameter n. Thus,

\begin{align*}\mathbb{P}[|N_n - n| \geq n^{1/2+\varepsilon}] \xrightarrow[n\to\infty]{} 0.\end{align*}

From this, it follows that, with high probability, the nodes V and V ^′ are not removed during the de-Poissonisation procedure. For the remainder of the proof, we work on the event that V and V ^′ are not removed. In the case where $N_n < n$ , conditionally on the position of V, the variable $\widetilde{\deg}(V) - \deg(V)$ follows a binomial distribution with $(n - N_n)$ trials and probability parameter $\mu_n(\mathcal{B}_{V}(R_n))$ . Thus, with high probability,

\begin{align*}\widetilde{\deg}(V) - \deg(V) \leq n^{1/2+2\varepsilon} \mu_n(\mathcal{B}_{r_n'}(R_n)) \leq K n^{2\beta + 2\varepsilon - 1/2} \mathrm{e}^{-w_n'/2} \delta_n = o(\delta_n),\end{align*}

where the last equality holds if $\varepsilon$ is chosen sufficiently small (because $\beta < \tfrac{1}{5}$ ). In the $N_n > n$ case, we can also prove with similar arguments that the de-Poissonisation reduces the degree of V ^′ by at most $o(\delta_n)$ . Therefore, we finally conclude from (8.20) that $\widetilde{\deg}(V) < \widetilde{\deg}(V')$ holds with high probability. In addition, straightforward estimates, using the fact that $t_n' \leq r(V) \leq r(V') \leq r_n'$ , show that in the de-Poissonised graph the ranks of the nodes V and V ^′ in the ranking of the nodes by increasing radii are in the interval $[n^{\beta},n^{\beta} a_n^{3\alpha}]$ , with high probability. It follows that (3.2) holds with high probability (the power $3\alpha$ is not problematic, as the sequence $(a_n)$ is an arbitrary sequence diverging to $+\infty$ ).

Let us conclude by proving Lemma 8.1.

Proof of Lemma 8.1. We recall the integral expression (5.1) of $\mu_n(\mathcal{B}_{r}(R_n))$ :

\begin{align*}\mu_n(\mathcal{B}_{r}(R_n)) = \frac{1}{\pi}\int_0^{R_n}\theta_r(y) \rho_n(y) \,\textrm{d}y.\end{align*}

Using the expression of $\theta_r(y)$ given by (5.4), we get

\begin{align*}\frac{\partial}{\partial r} \mu_n(\mathcal{B}_{r}(R_n)) = \frac{1}{\pi}\int_{R_n-r}^{R_n} f_{n,r}(y) \rho_n(y) \,\textrm{d}y,\end{align*}

where $f_{n,r}(y) \;:\!=\; ({\partial}/{\partial r}) \arccos(c_{n,r}(y))$ , with $c_{n,r}(y) \;:\!=\; {(\!\cosh(r)\cosh(y) -\cosh(R_n))}/{\sinh(r) \sinh(y)}$ . Since the angle $\theta_r(y)$ is decreasing in r (see Lemma 5.1), the integral above is always well defined in $\mathbb{R} \cup \{-\infty\}$ . The contribution of the integral over $(R_n-r,R_n-r+r_0)$ requires special treatment, because the quantity $f_{n,r}(y)$ diverges as y approaches $(R_n-r)$ . We show at the end of the proof that this contribution is $O(\mathrm{e}^{-\alpha r})$ (which can be incorporated into the first error term of the formula given in the statement). More precisely, we prove at the end of the proof that, for $r \in (r_0,R_n)$ ,

(8.21) \begin{align}\frac{\partial}{\partial r} \mu_n(\mathcal{B}_{r}(R_n)) = \frac{1}{\pi}\int_{R_n-r+r_0}^{R_n} f_{n,r}(y) \rho_n(y) \,\textrm{d}y + O(\mathrm{e}^{-\alpha r}).\end{align}

For now, let us estimate the integrand for $r \in (r_0,R_n)$ and $y \in (R_n-r+r_0,R_n)$ . A direct computation gives

\begin{align*}f_{n,r}(y)&= -\frac{\cosh(R_n)\cosh(r)-\cosh(y)}{\sinh(r)^2\sinh(y) \sqrt{1-c_{n,r}(y)^2}\,}.\end{align*}

Using $\cosh(x) = \mathrm{e}^x(1/2+O(\mathrm{e}^{-2x}))$ and $\sin(x) = \mathrm{e}^x(1/2+O(\mathrm{e}^{-2x}))$ , it follows that

(8.22) \begin{align}f_{n,r}(y)&= -\frac{2\mathrm{e}^{R_n-r-y}(1+O(\mathrm{e}^{-2r})+O(\mathrm{e}^{y-R_n-r}))}{(1+O(\mathrm{e}^{-2r} + \mathrm{e}^{-2y})) \sqrt{1-c_{n,r}(y)^2}}.\end{align}

If $r_0$ is chosen large enough, using $1/(1+x) = 1+O(x)$ for $|x| < \tfrac{1}{2}$ , we get

(8.23) \begin{align}f_{n,r}(y)&= -\frac{2\mathrm{e}^{R_n-r-y}(1+O(\mathrm{e}^{y-R_n-r}) + O(\mathrm{e}^{-2y}))}{\sqrt{1-c_{n,r}(y)^2} }.\end{align}

Note that we used the fact that $y-R_n-r \geq -2r$ to get rid of the $O(\mathrm{e}^{-2r})$ term. A similar computation gives, for $r \in (r_0,R_n)$ and $y \in (R_n-r+r_0,R_n)$ ,

\begin{align*}c_{n,r}(y)&= 1-2\mathrm{e}^{R_n-r-y}+O(\mathrm{e}^{-2r}+\mathrm{e}^{-2y}) .\end{align*}

Thus, if $r_0$ is chosen large enough, we have

\begin{align*}\frac{1}{\sqrt{1-c_{n,r}(y)^2}\,}&= \frac{\mathrm{e}^{-(R_n-r-y)/2}}{2}(1+O(\mathrm{e}^{R_n-r-y})).\end{align*}

Substituting this into (8.23) yields, for $r \in (r_0,R_n)$ and $y \in (R_n-r+r_0,R_n)$ ,

\begin{align*}f_{n,r}(y)&= -\mathrm{e}^{(R_n-r-y)/2}(1 + O(\mathrm{e}^{R_n-r-y})).\end{align*}

Estimating the hyperbolic terms in $\rho_n(y)$ , it follows that

(8.24) \begin{align}f_{n,r}(y)\rho_n(y)&= - \alpha \mathrm{e}^{(1/2-\alpha)R_n-r/2}\mathrm{e}^{(\alpha-1/2)y}(1 + O(\mathrm{e}^{R_n-r-y}) + O(\mathrm{e}^{-\alpha R_n})).\end{align}

Solving the integral in (8.21) using (8.24), without taking into account the error terms, gives

\begin{align*}-\frac{\alpha}{\pi} \int_{R_n-r+r_0}^{R_n} \mathrm{e}^{(1/2-\alpha)R_n-r/2}\mathrm{e}^{(\alpha-1/2)y} \,\textrm{d}y&= -\frac{C_{\alpha}}{2} \mathrm{e}^{-r/2} (1 + O(\mathrm{e}^{-(\alpha - 1/2)r})).\end{align*}

If $\alpha \neq \tfrac32$ then the integral over the first error term gives

\begin{align*}O\Bigg(\int_{R_n-r+r_0}^{R_n} \mathrm{e}^{(1/2-\alpha)R_n-r/2}\mathrm{e}^{(\alpha-1/2)y} \mathrm{e}^{R_n-r-y} \,\textrm{d}y \Bigg)&= O(\mathrm{e}^{-3r/2}) + O(\mathrm{e}^{-\alpha r}).\end{align*}

In the $\alpha = \tfrac32$ case, the addition of the error term $r\mathrm{e}^{-r}$ in the statement makes it correct. The integral over the second error term (corresponding to the $O(\mathrm{e}^{-\alpha R_n})$ term of (8.24)) is of order $O(\mathrm{e}^{-r/2 - \alpha R_n})$ , which can be incorporated into the first error term of the result. Thus, the computation of the integral appearing in (8.21) yields the correct estimate.

It remains to prove (8.21). This requires a bound on the speed of divergence of the function $f_{n,r}(y)$ , as y approaches $(R_n-r)$ . To obtain this bound, we first need to bound $c_{n,r}(y)$ , for y in the neighbourhood of $(R_n -r)$ . In the following, K stands for a positive constant whose value depends only on $\alpha, \nu$ , and $r_0$ and may change throughout the proof. For two functions f and g, we denote by $f \wedge g$ (respectively $f \vee g$ ) the minimum (respectively maximum) of f and g. Using the formula $\cosh(a+b) = \cosh(a)\cosh(b) + \sinh(a)\sinh(b)$ and the mean value theorem to bound the difference $\cosh(r+y) - \cosh(R_n)$ , we obtain, for $y \in (R_n-r,R_n-r+r_0)$ ,

\begin{align*}c_{n,r}(y)&= \frac{\cosh(r+y)- \sinh(r)\sinh(y) - \cosh(R_n)}{\sinh(r)\sinh(y)}\\&\geq K(y-(R_n-r))\,\mathrm{e}^{(R_n-r)-y} - 1.\end{align*}

Since $(R_n-r) - y \geq -r_0$ , we finally get

\begin{align*}c_{n,r}(y)&\geq (K(y-(R_n-r)) - 1) \wedge 0.\end{align*}

On the other hand, for $r_0$ and large enough n, approximating the hyperbolic terms by exponentials gives, for $y \in (R_n-r,R_n-r+r_0)$ ,

\begin{align*}c_{n,r}(y)&\leq 1 - K.\end{align*}

Thus, for $y \in (R_n-r,R_n-r+r_0)$ ,

\begin{align*}\frac{1}{\sqrt{1-c_{n,r}(y)^2}\,} \leq \frac{1}{\sqrt{1-(1-K)^2}\,} \vee \frac{1}{\sqrt{1-((K(y-(R_n-r)) - 1) \wedge 0)^2}\,}.\end{align*}

It follows that the function ${1}/{\sqrt{1-c_{n,r}(y)^2}\,}$ is integrable over the interval $(R_n-r,R_n-r+r_0)$ and the integral is bounded by a constant that does not depend on n. From this and the approximation of $f_{n,r}(y)$ given in (8.22) (which also holds for $y \in [R_n-r,R_n-r+r_0]$ ), we get

\begin{align*}\int_{R_n-r}^{R_n-r+r_0} f_{n,r}(y) \rho_n(y) \,\textrm{d}y = O(\rho_n(R_n-r+r_0)) = O(\mathrm{e}^{-\alpha r}).\end{align*}

This completes the proof.