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Largest eigenvalue of the configuration model
and breaking of ensemble equivalence

Pierfrancesco Dionigi Mathematical Institute, Leiden University, P.O. Box 9512, 2300 RA Leiden, The Netherlands. [email protected] Diego Garlaschelli Lorentz Institute for Theoretical Physics, Leiden University, P.O.  Box 9504, 2300 RA Leiden, The Netherlands & IMT School for Advanced Studies, Piazza S. Francesco 19, 55100 Lucca, Italy & INdAM-GNAMPA Istituto Nazionale di Alta Matematica, Italy [email protected] Rajat Subhra Hazra  and  Frank den Hollander Mathematical Institute, Leiden University, P.O. Box 9512, 2300 RA Leiden, The Netherlands [email protected] [email protected]
Abstract.

We analyse the largest eigenvalue of the adjacency matrix of the configuration model with large degrees, where the latter are treated as hard constraints. In particular, we compute the expectation of the largest eigenvalue for degrees that diverge as the number of vertices nn tends to infinity, uniformly on a scale between 11 and n\sqrt{n}, and show that a weak law of large numbers holds. We compare with what was derived in our earlier paper [14] for the Chung-Lu model, which in the regime considered represents the corresponding configuration model with soft constraints, and show that the expectation is shifted down by 11 asymptotically. This shift is a signature of breaking of ensemble equivalence between the hard and soft (also known as micro-canonical and canonical) versions of the configuration model. The latter result generalizes a previous finding [13] obtained in the case when all degrees are equal.

Key words and phrases:
Constrained random graphs; micro-canonical and canonical ensembles; adjacency matrix; largest eigenvalue; breaking of ensemble equivalence.
2000 Mathematics Subject Classification:
60B20,60C05, 60K35
PD, RSH, FdH and MM are supported by the Netherlands Organisation for Scientific Research (NWO) through the Gravitation-grant NETWORKS-024.002.003. DG is supported by the Dutch Econophysics Foundation (Stichting Econophysics, Leiden, The Netherlands) and by the European Union - NextGenerationEU - National Recovery and Resilience Plan (Piano Nazionale di Ripresa e Resilienza, PNRR), project ‘SoBigData.it - Strengthening the Italian RI for Social Mining and Big Data Analytics’ - Grant IR0000013 (n. 3264, 28/12/2021) (https://pnrr.sobigdata.it/).

1. Introduction and main results

Motivation.

Spectral properties of adjacency matrices in random graphs play a crucial role in various areas of network science. The largest eigenvalue is particularly sensitive to the graph architecture, making it a key focus. In this paper we focus on a random graph with a hard constraint on the degrees of nodes. In the homogeneous case (all degrees equal to dd), it reduces to a random dd-regular graph. In the heterogeneous case (different degrees), it is known as the configuration model. Our interest is characterizing the expected largest eigenvalue of the configuration model and comparing it with the same quantity for a corresponding random graph model where the degrees are treated as soft constraints.

The set of dd-regular graphs on nn vertices, with d=d(n)d=d(n), is non-empty when 1dn11\leq d\leq n-1 and dndn is even. Selecting a graph uniformly at random from this set results in what is known as the random dd-regular graph, denoted by Gn,dG_{n,d}. The spectral properties of Gn,dG_{n,d} are well-studied for d2d\geq 2 (for d=1d=1, the graph is trivially a set of disconnected edges). For instance, all eigenvalues of its adjacency matrix fall in the interval [d,d][-d,d], with the largest eigenvalue being dd. The computation of λ=max{|λ2|,|λn|}\lambda=\max\{|\lambda_{2}|,|\lambda_{n}|\}, where λ2\lambda_{2} and λn\lambda_{n} are the second-largest and the smallest eigenvalue, respectively, has been challenging. It is well known that for fixed dd the empirical distribution function of the eigenvalues of Gn,dG_{n,d} converges to the so-called Kesten-McKay law [22], the density of which is given by

fKM(x)={d4(d1)x22π(d2x2)if |x|2d1,0otherwise.f_{\rm KM}(x)=\begin{cases}\frac{d\sqrt{4(d-1)-x^{2}}}{2\pi(d^{2}-x^{2})}&\quad\text{if }|x|\leq 2\sqrt{d-1},\\ 0&\quad\text{otherwise}.\end{cases}

A lower bound of the type λ2d1\lambda\geq 2\sqrt{d-1} was expected from this, and was indeed proven in the celebrated Alon-Boppana theorem in [1], stating that λ2d1+Od(log2(n))\lambda\geq 2\sqrt{d-1}+O_{d}(\log^{-2}(n)), where the constant in the error term only depends on dd. In the same paper an upper bound of the type λ2d1+o(1)\lambda\leq 2\sqrt{d-1}+o(1) was conjectured . This conjecture was later proven in the pioneering work [17] (for a shorter proof, see [6]). When d=d(n)d=d(n)\to\infty, the spectral distribution converges to the semi-circular law [26]. It was proven in [10] that λ=O(d)\lambda=O(\sqrt{d}\,) with high probability when d=o(n)d=o(\sqrt{n}\,). This result was extended in [3] to d=o(n2/3)d=o(n^{2/3}). For recent results and open problems, we refer the reader to [30].

There has not been much work on the inhomogeneous setting where not all the degrees of the graph are the same. The natural extension of the regular random graph is the configuration model, where a degree sequence dn=(d1,,dn)\vec{d}_{n}=(d_{1},\ldots,d_{n}) is prescribed. Unless otherwise specified, we assume the degrees to be hard constraints, i.e. realized exactly on each configuration of the model (this is sometimes called the ‘microcanonical’ configuration model, as opposed to the ‘canonical’ one where the degrees are soft constraints realized only as ensemble averages [24, 20]). The empirical spectral distribution of the configuration model is known under some assumptions on the growth of the sum of the degrees. When i=1ndi=O(n)\sum_{i=1}^{n}d_{i}=O(n), the graph locally looks like a tree. It has been shown that the empirical spectral distribution exists and that the limiting measure is a functional of a size-biased Galton-Watson tree [7]. When i=1ndi\sum_{i=1}^{n}d_{i} grows polynomially with nn, the local geometry is no longer a tree. This case was recently studied in [12], where it is shown that the appropriately scaled empirical spectral distribution of the configuration model converges to the free multiplicative convolution of the semi-circle law with a measure that depends on the empirical degree sequence. The scaling of the largest and the second-largest eigenvalues has not yet been studied in full generality.

For the configuration model the behaviour of the largest eigenvalue is non-trivial. In the present paper, we consider the configuration model with large degrees, compute the expectation of the largest eigenvalue of its adjacency matrix, and prove a weak law of large numbers. When dd\to\infty, the empirical distribution of Gn,dG_{n,d} (appropriately scaled) converges to the semi-circle law. Also, if we look at Erdős-Rényi random graphs on nn vertices with connection probability d/nd/n, then the appropriately scaled empirical distribution converges to the semi-circle law. It is known that the latter two random graphs exhibit different behaviour for the expected largest eigenvalue [13]. This can be understood in the broader perspective of breaking ensemble equivalence [24, 20], which we discuss later. In the inhomogeneous setting, the natural graph to compare the configuration model with is the Chung-Lu model. For the case where the sum of the degrees grows with nn, it was shown in [11] that the empirical spectral distribution converges to the free multiplicative product of the semi-circle law with a measure that depends on the degree sequence, similar to [12]. In [14], we investigated the largest eigenvalue and its expectation for the Chung-Lu model (and even derived a central limit theorem). In the present paper we complete the picture by showing that, when the degrees are large, there is a gap between the expected largest eigenvalue of the Chung-Lu model and the configuration model. We refer the reader to the references listed in [13, 14] for more background.

Main theorem.

For nn\in\mathbb{N}, let CM(dn)\mathrm{CM}(\vec{d}_{n}) be the random graph on nn vertices generated according to the configuration model with degree sequence dn=(di)i[n]n\vec{d}_{n}=(d_{i})_{i\in[n]}\in\mathbb{N}^{n} [27, Chapter 7.2]. Define

m0=mini[n]di,m=maxi[n]di,mk=i[n](di)k,k.m_{0}=\min_{i\in[n]}d_{i},\qquad m_{\infty}=\max_{i\in[n]}d_{i},\qquad m_{k}=\sum_{i\in[n]}(d_{i})^{k},\quad k\in\mathbb{N}.

Throughout the paper we need the following assumptions on dn\vec{d}_{n} as nn\to\infty.

Assumption 1.1.


(D1) Bounded inhomogeneity: m0mm_{0}\asymp m_{\infty}.
(D2) Connectivity and sparsity: 1mn1\ll m_{\infty}\ll\sqrt{n}. \spadesuit

Under these assumptions, CM(dn)\mathrm{CM}(\vec{d}_{n}) is with high probability non-simple [27, Chapter 7.4]. We write \mathbb{P} and 𝔼\mathbb{E} to denote probability and expectation with respect to the law of CM(dn)\mathrm{CM}(\vec{d}_{n}) conditional on being simple, suppressing the dependence on the underlying parameters.

Let ACM(dn)A_{\mathrm{CM}(\vec{d}_{n})} be the adjacency matrix of CM(dn)\mathrm{CM}(\vec{d}_{n}). Let (λi)i[n](\lambda_{i})_{i\in[n]} be the eigenvalues of ACM(dn)A_{\mathrm{CM}(\vec{d}_{n})}, ordered such that λ1λn\lambda_{1}\geq\dots\geq\lambda_{n}. We are interested in the behaviour of λ1\lambda_{1} as nn\to\infty. Our main theorem reads as follows.

Theorem 1.2.

Subject to Assumption 1.1,

𝔼[λ1]=m2m1+m1m3m221+o(1),n,\mathbb{E}\left[\lambda_{1}\right]=\frac{m_{2}}{m_{1}}+\frac{m_{1}m_{3}}{m_{2}^{2}}-1+o(1),\qquad n\to\infty, (1.1)

and

λ1𝔼[λ1]1 in -probability. \frac{\lambda_{1}}{\mathbb{E}\left[\lambda_{1}\right]}\to 1\text{ in $\mathbb{P}$-probability. }

In [14] we looked at an alternative version of the configuration model, called the Chung-Lu model CMn(dn)\mathrm{CM}^{*}_{n}(\vec{d}_{n}), where the average degrees, rather than the degrees themselves, are fixed at dn\vec{d}_{n}. This is an ensemble with soft constraints; in the considered regime for the degrees, it coincides with a maximum-entropy ensemble, also called ‘canonical’ configuration model [24]. For this model we showed that, subject to Assumption 1.1,

𝔼[λ1]=m2m1+m1m3m22+o(1),n,\mathbb{E}^{*}\left[\lambda_{1}\right]=\frac{m_{2}}{m_{1}}+\frac{m_{1}m_{3}}{m_{2}^{2}}+o(1),\qquad n\to\infty, (1.2)

and λ1/𝔼[λ1]1\lambda_{1}/\mathbb{E}^{*}\left[\lambda_{1}\right]\to 1 in \mathbb{P}^{*}-probability, where \mathbb{P}^{*} and 𝔼\mathbb{E}^{*} denote expectation with respect to the law of CMn(dn)\mathrm{CM}^{*}_{n}(\vec{d}_{n}) and λ1\lambda_{1} is the largest eigenvalue of the ACMn(dn)A_{\mathrm{CM}^{*}_{n}(\vec{d}_{n})}. The notable difference between (1.1) and (1.2) is the shift by 1-1.

For the special case where all the degrees are equal to dd, we have m0=m=dm_{0}=m_{\infty}=d and mk=ndkm_{k}=nd^{k}, and so 𝔼[λ1]=d+o(1)\mathbb{E}[\lambda_{1}]=d+o(1) and 𝔼[λ1]=d+1+o(1)\mathbb{E}^{*}[\lambda_{1}]=d+1+o(1). In fact, (λ1=d)=1\mathbb{P}(\lambda_{1}=d)=1. Since in this model the degrees can fluctuate with the same law (soft constraint in the physics literature), this case reduces to the Erdős-Rényi random graph for which results on 𝔼[λ1]\mathbb{E}^{*}[\lambda_{1}] were already well known in [21, 19] and further analyzed in [15].

Breaking of ensemble equivalence.

The shift by 1-1 was proven in [13] for the homogeneous case with equal degrees and is a spectral signature of breaking of ensemble equivalence [24]. Indeed, a dd-regular graph is the ‘micro-canonical’ version of a random graph where all degrees are equal and ‘hard’, and the Erdős-Rényi random graph is the corresponding ‘canonical’ version where all degrees are equal and ‘soft’. More in general, CM(dn)\mathrm{CM}(\vec{d}_{n}) is the micro-canonical configuration model where the constraint on the degrees is ‘hard’, while CMn(dn)\mathrm{CM}^{*}_{n}(\vec{d}_{n}) is the canonical version where the constraint is ‘soft’. We refer the reader to [20] for the precise definition of these two configuration model ensembles and for the proof that they are not asymptotically equivalent in the measure sense [25]. This means that the relative entropy per node of \mathbb{P} with respect to \mathbb{P}^{*} has a strictly positive limit as nn\to\infty. This shows that the choice of constraint matters, not only on a microscopic scale but also on a macroscopic scale. Indeed, for non-equivalent ensembles one expects the existence of certain macroscopic properties that have different expectation values in the two ensembles (macrostate (in)equivalence [25]). The fact that the largest eigenvalue picks up this discrepancy is interesting. What is remarkable is that the shift by 1-1, under the hypotheses considered, holds true also in the case of heterogeneous degrees and remains the same irrespective of the scale of the degrees and of the distribution of the degrees on this scale.

Outline.

The remainder of this paper is organised as follows. In Section 2 we look at the issue of simplicity of the graph. In Section 3 we bound the spectral norm of the matrix

H=ACM(dn)𝔼[ACM(dn)].H=A_{\mathrm{CM}(\vec{d}_{n})}-\mathbb{E}[A_{\mathrm{CM}(\vec{d}_{n})}]. (1.3)

We use the proof of [10] to show that H=o(m)\|H\|=\mathrm{o}(m_{\infty}) with high probability. Using the latter we show that

λ1λ1(𝔼[ACM(dn)])m2m1\lambda_{1}\sim\lambda_{1}\left(\mathbb{E}[A_{\mathrm{CM}(\vec{d}_{n})}]\right)\sim\frac{m_{2}}{m_{1}}

with high probability. In Section 4 we use the estimates in Section 3 to prove Theorem 1.2.

2. Configuration model and simple graphs with a given degree sequence

In the context of random graphs with hard constraints on the degrees, we work with both the configuration model (a random multi-graph with a prescribed degree sequence) and with the conditional configuration model (a random simple graph with a prescribed degree sequence). We view the second as a special case of the first.

Configuration model.

The configuration model with degree sequence d=(d1,,dn)\vec{d}=(d_{1},\dots,d_{n}), CM(dn)\mathrm{CM}(\vec{d}_{n}), generates a graph through a perfect matching PP of the set of half-edges =i[n]{i}×[di]\mathcal{E}=\cup_{i\in[n]}\{i\}\times[d_{i}], i.e., P:P\colon\,\mathcal{E}\to\mathcal{E} such that PP is an isomorphism and an involution, and P(α)αP(\alpha)\neq\alpha for all α\alpha\in\mathcal{E}.

Refer to caption
(a)
Refer to caption
(b)

It turns out that a pairing scheme, such as matching half-edges from left to right or selecting pairs uniformly at random, is considered admissible as long as it yields the correct probability for a perfect matching (see, for example, [27, Lemma 7.6]). An important property of each admissible pairing scheme is that

CM(dn)(αβ)=1m11α,β.\mathbb{P}_{\mathrm{CM}(\vec{d}_{n})}\left(\alpha\sim\beta\right)=\frac{1}{m_{1}-1}\qquad\forall\alpha,\beta\in\mathcal{E}. (2.1)

We can therefore endow \mathcal{E} with the lexicographic order. An element α\alpha\in\mathcal{E}, α=(i,)\alpha=(i,\ell), is associated with a vertex v(α)=iv(\alpha)=i through the first component of α\alpha. An edge ee is an element of ×\mathcal{E}\times\mathcal{E} given by the pairing e={α,β}={α,P(α)}e=\{\alpha,\beta\}=\{\alpha,P(\alpha)\}. We define the configuration 𝒞\mathcal{C} as the set of all such ee. Given the lexicographic order, we may assume that α<β\alpha<\beta and identify arbitrarily the head and the tail of the edge: t(e)=v(α)t(e)=v(\alpha), h(e)=v(β)h(e)=v(\beta). We can then order the edges in their order of appearance via t(e)t(e), forming a list {ei,i=1,,m1/2}\{e_{i},\,i=1,\ldots,m_{1}/2\}. These properties of the configuration model will be used in Subsection 3.1. For the configuration model it is easy to check that111We adopt the convention that a self-loop contributes 22 to the degree, i.e., aiia_{ii} is twice the number of self-loops attached to vertex ii. This convention is useful because it yields jaij=di\sum_{j}a_{ij}=d_{i}.

𝔼[aii]=di(di1)m11,𝔼[aij]=didjm11,ij.\mathbb{E}[a_{ii}]=\frac{d_{i}(d_{i}-1)}{m_{1}-1},\qquad\mathbb{E}[a_{ij}]=\frac{d_{i}d_{j}}{m_{1}-1},\quad i\neq j. (2.2)

Indeed every matrix element aija_{ij}, can be expressed as

k=1dih=1dj𝟙(αβ,α=(i,k),β=(j,h)),\sum_{k=1}^{d_{i}}\sum^{d_{j}}_{h=1}\mathbbm{1}\big{(}\alpha\sim\beta,\alpha=(i,k),\beta=(j,h)\big{)},

where 𝟙(E)\mathbbm{1}(E) is the indicator function of the event EE. Taking expectations and using (2.1), we get 2.2.

Simple graphs with a given degree sequence.

A linked but different model is the one that samples uniformly at random a simple graph with a given degree sequence d\vec{d}, 𝒢(d)\mathcal{G}(\vec{d})222With a little abuse of notation we permit our simple graphs to have self-loops. We maintain the convention explained in the previous footnote.. An immediate question is whether there is any relation between CM(d)\mathrm{CM}(\vec{d}) and 𝒢(d)\mathcal{G}(\vec{d}). It turns out that the following is true (see [27] for reference):

Lemma 2.1.

Let be GnG_{n} a graph generated via the configuration model CM(dn)\mathrm{CM}(\vec{d}_{n}) with degree sequence d\vec{d}. Then

CM(dn)(Gn is simple)=exp[O(m12n2)]\mathbb{P}_{\mathrm{CM}(\vec{d}_{n})}(G_{n}\textnormal{ is simple})=\exp\left[-O\left(\frac{m^{2}_{1}}{n^{2}}\right)\right] (2.3)

and

CM(dn)(Gn=Gn is simple) is uniform.\mathbb{P}_{\mathrm{CM}(\vec{d}_{n})}\left(G_{n}=\cdot\mid G_{n}\textnormal{ is simple}\right)\textnormal{ is uniform}. (2.4)

Thus, we can identify the law of the uniform random graph with a given degree sequence with the law of the configuration model conditional on simplicity, i.e.,

𝒢(d)()=CM(d)(G is simple).\mathbb{P}_{\mathcal{G}(\vec{d})}(\cdot)=\mathbb{P}_{\mathrm{CM}(\vec{d})}(\cdot\mid\text{G is simple}).

It follows that the expectation under the uniform random graph with a given degree sequence can be expressed as a the expectation of the configuration model conditional on simplicity. As explained in [28, Remark 1.14], we have

𝔼𝒢(d)[aij]=𝒢(d)(ij)=(1+o(1))didjm1+didjdidjm1(1+O(1/m))=didjm1+O(1/m).\begin{split}\mathbb{E}_{\mathcal{G}(\vec{d})}[a_{ij}]&=\mathbb{P}_{\mathcal{G}(\vec{d})}(i\sim j)=(1+o(1))\frac{d_{i}d_{j}}{m_{1}+d_{i}d_{j}}\\ &\leq\frac{d_{i}d_{j}}{m_{1}(1+O(1/m_{\infty}))}=\frac{d_{i}d_{j}}{m_{1}}+O(1/m_{\infty}).\end{split} (2.5)

3. Bound on H\|H\|

In this section we derive the following bound on the spectral norm of the matrix HH defined in (1.3). This bound will play a crucial role in the proof of Theorem 1.2 in Section 4.

Theorem 3.1.

For every K>0K>0, there exists a constant C>0C>0 such that

HCm with probability 1o(nK),\|H\|\leq C\sqrt{m_{\infty}}\,\text{ with probability }1-o(n^{-K}),

where the constant CC depends on mm0\frac{m_{\infty}}{m_{0}} and KK only.

The proof is given in Section 3.1 and we follow the proof of Lemma 16 of [10]. In Sections 3.23.3 we derive some estimates that are needed in Section 3.1.

We have

H=max{λ1(H),|λn(H)|}.\|H\|=\max\{\lambda_{1}(H),\left|\lambda_{n}(H)\right|\}.

Recall (1.3). In Lemma 3.3 below we will see that 𝔼[ACM(dn)]\mathbb{E}[A_{\mathrm{CM}(\vec{d}_{n})}] is asymptotically rank 11 (more precisely, the other eigenvalues are of order O(n1)O(n^{-1}). This means that λi(ACM(dn))=O(n1)\lambda_{i}\left(A_{\mathrm{CM}(\vec{d}_{n})}\right)=O(n^{-1}), i1i\neq 1. Therefore, by Weyl’s inequality, it easy to see that λ1(H)\lambda_{1}(H) and λn(H)\lambda_{n}(H) are bounded above and below by λ2(AGn)\lambda_{2}(A_{G_{n}}) and λn(AGn)\lambda_{n}(A_{G_{n}}) with error O(n1)O(n^{-1}), while λ1(AGn)\lambda_{1}(A_{G_{n}}) is in a window of size m\sqrt{m_{\infty}} around m2m11\frac{m_{2}}{m_{1}-1}. We therefore have the following Corollary, which will be needed in Section 4:

Corollary 3.2.

For every K>0K>0,

λ1(AGn)λ1(𝔼[AGn])=O(m) with probability 1o(nK).\lambda_{1}(A_{G_{n}})-\lambda_{1}(\mathbb{E}[A_{G_{n}}])=O(\sqrt{m_{\infty}}\,)\text{ with probability }1-o(n^{-K}).

3.1. Spectral estimates

Notation.

Abbreviate Gn=CM(dn)G_{n}=\mathrm{CM}(\vec{d}_{n}). Let UU be uniformly distributed on [n][n], let dUd_{U} be the degree of a vertex that is picked uniformly at random, and let

ωn=𝔼[dU]\omega_{n}=\mathbb{E}[d_{U}]

be the average of the empirical degree distribution. Under Assumption 1.1, ωn\omega_{n}\to\infty and ωn=o(n)\omega_{n}=o(n). We define the normalised degree sequence (d^i)i[n](\hat{d}_{i})_{i\in[n]} as

d^i=diωn\hat{d}_{i}=\frac{d_{i}}{\omega_{n}} (3.1)

and the normalised adjacency matrix A^Gn\hat{A}_{G_{n}} as

A^Gn=AGnωn.\hat{A}_{G_{n}}=\frac{A_{G_{n}}}{\sqrt{\omega_{n}}}.

In the following we will need multiples of the vector

e~i=dim11,  1in,\tilde{e}_{i}=\frac{d_{i}}{\sqrt{m_{1}-1}},\,\,1\leq i\leq n, (3.2)

which is the eigenvector corresponding to the rank-11 approximation of 𝔼[AGn]\mathbb{E}[A_{G_{n}}], as it easy to check using (2.2).

Two lemmas.

The matrix 𝔼[AGn]\mathbb{E}[A_{G_{n}}] is asymptotically rank 1:

Lemma 3.3.

Define

e~=(e~1,,e~n)t\tilde{e}=\left(\tilde{e}_{1},\dots,\tilde{e}_{n}\right)^{t}

and

Ashift=𝔼[AGn]|e~e~|=1m11diag(d1,,dn).A_{\textnormal{shift}}=\mathbb{E}[A_{G_{n}}]-|\tilde{e}\rangle\langle\tilde{e}|=-\frac{1}{m_{1}-1}\,\textnormal{diag}\left(d_{1},\dots,d_{n}\right).

Then

Ashift=O(n1).\|A_{\textnormal{shift}}\|=O(n^{-1}).
Proof.

Using (2.2), we have AshiftA_{\textnormal{shift}} is a diagonal matrix and hence Ashiftmm11=O(n1)\|A_{\textnormal{shift}}\|\leq\frac{m_{\infty}}{m_{1}-1}=O(n^{-1}) under Assumption 1.1. Note that when all the degrees are equal to dd, then this bound is sharp, i.e., Ashift=ddn1\|A_{\textnormal{shift}}\|=\frac{d}{dn-1}, and 𝔼[AGn]\mathbb{E}[A_{G_{n}}] is exactly rank 1. ∎

It is shown in [12] that the empirical spectral distribution of AGnA_{G_{n}} has a deterministic limit given by μ=μscμD^\mu=\mu_{sc}\boxtimes\mu_{\hat{D}}, where μsc\mu_{sc} is the standard Wigner semicircle law, μD^\mu_{\hat{D}} is the distribution of d^U\hat{d}_{U}, and \boxtimes is the free product defined in [29, 4, 2].

Lemma 3.4.

Let (d^i)i[n](\hat{d}_{i})_{i\in[n]} be the normalised degree sequence, and suppose that Assumption 1.1 holds and

1ni=1nδd^iμD^.\frac{1}{n}\sum_{i=1}^{n}\delta_{\hat{d}_{i}}\Rightarrow\mu_{\hat{D}}. (3.3)

Then μD^\mu_{\hat{D}} is compactly supported.

Proof.

By Assumption 1.1 D(2),

0<clim infnminid^imaxid^ilim supnmaxid^iminid^iC<.0<c\leq\liminf_{n\to\infty}\frac{\min_{i}\hat{d}_{i}}{\max_{i}\hat{d}_{i}}\leq\limsup_{n\to\infty}\frac{\max_{i}\hat{d}_{i}}{\min_{i}\hat{d}_{i}}\leq C<\infty.

Hence the support of d^U\hat{d}_{U} is contained in a multiple of [c,C][c,C], and (3.3) implies that μD^\mu_{\hat{D}} is compactly supported. ∎

Key estimates.

From [12, Theorem 1] we know that, under Assumption 1.1, the empirical spectral distribution of A^Gn\hat{A}_{G_{n}}, written

μA^Gn,\mu_{\hat{A}_{G_{n}}},

converges weakly to μscμD^\mu_{\text{sc}}\boxtimes\mu_{\hat{D}}. The fact that μsc\mu_{sc} and μD^\mu_{\hat{D}} are compactly supported implies that μscμD^\mu_{\text{sc}}\boxtimes\mu_{\hat{D}} is compactly supported too. However, this still allows that HH has o(n)o(n) outliers, which possibly control H\|H\|. It is hard to analyse H\|H\| for sparse graphs, because it is related to expansion properties of the graph, mixing times of random walks on the graph, and more. To prove that H=O(m)\|H\|=O(\sqrt{m_{\infty}}\,), we will use the argument described in [10], which is an adaptation of the argument in [18]. We are only after the order of H\|H\|, not sharp estimates.

For random regular graphs with fixed degree, the problem of computing H\|H\| was solved in [16]. We are not aware of any proof concerning the second largest eigenvalue of configuration models with bounded degrees, but any sharp bound must come from techniques of the type employed in [16, 5]. In what follows we prove Theorem 3.1 based on [18] and its adaptation to the inhomogeneous setting in [10]. In the following, we present the proof as outlined in the latter paper, adapting it to our case to make our paper self-contained. This adjustment is necessary because the result we need is somewhat obscured in the context in which it appears in [10].

Let Dn=diag(d1,,dn)D_{n}=\text{diag}(d_{1},\dots,d_{n}). The transition kernel of a random walk on the graph GnG_{n} is given by Pn=Dn1AGnP_{n}=D_{n}^{-1}A_{G_{n}}. The random matrix PnP_{n} has as principal normalised eigenvector 1=(1,,1)/n\vec{1}=(1,\dots,1)/\sqrt{n} with eigenvalue 1. Define

λ=max{λ2(Pn),|λn(Pn)|}.\lambda^{*}=\max\{\lambda_{2}(P_{n}),|\lambda_{n}(P_{n})|\}.

Note that the matrices PnP_{n} and Sn=Dn1/2PnDn1/2S_{n}=D_{n}^{1/2}P_{n}D_{n}^{-1/2} have the same spectrum by a similarity transformation. Hence we can write the Rayleigh formula

λ=maxz:z,1=0|Pnz,z|z2=maxx:x,d=0|Snx,x|x2,\lambda^{*}=\max_{z\colon\,\langle z,\vec{1}\rangle=0}\frac{|\langle P_{n}z,z\rangle|}{\|z\|^{2}}=\max_{x\colon\,\langle x,\sqrt{\vec{d}}\rangle=0}\frac{|\langle S_{n}x,x\rangle|}{\|x\|^{2}},

where d=(d1,,dn)\sqrt{\vec{d}}=(\sqrt{d_{1}},\ldots,\sqrt{d_{n}}). Define Mn=Dn1AnDn1M_{n}=D_{n}^{-1}A_{n}D_{n}^{-1}. Let λ~\tilde{\lambda} be the second largest eigenvalue of MnM_{n}. Then

Snx,x=Dn1/2MnDn1/2x,x=MnDn1/2x,Dn1/2x.\langle S_{n}x,x\rangle=\langle D_{n}^{1/2}M_{n}D_{n}^{1/2}x,x\rangle=\langle M_{n}D_{n}^{1/2}x,D_{n}^{1/2}x\rangle.

Since

x,xx21mDn1/2x,Dn1/2xx2,\frac{\langle x,x\rangle}{\|x\|^{2}}\geq\frac{1}{m_{\infty}}\frac{\langle D_{n}^{1/2}x,D_{n}^{1/2}x\rangle}{\|x\|^{2}},

putting y=Dn1/2xy=D_{n}^{1/2}x we see that

λmmaxy,1=0|Mny,y|y2=mλ~,\lambda^{*}\leq m_{\infty}\max_{\langle y,\vec{1}\rangle=0}\frac{|\langle M_{n}y,y\rangle|}{\|y\|^{2}}=m_{\infty}\tilde{\lambda}, (3.4)

which gives a bound of the type H=O(m2λ~)\|H\|=O(m_{\infty}^{2}\tilde{\lambda}). Note that the matrix elements of MnM_{n} can be expressed as

(Mn)ij=aijdidj,(M_{n})_{ij}=\frac{a_{ij}}{d_{i}d_{j}}, (3.5)

where aija_{ij} counts the number of edges between vertices ii and jj, with the convention for the diagonal elements stated earlier. In view of (3.4), in order to obtain a bound on λ\lambda^{*} we must focus on λ~\tilde{\lambda}. In fact, we must show that

λ~=O(m3/2),\tilde{\lambda}=O(m_{\infty}^{-3/2}), (3.6)

in order to obtain the desired bound H=O(m)\|H\|=O(\sqrt{m_{\infty}}\,). To achieve this we proceed in steps:

  1. We reduce the computation of λ~\tilde{\lambda} to the analysis of two terms.

  2. In (3.7) below we identify the leading order of 𝔼[λ~]\mathbb{E}[\tilde{\lambda}] from these two terms, which turns out to be O(m3/2)O(m_{\infty}^{-3/2}).

  3. We show that the other terms are of higher order and therefore are negligible.

  4. We prove concentration around the mean through concentration of the leading order term in Lemma 3.8 below.

  5. Recalling the denominator of (3.5), we get a bound on H\|H\| after multiplying by m2m_{\infty}^{2}.

To prove (3.6), we reduce the problem to a maximisation problem in a simpler space. Namely, let ε(0,1)\varepsilon\in(0,1), and

T={x(εn)n:i[n]xi=0,i[n]xi21}.T=\left\{x\in\left(\frac{\varepsilon}{\sqrt{n}}\mathbb{Z}\right)^{n}\colon\,\sum_{i\in[n]}x_{i}=0,\,\sum_{i\in[n]}x_{i}^{2}\leq 1\right\}.

Then, using the formula for the volume of the nn-dimensional ball, we have

|T|((2+ε)n2ε)nπn/2Γ(n2+1)((2+ε)2πe2ε)n.|T|\leq\left(\frac{(2+\varepsilon)\sqrt{n}}{2\varepsilon}\right)^{n}\frac{\pi^{n/2}}{\Gamma(\tfrac{n}{2}+1)}\leq\left(\frac{(2+\varepsilon)\sqrt{2\pi e}}{2\varepsilon}\right)^{n}.

The maximisation over n\mathbb{R}^{n} can be reduced to over TT and this leads to an error that depends on ε\varepsilon.

Lemma 3.5.

Let

λ=max{|x,Mny|:x,yT}.\lambda=\max\{|\langle x,M_{n}y\rangle|\colon\,x,y\in T\}.

Then

λ~(1ε)2λ.\tilde{\lambda}\leq(1-\varepsilon)^{-2}\lambda.
Proof.

Let be 𝒮={xn:i[n]xi=0,x1}\mathcal{S}=\{x\in\mathbb{R}^{n}\colon\,\sum_{i\in[n]}x_{i}=0,\|x\|\leq 1\}. We want to show that for every x𝒮x\in\mathcal{S} there is a vector yTy\in T such that xyε\|x-y\|\leq\varepsilon and i[n](xiyi)=0\sum_{i\in[n]}(x_{i}-y_{i})=0. Let us write the components of xx as

xi=εmin+fi,i[n],x_{i}=\varepsilon\frac{m_{i}}{\sqrt{n}}+f_{i},\qquad i\in[n],

where mim_{i}\in\mathbb{Z} and fi[0,εn1/2)f_{i}\in[0,\varepsilon n^{-1/2}) is an error term. Because i[n]xi=0\sum_{i\in[n]}x_{i}=0, we choose mim_{i} such that i[n]fiv=vεfn1/2\sum_{i\in[n]}f_{i}v=v\varepsilon fn^{-1/2}, where ff is a non-negative integer smaller than nn. We relabel the indices in such a way that mimjm_{i}\leq m_{j} when iji\leq j. Now consider the vector yy given by

yi={εmi+1nif if,εminif i>f.y_{i}=\left\{\begin{array}[]{ll}\varepsilon\frac{m_{i}+1}{\sqrt{n}}&\text{if }i\leq f,\\ \varepsilon\frac{m_{i}}{\sqrt{n}}&\text{if }i>f.\end{array}\right.

It follows that i[n]yi=0\sum_{i\in[n]}y_{i}=0 and y1\|y\|\leq 1, and therefore yTy\in T. Furthermore, because |xiyi|εn1/2|x_{i}-y_{i}|\leq\varepsilon n^{-1/2} by construction, we have the required property xy1\|x-y\|\leq 1. Iterating the previous argument, we can express every vector x𝒮x\in\mathcal{S} in terms of a sequence of vectors (y(i))i[n](y^{(i)})_{i\in[n]} in TT such that

x=i[n]εiy(i).x=\sum_{i\in[n]}\varepsilon^{i}y^{(i)}.

Therefore, because (3.4) is maximized on 𝒮\mathcal{S}, we have

x,Mnx=i,j[n]×[n]εi+jx(i)Mnx(j)1(1ε)2max{|y,Mnz|:y,zT},\langle x,M_{n}x\rangle=\sum_{i,j\in[n]\times[n]}\varepsilon^{i+j}\langle x^{(i)}M_{n}x^{(j)}\rangle\leq\frac{1}{(1-\varepsilon)^{2}}\max\{|\langle y,M_{n}z\rangle|\colon\,y,z\in T\},

from which the claim follows. ∎

Our next goal is to show that |x,Mny|=o(m3/2)|\langle x,M_{n}y\rangle|=o(m_{\infty}^{-3/2}) for all x,yTx,y\in T with a suitably high probability. This can be done in the following way. Fix x,yTx,y\in T and define the random variable

X=i,j[n]×[n]xi(Mn)ijyj.X=\sum_{i,j\in[n]\times[n]}x_{i}(M_{n})_{ij}y_{j}.

Define the set of indices

={(i,j)[n]×[n]: 0<|xiyj|<mn}.\mathcal{B}=\left\{(i,j)\in[n]\times[n]\colon\,0<|x_{i}y_{j}|<\frac{\sqrt{m_{\infty}}}{n}\right\}.

Then we can rewrite X=X+X′′X=X^{\prime}+X^{\prime\prime} with

X=(i,j)xi(Mn)ijyj,X′′=(i,j)xi(Mn)ijyj.X^{\prime}=\sum_{(i,j)\in\mathcal{B}}x_{i}(M_{n})_{ij}y_{j},\qquad X^{\prime\prime}=\sum_{(i,j)\notin\mathcal{B}}x_{i}(M_{n})_{ij}y_{j}.

In Section 3.2 we show that 𝔼[X]\mathbb{E}[X^{\prime}] is of the correct order and that XX^{\prime} is well concentrated around its mean. In Section 3.3 we analyse X′′X^{\prime\prime}, which is of a different nature and requires that we exclude subgraphs in the configuration model that are too dense.

3.2. Estimate of first contribution

1. Use (2.2) and (3.5) to write out 𝔼[mij]=𝔼[(Mn)ij]\mathbb{E}[m_{ij}]=\mathbb{E}[(M_{n})_{ij}]. This gives

𝔼[X]=(i,j)xiyjm11(i,i)xiyidi2(m11).\mathbb{E}[X^{\prime}]=\sum_{(i,j)\in\mathcal{B}}\frac{x_{i}y_{j}}{m_{1}-1}-\sum_{(i,i)\in\mathcal{B}}\frac{x_{i}y_{i}}{d_{i}^{2}(m_{1}-1)}.

In view of the bound on |xiyi||x_{i}y_{i}|, the last term gives a contribution

|(i,i)xiyidi2(m11)|=O(1m1m3/2).\left|\sum_{(i,i)\in\mathcal{B}}\frac{x_{i}y_{i}}{d_{i}^{2}(m_{1}-1)}\right|=O\left(\frac{1}{m_{1}m_{\infty}^{3/2}}\right).

Since x,yTx,y\in T, we have that i[n]xi=0\sum_{i\in[n]}x_{i}=0 and j[n]yj=0\sum_{j\in[n]}y_{j}=0, and therefore (i,j)[n]×[n]xiyj=0\sum_{(i,j)\in[n]\times[n]}x_{i}y_{j}=0. Hence

|(i,j)xiyj|=|(i,j)xiyj|,\left|\sum_{(i,j)\in\mathcal{B}}x_{i}y_{j}\right|=\left|\sum_{(i,j)\notin\mathcal{B}}x_{i}y_{j}\right|,

where we can bound the right-hand side as

|(i,j)xiyj|(i,j):|xiyj|mn|xiyj|(i,j):|xiyj|mnxi2yj2|xiyj|nm(i,j)xi2yj2nm.\left|\sum_{(i,j)\notin\mathcal{B}}x_{i}y_{j}\right|\leq\sum_{(i,j)\colon|x_{i}y_{j}|\geq\frac{\sqrt{m_{\infty}}}{n}}\left|x_{i}y_{j}\right|\leq\sum_{(i,j)\colon|x_{i}y_{j}|\geq\frac{\sqrt{m_{\infty}}}{n}}\frac{x_{i}^{2}y_{j}^{2}}{|x_{i}y_{j}|}\leq\frac{n}{\sqrt{m_{\infty}}}\sum_{(i,j)}x^{2}_{i}y_{j}^{2}\leq\frac{n}{\sqrt{m_{\infty}}}.

We can therefore conclude that

|𝔼[X]|n(m11)m+O(1m1m3/2).\left|\mathbb{E}[X^{\prime}]\right|\leq\frac{n}{(m_{1}-1)\sqrt{m_{\infty}}}+O\left(\frac{1}{m_{1}m_{\infty}^{3/2}}\right). (3.7)

2. To prove that XX^{\prime} is concentrated around its mean, we use an argument originally developed in [9, 23] and used for the configuration model in [18, 10]. This argument employs the martingale structure of the configuration model conditional on partial pairings. Define

χ(x)={x,if |x|<mn,0,otherwise.\chi(x)=\begin{cases}x,\qquad\text{if }|x|<\frac{\sqrt{m_{\infty}}}{n},\\ 0,\qquad\text{otherwise.}\end{cases}

We can then express XX^{\prime} as

X=e𝒞χ(xt(e)yh(e))dt(e)dh(e)+e𝒞χ(xh(e)yt(e))dt(e)dh(e)=Xa+Xb.X^{\prime}=\sum_{e\in\mathcal{C}}\frac{\chi(x_{t(e)}y_{h(e)})}{d_{t(e)}d_{h(e)}}+\sum_{e\in\mathcal{C}}\frac{\chi(x_{h(e)}y_{t(e)})}{d_{t(e)}d_{h(e)}}=X_{a}^{\prime}+X_{b}^{\prime}.

We divide the set of pairings 𝒞\mathcal{C} into three sets

𝒞1={e𝒞:|xt(e)|>1εn},𝒞2={e𝒞:|yt(e)|>1εn,|xt(e)|1εn},𝒞3={e𝒞:|yt(e)|1εn,|xt(e)|1εn},\begin{split}&\mathcal{C}_{1}=\left\{e\in\mathcal{C}\colon\,|x_{t(e)}|>\frac{1}{\varepsilon\sqrt{n}}\right\},\\ &\mathcal{C}_{2}=\left\{e\in\mathcal{C}\colon\,|y_{t(e)}|>\frac{1}{\varepsilon\sqrt{n}},\,|x_{t(e)}|\leq\frac{1}{\varepsilon\sqrt{n}}\right\},\\ &\mathcal{C}_{3}=\left\{e\in\mathcal{C}\colon\,|y_{t(e)}|\leq\frac{1}{\varepsilon\sqrt{n}},\,|x_{t(e)}|\leq\frac{1}{\varepsilon\sqrt{n}}\right\},\end{split}

and write

Xa=X1+X2+X3,Xi=e𝒞iχ(xt(e)yh(e))dt(e)dh(e).X_{a}^{\prime}=X_{1}+X_{2}+X_{3},\qquad X_{i}=\sum_{e\in\mathcal{C}_{i}}\frac{\chi(x_{t(e)}y_{h(e)})}{d_{t(e)}d_{h(e)}}.

We can do a similar decomposition for XbX_{b}^{\prime}.

3. The following martingale lemma from [18, 10] is the core estimate that we want to apply to each of the XiX_{i}’s.

Lemma 3.6.

Let be G1G_{1} and G2G_{2} two graphs generated via the configuration model through perfect matchings P1P_{1} and P2P_{2}. Let {ei(1)}i1\{e^{(1)}_{i}\}_{i\geq 1} be the edges of G1G_{1} and {ei(2)}i1\{e^{(2)}_{i}\}_{i\geq 1} be the edges of G2G_{2}, ordered as above. Define an equivalence relation on the probability space Ω\Omega by setting G1kG2G_{1}\equiv_{k}G_{2} when {ei(1)}i=1k={ei(2)}i=1k\{e^{(1)}_{i}\}_{i=1}^{k}=\{e^{(2)}_{i}\}_{i=1}^{k}, i.e., the first kk pairings match. Let Ωk\Omega_{k} be the set of equivalence classes, and k\mathcal{F}_{k} the corresponding σ\sigma-algebra with 0=\mathcal{F}_{0}=\mathcal{F}. Consider a bounded measurable f:𝒢(d)f\colon\,\mathcal{G}(\vec{d})\to\mathbb{R}, and set Yk=𝔼[fk]Y_{k}=\mathbb{E}[f\mid\mathcal{F}_{k}]. Note that

  • (Yk)0km1/2(Y_{k})_{0\leq k\leq m_{1}/2} is a Doob martingale with 𝔼[Ykk1]=Yk1\mathbb{E}[Y_{k}\mid\mathcal{F}_{k-1}]=Y_{k-1}
    and Y0=𝔼[f]Y_{0}=\mathbb{E}[f], Ym1/2=fY_{m_{1}/2}=f.

Define Zk=YkYk1Z_{k}=Y_{k}-Y_{k-1}, and suppose that there exist functions (gk(ζ))1k12m1(g_{k}(\zeta))_{1\leq k\leq\tfrac{1}{2}m_{1}} such that

𝔼[eζ2Zk2k1]gk(ζ),1km1/2.\mathbb{E}\left[\mathrm{e}^{\zeta^{2}Z^{2}_{k}}\mid\mathcal{F}_{k-1}\right]\leq g_{k}(\zeta),\qquad 1\leq k\leq m_{1}/2.

Then, for all t0t\geq 0 and ζ>0\zeta>0,

(|f𝔼[f]|t)2eζtk=1m1/2gk(ζ).\mathbb{P}\left(\left|f-\mathbb{E}[f]\right|\geq t\right)\leq 2\,\mathrm{e}^{-\zeta t}\prod_{k=1}^{m_{1}/2}g_{k}(\zeta). (3.8)
Remark 3.7.

The condition on the existence gk(ζ)g_{k}(\zeta) can be rephrased as the existence of a random variable WkW_{k} that stochastically dominates ZkZ_{k} on (Ωk1,k1)(\Omega_{k-1},\mathcal{F}_{k-1}) (see [18]). \spadesuit

Lemma 3.8.

[10, Lemma 15] There exist constants B>0B_{\ell}>0, =1,2,3\ell=1,2,3, depending only on the ratio mm0\frac{m_{\infty}}{m_{0}}, such that

(|X𝔼[X]|tm3/2)2etn+Bn,=1,2,3.\mathbb{P}\left(|X_{\ell}-\mathbb{E}[X_{\ell}]|\geq\frac{t}{m_{\infty}^{3/2}}\right)\leq 2\,\mathrm{e}^{-tn+B_{\ell}n},\qquad\ell=1,2,3. (3.9)
Proof.

While the properties in 𝒞3\mathcal{C}_{3} allow us to apply standard martingale arguments to capture X3X_{3} (see below), the properties in 𝒞1\mathcal{C}_{1} and 𝒞2\mathcal{C}_{2} force us to use Lemma  3.6 to capture X1X_{1} and X2X_{2}.

i. From the definition of 𝒞1\mathcal{C}_{1} and 𝒞2\mathcal{C}_{2} it follows that a bound on X1X_{1} implies by symmetry a bound on X2X_{2} (with, possibly, different constants). We will therefore focus on X1X_{1}, the result for X2X_{2} carrying through trivially. Without loss of generality we may reorder the indices in such a way that |xi||xi+1||x_{i}|\geq|x_{i+1}| and for each ei={αi,βi}e_{i}=\{\alpha_{i},\beta_{i}\} use the lexicographic order αi+1>αi\alpha_{i+1}>\alpha_{i} and βi>αi\beta_{i}>\alpha_{i} (i.e., we perform the pairing sequentially from left to right; see [27, Lemma 7.6]). It follows that v(αi)v(αi+1)v(\alpha_{i})\leq v(\alpha_{i+1}) and |xv(αi)||xv(αi+1)||x_{v(\alpha_{i})}|\geq|x_{v(\alpha_{i+1})}|. Define

χ^(x,y)={xy,if |xy|<mn and |x|>1ϵn,0,otherwise.\hat{\chi}(x,y)=\begin{cases}xy,\quad&\text{if }|xy|<\frac{\sqrt{m_{\infty}}}{n}\text{ and }|x|>\frac{1}{\epsilon\sqrt{n}},\\ 0,\quad&\text{otherwise}.\end{cases}

For each e={α,β}e=\{\alpha,\beta\} in the configuration 𝒞\mathcal{C} we have

X1=e𝒞q(e)X_{1}=\sum_{e\in\mathcal{C}}q(e)

with q(e)=χ^(xv(α),yv(β))/dv(α)dv(β)q(e)=\hat{\chi}(x_{v(\alpha)},y_{v(\beta)})/d_{v(\alpha)}d_{v(\beta)}.

ii. Next, take Yk=𝔼[X1k]Y_{k}=\mathbb{E}[X_{1}\mid\mathcal{F}_{k}], with Y0=𝔼[X1]Y_{0}=\mathbb{E}[X_{1}] and Ym=X1Y_{m}=X_{1}. Then (Yk)k0(Y_{k})_{k\in\mathbb{N}_{0}} is a Doob martingale and, by the definition of the configuration model, we can write Zk=YkYk1Z_{k}=Y_{k}-Y_{k-1} as

Zk(𝒞)=212m1k(12m1k)!(m12k)!(𝒞k𝒞X1(𝒞)1m12k+1𝒞′′k1𝒞X1(𝒞′′)).Z_{k}(\mathcal{C})=\frac{2^{\tfrac{1}{2}m_{1}-k}(\tfrac{1}{2}m_{1}-k)!}{(m_{1}-2k)!}\left(\sum_{\mathcal{C}^{\prime}\equiv_{k}\mathcal{C}}X_{1}(\mathcal{C}^{\prime})-\frac{1}{m_{1}-2k+1}\sum_{\mathcal{C}^{\prime\prime}\equiv_{k-1}\mathcal{C}}X_{1}(\mathcal{C}^{\prime\prime})\right).

Now that we have an expression for ZkZ_{k}, we can use the method of switching (see, for example, [31]). Indeed, given a 𝒞k𝒞\mathcal{C}^{\prime}\equiv_{k}\mathcal{C}, we can define a quantity 𝒞η\mathcal{C}^{\prime}_{\eta} as follows. Given the first kk pairings of 𝒞\mathcal{C}, let II be the set of points already paired, and let {α,β}\{\alpha,\beta\} be the kk-th pair. Put ηI{β}\eta\notin I-\{\beta\} and {η,γ}𝒞\{\eta,\gamma\}\in\mathcal{C}^{\prime}. Then 𝒞η\mathcal{C}^{\prime}_{\eta} is the pairing obtained from 𝒞\mathcal{C}^{\prime} by mapping

{α,β},{η,γ}{α,η},{γ,β}.\{\alpha,\beta\},\{\eta,\gamma\}\to\{\alpha,\eta\},\{\gamma,\beta\}.

Is easy to see that 𝒞ηk1𝒞\mathcal{C}^{\prime}_{\eta}\equiv_{k-1}\mathcal{C} and that {{𝒞η:ηI{β}}𝒞k𝒞}\{\{\mathcal{C}^{\prime}_{\eta}\colon\,\eta\notin I-\{\beta\}\}\mid\mathcal{C}^{\prime}\equiv_{k}\mathcal{C}\} is a partition of {𝒞′′𝒞′′k1𝒞}\{\mathcal{C}^{\prime\prime}\mid\mathcal{C}^{\prime\prime}\equiv_{k-1}\mathcal{C}\}. We can therefore rewrite

Zk(𝒞)=212m1k(12m1k)!(m12k)!𝒞k𝒞ηI(X1(𝒞)X1(𝒞η))=ηIγI,γηq({α,β})+q({η,γ})q({α,η})q({γ,β})(2m2k+1)(2m2k1).\begin{split}Z_{k}(\mathcal{C})&=\frac{2^{\tfrac{1}{2}m_{1}-k}(\tfrac{1}{2}m_{1}-k)!}{(m_{1}-2k)!}\sum_{\mathcal{C}^{\prime}\equiv_{k}\mathcal{C}}\sum_{\eta\notin I}(X_{1}(\mathcal{C}^{\prime})-X_{1}(\mathcal{C}^{\prime}_{\eta}))\\ &=\sum_{\eta\notin I}\sum_{\gamma\notin I,\gamma\neq\eta}\frac{q(\{\alpha,\beta\})+q(\{\eta,\gamma\})-q(\{\alpha,\eta\})-q(\{\gamma,\beta\})}{(2m-2k+1)(2m-2k-1)}.\end{split}

Because ixu21\sum_{i}x_{u}^{2}\leq 1, there are at most ε2n\varepsilon^{2}n indices of xx such that |xi|>1/(εn)|x_{i}|>1/(\varepsilon\sqrt{n}). By the definition of X1X_{1}, the lexicographic ordering of {αi,βi}\{\alpha_{i},\beta_{i}\} and the ordering of |xj|>|xj+1||x_{j}|>|x_{j+1}|, there exists a k~\tilde{k} such that Zk~=0Z_{\tilde{k}}=0. Take k~=ε2mn\tilde{k}=\varepsilon^{2}m_{\infty}n. For k>k~k>\tilde{k}, we have that

m12k1m12ε2mn1m0n,m_{1}-2k-1\geq m_{1}-2\varepsilon^{2}m_{\infty}n-1\geq m_{0}n,

where the free parameter ε\varepsilon has to be fixed such that the last inequality holds. (Note that, because there exists a constant θ\theta such that mm0<θ\frac{m_{\infty}}{m_{0}}<\theta, we can always choose an ε\varepsilon small enough so that this holds). Hence we can bound

|Zk(𝒞)|1(m0n)2ηIγI,γη(|q({α,β})|+|q({η,γ})|+|q({α,η})|+|q({γ,β})|).|Z_{k}(\mathcal{C})|\leq\frac{1}{(m_{0}n)^{2}}\sum_{\eta\notin I}\sum_{\gamma\notin I,\gamma\neq\eta}\left(|q(\{\alpha,\beta\})|+|q(\{\eta,\gamma\})|+|q(\{\alpha,\eta\})|+|q(\{\gamma,\beta\})|\right).

iii. Define

yα=1|xv(α)|min{|yxv(α)|,mn}.y^{\alpha}=\frac{1}{|x_{v(\alpha)}|}\text{min}\left\{|yx_{v(\alpha)}|,\frac{\sqrt{m_{\infty}}}{n}\right\}.

Because α<β\alpha<\beta, we can bound

q({α,β})=χ^(xv(α),yv(β))dv(α)dv(β)yvβα|xv(α)|m02.q(\{\alpha,\beta\})=\frac{\hat{\chi}(x_{v(\alpha)},y_{v(\beta)})}{d_{v(\alpha)}d_{v(\beta)}}\leq\frac{y^{\alpha}_{v_{\beta}}|x_{v(\alpha)}|}{m_{0}^{2}}.

A similar bound holds for {α,η}\{\alpha,\eta\} for the same reason. For the other two edges, {η,γ}\{\eta,\gamma\} and {γ,β}\{\gamma,\beta\}, we need to upper bound with a symmetric term, because we do not know whether γ>η\gamma>\eta or γ<η\gamma<\eta. Thus, we have the upper bound

q({η,γ})1m02(yvηγ|xv(γ)|+yvγη|xv(η)|)q(\{\eta,\gamma\})\leq\frac{1}{m_{0}^{2}}\left(y^{\gamma}_{v_{\eta}}|x_{v(\gamma)}|+y^{\eta}_{v_{\gamma}}|x_{v(\eta)}|\right)

(the same bound holds for {γ,β}\{\gamma,\beta\}). Moreover, by the lexicographic order, xv(α)x_{v(\alpha)} bounds all the other components, and therefore yvβα|xv(α)yvβα|xv(α)y^{\alpha}_{v_{\beta}}|x_{v(\alpha)}\leq y^{\alpha}_{v_{\beta}}|x_{v(\alpha)}. Now note that i|yi|n\sum_{i}|y_{i}|\leq\sqrt{n} (because iyi21\sum_{i}y_{i}^{2}\leq 1), and so

ηIyv(η)αηI|yv(η)|mi|yi|mn.\sum_{\eta\notin I}y^{\alpha}_{v(\eta)}\leq\sum_{\eta\notin I}|y_{v(\eta)}|\leq m_{\infty}\sum_{i}|y_{i}|\leq m_{\infty}\sqrt{n}.

By the previous considerations, substituting into the expression for Z(𝒞)Z(\mathcal{C}), we have

|Zk(𝒞)|1m02(yvβα|xv(α)|+(yvηα|xv(α)|+yvγα|xv(α)|)+yvηα|xv(α)|+(yαvβ|xv(α)|+yvγα|xv(α)|))4m2m04|xv(α)|(yv(β)α+1n).\begin{split}|Z_{k}(\mathcal{C})|&\leq\frac{1}{m_{0}^{2}}\left(y^{\alpha}_{v_{\beta}}|x_{v(\alpha)}|+\left(y^{\alpha}_{v_{\eta}}|x_{v(\alpha)}|+y^{\alpha}_{v_{\gamma}}|x_{v(\alpha)}|\right)+y^{\alpha}_{v_{\eta}}|x_{v(\alpha)}|+\left(y^{\alpha}{v_{\beta}}|x_{v(\alpha)}|+y^{\alpha}_{v_{\gamma}}|x_{v(\alpha)}|\right)\right)\\ &\leq\frac{4m_{\infty}^{2}}{m_{0}^{4}}|x_{v(\alpha)}|\left(y^{\alpha}_{v(\beta)}+\frac{1}{\sqrt{n}}\right).\end{split}

iv. From the above bounds we are able to obtain an upper bound for 𝔼[exp(ζ2Zk2)k1]\mathbb{E}[\exp(\zeta^{2}Z_{k}^{2})\mid\mathcal{F}_{k-1}] and then use lemma 3.6. Indeed,

𝔼[eζ2Zk2k1]1m12k1ωI\βexp[16ζ2m4m08(xv(α))2](yv(ω)α+1n)2.\mathbb{E}\left[\mathrm{e}^{\zeta^{2}Z_{k}^{2}}\mid\mathcal{F}_{k-1}\right]\leq\frac{1}{m_{1}-2k-1}\sum_{\omega\notin I\backslash\beta}\exp\left[16\zeta^{2}m_{\infty}^{4}m_{0}^{-8}(x_{v(\alpha)})^{2}\right]\left(y^{\alpha}_{v(\omega)}+\frac{1}{\sqrt{n}}\right)^{2}.

Looking at (3.8), we see that we have to fix ζ=m3/2n\zeta=m_{\infty}^{3/2}n in order to achieve the required bound. Using that xv(α)m/(ϵn)x_{v(\alpha)}\leq\sqrt{m_{\infty}}/(\epsilon\sqrt{n}) (because xv(α)yv(β)αmnx_{v(\alpha)}y^{\alpha}_{v(\beta)}\leq\sqrt{m_{\infty}}{n} and yv(β)αε/ny^{\alpha}_{v(\beta)}\geq\varepsilon/\sqrt{n}), the exponent in the previous display is bounded by 64θ8ε264\theta^{8}\varepsilon^{-2}. Using that ex1+xex\mathrm{e}^{x}\leq 1+x\mathrm{e}^{x}, x0x\geq 0, and putting B=64θ8ϵ2B=64\theta^{8}\epsilon^{-2}, we have

𝔼[eζ2Zk2|k1]1+Bm12k1ωI\βζ2m4m08(xv(α))2(yv(ω)α+1n)21+Bζ2m4m09(xv(α))2ω(yv(ω)2+2yv(ω)1n+1n)exp[4Bζ2m5m09n(xv(α))2].\begin{split}\mathbb{E}\left[e^{\zeta^{2}Z_{k}^{2}}|\mathcal{F}_{k-1}\right]&\leq 1+\frac{B}{m_{1}-2k-1}\sum_{\omega\notin I\backslash\beta}\zeta^{2}m_{\infty}^{4}m_{0}^{-8}(x_{v(\alpha)})^{2}\left(y^{\alpha}_{v(\omega)}+\frac{1}{\sqrt{n}}\right)^{2}\\ &\leq 1+B\zeta^{2}m_{\infty}^{4}m_{0}^{-9}(x_{v(\alpha)})^{2}\sum_{\omega}\left(y^{2}_{v(\omega)}+2y_{v(\omega)}\frac{1}{\sqrt{n}}+\frac{1}{n}\right)\\ &\leq\exp\left[4B\frac{\zeta^{2}m_{\infty}^{5}}{m_{0}^{9}n}(x_{v(\alpha)})^{2}\right].\end{split}

Next, let us pick an index i(k)i(k) such that, for all 𝒞Ω\mathcal{C}\in\Omega,

𝔼[eζ2Zk2k1]exp[4Bζ2m5m09n(xi(k))2].\mathbb{E}\left[e^{\zeta^{2}Z_{k}^{2}}\mid\mathcal{F}_{k-1}\right]\leq\exp\left[4B\frac{\zeta^{2}m_{\infty}^{5}}{m_{0}^{9}n}(x_{i(k)})^{2}\right].

One possible choice is to take i(k)=k/mi(k)=\lceil k/m_{\infty}\rceil. We finally get

(|X1𝔼[X1]|tm3/2)2eζtm3/2ek=112m14Bζ2m5m09n(xi(k))22etn+4Bm9m09n,\mathbb{P}\left(|X_{1}-\mathbb{E}[X_{1}]|\geq\frac{t}{m_{\infty}^{3/2}}\right)\leq 2\,\mathrm{e}^{-\frac{\zeta t}{m_{\infty}^{3/2}}}\mathrm{e}^{\sum_{k=1}^{\tfrac{1}{2}m_{1}}4B\frac{\zeta^{2}m_{\infty}^{5}}{m_{0}^{9}n}(x_{i(k)})^{2}}\leq 2\,\mathrm{e}^{-tn+4B\frac{m_{\infty}^{9}}{m_{0}^{9}}n},

which proves what was said at the beginning of the proof (3.9) for =1,2\ell=1,2.

v. Finally, consider X3X_{3}. In view of the bounds in 𝒞3\mathcal{C}_{3}, this case can be dealt with via classical martingale arguments (see, for example, the McDiarmid inequality and its generalizations in [8]). Considering the variables Yk=𝔼[X3𝒞3]Y_{k}=\mathbb{E}[X_{3}\mid\mathcal{C}_{3}], we have that |YkYk1|4/(ε2nm02)|Y_{k}-Y_{k-1}|\leq 4/(\varepsilon^{2}nm_{0}^{2}). Thus, given our choice of ε=ε(θ)\varepsilon=\varepsilon(\theta) being constant, we have

(|X3𝔼[X3]|tm3/2)2et2ε4n2m0416m3m12eC(ε,θ)t2n,\mathbb{P}\left(|X_{3}-\mathbb{E}[X_{3}]|\geq tm_{\infty}^{-3/2}\right)\leq 2\,\mathrm{e}^{-\frac{t^{2}\varepsilon^{4}n^{2}m_{0}^{4}}{16m_{\infty}^{3}m_{1}}}\leq 2\,\mathrm{e}^{-C(\varepsilon,\theta)t^{2}n},

where C(ε,θ)>0C(\varepsilon,\theta)>0. ∎

4. Combining the results for =1,2,3\ell=1,2,3 in the above lemma, we get an exponential bound on XaX^{\prime}_{a} of the type

(|Xa𝔼[Xa]|tm3/2)2etn+Ban,\mathbb{P}\left(|X^{\prime}_{a}-\mathbb{E}[X^{\prime}_{a}]|\geq\frac{t}{m_{\infty}^{3/2}}\right)\leq 2\mathrm{e}^{-tn+B_{a}n},

where BaB_{a} is a suitable constant, possibly different from any of the BB_{\ell}. By symmetry, the same holds for XbX^{\prime}_{b}, which proves the concentration around m3/2m_{\infty}^{-3/2} of XX^{\prime}.

Remark 3.9.

Up to now we have worked with multi-graphs, so we have to pass to simple graphs with a prescribed degree sequence. Is easy to see that, because of (2.3), we can express the final result as saying that, conditional on the event that the graph is simple, there exist two constants ξ^<ξ\hat{\xi}<\xi in (0,1)(0,1) and a constant K(θ,ξ)>0K(\theta,\xi)>0 such that

(|X𝔼[X]|Km3/2)eO(d^2)ξ^nξn.\mathbb{P}\left(|X^{\prime}-\mathbb{E}[X^{\prime}]|\geq K{m_{\infty}^{-3/2}}\right)\leq\mathrm{e}^{O(\hat{d}^{2})}\hat{\xi}^{n}\leq\xi^{n}.

\spadesuit

3.3. Estimate of second contribution

It remains to show that the pairs with x,yx,y\notin\mathcal{B} give a bounded contribution of the order O(m3/2)O(m_{\infty}^{-3/2}) with sufficiently high probability. For this it suffices to show that a simple random graph GG with a prescribed degree sequence d\vec{d} cannot have too dense subgraphs (see [18, Lemma 2.5] and [10, Lemma 16]):

Lemma 3.10.

Let GG be a simple random graph of size nn drawn uniformly at random with a given degree sequence d\vec{d}. Let A,B[n]A,B\subseteq[n] be two subsets of the vertex set, and let e(A,B)e(A,B) be the set of edges e={α,β}e=\{\alpha,\beta\} such that either αA,βB\alpha\in A,\beta\in B or αB,βA\alpha\in B,\beta\in A. Since μ(A,B)=θ|A||B|mn\mu(A,B)=\theta|A||B|\frac{m_{\infty}}{n} with θ>m/m0\theta>m_{\infty}/m_{0} a sufficiently large constant, for any K>0K>0 there exist a constant C=C(θ,K)C=C(\theta,K) such that with probability 1o(nK)1-o(n^{-K}) any pair A,BA,B with |A||B||A|\leq|B| satisfies at least one of the following:

e(A,B)Cμ(A,B)\displaystyle e(A,B)\leq C\mu(A,B) (3.10)
e(A,B)log(e(A,B)μ(A,B))C|B|log(n|B|).\displaystyle e(A,B)\log\left(\frac{e(A,B)}{\mu(A,B)}\right)\leq C|B|\log\left(\frac{n}{|B|}\right). (3.11)

The above lemma has the following corollary.

Corollary 3.11.

For all x,yTx,y\in T,

X′′=O(1m3/2)with probability at least 1O(nK),X^{\prime\prime}=O\left(\frac{1}{m_{\infty}^{3/2}}\right)\quad\textnormal{with probability at least }1-O(n^{-K}),

where KK is the constant in Lemma 3.10.

Proof.

Fix x,yTx,y\in T and define

Si(x)={:ε2in|x|<ε1in},iI,Sj(y)={:ε2in|y|<ε1in},jJ,\begin{split}S_{i}(x)&=\left\{\ell\colon\,\frac{\varepsilon^{2-i}}{\sqrt{n}}\leq|x_{\ell}|<\frac{\varepsilon^{1-i}}{\sqrt{n}}\right\},\quad i\in I,\\ S_{j}(y)&=\left\{\ell\colon\,\frac{\varepsilon^{2-i}}{\sqrt{n}}\leq|y_{\ell}|<\frac{\varepsilon^{1-i}}{\sqrt{n}}\right\},\quad j\in J,\end{split}

where I={i|Si(x)}I=\{i|S_{i}(x)\neq\emptyset\} (JJ is defined similarly). Then, for xTx\in T, write

xu|S={xu,if uS,0,otherwise.x_{u}|_{S}=\begin{cases}x_{u},\quad\textnormal{if }u\in S,\\ 0,\quad\textnormal{otherwise.}\end{cases}

In order to apply Lemma 3.10, define Ai=Si(x)A_{i}=S_{i}(x) and Bj=Sj(y)B_{j}=S_{j}(y), and let aia_{i} and bib_{i} be their cardinality, respectively. Divide the set of indices into two groups:

\displaystyle\mathcal{E} ={(i,j):i,j>0,ε2ij>m,aibj},\displaystyle=\big{\{}(i,j)\colon\,i,j>0,\varepsilon^{2-i-j}>\sqrt{m_{\infty}},a_{i}\leq b_{j}\big{\}},
\displaystyle\mathcal{E}^{\prime} ={(i,j):i,j>0,ε2ij>m,ai>bj}.\displaystyle=\big{\{}(i,j)\colon\,i,j>0,\varepsilon^{2-i-j}>\sqrt{m_{\infty}},a_{i}>b_{j}\big{\}}.

By the definition of X′′X^{\prime\prime} and the set \mathcal{B}, we have

X′′=xiyj>m/nxiAijyj=(i,j)(x|Ai)tAy|Bj+(i,j)(x|Ai)tAy|Bj.X^{\prime\prime}=\sum_{x_{i}y_{j}>\sqrt{m_{\infty}}/n}x_{i}A_{ij}y_{j}=\sum_{(i,j)\in\mathcal{E}}(x|_{A_{i}})^{t}Ay|_{B_{j}}+\sum_{(i,j)\in\mathcal{E}^{\prime}}(x|_{A_{i}})^{t}Ay|_{B_{j}}.

It suffices to show that either of the contributions coming from \mathcal{E} or \mathcal{E}^{\prime} are O(m3/2)O(m_{\infty}^{-3/2}) (the other will follow by symmetry). Focus on \mathcal{E}. Putting ei,j=e(Ai,Bj)e_{i,j}=e(A_{i},B_{j}) and μi,j=μ(Ai,Bj)\mu_{i,j}=\mu(A_{i},B_{j}), we see that the bound can be rewritten as

1n(i,j)ei,jεi+j=O(m).\frac{1}{n}\sum_{(i,j)\in\mathcal{E}}\frac{e_{i,j}}{\varepsilon^{i+j}}=O(\sqrt{m_{\infty}}\,).

Divide \mathcal{E} into the union of a\mathcal{E}_{a} and b\mathcal{E}_{b}, where a\mathcal{E}_{a} satisfies (3.10) and b\mathcal{E}_{b} satisfies (3.11). Clearly =ab\mathcal{E}=\mathcal{E}_{a}\cup\mathcal{E}_{b} and

1n(i,j)ei,jεi+j1n(i,j)aei,jεi+j+1n(i,j)bei,jεi+j.\frac{1}{n}\sum_{(i,j)\in\mathcal{E}}\frac{e_{i,j}}{\varepsilon^{i+j}}\leq\frac{1}{n}\sum_{(i,j)\in\mathcal{E}_{a}}\frac{e_{i,j}}{\varepsilon^{i+j}}+\frac{1}{n}\sum_{(i,j)\in\mathcal{E}_{b}}\frac{e_{i,j}}{\varepsilon^{i+j}}.

If we are able to show that both contributions from a\mathcal{E}_{a} and b\mathcal{E}_{b} are O(m)O(\sqrt{m_{\infty}}\,), then the theorem follows. It is easy to see that a\mathcal{E}_{a} gives a bounded contribution. Indeed,

1n(i,j)aei,jεi+j1n2(i,j)aCaibjθmεi+jCmn2(i,j)aaibjε2(i+j)m=O(m),\frac{1}{n}\sum_{(i,j)\in\mathcal{E}_{a}}\frac{e_{i,j}}{\varepsilon^{i+j}}\leq\frac{1}{n^{2}}\sum_{(i,j)\in\mathcal{E}_{a}}\frac{Ca_{i}b_{j}\theta m_{\infty}}{\varepsilon^{i+j}}\leq\frac{C^{\prime}m_{\infty}}{n^{2}}\sum_{(i,j)\in\mathcal{E}_{a}}\frac{a_{i}b_{j}}{\varepsilon^{2(i+j)}\sqrt{m_{\infty}}}=O(\sqrt{m_{\infty}}\,),

where in the last step we use that, because ixi21\sum_{i}x_{i}^{2}\leq 1,

iIaiε2(i2)n,jJbjε2(i2)n.\sum_{i\in I}\frac{a_{i}}{\varepsilon^{2(i-2)}}\leq n,\qquad\sum_{j\in J}\frac{b_{j}}{\varepsilon^{2(i-2)}}\leq n.

It remains to show that

1n(i,j)bei,jεi+j=O(m).\frac{1}{n}\sum_{(i,j)\in\mathcal{E}_{b}}\frac{e_{i,j}}{\varepsilon^{i+j}}=O(\sqrt{m_{\infty}}\,).

In order to do so, we divide b\mathcal{E}_{b} into subsets b()\mathcal{E}_{b}^{(\ell)}, =1,,5\ell=1,\dots,5, having the following properties:

(1)εjεim.(2)ei,jμi,jεi+jm.(3)log(ei,jμi,j)14log(nbj).(4)nbje4j.(5)nbj>e4j.\begin{split}&(1)\quad\varepsilon^{j}\geq\varepsilon^{i}\sqrt{m_{\infty}}.\\ &(2)\quad e_{i,j}\leq\frac{\mu_{i,j}}{\varepsilon^{i+j}\sqrt{m_{\infty}}}.\\ &(3)\quad\log\left(\frac{e_{i,j}}{\mu_{i,j}}\right)\geq\frac{1}{4}\log\left(\frac{n}{b_{j}}\right).\\ &(4)\quad\frac{n}{b_{j}}\leq\mathrm{e}^{-4j}.\\ &(5)\quad\frac{n}{b_{j}}>\mathrm{e}^{-4j}.\end{split}

For j>ij>i we have that b()b()\mathcal{E}_{b}^{(\ell)}\nsubseteq\mathcal{E}_{b}^{(\ell)} and b=b()\mathcal{E}_{b}=\cup_{\ell}\mathcal{E}_{b}^{(\ell)}. Thus it suffices to show a bound of O(n)O(\sqrt{n}) for each of the quantities S=1/nb()ei,j/εi+jS_{\ell}=1/n\sum_{\mathcal{E}_{b}^{(\ell)}}e_{i,j}/\varepsilon^{i+j}.

\bullet For =1\ell=1 we note that, since ei,jaime_{i,j}\leq a_{i}m_{\infty},

S11nij|εjεimaimεi+j=bO(1niaimε2i)=O(m).S_{1}\leq\frac{1}{n}\sum_{i}\sum_{j|\varepsilon^{j}\geq\varepsilon^{i}\sqrt{m_{\infty}}}\frac{a_{i}m_{\infty}}{\varepsilon^{i+j}}=bO\left(\frac{1}{n}\sum_{i}\frac{a_{i}\sqrt{m_{\infty}}}{\varepsilon^{2i}}\right)=O(\sqrt{m_{\infty}}\,).

\bullet For =2\ell=2 we obtain

S21nijμijε2(i+j)m=O(mn2ijaibjε2(i+j))=O(n).S_{2}\leq\frac{1}{n}\sum_{ij}\frac{\mu_{ij}}{\varepsilon^{2(i+j)}\sqrt{m_{\infty}}}=O(\frac{\sqrt{m_{\infty}}}{n^{2}}\sum_{ij}\frac{a_{i}b_{j}}{\varepsilon^{2(i+j)}})=O(\sqrt{n}).

\bullet For =3\ell=3, because the pairs (i,j)b(i,j)\in\mathcal{E}_{b} have property (3.11), it follows easily that eij=O(bj)e_{ij}=O(b_{j}). Furthermore, because b(3)b(1)\mathcal{E}^{(3)}_{b}\nsubseteq\mathcal{E}_{b}^{(1)}, we have that (i,j)b(3)\forall(i,j)\in\mathcal{E}_{b}^{(3)}, ej<eime^{j}<e_{i}\sqrt{m_{\infty}}. It follows that

S3=O(1nji|εi>ej/mbjεi+j)=O(1njmbjε2j)=O(m).S_{3}=O\left(\frac{1}{n}\sum_{j}\sum_{i|\varepsilon^{i}>e^{j}/\sqrt{m_{\infty}}}\frac{b_{j}}{\varepsilon^{i+j}}\right)=O\left(\frac{1}{n}\sum_{j}\frac{\sqrt{m_{\infty}}\,b_{j}}{\varepsilon^{2j}}\right)=O(\sqrt{m_{\infty}}\,).

\bullet For =4\ell=4 we take advantage of the fact that (i,j)b(4)(i,j)\in\mathcal{E}_{b}^{(4)} do not belong to b(3)\mathcal{E}_{b}^{(3)} and b(2)\mathcal{E}_{b}^{(2)}. This implies that

eijmij1ejeijμij1εi+jm.\frac{e_{ij}}{m_{ij}}\leq\frac{1}{e^{j}}\qquad\frac{e_{ij}}{\mu_{ij}}\geq\frac{1}{\varepsilon^{i+j}\sqrt{m_{\infty}}}.

Hence we have εim\varepsilon^{-i}\leq\sqrt{m_{\infty}} and, by (3.11), also eij=O(jbj)e_{ij}=O(jb_{j}). We can therefore conclude that

S4=O(1nji|εimjbjεi+j)=O(mnjjbjεj)=O(m),S_{4}=O\left(\frac{1}{n}\sum_{j}\sum_{i|\varepsilon^{-i}\leq\sqrt{m_{\infty}}}\frac{jb_{j}}{\varepsilon^{i+j}}\right)=O\left(\frac{\sqrt{m_{\infty}}}{n}\sum_{j}\frac{jb_{j}}{\varepsilon^{j}}\right)=O(\sqrt{m_{\infty}}\,),

where in the last equality we use that jJbj/(nε2)=O(1)\sum_{j\in J}b_{j}/(n\varepsilon^{2})=O(1).

\bullet For =5\ell=5, using the property in (5) and (3.11), we have

eijCnε4jlogε4j=O(njε4j).e_{ij}\leq Cn\varepsilon^{4j}\log\varepsilon^{-4j}=O(nj\varepsilon^{4j}).

Also, using that b(4)b(4)\mathcal{E}_{b}^{(4)}\nsubseteq\mathcal{E}_{b}^{(4)}, we have εj<εim\varepsilon^{j}<\varepsilon^{i}\sqrt{m_{\infty}}, from which we conclude that

S5=O(ji|εi>εj/mjε3ji)=O(mjjε2j)=O(m).S_{5}=O\left(\sum_{j}\sum_{i|\varepsilon^{i}>\varepsilon^{j}/\sqrt{m_{\infty}}}j\varepsilon^{3j-i}\right)=O\left(\sqrt{m_{\infty}}\sum_{j}j\varepsilon^{2j}\right)=O(\sqrt{m_{\infty}}\,).

This completes the proof. ∎

4. Proof of the main theorem

Expansion.

Throughout this section we abbreviate A=AGnA=A_{G_{n}} and condition on the event that GnG_{n} is simple. Recall Lemma 3.3. Compute

Av1=λ1v1,(H+|e~e~|+(𝔼[A]|e~e~|))v1=λ1v1,(H+|e~e~|diag(d1m11,,dnm11))v1=λ1v1.\begin{split}&Av_{1}=\lambda_{1}v_{1},\\ &\left(H+|\tilde{e}\rangle\langle\tilde{e}|+\left(\mathbb{E}[A]-|\tilde{e}\rangle\langle\tilde{e}|\right)\right)v_{1}=\lambda_{1}v_{1},\\ &\left(H+|\tilde{e}\rangle\langle\tilde{e}|-\textnormal{diag}\left(\frac{d_{1}}{m_{1}-1},\dots,\frac{d_{n}}{m_{1}-1}\right)\right)v_{1}=\lambda_{1}v_{1}.\end{split}

Rewriting the equation we have,

e~,v1e~=(λ1𝕀+diag(d1m11,,dnm11)H)v1.\begin{split}&\langle\tilde{e},v_{1}\rangle\tilde{e}=\left(\lambda_{1}\mathbb{I}+\textnormal{diag}\left(\frac{d_{1}}{m_{1}-1},\dots,\frac{d_{n}}{m_{1}-1}\right)-H\right)v_{1}.\end{split}

Therefore componentwise we have the following inequality,

(λ1+m0m11)(𝕀Hλ1+m0m11)v1e~,v1e~(λ1+mm11)(𝕀Hλ1+mm11)v1,\begin{split}&\left(\lambda_{1}+\frac{m_{0}}{m_{1}-1}\right)\left(\mathbb{I}-\frac{H}{\lambda_{1}+\frac{m_{0}}{m_{1}-1}}\right)v_{1}\leq\langle\tilde{e},v_{1}\rangle\tilde{e}\leq\left(\lambda_{1}+\frac{m_{\infty}}{m_{1}-1}\right)\left(\mathbb{I}-\frac{H}{\lambda_{1}+\frac{m_{\infty}}{m_{1}-1}}\right)v_{1},\end{split}

If xx, yy and ee are non-negative vector (the non-negativity of v1v_{1} follows from Perron-Frobenius theory) with xyx\geq y, then e,xe,y\langle e,x\rangle\geq\langle e,y\rangle, we can use Corollary 3.2 and invert the matrix multiplying v1v_{1}. Indeed, given that H=O(m)\|H\|=O(\sqrt{m_{\infty}}\,) and λ1m2/m1\lambda_{1}\sim m_{2}/m_{1} on the event with probability 1o(nK)1-o(n^{-K}) of Corollary 3.2 and Theorem 3.1, we can invert and expand

(𝕀Hλ1+m0m11)1=k0Hk(λ1+m0m11)k,\left(\mathbb{I}-\frac{H}{\lambda_{1}+\frac{m_{0}}{m_{1}-1}}\right)^{-1}=\sum_{k\in\mathbb{N}_{0}}\frac{H^{k}}{(\lambda_{1}+\frac{m_{0}}{m_{1}-1})^{k}},

and similarly for mm_{\infty}. Thus,

λ1m2m11m0m11+ke~,Hke~(λ1+m0m11)kλ1m2m11mm11+ke~,Hke~(λ1+mm11)k.\begin{split}&\lambda_{1}\leq\frac{m_{2}}{m_{1}-1}-\frac{m_{0}}{m_{1}-1}+\sum_{k\in\mathbb{N}}\frac{\langle\tilde{e},H^{k}\tilde{e}\rangle}{(\lambda_{1}+\frac{m_{0}}{m_{1}-1})^{k}}\\ &\lambda_{1}\geq\frac{m_{2}}{m_{1}-1}-\frac{m_{\infty}}{m_{1}-1}+\sum_{k\in\mathbb{N}}\frac{\langle\tilde{e},H^{k}\tilde{e}\rangle}{(\lambda_{1}+\frac{m_{\infty}}{m_{1}-1})^{k}}.\end{split}

Our final goal is to determine the expectation 𝔼[λ1]\mathbb{E}[\lambda_{1}], which splits as

𝔼[λ1]=𝔼[λ1|]()+𝔼[λ1|c](c).\mathbb{E}[\lambda_{1}]=\mathbb{E}[\lambda_{1}|\mathcal{E}]\,\mathbb{P}(\mathcal{E})+\mathbb{E}[\lambda_{1}|\mathcal{E}^{c}]\,\mathbb{P}(\mathcal{E}^{c}). (4.1)

The event c\mathcal{E}^{c} has probability at most nKn^{-K}, where KK is a large arbitrary constant. Thus, given the deterministic bound λ1n\lambda_{1}\leq n, we may focus on 𝔼[λ1|]()\mathbb{E}[\lambda_{1}|\mathcal{E}]\,\mathbb{P}(\mathcal{E}). In order to do this, we need to be able to handle terms of the type

m2m11+𝔼[e~,He~]m2m11(1+o(1))+𝔼[e~,H2e~](m2m11)2(1+o(1))+k{1,2}𝔼[e~,Hke~](m2m11)k(1+o(1))\displaystyle\frac{m_{2}}{m_{1}-1}+\frac{\mathbb{E}[\langle\tilde{e},H\tilde{e}\rangle]}{\frac{m_{2}}{m_{1}-1}(1+o(1))}+\frac{\mathbb{E}[\langle\tilde{e},H^{2}\tilde{e}\rangle]}{\left(\frac{m_{2}}{m_{1}-1}\right)^{2}(1+o(1))}+\sum_{k\in\mathbb{N}\setminus\{1,2\}}\frac{\mathbb{E}[\langle\tilde{e},H^{k}\tilde{e}\rangle]}{\left(\frac{m_{2}}{m_{1}-1}\right)^{k}(1+o(1))}

Since

𝔼[e~,Hke~](m2m11)k(1+o(1))mk/2(m2m11)k1=o(1m0k/21),\frac{\mathbb{E}[\langle\tilde{e},H^{k}\tilde{e}\rangle]}{\left(\frac{m_{2}}{m_{1}-1}\right)^{k}(1+o(1))}\leq\frac{m_{\infty}^{k/2}}{\left(\frac{m_{2}}{m_{1}-1}\right)^{k-1}}=o\left(\frac{1}{m_{0}^{k/2-1}}\right),

the last sum is o(1/m)o(1/\sqrt{m_{\infty}}), which is an error term. It therefore remains to study e~,Hke~\langle\tilde{e},H^{k}\tilde{e}\rangle, k=1,2k=1,2. The study of these moments for the configuration model is more involved than for random regular graphs.

Case k=1k=1.

Compute

e~,He~=e~,(A𝔼[A])e~=1m11(ijdidjaij1m11ijdi2dj2)=jdj(ijdi)m11m22(m11)2.\begin{split}\langle\tilde{e},H\tilde{e}\rangle&=\langle\tilde{e},(A-\mathbb{E}[A])\,\tilde{e}\rangle\\ &=\frac{1}{m_{1}-1}\left(\sum_{ij}d_{i}d_{j}a_{ij}-\frac{1}{m_{1}-1}\sum_{ij}d_{i}^{2}d_{j}^{2}\right)=\frac{\sum_{j}d_{j}(\sum_{i\sim j}d_{i})}{m_{1}-1}-\frac{m_{2}^{2}}{(m_{1}-1)^{2}}.\end{split}

Since

𝔼[jdjijdi]=ijdjdi𝔼aij=m22m11m3m11,\mathbb{E}\left[\sum_{j}d_{j}\sum_{i\sim j}d_{i}\right]=\sum_{ij}d_{j}d_{i}\mathbb{E}{a_{ij}}=\frac{m_{2}^{2}}{m_{1}-1}-\frac{m_{3}}{m_{1}-1},

where the last term comes from the presence of selfloops. It follows that 𝔼[e~,Hke~]=O(1/n)\mathbb{E}[\langle\tilde{e},H^{k}\tilde{e}\rangle]=O(1/\sqrt{n}).

Case k=2k=2.

Compute

e~,H2e~\displaystyle\langle\tilde{e},H^{2}\tilde{e}\rangle =e~,(A𝔼[A])2e~\displaystyle=\langle\tilde{e},(A-\mathbb{E}[A])^{2}\,\tilde{e}\rangle
=1m11(ijkdidkaijajk1m11ijkdidk2aijdj\displaystyle=\frac{1}{m_{1}-1}\bigg{(}\sum_{ijk}d_{i}d_{k}a_{ij}a_{jk}-\frac{1}{m_{1}-1}\sum_{ijk}d_{i}d^{2}_{k}a_{ij}d_{j}
1m11ijkdi2djdkajk+1(m11)2ijkdi2dj2dk2).\displaystyle\qquad-\frac{1}{m_{1}-1}\sum_{ijk}d^{2}_{i}d_{j}d_{k}a_{jk}+\frac{1}{(m_{1}-1)^{2}}\sum_{ijk}d_{i}^{2}d_{j}^{2}d_{k}^{2}\bigg{)}.

Write

1m11ijkdidk2aijdj=m2m11ijkdiaijdj=m2m11idi(ijdj)\frac{1}{m_{1}-1}\sum_{ijk}d_{i}d^{2}_{k}a_{ij}d_{j}=\frac{m_{2}}{m_{1}-1}\sum_{ijk}d_{i}a_{ij}d_{j}=\frac{m_{2}}{m_{1}-1}\sum_{i}d_{i}\left(\sum_{i\sim j}d_{j}\right)

(by symmetry the third term is equal) and

ijkdidkaijajk=j(ijdi)2=jijdi2+ki,jkijdidj=m3+ki,jkijdidj.\sum_{ijk}d_{i}d_{k}a_{ij}a_{jk}=\sum_{j}\left(\sum_{i\sim j}d_{i}\right)^{2}=\sum_{j}\sum_{i\sim j}d_{i}^{2}+\sum_{k}\sum_{\begin{subarray}{c}i,j\sim k\\ i\neq j\end{subarray}}d_{i}d_{j}=m_{3}+\sum_{k}\sum_{\begin{subarray}{c}i,j\sim k\\ i\neq j\end{subarray}}d_{i}d_{j}.

Indeed in jijdi2\sum_{j}\sum_{i\sim j}d_{i}^{2}, the summand di2d_{i}^{2} appears exactly did_{i} times, because the node ii has exactly did_{i} neighbours, and so jijdi2=m3\sum_{j}\sum_{i\sim j}d_{i}^{2}=m_{3}. Putting the terms together, we get

e~,H2e~=1m11(m3+ki,jkijdidj2m2m11idi(ijdj)+m23(m11)2).\langle\tilde{e},H^{2}\tilde{e}\rangle=\frac{1}{m_{1}-1}\left(m_{3}+\sum_{k}\sum_{\begin{subarray}{c}i,j\sim k\\ i\neq j\end{subarray}}d_{i}d_{j}-2\frac{m_{2}}{m_{1}-1}\sum_{i}d_{i}\left(\sum_{i\sim j}d_{j}\right)+\frac{m_{2}^{3}}{(m_{1}-1)^{2}}\right).

Taking expectations, we get

𝔼[e~,H2e~]=1m11(m3+𝔼[ki,jkijdidjm23(m11)2)].\mathbb{E}[\langle\tilde{e},H^{2}\tilde{e}\rangle]=\frac{1}{m_{1}-1}\left(m_{3}+\mathbb{E}\left[\sum_{k}\sum_{\begin{subarray}{c}i,j\sim k\\ i\neq j\end{subarray}}d_{i}d_{j}-\frac{m^{3}_{2}}{(m_{1}-1)^{2}}\right)\right].

Note that 𝔼[ki,jk,ijdidj]\mathbb{E}[\sum_{k}\sum_{i,j\sim k,\,i\neq j}d_{i}d_{j}] is a sum over the wedges centered at vertex kk, summed all kk. We can swap the summation over pairs of vertices, and choose a third neighbour to form a wedge, which gives

ki,jkijdidj=i,jijdidjk𝟙(ki,kj).\sum_{k}\sum_{\begin{subarray}{c}i,j\sim k\\ i\neq j\end{subarray}}d_{i}d_{j}=\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d_{i}d_{j}\sum_{k}\mathbbm{1}_{(k\sim i,k\sim j)}.

Compute

𝔼[i,jijdidjk𝟙(ki,kj)]=i,jijdidjk𝔼[𝟙(ki,kj)]=i,jijdidjk(didjdk(dk1)(m11)(m12)𝟙ki,kj+didj(dk1)(dk2)(m11)(m12)𝟙k=i or k=j)=i,jijdidjk(didjdk(dk1)(m11)(m12)2didj(dk1)(m11)(m12)𝟙k=i or k=j)=1(m11)(m12)i,jijdi2dj2k(dk(dk1)2(dk1)𝟙k=i or k=j)=1(m11)(m12)((m2m1)(i,jdi2dj2idi4)2i,jijdi2dj32i,jijdi3dj2+4i,jijdi2dj2)=1(m11)(m12)((m2m1)(m22m4)4(i,jdi2dj3idi5)+4(i,jdi2dj2idi4))=1(m11)(m12)((m2m1)(m22m4)4(m3m2m5)+4(m22m4))=m23m2m4m1m22+m1m44m3m2+4m5+4m224m4(m11)(m12).\begin{split}&\mathbb{E}\left[\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d_{i}d_{j}\sum_{k}\mathbbm{1}_{(k\sim i,k\sim j)}\right]=\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d_{i}d_{j}\sum_{k}\mathbb{E}\left[\mathbbm{1}_{(k\sim i,k\sim j)}\right]\\ &=\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d_{i}d_{j}\sum_{k}\left(\frac{d_{i}d_{j}d_{k}(d_{k}-1)}{(m_{1}-1)(m_{1}-2)}\mathbbm{1}_{k\neq i,k\neq j}+\frac{d_{i}d_{j}(d_{k}-1)(d_{k}-2)}{(m_{1}-1)(m_{1}-2)}\mathbbm{1}_{k=i\textnormal{ or }k=j}\right)\\ &=\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d_{i}d_{j}\sum_{k}\left(\frac{d_{i}d_{j}d_{k}(d_{k}-1)}{(m_{1}-1)(m_{1}-2)}-\frac{2d_{i}d_{j}(d_{k}-1)}{(m_{1}-1)(m_{1}-2)}\mathbbm{1}_{k=i\textnormal{ or }k=j}\right)\\ &=\frac{1}{(m_{1}-1)(m_{1}-2)}\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d^{2}_{i}d^{2}_{j}\sum_{k}\left(d_{k}(d_{k}-1)-2(d_{k}-1)\mathbbm{1}_{k=i\textnormal{ or }k=j}\right)\\ &=\frac{1}{(m_{1}-1)(m_{1}-2)}\left((m_{2}-m_{1})\left(\sum_{i,j}d_{i}^{2}d_{j}^{2}-\sum_{i}d_{i}^{4}\right)-2\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d^{2}_{i}d^{3}_{j}-2\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d^{3}_{i}d^{2}_{j}+4\sum_{\begin{subarray}{c}i,j\\ i\neq j\end{subarray}}d_{i}^{2}d_{j}^{2}\right)\\ &=\frac{1}{(m_{1}-1)(m_{1}-2)}\left((m_{2}-m_{1})\left(m_{2}^{2}-m_{4}\right)-4\left(\sum_{i,j}d^{2}_{i}d^{3}_{j}-\sum_{i}d_{i}^{5}\right)+4\left(\sum_{i,j}d_{i}^{2}d_{j}^{2}-\sum_{i}d_{i}^{4}\right)\right)\\ &=\frac{1}{(m_{1}-1)(m_{1}-2)}\left((m_{2}-m_{1})\left(m_{2}^{2}-m_{4}\right)-4\left(m_{3}m_{2}-m_{5}\right)+4\left(m_{2}^{2}-m_{4}\right)\right)\\ &=\frac{m_{2}^{3}-m_{2}m_{4}-m_{1}m_{2}^{2}+m_{1}m_{4}-4m_{3}m_{2}+4m_{5}+4m_{2}^{2}-4m_{4}}{(m_{1}-1)(m_{1}-2)}.\end{split}

Hence

𝔼[e~,H2e~]\displaystyle\mathbb{E}[\langle\tilde{e},H^{2}\tilde{e}\rangle]
=1m11(m3+m23m2m4m1m22+m1m44m3m2+4m5+4m224m4(m11)(m12)m23(m11)2).\displaystyle=\frac{1}{m_{1}-1}\bigg{(}m_{3}+\frac{m_{2}^{3}-m_{2}m_{4}-m_{1}m_{2}^{2}+m_{1}m_{4}-4m_{3}m_{2}+4m_{5}+4m_{2}^{2}-4m_{4}}{(m_{1}-1)(m_{1}-2)}-\frac{m^{3}_{2}}{(m_{1}-1)^{2}}\bigg{)}.

Given the event \mathcal{E}, by (4.1) and Corollary 3.2, we have that 𝔼[λ1]\mathbb{E}[\lambda_{1}] concentrates around m2/m1m_{2}/m_{1} with an O(m)O(\sqrt{m_{\infty}}) error, and so we can write

𝔼[e~,H2e~λ12]=𝔼[e~,H2e~](m2m11)2(1+o(1))+o(1).\mathbb{E}\left[\frac{\langle\tilde{e},H^{2}\tilde{e}\rangle}{\lambda_{1}^{2}}\right]=\frac{\mathbb{E}\left[\langle\tilde{e},H^{2}\tilde{e}\rangle\right]}{(\frac{m_{2}}{m_{1}-1})^{2}}(1+o(1))+o(1).

We run through the various contributions separately (using Assumption 1.1). Noting that nm0kmknmknm_{0}^{k}\leq m_{k}\leq nm_{\infty}^{k} and that. there are positive constants c,Cc,C such that cmm0Cc\leq\frac{m_{\infty}}{m_{0}}\leq C, we have

m23m22(m11)(m12)=Θ(1n),\displaystyle\frac{m_{2}^{3}}{m_{2}^{2}(m_{1}-1)(m_{1}-2)}=\Theta\left(\frac{1}{n}\right),
m2m4m22(m12)=o(1n),m1m4m22(m12)=Θ(1n),4m3m2m22(m12)=o(1n),\displaystyle\frac{m_{2}m_{4}}{m_{2}^{2}(m_{1}-2)}=o\left(\frac{1}{\sqrt{n}}\right),\quad\frac{m_{1}m_{4}}{m_{2}^{2}(m_{1}-2)}=\Theta\left(\frac{1}{n}\right),\quad\frac{4m_{3}m_{2}}{m_{2}^{2}(m_{1}-2)}=o\left(\frac{1}{n}\right),
4m5m22(m12)=o(1n3/2),4m22m22(m12)=o(1n),4m4m22(m12)=o(1n2).\displaystyle\frac{4m_{5}}{m_{2}^{2}(m_{1}-2)}=o\left(\frac{1}{n^{3/2}}\right),\quad\frac{4m_{2}^{2}}{m_{2}^{2}(m_{1}-2)}=o\left(\frac{1}{n}\right),\quad\frac{4m_{4}}{m_{2}^{2}(m_{1}-2)}=o\left(\frac{1}{n^{2}}\right).

Therefore

𝔼[e~,H2e~](m2/(m11))2=m3m1m221+o(1n),\frac{\mathbb{E}\left[\langle\tilde{e},H^{2}\tilde{e}\rangle\right]}{(m_{2}/(m_{1}-1))^{2}}=\frac{m_{3}m_{1}}{m^{2}_{2}}-1+o(\frac{1}{\sqrt{n}}),

which settles (1.1).

Weak law of large numbers.

We want to show that

λ1𝔼λ11\frac{\lambda_{1}}{\mathbb{E}{\lambda_{1}}}\to 1

in \mathbb{P}-probability. Using Corollary 3.2 and the Weyl interlacing inequality, we have that, with probability 1nK1-n^{-K} for K>0K>0,

m2m1O(m)λ1m2m1+O(m).\frac{m_{2}}{m_{1}}-O\left(\sqrt{m_{\infty}}\,\right)\leq\lambda_{1}\leq\frac{m_{2}}{m_{1}}+O(\sqrt{m_{\infty}}\,).

By (1.1),

m2m1O(m)m2m1(1+o(1))λ1𝔼[λ1]m2m1+O(m)m2m1(1+o(1)).\frac{\frac{m_{2}}{m_{1}}-O(\sqrt{m_{\infty}}\,)}{\frac{m_{2}}{m_{1}}(1+o(1))}\leq\frac{\lambda_{1}}{\mathbb{E}[\lambda_{1}]}\leq\frac{\frac{m_{2}}{m_{1}}+O(\sqrt{m_{\infty}}\,)}{\frac{m_{2}}{m_{1}}(1+o(1))}.

It follows that

1O(1m))λ1𝔼[λ1]1+O(1m)1-O\left(\frac{1}{\sqrt{m_{\infty}}}\right))\leq\frac{\lambda_{1}}{\mathbb{E}[\lambda_{1}]}\leq 1+O\left(\frac{1}{\sqrt{m_{\infty}}}\right)

with probability 1nK1-n^{-K}, and hence the claim follows.

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