Optimal bounds for convergence of expected spectral distributions to the semi-circular law

Abstract

Let $\mathbf {X}=(X_{jk})_{j,k=1}^n$ denote a Hermitian random matrix with entries $X_{jk}$, which are independent for $1\le j\le k\le n$. We consider the rate of convergence of the empirical spectral distribution function of the matrix $\mathbf {X}$ to the semi-circular law assuming that $\mathbf{E}X_{jk}=0$, $\mathbf{E}X_{jk}^2=1$ and that

$$\begin{aligned} \sup _{n\ge 1}\sup _{1\le j,k\le n}\mathbf{E}|X_{jk}|^4=:\mu _4<\infty , \end{aligned}$$

and

$$\begin{aligned} \sup _{1\le j,k\le n}|X_{jk}|\le D_0n^{\frac{1}{4}}. \end{aligned}$$

By means of a recursion argument it is shown that the Kolmogorov distance between the expected spectral distribution of the Wigner matrix $\mathbf {W}=\frac{1}{\sqrt{n}}\mathbf {X}$ and the semicircular law is of order $O(n^{-1})$.

1 Introduction

Consider a family $\mathbf {X} = \{X_{jk}\}$, $1 \le j \le k \le n$, of independent real random variables defined on some probability space , for any $n\ge 1$. Assume that $X_{jk} = X_{kj}$, for $1 \le k < j \le n$, and introduce the symmetric matrices

$$\begin{aligned} \mathbf {W} = \ \frac{1}{\sqrt{n}} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} X_{11} &{} X_{12} &{} \cdots &{} X_{1n} \\ X_{21} &{} X_{22} &{} \cdots &{} X_{2n} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ X_{n1} &{} X_{n2} &{} \cdots &{} X_{nn} \\ \end{array} \right) . \end{aligned}$$

The matrix $\mathbf {W}$ has a random spectrum $\{\lambda _1,\dots ,\lambda _n\}$ and an associated spectral distribution function $\mathcal {F}_{n}(x) = \frac{1}{n}\ \mathrm{card}\,\{j \le n: \lambda _j \le x\}, x \in {\mathbb {R}}$. Averaging over the random values $X_{ij}(\omega )$, define the expected (non-random) empirical distribution functions $F_{n}(x) = \mathbf{E}\,\mathcal {F}_{n}(x)$. Let $G(x)$ denote the semi-circular distribution function with density $g(x)=G'(x)=\frac{1}{2\pi }\sqrt{4-x^2}\mathbb {I}_{[-2,2]}(x)$, where $\mathbb {I}_{[a,b]}(x)$ denotes the indicator-function of the interval $[a,b]$. The rate of convergence to the semi-circular law has been studied by several authors. We proved in [13] that the Kolmogorov distance between $\mathcal {F}_n(x)$ and the distribution function $G(x)$, $\Delta _n^*:=\sup _x|\mathcal {F}_n(x)-G(x)|$ is of order $O_P(n^{-\frac{1}{2}})$ (i.e. $n^{\frac{1}{2}}\Delta _n^*$ is bounded in probability). Bai et al. [1, 2] and Girko [8] showed that $\Delta _n:=\sup _x| F_n(x)-G(x)|=O(n^{-\frac{1}{2}})$. Bobkov, Götze and Tikhomirov [4] proved that $\Delta _n$ and $\mathbf{E}\Delta _n^*$ have order $O(n^{-\frac{2}{3}})$ assuming a Poincaré inequality for the distribution of the matrix elements. For the Gaussian Unitary Ensemble respectively for the Gaussian Orthogonal Ensemble, see [12] respectively [21], it has been shown that $\Delta _n=O(n^{-1})$. Denote by $\gamma _{n1}\le \cdots \le \gamma _{nn}$, the quantiles of $G$, i.e. $G(\gamma _{nj})=\frac{j}{n}$, and introduce the notation ${\text {llog}}_n:=\log \log n$. Erdös et al. [6, 7] showed, for matrices with elements $X_{jk}$ which have a uniformly sub exponential decay, i.e.

$$\begin{aligned} \Pr \{|X_{jk}|>t\}\le A\exp \{-t^{\varkappa }\}, \end{aligned}$$

(1.1)

for some $\varkappa >0$, $A>0$ and for any $t\ge 1$, the following result

$$\begin{aligned}&\Pr \left\{ \exists \,\, j:\,|\lambda _j-\gamma _{nj}|\ge (\log n)^{C {\text {llog}}_n} \left[ \min (j,N-j+1)\right] ^{-\frac{1}{3}}n^{-\frac{2}{3}} \right\} \nonumber \\&\quad \le C\exp \{-(\log n)^{c {\text {llog}}_n}\}, \end{aligned}$$

(1.2)

for $n$ large enough. It is straightforward to check that this bound implies that

$$\begin{aligned} \Pr \left\{ \sup _x|\mathcal {F}_n(x)-G(x)|\le Cn^{-1}(\log n)^{C {\text {llog}}_n}\right\} \ge 1- C\exp \{-(\log n)^{c {\text {llog}}_n}\}. \end{aligned}$$

(1.3)

From the last inequality it follows that $\mathbf{E}\Delta _n^*\le Cn^{-1}(\log n)^{C {\text {llog}}_n}$. Similar results were obtained in [20], Theorem 32], assuming additionally that the distributions of the entries of matrices have vanishing third moment.

In this paper we derive the optimal bound for the rate of convergence of the expected spectral distribution to the semi-circular law. Using arguments similar to those used in [17] we provide a self-contained proof based on recursion methods developed in the papers of Götze and Tikhomirov [9, 13] and [22]. It follows from the results of Gustavsson [14] that the best possible bound in the Gaussian case for the rate of convergence in probability is $O(n^{-1}\sqrt{\log n})$. The best possible bound for $\Delta _n$ is of order $O(n^{-1})$. For Gaussian matrices such bounds were obtained in [12] and [21]. Our setup includes the case that the distributions of $X_{jk}=X_{jk}^{(n)}$ may depend on $n$. In the following we shall investigate the rate of convergence of expected spectral distribution function $F_n(x)=\mathbf{E}\mathcal {F}_n(x)$ to the semi-circular distribution function by estimating the quantity $\Delta _n$. The main result of this paper is the following

Theorem 1.1

Let $\mathbf{E}X_{jk}=0$, $\mathbf{E}X_{jk}^2=1$. Assume that

$$\begin{aligned} \sup _{n\ge 1}\sup _{1\le j,k\le n}\mathbf{E}|X_{jk}|^4=: \mu _4<\infty . \end{aligned}$$

(1.4)

Assume as well that there exists a constant $D_0$ such that for all $n\ge 1$

$$\begin{aligned} \sup _{1\le j,k\le n}|X_{jk}|\le D_0n^{\frac{1}{4}}. \end{aligned}$$

(1.5)

Then, there exists a positive constant $C=C(D_0,\mu _4)$ depending on $D_0$ and $\mu _4$ only such that

$$\begin{aligned} \Delta _n=\sup _x|F_n(x)-G(x)|\le Cn^{-1}. \end{aligned}$$

(1.6)

Corollary 1.1

Let $\mathbf{E}X_{jk}=0$, $\mathbf{E}X_{jk}^2=1$. Assume that

$$\begin{aligned} \sup _{n\ge 1}\sup _{1\le j,k\le n}\mathbf{E}|X_{jk}|^8=:\mu _8<\infty . \end{aligned}$$

(1.7)

Then, there exists a positive constant $C=C(\mu _8)$ depending on $\mu _8$ only such that

$$\begin{aligned} \Delta _n\le Cn^{-1}. \end{aligned}$$

(1.8)

Remark 1.2

Note that the bound (1.6) in Theorem 1.1 and the bound (1.8) in the Corollary 1.1 are not improvable and coincide with the corresponding bounds in the Gaussian case.

We state here as well the results for the Stieltjes transform of the expected spectral distribution of the matrix $\mathbf {W}$. Let $\mathbf {R}$ denote the resolvent matrix of the matrix $\mathbf {W}$,

$$\begin{aligned} \mathbf {R}:=\mathbf {R}(z)=(\mathbf {W}-z\mathbf {I})^{-1}. \end{aligned}$$

Here and in what follows $\mathbf {I}$ denotes the unit matrix of corresponding dimension. For any distribution function $F(x)$ we define the Stieltjes transform $s_F(z)$, for $z=u+iv$ with $v>0$, via formula

$$\begin{aligned} s_F(z)=\int _{-\infty }^{\infty }\frac{1}{x-z}dF(x). \end{aligned}$$

Denote by $m_n(z)$ the Stieltjes transform of the distribution function $\mathcal {F}_n(x)$. It is a well-known fact that

$$\begin{aligned} m_n(z)=\frac{1}{n}\sum _{j=1}^n\frac{1}{\lambda _j-z}=\frac{1}{n}\mathrm{Tr}\,\mathbf {R}. \end{aligned}$$

By $s(z)$ we denote the Stieltjes transform of the semi-circular law,

$$\begin{aligned} s(z)=\frac{-z+\sqrt{z^2-4}}{2}. \end{aligned}$$

The Stieltjes transform of the semi-circular distribution satisfies the equation

$$\begin{aligned} s^2(z)+zs(z)+1=0 \end{aligned}$$

(1.9)

(see, for example, [13], equality (4.20)]).

Introduce for $z=u+iv$ and a positive constant $A_0>0$

$$\begin{aligned} v_0:= A_0n^{-1}, \quad \text {and} \quad \gamma :=\gamma (z):=|2-|u||. \end{aligned}$$

(1.10)

For any $0<\varepsilon <\frac{1}{2}$, and $A_0>0$, define a region $\mathbb {G}=\mathbb {G}(A_0,n,\varepsilon )\subset \mathbb {C}_+$, by

$$\begin{aligned} \mathbb {G}:=\left\{ z=u+iv\in \mathbb {C}_+: -2+\varepsilon \le u\le 2-\varepsilon ,\, v\ge v_0/\sqrt{\gamma (z)}\right\} . \end{aligned}$$

(1.11)

Let $a>0$ be a positive number such that

$$\begin{aligned} \frac{1}{\pi }\int _{|u|\le a}\frac{1}{u^2+1}du=\frac{3}{4}. \end{aligned}$$

(1.12)

We prove the following result.

Theorem 1.3

Let $\frac{1}{2}>\varepsilon >0$ be a sequence of positive numbers in (1.11) such that

$$\begin{aligned} \varepsilon ^{\frac{3}{2}} =2v_0a. \end{aligned}$$

(1.13)

Assuming the conditions of Theorem 1.1, there exists a positive constant $C=C(D_0,A_0,\mu _4)$ depending on $D$, $A_0$ and $\mu _4$ only, such that, for $z\in \mathbb {G}$

$$\begin{aligned} |\mathbf{E}m_n(z)-s(z)|\le \frac{C}{n v^{\frac{3}{4}}}+\frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

1.1 Sketch of the proof

1. We start with an estimate of the Kolmogorov-distance to the Wigner distribution via an integral over the difference of the corresponding Stieltjes transforms along a contour in the upper half-plane using a smoothing inequality (2.1) and Cauchy’s formula developed by the authors in [11]. The resulting bound (2.3) involves an integral over a segment at a fixed distance, say $V=4$, from the real axis and a segment $u+i A_0 n^{-1}(2-\left| u\right| )^{-\frac{1}{2}}, \; |u|\le 2$ at a distance of order $n^{-1}$ but avoiding to come close to the endpoints $\pm 2$ of the support. These segments are part of the boundary of an $n$-dependent region $\mathbb {G}$ where bounds of Stieltjes transforms are needed. Since the Stieltjes-transform and the diagonal elements $R_{jj}(z)$ of the resolvent of the Wigner-matrix $\mathbf {W}$ are uniformly bounded on the segment with $\mathrm{Im}\,z=V$ by $1{/}V$ (see Sect. 3.1) proving a bound of order $O(n^{-1})$ for the latter segment near the $x$-axis is the essential problem.

2. In order to investigate this crucial part of the error we start with the 2nd resolvent or self-consistency equation for the expected Stieltjes transform resp. the quantities $R_{jj}(z)$ of $\mathbf {W}$ (see (3.2) below) based on the difference of the resolvent of $\mathbf {W}^{(j)}$ ($j$th row and column removed) and $\mathbf {W}$. For the equivalent representation for the difference of Stieltjes transforms (see (3.3)) we have to show an error bound of order $O((nv)^{-\frac{3}{2}}|z^2-4|^{-\frac{1}{4}})$ for $z\in \mathbb {G}$. To prove this bound we use a recursive version of this representations as in (5.32). Obviously bounds for $\mathbf{E}|R_{jj}|^p$ for $z=u+iv$ close to the real line are needed for the sufficiently large $p$, which follow once the error terms $\varepsilon _j$ are small in the region $\mathbb {G}$. But proving that $\mathbf{E}|\varepsilon _j|^p$ is small requires in turn again bounds of $\mathbf{E}|R_{jj}|^{2p}$.

An approach suggested recently in [17] turns out to be very fruitful in dealing with this recursion problem. Assuming that $\mathbf{E}|R_{jj}|^{2p}\le C_0^{2p}$ for some $z=u+iv$, we can show that $\mathbf{E}|R_{jj}|^p\le C_0^p$ with $z=u+iv/s_0$ with some fixed scale factor $s_0>1$. This allows us to prove by induction a bound of type $\mathbf{E}|R_{jj}|^q\le C_0^q$ for some fixed $q$ (independent of $n$) and $z=u+iv$ with $v\ge Cn^{-1}$ starting with $\mathbf{E}|R_{jj}|^p\le C_0^p$ for $p=s_0^q$ and $z=u+iv$ for fixed $v=4$, say. The latter assumption can be easily verified.

Note that one of the errors, that is $\varepsilon _{j2}$, in (3.2) is a quadratic form in independent random variables. Thus, in case that $W$ has entries with exponential or even sub-Gaussian tails, inequalities for quadratic forms of independent random variables, like [11], Lemma 3.8] or [17], Proposition A.1], could be applied.

Assuming eight moments in Corollary 1.1 or four moments and a truncation condition in Theorem 1.1 only, we can’t use these strong tail estimates for quadratic forms anymore. Our solution is an recursive application of Burkholder’s inequality for the $p$th moment resulting in a bound involving moments of order $p/2$ of another quadratic form in independent variables in each step. This is the crucial part of the moment recursion for $R_{jj}$ described above. Details of this procedure are described in Sects. 5.1 and 5.2.

3. In Sect. 6 we prove a bound for the error $\mathbf{E}\Lambda _n:=\mathbf{E}(m_n(z)-s(z))$ of the form ${n^{-1} v^{-\frac{3}{4}}}+(n v)^{-\frac{3}{2}}|z^2-4|^{-\frac{1}{4}}$ which suffices to prove the rate $O(n^{-1})$ in Theorem 1.1. Here we use a series of martingale-type decompositions to evaluate the expectation $\mathbf{E}m_n(z)$ combined with the bound $\mathbf{E}|\Lambda _n|^2 \le C (nv)^{-2}$ of Lemma 7.24 in the Appendix which is again based on a recursive inequality for $\mathbf{E}|\Lambda _n|^2$ in (7.71). A direct application of this bound to estimate the error terms $\varepsilon _{j4}$ would result in a less precise bound of order $O(n^{-1}\log n)$ in Theorem 1.1. Bounds of such type will be shown for the Kolmogorov distance of the random spectral distribution to Wigner’s law in a separate paper. For the expectation we provide sharper bounds in Sect. 6.2 involving $m'_n(z)$.

4. The necessary auxiliary bounds for all these steps are collected in the Appendix.

2 Bounds for the Kolmogorov distance of spectral distributions via Stieltjes transforms

To bound the error $\Delta _n$ we shall use an approach developed in previous work of the authors, see [13].

We modify the bound of the Kolmogorov distance between an arbitrary distribution function and the semi-circular distribution function via their Stieltjes transforms obtained in [13], Lemma 2.1]. For $x\in [-2,2]$ define $\gamma (x):=2-|x|$. Given $\frac{1}{2}>\varepsilon >0$ introduce the interval $\mathbb {J}_{\varepsilon }=\{x\in [-2,2]:\, \gamma (x)\ge \varepsilon \}$ and $\mathbb {J}'_{\varepsilon }=\mathbb {J}_{\varepsilon /2}$. For a distribution function $F$ denote by $S_F(z)$ its Stieltjes transform.

Proposition 2.1

Let $v>0$ and $a>0$ and $\frac{1}{2}>\varepsilon >0$ be positive numbers such that

$$\begin{aligned} \frac{1}{\pi }\int _{|u|\le a}\frac{1}{u^2+1}du=\frac{3}{4}, \end{aligned}$$

(2.1)

and

$$\begin{aligned} 2va\le \varepsilon ^{\frac{3}{2}}. \end{aligned}$$

(2.2)

If $G$ denotes the distribution function of the standard semi-circular law, and $F$ is any distribution function, there exist some absolute constants $C_1$ and $C_2$ such that

$$\begin{aligned} \Delta (F,G):= & {} \sup _{x}|F(x)-G(x)|\\\le & {} 2 \sup _{x\in \mathbb {J}'_{\varepsilon }}\left| \mathrm {Im}\;\!\int _{-\infty }^x\left( S_F\left( u+i\frac{v}{\sqrt{\gamma }}\right) -S_G\left( u+i\frac{v}{\sqrt{\gamma }}\right) \right) du\right| +C_1v +C_2\varepsilon ^{\frac{3}{2}}. \end{aligned}$$

Remark 2.2

For any $x\in \mathbb {J}_{\varepsilon }$ we have $\gamma =\gamma (x)\ge \varepsilon$ and according to condition (2.2), $\frac{av}{\sqrt{\gamma }}\le \frac{\varepsilon }{2}$.

For a proof of this Proposition see [11], Proposition 2.1].

Lemma 2.1

Under the conditions of Proposition 2.1, for any $V>v$ and $0<v\le \frac{\varepsilon ^{3/2}}{2a}$ and $v'=v/\sqrt{\gamma }, \gamma = 2-|x|$, $x\in {\mathbb {J}}'_{\varepsilon }$ as above, the following inequality holds

$$\begin{aligned}&\sup _{x\in \mathbb {J}'_{\varepsilon }}\left| \int _{-\infty }^x(\mathrm {Im}\;\!(S_F(u+iv')-S_G(u+iv'))du\right| \\&\quad \le \int _{-\infty }^{\infty }|S_F(u+iV)-S_G(u+iV)|du\\&\qquad + \sup _{x\in \mathbb {J}'_{\varepsilon }}\left| \int _{v'}^V \left( S_F(x+iu)-S_G(x+iu)\right) du\right| . \end{aligned}$$

Proof

Let $x\in \mathbb {J}'_{\varepsilon }$ be fixed. Let $\gamma =\gamma (x)$. Put $z=u+iv'$. Since $v'=\frac{v}{\sqrt{\gamma }}\le \frac{\varepsilon }{2a}$, see (2.2), we may assume without loss of generality that $v'\le 4$ for $x\in \mathbb {J}'_{\varepsilon }$. Since the functions $S_F(z)$ and $S_G(z)$ are analytic in the upper half-plane, it is enough to use Cauchy’s theorem. We can write for $x\in \mathbb {J}'_{\varepsilon }$

$$\begin{aligned} \int _{-\infty }^{x}\mathrm {Im}\;\!(S_F(z)-S_G(z))du=\mathrm {Im}\;\!\left\{ \lim _{L\rightarrow \infty } \int _{-L}^x(S_F(u+iv')-S_G(u+iv'))du\right\} . \end{aligned}$$

By Cauchy’s integral formula, we have

$$\begin{aligned} \int _{-L}^x(S_F(z)-S_G(z))du= & {} \int _{-L}^x(S_F(u+iV)-S_G(u+iV))du\\&+\int _{v'}^V(S_F(-L+iu)-S_G(-L+iu))du\\&-\int _{v'}^V(S_F(x+iu)-S_G(x+iu))du. \end{aligned}$$

Denote by $\xi \text { (resp. }\eta )$ a random variable with distribution function $F(x)$ (resp. $G(x)$). Then we have

$$\begin{aligned} |S_F(-L+iu)|=\left| \mathbf{E}\frac{1}{\xi +L-iu}\right| \le {v'}^{-1}\Pr \{|\xi |>L/2\}+\frac{2}{L}, \end{aligned}$$

for any $v'\le u\le V$. Similarly,

$$\begin{aligned} |S_G(-L+iu)|\le {v'}^{-1}\Pr \{|\eta |>L/2\}+\frac{2}{L}. \end{aligned}$$

These inequalities imply that

$$\begin{aligned} \left| \int _{v'}^V(S_F(-L+iu)-S_G(-L+iu))du\right| \rightarrow 0\quad \text {as}\quad L\rightarrow \infty , \end{aligned}$$

which completes the proof. $\square$

Combining the results of Proposition 2.1 and Lemma 2.1, we get

Corollary 2.2

Under the conditions of Proposition 2.1 the following inequality holds

$$\begin{aligned} \Delta (F,G)\le & {} 2\int _{-\infty }^{\infty }|S_F(u+iV)-S_G(u+iV)|du+ C_1v_0+C_2\varepsilon ^{\frac{3}{2}}\nonumber \\&+ \,2 \sup _{x\in \mathbb {J}'_{\varepsilon }}\int _{v'}^V|S_F(x+iu)-S_G(x+iu)|du, \end{aligned}$$

(2.3)

where $v'=\frac{v_0}{\sqrt{\gamma }}$ with $\gamma =2-|x|$ and $C_1,C_2 >0$ denote absolute constants.

3 Proof of Theorem 1.1

Proof

We shall apply Corollary 2.2 to prove the Theorem 1.1. We choose $V=4$ and $v_0$ as defined in (1.10) and use the quantity $\varepsilon =(2av_0)^{\frac{2}{3}}$.

3.1 Estimation of the first integral in (2.3) for $V=4$

Denote by $\mathbb {T}=\{1,\ldots ,n\}$. In the following we shall systematically use for any $n\times n$ matrix $\mathbf {W}$ together with its resolvent $\mathbf {R}$, its Stieltjes transform $m_n$ etc. the corresponding quantities $\mathbf {W}^{(\mathbb {A})}$, its resolvent $\mathbf {R}^{(\mathbb {A})}$ and its Stieltjes transform $m_n^{(\mathbb {A})}$ for the corresponding sub matrix with entries $X_{jk}, j, k \not \in \mathbb {A}$, $\mathbb {A} \subset \mathbb {T}=\{1,\ldots ,n\}$. Let $\mathbb {T}_{\mathbb {A}}=\mathbb {T}{\setminus }\mathbb {A}$. Observe that

$$\begin{aligned} m_n^{(\mathbb {A})}(z)=\frac{1}{n}\sum _{j\in \mathbb {T}_{\mathbb {A}}}\frac{1}{\lambda ^{(\mathbb {A})}_j-z}. \end{aligned}$$

By $\mathfrak {M}^{(\mathbb {A})}$ we denote the $\sigma$-algebra generated by $X_{lk}$ with $l,k\in \mathbb {T}_{\mathbb {A}}$. If $\mathbb {A}=\emptyset$ we shall omit the set $\mathbb {A}$ as exponent index.

We shall use the representation

$$\begin{aligned} R_{jj}=\frac{1}{-z+\frac{1}{\sqrt{n}}X_{jj}-\frac{1}{n}{{\sum _{k,l\in \mathbb {T}_j}}}X_{jk}X_{jl}R^{(j)}_{kl}}, \end{aligned}$$

(3.1)

(see, for example, [13], equality (4.6)]). We may rewrite it as follows

$$\begin{aligned} R_{jj}=-\frac{1}{z+m_n(z)}+\frac{1}{z+m_n(z)}\varepsilon _jR_{jj}, \end{aligned}$$

(3.2)

where $\varepsilon _j:=\varepsilon _{j1}+\varepsilon _{j2}+ \varepsilon _{j3}+ \varepsilon _{j4}$ with

$$\begin{aligned} \varepsilon _{j1}:= & {} \frac{1}{\sqrt{n}}X_{jj},\quad \varepsilon _{j2}:=-\frac{1}{n}{\sum _{k\ne l \in \mathbb {T}_j}}X_{jk}X_{jl}R^{(j)}_{kl},\\ \varepsilon _{j3}:= & {} -\frac{1}{n}{\sum _{k\in \mathbb {T}_j}}(X_{jk}^2-1)R^{(j)}_{kk},\quad \varepsilon _{j4}:=\frac{1}{n}(\mathrm{Tr}\,\mathbf {R}-\mathrm{Tr}\,\mathbf {R}^{(j)}).\quad \end{aligned}$$

Let

$$\begin{aligned} \Lambda _n:=\Lambda _n(z):=m_n(z)-s(z)=\frac{1}{n}\mathrm{Tr}\,\mathbf {R}-s(z). \end{aligned}$$

Summing equality (3.2) in $j=1,\ldots ,n$ and solving with respect $\Lambda _n$, we get

$$\begin{aligned} \Lambda _n= m_n(z)-s(z)=\frac{T_n}{z+m_n(z)+s(z)}, \end{aligned}$$

(3.3)

where

$$\begin{aligned} T_n=\frac{1}{n}\sum _{j=1}^n\varepsilon _jR_{jj}. \end{aligned}$$

Obvious bounds like $|z+s(z)|\ge 1$, $|\lambda _j-z|^{-1}\le v^{-1}$, $\max \{|R_{jj}^{(\mathbb {J})}(z)|,\,|m_n^{(\mathbb {J})}(z)|\}\le v^{-1}$, imply that for $V=4$ and for any $\mathbb {J}\subset \mathbb {T}$,

$$\begin{aligned} |m_n^{(\mathbb {J})}(z)|\le & {} \frac{1}{4}\le \frac{1}{2}|z+s(z)|,\\ |s(z)-m_n^{(\mathbb {J})}(z)|\le & {} \frac{1}{2} \le \frac{1}{2}|z+s(z)|, \quad \text {a.s.} \end{aligned}$$

and therefore,

$$\begin{aligned} |z+m_n^{(\mathbb {J})}(z)+s(z)|\ge \frac{1}{2}|z+s(z)|,\quad |z+m_n^{(\mathbb {J})}(z)|\ge \frac{1}{2}|s(z)+z|. \end{aligned}$$

(3.4)

Using equality (3.3), we may write

$$\begin{aligned} \mathbf{E}\Lambda _n= & {} \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j}R_{jj}}{z+m_n(z)+s(z)}\\= & {} \sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }s(z)}{z+m_n(z)+s(z)}+ \sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }(R_{jj}-s(z)) }{z+m_n(z)+s(z)}. \end{aligned}$$

We use that $\mathbf{E}\{\varepsilon _{j\nu }|\mathfrak {M}^{(j)}\}=0$, for $\nu =1,2,3$ and obtain

$$\begin{aligned} \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }s(z)}{z+m_n(z)+s(z)}=- \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }\varepsilon _{j4}s(z)}{ (z+m_n^{(j)}(z)+s(z))(z+m_n(z)+s(z))}. \end{aligned}$$

Thus, according to inequalities (3.4), Lemmas 7.8, 7.9, 7.10, 7.12 in the Appendix and Eq. (1.9), we obtain

$$\begin{aligned} \left| \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }s(z)}{z+m_n(z)+s(z)}\right| \le 4|s(z)|^3\frac{1}{n}\sum _{j=1}^n\mathbf{E}|\varepsilon _{j4}\varepsilon _{j\nu }| \le \frac{C|s(z)|^2}{n^{\frac{3}{2}}}, \end{aligned}$$

where $C$ depends on $\mu _4$ only. For $\nu =4$, Lemma 7.12 in the Appendix, inequality (3.4) and relation (1.9) yield

$$\begin{aligned} \frac{1}{n}\sum _{j=1}^n\frac{|s(z)||\varepsilon _{j4}|}{|z+m_n(z)+s(z)|}\le \frac{C}{n}|s(z)|^2 \end{aligned}$$

(3.5)

with some absolute constant $C$. Furthermore, applying the Cauchy–Schwartz inequality and inequality (3.4) and relation (1.9), we get

$$\begin{aligned} \left| \sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }(R_{jj}-s(z)) }{z+m_n(z)+s(z)} \right| \le C|s(z)|\sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}| \varepsilon _{j\nu }|^2\mathbf{E}^{\frac{1}{2}}|R_{jj}-s(z)|^2. \end{aligned}$$

(3.6)

We may rewrite the representation (3.2) using $\Lambda _n=m_n(z)-s(z)$ and (1.9) as (compare (3.1))

$$\begin{aligned} R_{jj}=s(z)-s(z)\varepsilon _jR_{jj}-s(z)\Lambda _nR_{jj}. \end{aligned}$$

(3.7)

Applying representations (3.3) and (3.7) together with (3.4) and $|R_{jj}|\le \frac{1}{4}$, we obtain

$$\begin{aligned} \mathbf{E}|R_{jj}(z)-s(z)|^2\le C|s(z)|^2\frac{1}{n}\sum _{l=1}^n\mathbf{E}|\varepsilon _l|^2. \end{aligned}$$

(3.8)

Combining inequalities (3.6) and (3.8), we get

$$\begin{aligned} \left| \sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }(R_{jj}-s(z)) }{z+m_n(z)+s(z)}\right| \le C|s(z)|^2\sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}|\varepsilon _{j\nu }|^2. \end{aligned}$$

(3.9)

Applying now Lemmas 7.8, 7.9, 7.10 and 7.12, we get

$$\begin{aligned} \left| \sum _{\nu =1}^4\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }(R_{jj}-s(z)) }{z+m_n(z)+s(z)}\right| \le \frac{C|s(z)|^2}{n}. \end{aligned}$$

(3.10)

Inequality (3.5) and (3.10) together imply

$$\begin{aligned} |\mathbf{E}\Lambda _n|\le \frac{C}{n}|s(z)|^2. \end{aligned}$$

(3.11)

Consider now the integral

$$\begin{aligned} Int(V)=\int _{-\infty }^{\infty }|\mathbf{E}m_n(u+iV)-s(u+iV)|du \end{aligned}$$

for $V=4$. Using inequality (3.11), we have

$$\begin{aligned} |Int(V)|\le \frac{C}{n}\int _{-\infty }^{\infty }|s(u+Vi)|^2du. \end{aligned}$$

Finally, we note that

$$\begin{aligned} \int _{-\infty }^{\infty }|s(z)|^2dx\le \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \frac{1}{(x-u)^2+V^2}dudF_n(x)\le \frac{\pi }{V}. \end{aligned}$$

(3.12)

Therefore,

$$\begin{aligned} \int _{-\infty }^{\infty }|\mathbf{E}m_n(u+iV)-s(u+iV)|du\le \frac{C}{n}. \end{aligned}$$

(3.13)

3.2 Estimation of the second integral in (2.3)

To finish the proof of Theorem 1.1 we need to bound the second integral in (2.3) for $z\in \mathbb {G}$ and $v_0=A_0n^{-1}$, where $\varepsilon =(2av_0)^{\frac{2}{3}}$ is defined in such a way that condition (2.2) holds. We shall use the results of Theorem 1.3. According to these results we have, for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathbf{E}m_n(z)-s(z)|\le \frac{C}{n v^{\frac{3}{4}}}+\frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

(3.14)

We have

$$\begin{aligned} \int _{v_0/\sqrt{\gamma }}^V|\mathbf{E}(m_n(x+iv)-s(x+iv))|dv\le \frac{C}{n}\int _{\frac{v_0}{\sqrt{\gamma }}}^V\frac{dv}{v^{\frac{3}{4}}}+\frac{C}{n\sqrt{n}\gamma ^{\frac{1}{4}}}\int _{\frac{v_0}{\sqrt{\gamma }}}^V\frac{dv}{v^{\frac{3}{2}}}. \end{aligned}$$

After integrating we get

$$\begin{aligned} \int _{v_0/\sqrt{\gamma }}^V|\mathbf{E}(m_n(x+iv)-s(x+iv))|dv\le \frac{C}{n}+\frac{ C\gamma ^{\frac{1}{4}}}{n\sqrt{n}\gamma ^{\frac{1}{4}}v_0^{\frac{1}{2}}}\le \frac{C}{n}. \end{aligned}$$

(3.15)

Inequalities (3.13) and (3.15) complete the proof of Theorem 1.1. Thus Theorem 1.1 is proved. $\square$

4 The proof of Corollary 1.1

To prove the Corollary 1.1 we consider truncated random variables $\widehat{X}_{jl}$ defined by

$$\begin{aligned} \widehat{X}_{jl}:=X_{jl}\mathbb {I}\{|X_{jl}|\le cn^{\frac{1}{4}} \}. \end{aligned}$$

(4.1)

Let $\widehat{\mathcal {F}}_n(x)$ denote the empirical spectral distribution function of the matrix $\widehat{\mathbf {W}}=\frac{1}{\sqrt{n}}(\widehat{X}_{jl})$.

Lemma 4.1

Assuming the conditions of Theorem 1.1 there exists a constant $C>0$ depending on $\mu _8$ only such that

$$\begin{aligned} \mathbf{E}\left\{ \sup _x|\mathcal {F}_n({x})-\widehat{\mathcal {F}}_n(x)|\right\} \le \frac{C}{n}. \end{aligned}$$

Proof

We shall use the rank inequality of Bai. See [3], Theorem A.43, p. 503]. According this inequality

$$\begin{aligned} \mathbf{E}\left\{ \sup _x|{\mathcal {F}}_n(x)-\widehat{\mathcal {F}}_n(x)|\right\} \le \frac{1}{n}\mathbf{E}\left\{ \text {rank}(\mathbf {X}-\widehat{\mathbf {X}})\right\} . \end{aligned}$$

Observing that the rank of a matrix is not larger then numbers of its non-zero entries, we may write

$$\begin{aligned} \mathbf{E}\left\{ \sup _x|{\mathcal {F}}_n(x)-\widehat{\mathcal {F}}_n(x)|\right\} {\le } \frac{1}{n}\sum _{j,k=1}^n\mathbf{E}\mathbb {I}\left\{ |X_{jk}|\ge Cn^{\frac{1}{4}}\right\} {\le } \frac{1}{n^3}\sum _{j,k=1}^n\mathbf{E}|X_{jk}|^8\le \frac{C\mu _8}{n}. \end{aligned}$$

Thus, the Lemma is proved. $\square$

Note that in the bound of the first integral in (2.3) we used the condition (1.4) only. We shall compare the Stieltjes transform of the matrix $\widehat{\mathbf {W}}$ and the matrix obtained from $\widehat{\mathbf {W}}$ by centralizing and normalizing its entries. Introduce $\widetilde{X}_{jk}=\widehat{X}_{jk}-\mathbf{E}\widehat{X}_{jk}$ and $\widetilde{\mathbf {W}}=\frac{1}{\sqrt{n}}(\widetilde{X}_{jk})_{j,k=1}^n$. We normalize the r.v.’s $\widetilde{X}_{jk}$. Let $\sigma _{jk}^2=\mathbf{E}|\widetilde{X}_{jk}|^2$. We define the r.v.’s $\breve{X}_{jk}=\sigma _{jk}^{-1}\widetilde{X}_{jk}$. Finally, let $\breve{m}_n(z)$ (resp. $\widehat{m}_n(z)$, $\widetilde{m}_n(z)$) denote the Stieltjes transform of the empirical spectral distribution function of the matrix $\breve{\mathbf {W}}= \frac{1}{\sqrt{n}}(\breve{X}_{jk})_{j,k=1}^n$ (resp. $\widehat{\mathbf {W}}$, $\widetilde{\mathbf {W}}$).

Remark 4.1

Note that

$$\begin{aligned} |\breve{X}_{jl}|\le D_1n^{\frac{1}{4}}, \quad \mathbf{E}\breve{X}_{jl}=0 \quad \text {and}\quad \mathbf{E}{\breve{X}}_{jk}^2=1, \end{aligned}$$

(4.2)

for some absolute constant $D_1$. That means that the matrix $\breve{\mathbf {W}}$ satisfies the conditions of Theorem 1.3.

Lemma 4.2

There exists some absolute constant $C$ depending on $\mu _8$ such that

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-\breve{m}_n(z)|\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}}. \end{aligned}$$

Proof

Note that

$$\begin{aligned} \breve{m}_n(z)=\frac{1}{n}\mathrm{Tr}\,(\breve{\mathbf {W}}-z\mathbf {I})^{-1}=: \frac{1}{n}\mathrm{Tr}\,\breve{\mathbf {R}},\, \widetilde{m}_n(z)=\frac{1}{n}\mathrm{Tr}\,(\widetilde{\mathbf {W}}- z\mathbf {I})^{-1}=:\frac{1}{n}\mathrm{Tr}\,\widetilde{\mathbf {R}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \widetilde{m}_n(z)-\breve{m}_n(z)=\frac{1}{n}\mathrm{Tr}\,(\widetilde{\mathbf {R}}-\breve{\mathbf {R}}) =\frac{1}{n}\mathrm{Tr}\,(\widetilde{\mathbf {W}}-\breve{\mathbf {W}}) \widetilde{\mathbf {R}}\widehat{\mathbf {R}}. \end{aligned}$$

(4.3)

Using the simple inequalities $|\mathrm{Tr}\,\mathbf {A}\mathbf {B}|\le \Vert \mathbf {A}\Vert _2\Vert \mathbf {B}\Vert _2$ and $\Vert \mathbf {A}\mathbf {B}\Vert _2\le \Vert \mathbf {A}\Vert \Vert \mathbf {B}\Vert _2$, we get

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-\breve{m}_n(z)|\le n^{-1}\mathbf{E}^{\frac{1}{2}}\Vert \widetilde{\mathbf {R}}\Vert ^2\Vert \breve{\mathbf {R}}\Vert _2^2\mathbf{E}^{\frac{1}{2}}\Vert \widetilde{\mathbf {W}}-\breve{\mathbf {W}}\Vert _2^2. \end{aligned}$$

(4.4)

Furthermore, we note that,

$$\begin{aligned} \widetilde{\mathbf {W}}-\breve{\mathbf {W}}=\frac{1}{\sqrt{n}}((1-\sigma _{jk})\breve{X}_{jk}), \end{aligned}$$

(4.5)

and

$$\begin{aligned} \Vert \widetilde{\mathbf {W}}-\breve{\mathbf {W}}\Vert _2\le \max _{1\le j,k\le n}\{1-\sigma _{jk}\}\Vert \breve{\mathbf {W}}\Vert _2. \end{aligned}$$

Since

$$\begin{aligned} 0<1-\sigma _{jk}\le 1-\sigma _{jk}^2\le Cn^{-\frac{3}{2}}\mu _8, \end{aligned}$$

therefore

$$\begin{aligned} \mathbf{E}\Vert \widetilde{\mathbf {W}}-\breve{\mathbf {W}}\Vert _2^2\le C\mu _8^2n^{-2}. \end{aligned}$$

(4.6)

Applying Lemma 7.6 inequality (7.11) in the Appendix we get

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\Vert \widetilde{\mathbf {R}}\Vert ^2\Vert \breve{\mathbf {R}}\Vert _2^2\le v^{-1}\mathbf{E}^{\frac{1}{2}}\Vert \breve{\mathbf {R}}\Vert _2^2\le v^{-1}\left( \sum _{j,k}\mathbf{E}|\breve{R}_{jk}|^2\right) ^{\frac{1}{2}}\le v^{-\frac{3}{2}}\sqrt{n}(\mathrm {Im}\;\!{\breve{m}_n(z)})^{\frac{1}{2}}. \end{aligned}$$

(4.7)

Using now inequalities (4.6) and (4.7), we obtain

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-{\breve{m}_n(z)}|\le Cn^{-\frac{3}{2}}v^{-\frac{3}{2}}\left( \frac{1}{n}\sum _{j=1}^n\mathbf{E}|\breve{\mathbf {R}}_{jj}|\right) ^{\frac{1}{2}}. \end{aligned}$$

According Remark 4.1, we may apply Corollary 5.14 in Sect. 5 with $q=1$ to prove the claim. Thus, Lemma 4.2 is proved. $\square$

Lemma 4.3

For some absolute constant $C>0$ we have

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-\widehat{m}_n(z)|\le \frac{C\mu _8}{n^{\frac{3}{2}}v^{\frac{3}{2}}}. \end{aligned}$$

Proof

Similar to (4.3), we write

$$\begin{aligned} \widetilde{m}_n(z)-\widehat{m}_n(z)=\frac{1}{n}\mathrm{Tr}\,(\widetilde{\mathbf {R}}- \widehat{\mathbf {R}}) =\frac{1}{n}\mathrm{Tr}\,(\widetilde{\mathbf {W}}-\widehat{ \mathbf {W}})\widetilde{\mathbf {R}}\widehat{\mathbf {R}}. \end{aligned}$$

This yields

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-\widehat{m}_n(z)|\le n^{-1}\mathbf{E}\Vert \widehat{\mathbf {R}}\Vert \Vert \widetilde{\mathbf {R}}\Vert _2\Vert \mathbf{E}\widehat{\mathbf {W}}\Vert _2. \end{aligned}$$

(4.8)

Furthermore, we note that, by definition (4.1) and condition (1.7), we have

$$\begin{aligned} |\mathbf{E}\widehat{X}_{jk}|\le Cn^{-\frac{7}{4}}\mu _8. \end{aligned}$$

(4.9)

Applying (Lemma 7.6 inequality (7.11) in the Appendix) and inequality (4.9), we obtain using $\Vert \widehat{\mathbf {R}}\Vert \le v^{-1}$,

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-\widehat{m}_n(z)|\le n^{-\frac{7}{4}}v^{-\frac{3}{2}}\mathbf{E}^{\frac{1}{2}}|\widetilde{m}_n(z)|. \end{aligned}$$

By Lemma 4.2,

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)|\le \mathbf{E}|\breve{m}_n(z)|+C, \end{aligned}$$

for some constant $C$ depending on $\mu _8$ and $A_0$. According to Corollary 5.14 in Sect. 5.2 with $q=1$

$$\begin{aligned} \mathbf{E}|\breve{m}_n(z)|\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}|\breve{R}_{jj}|\le C, \end{aligned}$$

with a constant $C$ depending on $\mu _4$, $D_0$. Using these inequalities, we get

$$\begin{aligned} \mathbf{E}|\widetilde{m}_n(z)-\widehat{m}_n(z)|\le \frac{C\mu _8}{n^{\frac{7}{4}}v^{\frac{3}{2}}}\le \frac{C\mu _8}{n^{\frac{3}{2}}v^{\frac{3}{2}}}. \end{aligned}$$

Thus Lemma 4.3 is proved. $\square$

Corollary 4.4

Assuming the conditions of Corollary 1.1, we have for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathbf{E}\widehat{m}_n(z)-s(z)|\le \frac{C}{(nv)^{\frac{3}{2}}}+\frac{C}{n^2v^2\sqrt{\gamma }}. \end{aligned}$$

Proof

The proof immediately follows from the inequality

$$\begin{aligned} |\mathbf{E}\widehat{m}_n(z)-s(z)|\le |\mathbf{E}(\widehat{m}_n(z)-\breve{m}_n(z))|+|\mathbf{E}\breve{m}_n(z)-s(z)|, \end{aligned}$$

Lemmas 4.2 and 4.3 and Theorem 1.3. $\square$

The proof of Corollary 1.1 follows now from Lemma 4.1, Corollary 2.2, inequality (3.13) and inequality

$$\begin{aligned} \sup _{x\in \mathbb {J}_{\varepsilon }}\int _{v_0/\sqrt{\gamma }}^V|\mathbf{E}\widehat{m}_n(x+iv)-s(x+iv)|dv\le \frac{C}{n}. \end{aligned}$$

5 Resolvent matrices and quadratic forms

The crucial problem in the proof of Theorem 1.3 is the following bound for any $z\in \mathbb {G}$

$$\begin{aligned} \mathbf{E}|R_{jj}|^p\le C^p, \end{aligned}$$

for $j=1,\ldots ,n$ and some absolute constant $C>0$. To prove this bound we use an approach similar to the proof of Lemma 3.4 in [17]. In order to arrive at our goal we need additional bounds of quadratic forms of type

$$\begin{aligned} \mathbf{E}\left| \frac{1}{n}\sum _{l\ne k}X_{jl}X_{jk}R^{(j)}_{kl}\right| ^p\le \left( \frac{Cp}{\sqrt{nv}}\right) ^p. \end{aligned}$$

To prove this bound we recurrently use Rosenthal’s and Burkholder’s inequalities.

5.1 The key lemma

In this Section we provide auxiliary lemmas needed for the proof of Theorem 1.1.

For any $\mathbb {J}\subset \mathbb {T}$ introduce $\mathbb {T}_{\mathbb {J}}=\mathbb {T}{\setminus }\mathbb {J}$. We introduce the quantity, for some $\mathbb {J}\subset \mathbb {T}$,

$$\begin{aligned} B_p^{(\mathbb {J})}:=\left[ \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J}}}\left( \sum _{r\in \mathbb {T}_{\mathbb {J}}}|R_{qr}^{(\mathbb {J})}|^2\right) ^p\right] . \end{aligned}$$

By Lemma 7.6, inequality (7.12) in the Appendix, we have

$$\begin{aligned} \mathbf{E}B_p^{(\mathbb {J})}\le v^{-p}\frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J}}}\mathbf{E}|R_{qq}^{(\mathbb {J})}|^p. \end{aligned}$$

(5.1)

Furthermore, introduce the quantities

$$\begin{aligned} Q^{(\mathbb {J},k)}_{\nu }= & {} \sum _{l\in \mathbb {T}_{\mathbb {J},k}} \left| \sum _{r\in \mathbb {T}_{\mathbb {J},k}\cap \{1,\ldots ,l-1\}}X_{kr} a_{lr}^{(\mathbb {J},k,\nu )}\right| ^2,\nonumber \\ Q^{(\mathbb {J},k)}_{\nu 1}= & {} \sum _{r\in \mathbb {T}_{\mathbb {J},k}} a_{rr}^{(\mathbb {J},k,\nu +1)},\nonumber \\ Q^{(\mathbb {J},k)}_{\nu 2}= & {} \sum _{r\in \mathbb {T}_{\mathbb {J},k}}(X_{kr}^2-1) a_{rr}^{(\mathbb {J},k,\nu +1)},\nonumber \\ Q^{(\mathbb {J},k)}_{\nu 3}= & {} \sum _{ r\ne q\in \mathbb {T}_{\mathbb {J},k}}X_{kr} X_{kq}a_{qr}^{(\mathbb {J},k,\nu +1)}, \end{aligned}$$

(5.2)

where, $a^{(\mathbb {J},k,0)}_{qr}$ are defined recursively via

$$\begin{aligned} a_{qr}^{(\mathbb {J},k,0)}= & {} \frac{1}{\sqrt{n}} R^{(\mathbb {J},k)}_{qr},\nonumber \\ a_{qr}^{(\mathbb {J},k,\nu +1)}= & {} \sum _{l\in \{\max \{q,r\}+1,\ldots ,n\}\cap \mathbb {T}_{\mathbb {J},k}}a_{rl}^{(\mathbb {J},k,\nu )}\overline{a}_{lq}^{(\mathbb {J},k,\nu )},\quad \text {for} \quad \nu =0,\ldots ,L. \end{aligned}$$

(5.3)

Using these notations we have

$$\begin{aligned} Q^{(\mathbb {J},k)}_{\nu }=Q^{(\mathbb {J},k)}_{\nu 1}+Q^{( \mathbb {J},k)}_{\nu 2}+Q^{(\mathbb {J},k)}_{\nu 3}. \end{aligned}$$

(5.4)

Lemma 5.1

Under the conditions of Theorem 1.1 we have

$$\begin{aligned} \sum _{r\in \mathbb {T}_{\mathbb {J},k}} |a_{qr}^{(\mathbb {J},k,\nu +1)}|^2\le \left( \sum _{l,r\in \mathbb {T}_{\mathbb {J},k}}^n|a_{lr}^{(\mathbb {J},k,\nu )}|^2\right) \left( \sum _{l\in \mathbb {T}_{\mathbb {J},k}}^n|a_{ql}^{(\mathbb {J},k,\nu )}|^2\right) . \end{aligned}$$

(5.5)

Moreover,

$$\begin{aligned} \sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,\nu +1)}|^2\le \left( \sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,\nu )}|^2\right) ^2. \end{aligned}$$

(5.6)

Proof

We apply Hölder’s inequality and obtain

$$\begin{aligned} |a_{q,r}^{(\mathbb {J},k,\nu +1)}|^2\le \sum _{l\in \mathbb {T}_{\mathbb {J},k}} |a_{ql}^{(\mathbb {J},k,\nu )}|^2\sum _{l\in \mathbb {T}_{\mathbb {J},k}}| a_{lr}^{(\mathbb {J},k,\nu )}|^2. \end{aligned}$$

Summing in $q$ and $r$, (5.5) and (5.6) follow. $\square$

Corollary 5.2

Under the conditions of Theorem 1.1 we have

$$\begin{aligned} \sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,\nu )}|^2\le \left( \left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+ \frac{1}{nv}\right) v^{-1}\right) ^{2^{\nu }} \end{aligned}$$

and

$$\begin{aligned} \sum _{r\in \mathbb {T}_k} |a_{qr}^{(\mathbb {J},k,\nu )}|^2\le \left( \left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+ \frac{1}{nv}\right) v^{-1}\right) ^{2^{\nu }-1}{n^{-1}}v^{-1}\mathrm {Im}\;\!R_{qq}^{(\mathbb {J},k)}. \end{aligned}$$

Proof

By definition of $a_{qr}^{(\mathbb {J},k,0)}$, see (5.3), applying (Lemma 7.6 inequality (7.11) in the Appendix), we get

$$\begin{aligned} \sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,0)}|^2\le \frac{1}{n}\sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|R_{qr}^{(\mathbb {J},k)}|^2 \le \left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+ \frac{1}{nv}\right) v^{-1}, \end{aligned}$$

(5.7)

and by definition (5.3),

$$\begin{aligned} \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,0)}|^2\le \frac{1}{n}\sum _{r\in \mathbb {T}_{\mathbb {J},k}}|R_{qr}^{(\mathbb {J},k)}|^2. \end{aligned}$$

(5.8)

The general case follows now by induction in $\nu$, Lemma 5.1, and Lemma 7.6 inequality (7.12) in the Appendix. $\square$

Corollary 5.3

Under the conditions of Theorem 1.1 we have

$$\begin{aligned} a_{rr}^{(\mathbb {J},k,\nu +1)}\le \left( \left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+ \frac{1}{nv}\right) v^{-1}\right) ^{2^{\nu }-1}{n^{-1}}v^{-1}\mathrm {Im}\;\!R_{rr}^{(\mathbb {J},k)}. \end{aligned}$$

(5.9)

Proof

The result immediately follows from the definition of $a_{rr}^{(k,\nu )}$ and Corollary 5.2.

$\square$

In what follows we shall use the notations

$$\begin{aligned} \Psi ^{(\mathbb {J})}= & {} \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+\frac{1}{nv}, \quad (A_{\nu ,p}^{(\mathbb {J})})^2=\mathbf{E}(\Psi ^{(\mathbb {J})})^{(2^{\nu }-1)2 p}, \quad T_{\nu ,p}^{(\mathbb {J},k)}=\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu }|^p,\nonumber \\ A_p^{(\mathbb {J})}:= & {} 1+\mathbf{E}^{\frac{1}{4}}|\Psi ^{(\mathbb {J})}|^{4p}. \end{aligned}$$

(5.10)

Let $s_0$ denote some fixed number (for instance $s_0=2^8$). Let $A_1$ be a constant (to be chosen later) and $0<v_1\le 4$ a constant such that $v_0=A_0n^{-1}\le v_1$ for all $n\ge 1$.

Lemma 5.4

Assuming the conditions of Theorem 1.1 and for $p\le A_1(nv)^{\frac{1}{4}}$

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le C_0^p,\quad \text {for} \quad v\ge v_1,\quad \text {for all}\quad j=1,\ldots ,n, \end{aligned}$$

(5.11)

we have for $v\ge v_1/s_0$ and $p\le A_1(nv)^{\frac{1}{4}}$, and $k\in \mathbb {T}_{\mathbb {J}}$

$$\begin{aligned} \mathbf{E}\left( Q^{(\mathbb {J},k)}_0\right) ^p\le 6\left( \frac{C_3p}{\sqrt{2}}\right) ^{2p}v^{-p}A_p^{(\mathbb {J})}. \end{aligned}$$

(5.12)

Proof

Using the representation (5.4) and the triangle inequality, we get

$$\begin{aligned} \mathbf{E}| Q^{(\mathbb {J},k)}_{\nu }|^p\le 3^p\left( \mathbf{E}|Q^{(\mathbb {J},k)}_{\nu 1} |^p+\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu 2}|^p+\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu 3}|^p\right) . \end{aligned}$$

(5.13)

Let $\mathfrak {M}^{(\mathbb {A})}$ denote the $\sigma$-algebra generated by r.v.’s $X_{j,l}$ for $j,l\in \mathbb {T}_{\mathbb {A}}$, for any set $\mathbb {A}$. Conditioning on $\mathfrak {M}^{(\mathbb {J},k)}$ ($\mathbb {A}=\mathbb {J}\cup \{k\}$) and applying Rosenthal’s inequality (see Lemma 7.1), we get

$$\begin{aligned} \mathbf{E}|Q^{(\mathbb {J},k)}_{\nu 2}|^p\le C_1^pp^p\left( \mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a_{rr}^{(\mathbb {J},k,\nu +1)}|^2\right) ^{\frac{p}{2}}+\sum _{r\in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}|a_{rr}^{(\mathbb {J},k,\nu +1)}|^{p} \mathbf{E}|X_{kr}|^{2p}\right) , \end{aligned}$$

(5.14)

where $C_1$ denotes the absolute constant in Rosenthal’s inequality. By Remark 4.1, we get

$$\begin{aligned} \mathbf{E}|Q^{(\mathbb {J},k)}_{\nu 2}|^p\le C_1^pp^p\left( \mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a_{rr}^{(\mathbb {J},k,\nu +1)}|^2\right) ^{\frac{p}{2}}+n^{\frac{p}{2}}\frac{1}{n}\sum _{r\in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}|a_{rr}^{(\mathbb {J},k,\nu +1)}|^{p}\right) .\nonumber \\ \end{aligned}$$

(5.15)

Analogously conditioning on $\mathfrak {M}^{(\mathbb {J},k)}$ and applying Burkholder’s inequality (see Lemma 7.3), we get

$$\begin{aligned} \mathbf{E}| Q^{(\mathbb {J},k)}_{\nu 3}|^p\le & {} C_2^pp^p\left( \mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}\Bigg |\sum _{q\in \mathbb {T}_{\mathbb {J},k}\cap \{1,\ldots ,r-1\}}X_{kq} a^{(\mathbb {J},k,\nu +1)}_{rq}|^2\right) ^{\frac{p}{2}} \right. \nonumber \\&\left. +\sum _{q=1}^{n-1}\mathbf{E}\left| \sum _{r=1}^{q-1}X_{kr}a^{(\mathbb {J},k,\nu +1)}_{rq}\right| ^p\mathbf{E}|X_{kq}|^p\right) , \end{aligned}$$

(5.16)

where $C_2$ denotes the absolute constant in Burkholder’s inequality. Conditioning again on $\mathfrak {M}^{(\mathbb {J},k)}$ and applying Rosenthal’s inequality, we obtain

$$\begin{aligned} \mathbf{E}\left| \sum _{r\in \mathbb {T}_{\mathbb {J},k}}X_{kr}a^{(\mathbb {J},k,\nu +1)}_{rq}\right| ^p\le & {} C_1^pp^p\left( \mathbf{E}\left( \sum _{r=1}^{q-1}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^2\right) ^{\frac{p}{2}}\right. \nonumber \\&\left. +\sum _{r\in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}|a^{(\mathbb {J},k, \nu +1)}_{rq}|^p\mathbf{E}|X_{kr}|^p\right) . \end{aligned}$$

(5.17)

Combining inequalities (5.16) and (5.17), we get

$$\begin{aligned} \mathbf{E}| Q^{(\mathbb {J},k)}_{\nu 3}|^p\le & {} C_2^pp^p\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu +1}|^{\frac{p}{2}} \!+\! C_1^pC_2^pp^{2p}\sum _{q\in \mathbb {T}_{\mathbb {J},k}} \mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^2\right) ^{\frac{p}{2}}\mathbf{E}|X_{kq}|^p\nonumber \\&+ \,C_1^pC_2^p p^{2p}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}\sum _{r\in \mathbb {T}_{\mathbb {J},k}} \mathbf{E}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^p\mathbf{E}|X_{kq}|^p\mathbf{E}|X_{kr}|^p. \end{aligned}$$

(5.18)

Using Remark 4.1, this implies

$$\begin{aligned} \mathbf{E}| Q^{(\mathbb {J},k)}_{\nu 3}|^p\le & {} C_2^pp^p\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu +1}|^{\frac{p}{2}}+C_1^pC_2^pp^{2p}n^{\frac{p}{4}} \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}} \mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^2\right) ^{\frac{p}{2}}\nonumber \\&+ \,C_1^pC_2^pp^{2p}n^{\frac{p}{2}}\frac{1}{n^2}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}\sum _{r\in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^p. \end{aligned}$$

(5.19)

Using the definition (5.2) of $Q^{(\mathbb {J},k)}_{\nu 1}$ and the definition (5.3) of coefficients $a^{(\mathbb {J},k,\nu +1)}_{rr}$, it is straightforward to check that

$$\begin{aligned} \mathbf{E}|Q^{(\mathbb {J},k)}_{\nu 1}|^p\le \mathbf{E}\left[ \sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,\nu )}|^2\right] ^p. \end{aligned}$$

(5.20)

Combining (5.15), (5.19) and (5.20), we get by (5.13)

$$\begin{aligned} 3^{-p}\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu }|^p\le & {} C_2^pp^p\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu +1}|^{\frac{p}{2}} +C_1^pC_2^pp^{2p}n^{\frac{p}{4}}\frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^2\right) ^{\frac{p}{2}}\nonumber \\&+\, C_1^pC_2^pp^{2p}n^{\frac{p}{2}}\frac{1}{n^2}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}\sum _{r \in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}|a^{(\mathbb {J},k,\nu +1)}_{rq}|^p\nonumber \\&+\,C_1^pC_2^p\mathbf{E}\left[ \sum _{q,r\in \mathbb {T}_{\mathbb {J},k}}|a_{qr}^{(\mathbb {J},k,\nu )}|^2\right] ^p\nonumber \\&+\,C_1^pC_2^pp^p\left( \mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},k}}|a_{rr}^{(\mathbb {J},k,\nu +1)}|^2\right) ^{\frac{p}{2}} +n^{\frac{p}{2}}\frac{1}{n}\sum _{r\in \mathbb {T}_{\mathbb {J},k}}\mathbf{E}|a_{rr}^{(\mathbb {J},k,\nu +1)}|^{p}\right) .\nonumber \\ \end{aligned}$$

(5.21)

Applying now Lemma 5.1 and Corollaries 5.2 and 5.3, we obtain

$$\begin{aligned} \mathbf{E}|Q^{(\mathbb {J},k)}_{\nu }|^p\le & {} C_3^pp^p\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu +1}|^{\frac{p}{2}}+ C_3^p\mathbf{E}\left( \mathrm {Im}\;\!m_n(z)+\frac{1}{nv}\right) ^{(2^{\nu }-1)p}v^{-(2^{\nu }-1) p}\nonumber \\&+\, C_3^pp^{2p}v^{-2^{\nu }p}n^{-\frac{p}{4}}\mathbf{E}\left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z) +\frac{1}{nv}\right) ^{(2^{\nu }-1)\frac{2}{p}} \left( \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}} |R_{qq}^{(\mathbb {J},k)}|^{\frac{p}{2}}\right) \nonumber \\&+\, C_3^pp^{2p}n^{-\frac{p}{2}}v^{-2^{\nu }p}\mathbf{E}\left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+\frac{1}{nv}\right) ^{(2^{\nu }-1)p} \left( \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}|R_{qq}^{(\mathbb {J},k)}|^{p}\right) ,\nonumber \\ \end{aligned}$$

(5.22)

where $C_3=3C_1C_2$. Applying Cauchy–Schwartz inequality, we may rewrite the last inequality in the form

$$\begin{aligned} \mathbf{E}|Q^{(\mathbb {J},k)}_{\nu }|^p\le & {} C_3^pp^p\mathbf{E}|Q^{(\mathbb {J},k)}_{\nu +1}|^{\frac{p}{2}}+ C_3^p\mathbf{E}\left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+\frac{1}{nv}\right) ^{(2^{\nu }-1) p}v^{-(2^{\nu }-1) p}\nonumber \\&+\,C_3^pp^{2p}v^{-2^{\nu }p}n^{-\frac{p}{4}}\mathbf{E}^{\frac{1}{2}}\left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z) +\frac{1}{nv}\right) ^{(2^{\nu }-1)p}\mathbf{E}^{\frac{1}{4}}\left( \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}|R_{qq}^{(\mathbb {J},k)}|^{2p}\right) \nonumber \\&+\,C_3^pp^{2p}n^{-\frac{p}{2}}v^{-2^{\nu }p}\mathbf{E}^{\frac{1}{2}}\left( \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+\frac{1}{nv}\right) ^{(2^{\nu }-1)2p} \mathbf{E}^{\frac{1}{2}}\left( \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}|R_{qq}^{(\mathbb {J},k)}|^{2p}\right) .\nonumber \\ \end{aligned}$$

(5.23)

Introduce the notation

$$\begin{aligned} \Gamma _p(z):=\mathbf{E}^{\frac{1}{2}}\left( \frac{1}{n}\sum _{q\in \mathbb {T}_{\mathbb {J},k}}|R_{qq}^{(\mathbb {J},k)}|^{2p}\right) . \end{aligned}$$

We rewrite the inequality (5.23) using $\Gamma _p(z)$ and the notations of (5.10) as follows

$$\begin{aligned} T_{\nu ,p}^{(\mathbb {J},k)}\le & {} (C_3p)^pT_{\nu +1,p/2}^{(\mathbb {J},k)}+C_3^p A^{(\mathbb {J})}_{\nu ,p}v^{-(2^{\nu }-1) p}\nonumber \\&+\, (C_3p^2)^{p}\left( v^{-2^{\nu }p}n^{-\frac{p}{4}}(A^{(\mathbb {J})}_{\nu ,p/2})^{\frac{1}{2}}\Gamma _p^{\frac{1}{2}}(z)+v^{-2^{\nu }p}n^{-\frac{p}{2}}A^{(\mathbb {J})}_{\nu ,p} \Gamma _p(z)\right) .\nonumber \\ \end{aligned}$$

(5.24)

Note that

$$\begin{aligned} A_{0,p}^{(\mathbb {J})}=1,\quad A_{\nu ,p/2^{\nu }}^{(\mathbb {J})}\le \sqrt{1+\mathbf{E}(\Psi ^{(\mathbb {J})})^{2p}}\le 1+\mathbf{E}^{\frac{1}{4}}(\Psi ^{(\mathbb {J})})^{4p}, \end{aligned}$$

where $\Psi ^{(\mathbb {J})}=\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)+\frac{1}{nv}$. Furthermore,

$$\begin{aligned} \Gamma _{p/2^{\nu }}\le \Gamma _{p}^{\frac{1}{2^{\nu }}}. \end{aligned}$$

Without loss of generality we may assume $p=2^L$ and $\nu =0,\ldots ,L$. We may write

$$\begin{aligned} T_{0,p}^{(\mathbb {J},k)}\le (C_3p)^pT_{1,p/2}^{(\mathbb {J},k)}+C_3^p+ (C_3p^2)^{p}v^{- p}\left( n^{-\frac{p}{4}}\Gamma _p^{\frac{1}{2}}(z)+ n^{-\frac{p}{2}}\Gamma _p(z)\right) . \end{aligned}$$

By induction we get

$$\begin{aligned} T_{0,p}^{(\mathbb {J},k)}\le & {} \prod _{\nu =0}^L(C_3p/2^{\nu })^{p/2^{\nu }} T_{L,1}^{(\mathbb {J},k)}+A_{p}^{(\mathbb {J})}\sum _{l=1}^L \left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) v^{-(2^{l}-1)p/2^{l}}\nonumber \\&+ A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }} \right) (n^{-p}\Gamma _p^2)^{\frac{1}{2^{l+1}}}\nonumber \\&+ A_{p}^{(\mathbb {J})}\sum _{l=1}^L\left( \prod _{\nu =0}^{l-1} (C_3p/2^{\nu })^{p/2^{\nu }}\right) (n^{-p}{\Gamma _p}^2)^{\frac{1}{2^{l}}}. \end{aligned}$$

(5.25)

It is straightforward to check that

$$\begin{aligned} \sum _{\nu =1}^{l-1}\frac{\nu }{2^{\nu }}=2\left( 1-\frac{l+1}{2^{l}}\right) . \end{aligned}$$

Note that, for $l\ge 1$,

$$\begin{aligned} \prod _{\nu =0}^{l-1}\left( C_3(p/2^{\nu })\right) ^{p/2^{\nu }}= \frac{(C_3p)^{2p(1-2^{-l})}}{2^{2p(1-\frac{2l+1}{2^{l}})}}=2^{2p\frac{l}{2^{l}}} \left( \frac{C_3p}{2}\right) ^{2p(1-2^{-l})}. \end{aligned}$$

(5.26)

Applying this relation, we get

$$\begin{aligned} A_{p}^{(\mathbb {J})}\sum _{l=0}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) v^{-(2^{l}-1)p/2^{l}}\le A_p^{(\mathbb {J})}\left( \frac{C_3p}{2}\right) ^{2p}v^{-p} \sum _{l=0}^{L-1}2^{2p\frac{l}{2^l}}\left( \frac{4v}{C_3^2p^2}\right) ^{\frac{p}{2^l}}. \end{aligned}$$

Note that for $l\ge 0$, $\frac{l}{2^l}\le \frac{1}{2}$ and recall that $p=2^L$. Using this observation, we get

$$\begin{aligned} A_{p}^{(\mathbb {J})}\sum _{l=0}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) v^{-(2^{l}-1)p/2^{l}}\le A_p^{(\mathbb {J})}\left( \frac{C_3p}{2}\right) ^{2p}v^{-p}2^{p} \sum _{l=0}^{L-1}\left( \frac{4v}{C_3^2p^2}\right) ^{2^{L-l}}. \end{aligned}$$

This implies that for $\frac{4v}{C_3^2p^2}\le \frac{1}{2}$,

$$\begin{aligned} A_{p}^{(\mathbb {J})}\sum _{l=1}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }} \right) v^{-(2^{l}-1)p/2^{l}}\le (C_3p)^{2p}A_p^{(\mathbb {J})}v^{-p}. \end{aligned}$$

Furthermore, by definition of $T_{\nu ,p}$, we have

$$\begin{aligned} T_{L,1}^{(\mathbb {J},k)}=\mathbf{E}Q^{(\mathbb {J},k)}_L\le \mathbf{E}\sum _{q,r\in \mathbb {T}_k}(a_{qr}^{(\mathbb {J},k,L)})^2. \end{aligned}$$

Applying Corollary 5.2 and Hölder’s inequality, we get

$$\begin{aligned} T_{L,1}^{(\mathbb {J},k)}\le E(v^{-1}\Psi ^{(\mathbb {J})})^p\le v^{-p}A_p^{(\mathbb {J})}. \end{aligned}$$

(5.27)

By condition (5.11), we have

$$\begin{aligned} \Gamma _p:=\Gamma _p(u+iv)\le s_0^{2p}C_0^{2p}. \end{aligned}$$

Using this inequality, we get,

$$\begin{aligned}&A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) n^{-\frac{p}{2^{l+1}}}\Gamma _p^{\frac{2}{2^{l+1}}}\nonumber \\&\quad \le A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=0}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) (s_0^4C_0^4n^{-1})^{\frac{p}{2^{l+1}}}. \end{aligned}$$

(5.28)

Applying relation (5.26), we obtain

$$\begin{aligned}&A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) n^{-\frac{p}{2^{l+1}}}\Gamma _p^{\frac{1}{2^{l}}}\\&\quad \le \left( \frac{C_3p}{2}\right) ^{2p}A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L2^{2p\frac{l}{2^l}} \left( \frac{C_3p}{2}\right) ^{-\frac{2p}{2^{l}}}(s_0^4C_0^4n^{-1})^{\frac{p}{2^{l+1}}}\\&\quad =\left( \frac{C_3p}{2}\right) ^{2p}A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L2^{p\frac{l}{2^{l-1}}} \left( \frac{C_3p}{2}\right) ^{-\frac{2p}{2^{l-1}}} ((s_0^4C_0^4n^{-1})^{\frac{1}{4}})^{\frac{p}{2^{l-1}}}\\&\quad =\left( \frac{C_3p}{\sqrt{2}}\right) ^{2p}A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L \left( \frac{(s_0C_0)}{C_3pn^{\frac{1}{4}}}\right) ^{{2^{L-l+1}}}. \end{aligned}$$

Without loss of generality we may assume that $C_3\ge 2(C_0s_0)$. Then we get

$$\begin{aligned} A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=0}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }}\right) n^{-\frac{p}{2^{l+1}}}\Gamma _p^{\frac{1}{2^{l}}}\le (C_3p)^{2p}A_{p}^{(\mathbb {J})}v^{-p}. \end{aligned}$$

Analogously we get

$$\begin{aligned} A_{p}^{(\mathbb {J})}v^{-p}\sum _{l=1}^L\left( \prod _{\nu =0}^{l-1}(C_3p/2^{\nu })^{p/2^{\nu }} \right) n^{-\frac{p}{2^{l}}}\Gamma _p^{\frac{1}{2^{l-1}}}\le ({C_3p})^{2p}v^{-p}A_p^{(\mathbb {J})}. \end{aligned}$$

(5.29)

Combining inequalities (5.25), (5.27), (5.28), (5.29), we finally arrive at

$$\begin{aligned} T_{0,p}^{(\mathbb {J},k)}\le 6(C_3p)^{2p}v^{-p}A_p^{(\mathbb {J})}. \end{aligned}$$

(5.30)

Thus, Lemma 5.4 is proved. $\square$

5.2 Diagonal entries of the resolvent matrix

We shall use the representation, for any $j\in \mathbb {T}_{\mathbb {J}}$,

$$\begin{aligned} R^{(\mathbb {J})}_{jj}=\frac{1}{-z+\frac{1}{\sqrt{n}}X_{jj}-\frac{1}{n}{{\sum _{k,l\in \mathbb {T}_{\mathbb {J},j}}}}X_{jk}X_{jl}R^{(\mathbb {J},j)}_{kl}}\,, \end{aligned}$$

(see, for example, [13], equality (4.6)]). Recall the following relations (compare (3.2), (3.7))

$$\begin{aligned} R^{(\mathbb {J})}_{jj}=-\frac{1}{z+m_n^{(\mathbb {J})}(z)}+ \frac{1}{z+m_n^{(\mathbb {J})}(z)}\varepsilon _j^{(\mathbb {J})}R^{(\mathbb {J})}_{jj}, \end{aligned}$$

(5.31)

or

$$\begin{aligned} R^{(\mathbb {J})}_{jj}=s(z)-s(z)\Lambda _n^{(\mathbb {J})}R^{(\mathbb {J})}_{jj}- s(z)\varepsilon _j^{(\mathbb {J})}R^{(\mathbb {J})}_{jj}, \end{aligned}$$

(5.32)

where $\varepsilon _j^{(\mathbb {J})}:=\varepsilon _{j1}^{(\mathbb {J})}+\varepsilon _{j2}^{(\mathbb {J})}+\varepsilon _{j3}^{(\mathbb {J})}+ \varepsilon _{j4}^{(\mathbb {J})}$ with

$$\begin{aligned} \varepsilon _{j1}^{(\mathbb {J})}&:=\frac{1}{\sqrt{n}}X_{jj},\, \varepsilon _{j2}^{(\mathbb {J})}:=-\frac{1}{n}{\sum _{k\ne l\in \mathbb {T}_{\mathbb {J},j}}}X_{jk}X_{jl}R^{(\mathbb {J},j)}_{kl},\, \varepsilon _{j3}^{(\mathbb {J})}:=-\frac{1}{n}{\sum _{k\in \mathbb {T}_{\mathbb {J},j}}}(X_{jk}^2-1)R^{(\mathbb {J},j)}_{kk}, \nonumber \\ \varepsilon _{j4}^{(\mathbb {J})}&:=\frac{1}{n}(\mathrm{Tr}\,\mathbf {R}^{(\mathbb {J})}-\mathrm{Tr}\,\mathbf {R}^{(\mathbb {J},j)}), \Lambda _n^{(\mathbb {J})}:=m_n^{(\mathbb {J})}(z)-s(z)=\frac{1}{n}\mathrm{Tr}\,\mathbf {R}^{(\mathbb {J})}-s(z). \end{aligned}$$

(5.33)

Since $|s(z)|\le 1$, the representation (5.32) yields, for any $p\ge 1$,

$$\begin{aligned} |R_{jj}^{(\mathbb {J})}|^p\le 3^p+3^p|\varepsilon _j^{(\mathbb {J})}|^p|R_{jj}^{(\mathbb {J})} |^p+3^p|\Lambda _n^{(\mathbb {J})}|^p|R_{jj}^{(\mathbb {J})}|^p. \end{aligned}$$

(5.34)

Applying the Cauchy–Schwartz inequality, we get

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le 3^p+3^p\mathbf{E}^{\frac{1}{2}}|\varepsilon _j^{(\mathbb {J})}|^{2p}\mathbf{E}^{\frac{1}{2}}|R_{jj}^{(\mathbb {J})}|^{2p}+3^p\mathbf{E}^{\frac{1}{2}}|\Lambda _n^{(\mathbb {J})}|^{2p} \mathbf{E}^{\frac{1}{2}}|R_{jj}^{(\mathbb {J})}|^{2p}. \end{aligned}$$

(5.35)

We shall investigate now the behavior of $\mathbf{E}|\varepsilon _j^{(\mathbb {J})}|^{2p}$ and $\mathbf{E}|\Lambda _n^{(\mathbb {J})}|^{2p}$. First we note,

$$\begin{aligned} \mathbf{E}|\varepsilon _j^{(\mathbb {J})}|^{2p}\le 4^{2p}\sum _{\nu =1}^4\mathbf{E}|\varepsilon _{j\nu }^{(\mathbb {J})}|^{2p}. \end{aligned}$$

Lemma 5.5

Assuming the conditions of Theorem 1.1 we have, for any $p\ge 1$, and for any $z=u+iv\in \mathbb {C}_+$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j1}^{(\mathbb {J})}|^{2p}\le \frac{\mu _4}{n^{\frac{p}{2}+1}}. \end{aligned}$$

Proof

The proof follows immediately from the definition of $\varepsilon _{j1}$ and condition (1.4). $\square$

Lemma 5.6

Assuming the conditions of Theorem 1.3 we have, for any $p\ge 1$, and for any $z=u+iv\in \mathbb {C}_+$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j4}^{(\mathbb {J})}|^{2p}\le \frac{1}{n^{2p}v^{2p}}. \end{aligned}$$

Proof

For a proof of this Lemma see [13], Lemma 4.1]. $\square$

Let $A_1>0$ and $0\le v_1\le 4$ be a fixed.

Lemma 5.7

Assuming the conditions of Theorem 1.1, and assuming for all $\mathbb {J}\subset T$ with $|\mathbb {J}|\le L$ and all $l\in \mathbb {T}_{\mathbb {J}}$

$$\begin{aligned} \mathbf{E}|R_{ll}^{(\mathbb {J})}|^{q}\le C_0^{q}, \quad \text {for} \quad 1\le q\le A_1(nv)^{\frac{1}{4}}\quad \text {and for} \quad v\ge v_1, \end{aligned}$$

(5.36)

we have, for all $v\ge v_1/s_0$, and for all $\mathbb {J}\subset \mathbb {T}$ with $|\mathbb {J}|\le L-1$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j3}^{(\mathbb {J})}|^{2p}\le (C_1p)^{2p}n^{-p}s_0^{2p}C_0^{4p}, \quad \text {for} \quad 1\le p\le A_1(nv)^{\frac{1}{4}}. \end{aligned}$$

Proof

Recall that $s_0=2^4$ and note that if $p\le A_1(nv)^{\frac{1}{4}}$ for $v\ge v_1/s_0$ then $q=2p\le A_1(nv)^{\frac{1}{4}}$ for $v\ge v_1$. Let $v':=vs_0$. If $v\ge v_1/s_0$ then $v'\ge v_1$. We have

$$\begin{aligned} q=2p\le 2A_1(nv)^{\frac{1}{4}}= 2A_1(nv's_0^{-1})^{\frac{1}{4}}=A_1(nv')^{\frac{1}{4}}. \end{aligned}$$

(5.37)

We apply now Rosenthal’s inequality for the moments of sums of independent random variables and get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j3}^{(\mathbb {J})}|^{2p}\le (C_1p)^{2p}n^{-2p}\left( \mathbf{E}\left( \sum _{l\in \mathbb {T}_{\mathbb {J},j}}|R_{ll}^{(\mathbb {J},j)}|^{2}\right) ^{p} +\mathbf{E}|X_{jl}|^{4p}\sum _{l\in \mathbb {T}_{\mathbb {J},j}}\mathbf{E}|R^{(\mathbb {J},j)}_{ll}|^{2p}\right) . \end{aligned}$$

According to inequality (5.37) we may apply Lemma 7.7 and condition (5.36) for $q=2p$. We get, for $v\ge v_1/s_0$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j3}^{(\mathbb {J})}|^{2p}\le (C_1p)^{2p}n^{-p}s_0^{2p}C_0^{2p}. \end{aligned}$$

We use as well that by the conditions of Theorem 1.1, $\mathbf{E}|X_{jl}|^{4p}\le D_0^{4p-4}n^{p-1}\mu _4$, and by Jensen’s inequality, $(\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R_{ll}^{(j)}|^{2})^{p}\le \frac{1}{n}\sum _{l\in \mathbb {T}_j}|R_{ll}^{(j)}|^{2p}$. Thus, Lemma 5.7 is proved. $\square$

Lemma 5.8

Assuming the conditions of Theorem 1.1, condition (5.36), for $v\ge v_1$ and $p\le A_1(nv)^{\frac{1}{4}}$, we have, for any $v\ge v_1/s_0$ and $p\le A_1(nv)^{\frac{1}{4}}$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}^{(\mathbb {J})}|^{2p}\le 6(C_3p)^{4p}n^{-p}v^{-p}A_p^{(\mathbb {J})}+ 2(C_3p)^{4p}n^{-p}v^{-p}(C_0s_0)^p. \end{aligned}$$

Proof

We apply Burkholder’s inequality for quadratic forms. See Lemma 7.3 in the Appendix. We obtain

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}^{(\mathbb {J})}|^{2p}&\le (C_1p)^{2p}n^{-2p}\left( \mathbf{E}\left( \sum _{l\in \mathbb {T}_{\mathbb {J},j}}\left| \sum _{r\in \mathbb {T}_{\mathbb {J},j}\cap \{1,\ldots ,l-1\}}X_{jr} R^{(\mathbb {J},j)}_{lr}\right| ^2\right) ^p\right. \\&\quad \left. +\max _{j,k}\mathbf{E}|X_{jk}|^{2p} \sum _{l\in \mathbb {T}_{\mathbb {J},j}} \mathbf{E}\left| \sum _{r\in \mathbb {T}_{\mathbb {J},j}\cap \{1,\ldots ,l-1\}}X_{jr} R^{(\mathbb {J},j)}_{lr}\right| ^{2p}\right) . \end{aligned}$$

Using now the quantity $Q_0^{(\mathbb {J},j)}$ for the first term and Rosenthal’s inequality and condition (1.4) for the second term, we obtain with Lemma 7.6, inequality (7.12), in the Appendix and $C_3=C_1C_2$

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}^{(\mathbb {J})}|^{2p}&\le (C_2p)^{2p}n^{-p}\mathbf{E}|Q^{(\mathbb {J},j)}_{0}|^{p}\nonumber \\&\quad +(C_3p)^{4p}n^{-\frac{3p}{2}}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J},j}}\mathbf{E}\left( \sum _{r\in \mathbb {T}_{\mathbb {J},j}\cap \{1,\ldots ,l-1\}}| R^{(\mathbb {J},j)}_{lr}|^2\right) ^{p}\nonumber \\&\quad +(C_3p)^{4p}n^{-p}\frac{1}{n^2} \sum _{l\in \mathbb {T}_{\mathbb {J},j}} \sum _{r\in \mathbb {T}_{\mathbb {J},j}\cap \{1,\ldots ,l-1\}}\mathbf{E}|R^{ (\mathbb {J},j)}_{lr}|^{2p}\nonumber \\&\le (C_2p)^{2p}n^{-p}\mathbf{E}|Q^{(\mathbb {J},j)}_{0}|^{p}+(C_3p)^{4p}n^{- \frac{3p}{2}}v^{-p}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J},j}}\mathbf{E}|R_{ll }^{(\mathbb {J},j)}|^p\nonumber \\&\quad +(C_3p)^{4p}n^{-p}v^{-p}\frac{1}{n^2} \sum _{l\in \mathbb {T}_{\mathbb {J},j}}\mathbf{E}|R_{ll}^{(\mathbb {J},j)}|^p. \end{aligned}$$

By Lemma 7.7 and condition (5.36), we get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}^{(\mathbb {J})}|^{2p}\le (C_3p)^{2p}n^{-p}\mathbf{E}|Q^{(\mathbb {J},j)}_{0}|^{p} +2(C_3p)^{3p}n^{-p}v^{-p}(C_0s_0)^p. \end{aligned}$$

Applying now Lemma 5.4, we get the claim. Thus, Lemma 5.8 is proved. $\square$

Recall that

$$\begin{aligned} \Lambda _n^{(\mathbb {J})}=\Lambda _n^{(\mathbb {J})}(z)=\frac{1}{n}\mathrm{Tr}\,\mathbf {R}^{(\mathbb {J})}-s(z),\quad \text {and} \quad T_n^{(\mathbb {J})}(z)=\frac{1}{n}\sum _{j\in \mathbb {T}_{\mathbb {J}}}\varepsilon _j^{(\mathbb {J})}R_{jj}^{(\mathbb {J})}. \end{aligned}$$

Lemma 5.9

Assuming the conditions of Theorem 1.1, we have, for $z=u+iv$ with $u\in [-2,2]$,

$$\begin{aligned} |\Lambda _n^{(\mathbb {J})}|\le C\left( \sqrt{|T_n^{(\mathbb {J})}(z)|}+\frac{\sqrt{| \mathbb {J}|}}{\sqrt{n}}\right) . \end{aligned}$$

Proof

See e. g. [17], inequality (2.10)]. For completeness we include short proof here. Obviously

$$\begin{aligned} \Lambda _n^{(\mathbb {J})}= \frac{s(z)\left( T_n^{(\mathbb {J})}(z)-\frac{|\mathbb {J}|}{n}\right) }{z+2s(z)+ \Lambda _n^{(\mathbb {J})}}. \end{aligned}$$

(5.38)

Denote by

$$\begin{aligned} \widetilde{T}_n(z)=s(z)\left( T_n^{(\mathbb {J})}(z)-\frac{|\mathbb {J}|}{n}\right) . \end{aligned}$$

First we assume that $|z+2s(z)+\Lambda _n^{(\mathbb {J})}|> \sqrt{|\widetilde{T}_n(z)|}$. Then the claim of Lemma 5.9 holds. In the case $|z+2s(z)+\Lambda _n^{(\mathbb {J})}|\le \sqrt{|\widetilde{T}_n(z)|}$, we assume $|\Lambda _n^{(\mathbb {J})}|>2\sqrt{|\widetilde{T}_n(z)|}$. Otherwise the Lemma is proved. Under these assumptions we have

$$\begin{aligned} |z+2s(z)|\ge |\Lambda _n^{(\mathbb {J})}|-|z+2s(z)+\Lambda _n^{(\mathbb {J})}|\ge \sqrt{|\widetilde{T}_n(z)|}. \end{aligned}$$

(5.39)

On the other hand

$$\begin{aligned} |z+s(z)+m_n^{(\mathbb {J})}|\ge \mathrm {Im}\;\!\{z+s(z)\}\ge \frac{1}{2}\mathrm {Im}\;\!\{z+2s(z)\}=\frac{1}{2}\mathrm {Im}\;\!\sqrt{z^2-4}. \end{aligned}$$

(5.40)

We take here the branch of $\sqrt{z}$ such that $\mathrm {Im}\;\!\sqrt{z}\ge 0$. Note that, for $z\in [-2,2]$ we have $\mathrm {Re}\;\!\{z^2-4\}\le 0$. This implies that

$$\begin{aligned} \mathrm {Im}\;\!\sqrt{z^2-4}\ge \frac{\sqrt{2}}{2}|z^2-4|^{\frac{1}{2}}=\frac{\sqrt{2}}{2}|z+2s(z)|. \end{aligned}$$

(5.41)

Inequalities (5.39), (5.40) and (5.41) together imply

$$\begin{aligned} |z+s(z)+m_n^{(\mathbb {J})}|\ge \frac{1}{2\sqrt{2}}|z+2s(z)|\ge \frac{\sqrt{2}}{2}\sqrt{|\widetilde{T}_n(z)|}. \end{aligned}$$

(5.42)

The last inequality and Eq. (5.38) complete the proof of Lemma 5.9. $\square$

Lemma 5.10

Assuming the conditions of Theorem 1.1 and condition (5.36), we obtain, for $|\mathbb {J}|\le Cn^{\frac{1}{2}}$

$$\begin{aligned} \mathbf{E}|\Lambda _n^{(\mathbb {J})}|^{2p}&\le \frac{C^p}{n^{\frac{p}{4}}}+ \left( \frac{\mu _4}{n^{\frac{p}{2}+1}}+\frac{1}{n^{2p}v^{2p}}+(C_1p)^{2p}n^{-p}s_0^{2p}C_0^{4p}\right. \\&\quad \left. +\, 6(C_3p)^{4p}n^{-p}v^{-p}A_{p}^{(\mathbb {J})}+ 2(C_3p)^{4p}n^{-p}v^{-p}(C_0s_0)^p\right) ^{\frac{1}{2}}(C_0s_0)^p. \end{aligned}$$

Proof

By Lemma 5.8, we have

$$\begin{aligned} \mathbf{E}|\Lambda _n^{(\mathbb {J})}|^{2p}\le C^p\mathbf{E}|T_n^{(\mathbb {J})}(z)|^p+\frac{|\mathbb {J}|^{\frac{p}{2}}}{n^{\frac{p}{2}}}\le C^p\mathbf{E}|T_n^{(\mathbb {J})}(z)|^p+\frac{C}{n^{\frac{p}{4}}}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \mathbf{E}|T_n^{(\mathbb {J})}(z)|^p\le \left( \frac{1}{n}\sum _{j\in \mathbb {T}}\mathbf{E}|\varepsilon _{j}^{(\mathbb {J})}|^{2p}\right) ^{\frac{1}{2}} \left( \frac{1}{n}\sum _{j\in \mathbb {T}}\mathbf{E}|R_{jj}^{(\mathbb {J})}|^{2p}\right) ^{\frac{1}{2}}. \end{aligned}$$

Lemmas 5.5–5.8 together with Lemma 7.7 imply

$$\begin{aligned} \frac{1}{n}\sum _{j\in \mathbb {T}}\mathbf{E}|\varepsilon _{j}^{(\mathbb {J})}|^{2p}&\le 4^{2p-1}\left( \frac{\mu _4}{n^{\frac{p}{2}+1}}+\frac{1}{n^{2p}v^{2p}}+(C_1 p)^{2p}n^{-p}s_0^{2p}C_0^{4p}\right. \nonumber \\&\quad \left. +\,6(C_3p)^{4p}n^{-p}v^{-p}A_{p}^{(\mathbb {J})}+2(C_3p)^{4p}n^{-p}v^{-p}(C_0s_0)^p\right) . \end{aligned}$$

(5.43)

Thus, Lemma 5.10 is proved. $\square$

Lemma 5.11

Assuming the conditions of Theorem 1.1 and condition (5.36), there exists an absolute constant $C_4$ such that, for $p\le A_1(nv)^{\frac{1}{4}}$ and $v\ge v_1/s_0$, we have, uniformly in $\mathbb {J}\subset \mathbb {T}$ such that $|\mathbb {J}|\le Cn^{\frac{1}{2}}$,

$$\begin{aligned} A_p^{(\mathbb {J})}\le C_4^p. \end{aligned}$$

Proof

We start from the obvious inequality, using $|s(z)|\le 1$,

$$\begin{aligned} \mathbf{E}|\Psi ^{(\mathbb {J})}|^{2p}\le 3^{2p}(1+(nv)^{-2p}+\mathbf{E}|\Lambda _n^{(\mathbb {J})}(z)|^{2p}). \end{aligned}$$

Furthermore, applying Lemma 5.10, we get

$$\begin{aligned} \mathbf{E}|\Psi ^{(\mathbb {J})}|^{2p}&\le 3^{2p}\left( 1+(nv)^{-p}+\left( \frac{\mu _4}{n^{\frac{p}{2}+1}}+\frac{1}{n^{2p}v^{2p}}+(C_1p)^{2p}n^{-p}s_0^{2p}C_0^{4p}\right. \right. \nonumber \\&\quad \left. \left. +\,6(C_3p)^{4p}n^{-p}v^{-p}A_{p}^{(\mathbb {J})}+ 2(C_3p)^{4p}n^{-p}v^{-p}(C_0s_0)^p\right) ^{\frac{1}{2}}(C_0s_0)^p\right) . \end{aligned}$$

(5.44)

By definition,

$$\begin{aligned} A_p^{(\mathbb {J})}\le 1+\mathbf{E}^{\frac{1}{2}}(\Psi ^{(\mathbb {J})})^{2p}. \end{aligned}$$

(5.45)

Inequalities (5.45) and (5.44) together imply

$$\begin{aligned} A_p^{(\mathbb {J})}&\le 1+3^{p}\left( 1+(nv)^{-\frac{p}{2}}+(C_0s_0)^{\frac{p}{2}} \left( \mu _4^{\frac{1}{4}}n^{-\frac{p}{8}}+\frac{1}{n^{\frac{p}{2}}v^{\frac{p}{2}}} +(C_1p)^{\frac{p}{2}}n^{-\frac{p}{4}}s_0^{\frac{p}{2}}C_0^{p}\right. \right. \\&\quad \left. \left. +\,3(C_3p)^{p}n^{-\frac{p}{4}}v^{-\frac{p}{4}}(A_p^{(\mathbb {J})})^{\frac{1}{4}}+ 2(C_3p)^{p}n^{-\frac{p}{4}}v^{-\frac{p}{4}}(C_0s_0)^{\frac{p}{4}}\right) \right) . \end{aligned}$$

Let $C'{=}s_0\max \{9,C_3^{\frac{1}{2}}, C_0^{\frac{3}{2}}C_1^{\frac{1}{2}}, 3C_3C_0^{\frac{1}{2}}, C_3C_0^{\frac{3}{4}}\}$. Using Lemma 7.4 with $x=(A_p^{(\mathbb {J})})^{\frac{1}{4}}$, $t=4$, $r=1$, we get

$$\begin{aligned} A_p^{(\mathbb {J})}&\le {C'}^p\left( 1+(nv)^{-\frac{p}{2}}+\mu _4^{\frac{1}{2}} n^{-\frac{p}{8}}+\frac{1}{n^{\frac{p}{2}}v^{\frac{p}{2}}}\right. \\&\qquad \left. +\,p^{\frac{p}{2}}n^{-\frac{p}{4}}+p^{p}n^{-\frac{p}{4}}v^{-\frac{p}{4}}+ p^{\frac{4p}{3}}n^{-\frac{p}{3}}v^{-\frac{p}{3}}\right) . \end{aligned}$$

For $p\le A_1(nv)^{\frac{1}{4}}$, we get, for $z\in \mathbb {G}$,

$$\begin{aligned} A_p^{(\mathbb {J})}\le C_4^p, \end{aligned}$$

where $C_4$ is some absolute constant. We may take $C_4=2C'$. $\square$

Corollary 5.12

Assuming the conditions of Theorem 1.1 and condition (5.36), we have , for $v\ge v_1/s_0$, and for any $\mathbb {J}\subset \mathbb {T}$ such that $|\mathbb {J}|\le \sqrt{n}$

$$\begin{aligned} \mathbf{E}|\Lambda _n^{(\mathbb {J})}|^{2p}\le C_0^{2p}\left( \frac{4^{\frac{p}{4}}\mu _4^{\frac{1}{2}}s_0^{p}}{n^{\frac{p}{4}}v^{\frac{p}{4}}}+\frac{s_0^{\frac{p}{2}}}{n^pv^p} +\frac{C_5^pp^{2p}}{n^{\frac{p}{2}}v^{\frac{p}{2}}}\right) , \end{aligned}$$

(5.46)

where

$$\begin{aligned} C_5:=4C_1^2s_0^{4}+6^{\frac{1}{p}}C_3^4C_4+2^{\frac{1}{p}}C_3^4s_0^3. \end{aligned}$$

Proof

Without loss of generality we may assume that $C_0>1$. The bound (5.46) follows now from Lemmas 5.10 and 5.11. $\square$

Lemma 5.13

Assuming the conditions of Theorem 1.1 and condition (5.36) for $\mathbb {J}\subset \mathbb {T}$ such that $|\mathbb {J}|\le L\le \sqrt{n}$, there exist positive constant $A_0, C_0, A_1$ depending on $\mu _4, D_0$ only , such that we have, for $p\le A_1(nv)^{\frac{1}{4}}$ and $v\ge v_1/s_0$ uniformly in $\mathbb {J}$ and $v_1$

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le C_0^p \end{aligned}$$

with $|\mathbb {J}|\le L-1$.

Proof

According to inequality (5.35), we have

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le 4^p\left( 1+\left( \mathbf{E}^{\frac{1}{2}}|\Lambda _n^{(\mathbb {J})}|^{2p}+ \mathbf{E}^{\frac{1}{2}}|\varepsilon _j^{(\mathbb {J})}|^{2p}\right) \mathbf{E}^{\frac{1}{2}}|R_{jj}^{(\mathbb {J})}|^{2p}\right) . \end{aligned}$$

Applying condition (5.36), we get

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le 4^p\left( 1+\left( \mathbf{E}^{\frac{1}{2}}|\Lambda _n^{(\mathbb {J})}|^{2p}+ \mathbf{E}^{\frac{1}{2}}|\varepsilon _{j1}^{(\mathbb {J})}|^{2p} +\cdots +\mathbf{E}^{\frac{1}{2}}|\varepsilon _{j4}^{(\mathbb {J})}|^{2p}\right) s_0^pC_0^p\right) . \end{aligned}$$

Combining results of Lemmas 5.5–5.8 and Corollary 5.12, we obtain

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p&\le 5^p\left( 1+s_0^pC_0^{2p}\left( \frac{4^{\frac{p}{4}} \mu _4^{\frac{1}{4}}s_0^{p}}{n^{\frac{p}{4}}v^{\frac{p}{4}}}+\frac{s_0^{p}}{n^pv^p} +\frac{C_5^pp^{2p}}{n^{\frac{p}{2}}v^{\frac{p}{2}}}\right) ^{\frac{1}{2}} \right. \\&\qquad \left. +\, s_0^pC_0^{3p}\left( \frac{4^{\frac{p}{4}}\mu _4^{\frac{1}{4}}}{n^{\frac{p}{4}}v^{\frac{p}{4}}} +\frac{s_0^{p}}{n^pv^p} +\frac{C_5^pp^{2p}}{n^{\frac{p}{2}}v^{\frac{p}{2}}} \right) \right) . \end{aligned}$$

We may rewrite the last inequality as follows

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le C_0^p\left( \frac{5^p}{C_0^p}+\frac{{\widehat{C}}_1^{\frac{p}{8}}}{(nv)^{\frac{p}{8}}} +\frac{({\widehat{C}}_2p^4)^{\frac{p}{4}}}{(nv)^{\frac{p}{4}}}+ \frac{({\widehat{C}}_3p^4)^{\frac{p}{2}}}{(nv)^{\frac{p}{2}}}+ \frac{{\widehat{C}}_4^{ p}}{(nv)^{p}}\right) , \end{aligned}$$

where

$$\begin{aligned} {\widehat{C}}_1&=5^8s_0^{12}C_0^8\mu _4^{\frac{1}{p}},\\ {\widehat{C}}_2&=5^4s_0^4C_0^4C_5^2\left( 1+2C_0^4\mu _4^{\frac{2}{p}}\right) ,\\ {\widehat{C}}_3&=5^2s_0^2C_0^2\left( s_0+C_5^2\right) ,\\ {\widehat{C}}_4&=5C_0^2s_0^2. \end{aligned}$$

Note that for

$$\begin{aligned} A_0\ge 2^8A_1^4\max \{{\widehat{C}}_1,\ldots ,{\widehat{C}}_4\} \end{aligned}$$

(5.47)

and $C_0\ge 25$, we obtain that

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le C_0^p. \end{aligned}$$

Thus Lemma 5.13 is proved. $\square$

Corollary 5.14

Assuming the conditions of Theorem 1.1, we have, for $p\le 8$ and $v\ge v_0= A_0n^{-1}$ there exist a constant $C_0>0$ depending on $\mu _4$ and $D_0$ only such that for all $1\le j\le n$ and all $z\in \mathbb {G}$

$$\begin{aligned} \mathbf{E}|R_{jj}|^p\le C_0^p, \end{aligned}$$

(5.48)

and

$$\begin{aligned} \mathbf{E}\frac{1}{|z+m_n(z)|^p}\le C_0^p. \end{aligned}$$

(5.49)

Proof

Let $L=[-\log _{s_0}v_0]+1$. Note that $s_0^{-L}\le v_0$ and $A_1\frac{n^{\frac{1}{4}}}{s_0^{\frac{L}{4}}}\ge A_1(nv_0)^{\frac{1}{4}}$. We may choose $C_0=25$ and $A_0, A_1$ such that (5.47) holds and

$$\begin{aligned} A_1(nv)^{\frac{1}{4}}\ge 8. \end{aligned}$$

Then, for $v=1$, and for any $p\ge 1$, for any set $\mathbb {J}\subset \mathbb {T}$ such that $|\mathbb {J}|\le L$

$$\begin{aligned} \mathbf{E}|R_{jj}^{(\mathbb {J})}|^p\le C_0^p. \end{aligned}$$

(5.50)

By Lemma 5.13, inequality (5.50) holds for $v\ge 1/s_0$ and for $p\le A_1n^{\frac{1}{4}}/s_0^{\frac{1}{4}}$ and for $\mathbb {J}\subset \mathbb {T}$ such that $|\mathbb {J}|\le L-1$. After repeated application of Lemma 5.13 (with (5.50) as assumption valid for $v \ge 1/s_0$) we arrive at the conclusion that the inequality (5.50) holds for $v\ge 1/s_0^{2}$, $p\le A_1n^{\frac{1}{4}}/s_0^{\frac{1}{2}}$ and all $\mathbb {J}\subset \mathbb {T}$ such that $|\mathbb {J}|\le L-2$. Continuing this iteration inequality (5.50) finally holds for $v\ge A_0n^{-1}$, $p\le 8$ and $\mathbb {J}=\emptyset$.

The proof of inequality of (5.49) is similar: We have by (1.9)

$$\begin{aligned} \frac{1}{|z+m_n(z)|}\le \frac{1}{|s(z)+z|}+\frac{|\Lambda _n|}{|z+m_n(z)||z+s(z)|}\le |s(z)|\left( 1+\frac{|\Lambda _n|}{|z+m_n(z)|}\right) .\nonumber \\ \end{aligned}$$

(5.51)

Furthermore, using that $|m_n'(z)|\le |\frac{1}{n}\mathrm{Tr}\,\mathbf {R}^2|\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}|R_{jj}|^2$ and Lemma 7.6, inequality (7.11) in the Appendix, we get

$$\begin{aligned} \left| \frac{d}{dz}\log (z+m_n(z))\right| \le \frac{|1+m_n'(z)|}{|z+m_n(z)|}\le \frac{1}{v}\frac{v+\mathrm {Im}\;\!m_n(z)}{|z+m_n(z)|}\le \frac{1}{v}. \end{aligned}$$

By integration, this implies that (see the proof of Lemma 7.7)

$$\begin{aligned} \frac{1}{|(u+iv/s_0)+m_n(u+iv/s_0)|}\le \frac{s_0}{|(u+iv)+m_n(u+iv)|}. \end{aligned}$$

(5.52)

Inequality (5.51) and the Cauchy–Schwartz inequality together imply

$$\begin{aligned} \mathbf{E}\frac{1}{|z+m_n(z)|^p}\le 2^p|s(z)|^p\left( 1+\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^{2p}\mathbf{E}^{\frac{1}{2}}\frac{1}{|z+m_n(z)|^{2p}}\right) . \end{aligned}$$

Applying inequality (5.52), we obtain

$$\begin{aligned} \mathbf{E}\frac{1}{|z+m_n(z)|^p}\le 2^p|s(z)|^p(1+\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^{2p}s_0^pC_0^p). \end{aligned}$$

Using Corollary 5.12, we get, for $v\ge 1/s_0$

$$\begin{aligned} \mathbf{E}\frac{1}{|z+m_n(z)|^p}\le 2^p|s(z)|^p\left( 1+\left( \frac{4^{\frac{p}{8}}\mu _4^{\frac{1}{4}}s_0^{\frac{p}{2}}}{n^{\frac{p}{8}}v^{\frac{p}{8}}} +\frac{s_0^{\frac{p}{4}}}{n^{\frac{p}{2}}v^{\frac{p}{2}}} +\frac{C_5^pp^{p}}{n^{\frac{p}{4}}v^{\frac{p}{4}}}\right) s_0^pC_0^{2p}\right) . \end{aligned}$$

Thus inequality (5.49) holds for $v\ge 1/s_0$ as well. Repeating this argument inductively with $A_0,A_1,C_)$ satisfying (5.47) for the regions $v \ge s_0^{-\nu }$, for $\nu =1,\ldots ,L$ and $z \in \mathbb {G}$, we get the claim. Thus, Corollary 5.14 is proved. $\square$

6 Proof of Theorem 1.3

We return now to the representation (5.32) which implies that

$$\begin{aligned} s_n(z)=\frac{1}{n}\sum _{j=1}^n\mathbf{E}R_{jj}=s(z)+\mathbf{E}\Lambda _n=s(z)+\mathbf{E}\frac{T_n(z)}{z+s(z)+m_n(z)}. \end{aligned}$$

(6.1)

We may continue the last equality as follows

$$\begin{aligned} s_n(z)=s(z)+\mathbf{E}\frac{\frac{1}{n}\sum _{j=1}^n\varepsilon _{j4}R_{jj}}{z+s(z)+m_n(z)}+ \mathbf{E}\frac{\widehat{T}_n(z)}{z+s(z)+m_n(z)}, \end{aligned}$$

(6.2)

where

$$\begin{aligned} \widehat{T}_n=\sum _{\nu =1}^3\frac{1}{n}\sum _{j=1}^n\varepsilon _{j\nu }R_{jj}. \end{aligned}$$

Note that the definition of $\varepsilon _{j4}$ in (5.32) and equality (7.34) together imply

$$\begin{aligned} \frac{1}{n}\sum _{j=1}^n\varepsilon _{j4}R_{jj}=\frac{1}{n^{2}}\mathrm{Tr}\,\mathbf {R}^2=\frac{1}{n}\frac{d m_n(z)}{dz}. \end{aligned}$$

(6.3)

Thus we may rewrite (6.2) as

$$\begin{aligned} s_n(z)=s(z)+\frac{1}{n}\mathbf{E}\frac{ m_n'(z)}{z+s(z)+m_n(z)}+ \mathbf{E}\frac{\widehat{T}_n(z)}{z+s(z)+m_n(z)}. \end{aligned}$$

(6.4)

Denote by

$$\begin{aligned} \mathfrak {T}=\mathbf{E}\frac{\widehat{T}_n(z)}{z+s(z)+m_n(z)}. \end{aligned}$$

(6.5)

6.1 Estimation of $\mathfrak {T}$

We represent $\mathfrak {T}$

$$\begin{aligned} \mathfrak {T}=\mathfrak {T}_{1}+\mathfrak {T}_{2}, \end{aligned}$$

where

$$\begin{aligned} \mathfrak {T}_{1}&=-\frac{1}{n}\sum _{j=1}^n\sum _{\nu =1}^3\mathbf{E}\frac{\varepsilon _{j\nu } \frac{1}{z+m_n^{(j)}(z)}}{z+m_n(z)+s(z)},\\ \mathfrak {T}_{2}&=\frac{1}{n}\sum _{j=1}^n\sum _{\nu =1}^3 \mathbf{E}\frac{\varepsilon _{j\nu }(R_{jj}+\frac{1}{z+m_n^{(j)}(z)})}{z+m_n(z)+s(z)}. \end{aligned}$$

6.1.1 Estimation of $\mathfrak {T}_{1}$

We may decompose $\mathfrak {T}_{1}$ as

$$\begin{aligned} \mathfrak {T}_{1}=\mathfrak {T}_{11}+\mathfrak {T}_{12}, \end{aligned}$$

(6.6)

where

$$\begin{aligned} \mathfrak {T}_{11}&=-\frac{1}{n}\sum _{j=1}^n\sum _{\nu =1}^3\mathbf{E}\frac{\varepsilon _{j\nu } \frac{1}{z+m_n^{(j)}(z)}}{z+m_n^{(j)}(z)+s(z)},\\ \mathfrak {T}_{12}&=-\frac{1}{n}\sum _{j=1}^n\sum _{\nu =1}^3\mathbf{E}\frac{\varepsilon _{j\nu }\varepsilon _{j4} \frac{1}{z+m_n^{(j)}(z)}}{(z+m_n^{(j)}(z)+s(z))(z+m_n(z)+s(z))}. \end{aligned}$$

It is easy to see that, by conditional expectation

$$\begin{aligned} \mathfrak {T}_{11}=0. \end{aligned}$$

(6.7)

Applying the Cauchy–Schwartz inequality, for $\nu =1,2,3$, we get

$$\begin{aligned}&\left| \mathbf{E}\frac{\varepsilon _{j\nu }\varepsilon _{j4} \frac{1}{z+m_n^{(j)}(z)}}{(z+m_n^{(j)}(z)+s(z))(z+m_n(z)+s(z))}\right| \nonumber \\&\quad \le \mathbf{E}^{\frac{1}{2}}\left| \frac{\varepsilon _{j\nu }}{(z+m_n^{(j)}(z))(z+m_n^{(j)}(z) +s(z))}\right| ^2\mathbf{E}^{\frac{1}{2}}\left| \frac{\varepsilon _{j4}}{z+m_n(z)+s(z)}\right| ^2. \end{aligned}$$

(6.8)

Applying the Cauchy–Schwartz inequality again, we get

$$\begin{aligned}&\mathbf{E}^{\frac{1}{2}}\left| \frac{\varepsilon _{j\nu }}{(z+m_n^{(j)}(z))(z+m_n^{(j)}(z)+s(z))}\right| ^2\nonumber \\&\quad \le \mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^4}\mathbf{E}^{\frac{1}{4}}\frac{1}{|z+m_n^{(j)}(z)|^4}. \end{aligned}$$

(6.9)

Inequalities (6.8), (6.9), Corollaries 7.23 and 5.14 together imply

$$\begin{aligned} \left| \mathbf{E}\frac{\varepsilon _{j\nu }\varepsilon _{j4} \frac{1}{z+m_n^{(j)}(z)}}{(z+m_n^{(j)}(z)+s(z))(z+m_n(z)+s(z))}\right| \le \frac{C}{nv}\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^4}. \end{aligned}$$

(6.10)

By Lemmas 7.14 and 7.5 we have for $\nu =1$

$$\begin{aligned} \mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^4}\le \frac{C}{\sqrt{n}\sqrt{|z^2-4|}}. \end{aligned}$$

(6.11)

By Corollary 7.17, inequality (7.31) with $\alpha =0$ and $\beta =4$ in the Appendix we have for $\nu =2,3$

$$\begin{aligned} \mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^4}\le \frac{C}{\sqrt{nv}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

(6.12)

with some constant $C>0$ depending on $\mu _4$ and $D_0$ only. Using that, by Lemma 7.13, for $z\in \mathbb {G}$,

$$\begin{aligned} \frac{1}{\sqrt{n}\sqrt{|z^2-4|}}\le \frac{\sqrt{v}}{\sqrt{nv}\sqrt{|z^2-4|}}\le \frac{C}{\sqrt{nv}|z^2-4|^{\frac{1}{4}}} \end{aligned}$$

(6.13)

we get from (6.10), (6.11) and (6.12) that for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathfrak {T}_{1}|\le \frac{C}{(nv)^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}} \end{aligned}$$

(6.14)

with some constant $C>0$ depending on $\mu _4$ and $D_0$ only.

6.1.2 Estimation of $\mathfrak {T}_{2}$

Using the representation (5.32), we write

$$\begin{aligned} \mathfrak {T}_{2}=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\widetilde{\varepsilon }_j^2 R_{jj}}{(z+m^{(j)}(z))(z+s(z)+m_n(z))}. \end{aligned}$$

Furthermore we note that

$$\begin{aligned} \widetilde{\varepsilon }_j^2=\varepsilon _{j2}^2+\eta _j, \end{aligned}$$

where

$$\begin{aligned} \eta _j=(\varepsilon _{j1}+\varepsilon _{j3})^2+2(\varepsilon _{j1}+\varepsilon _{j3}) \varepsilon _{j2}. \end{aligned}$$

We now decompose $\mathfrak {T}_{2}$ as follows

$$\begin{aligned} \mathfrak {T}_{2}=\mathfrak {T}_{21}+\mathfrak {T}_{22}+\mathfrak {T}_{23}, \end{aligned}$$

(6.15)

where

$$\begin{aligned} \mathfrak {T}_{21}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{{\varepsilon }_{j2}^2R_{jj}}{(z+m^{(j)}(z))(z+s(z)+m_n(z))},\\ \mathfrak {T}_{22}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\eta _jR_{jj}}{(z+m^{(j)}(z))(z+s(z)+m_n^{(j)}(z))},\\ \mathfrak {T}_{23}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\eta _jR_{jj}\varepsilon _{j4}}{(z+m^{(j)}(z))(z+s(z)+m_n^{(j)}(z))(z+s(z)+m_n(z))}. \end{aligned}$$

Applying the Cauchy–Schwartz inequality, we obtain

$$\begin{aligned} |\mathfrak {T}_{22}|\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}\frac{|\eta _j|^2}{|z+ m^{(j)}(z)|^2|z+s(z)+m_n^{(j)}(z)|^2}\mathbf{E}^{\frac{1}{2}}|R_{jj}|^2. \end{aligned}$$

(6.16)

In what follows we denote by $C$ a generic constant depending on $\mu _4$ and $D_0$ only. We note that

$$\begin{aligned} \mathbf{E}\{|\eta _j|^2|\mathfrak {M}^{(j)}\}&\le C(\mathbf{E}\{|\varepsilon _{j1}|^4| \mathfrak {M}^{(j)}\}+\mathbf{E}\{|\varepsilon _{j3}|^4|\mathfrak {M}^{(j)}\})\\&\quad +C\left( \mathbf{E}\{|\varepsilon _{j1}|^4|\mathfrak {M}^{(j)}\}+ \mathbf{E}\{|\varepsilon _{j3}|^4|\mathfrak {M}^{(j)}\}\right) ^{\frac{1}{2}} \mathbf{E}^{\frac{1}{2}}\{|\varepsilon _{j2}|^4|\mathfrak {M}^{(j)}\}. \end{aligned}$$

Using Lemmas 7.14, 7.15, 7.16, we get

$$\begin{aligned} \mathbf{E}\{|\eta _j|^2\Big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^2}+\frac{C}{n^2}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R_{ll}^{(j)}|^4\nonumber \\&\quad + \left( \frac{C}{n}+\frac{C}{n}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}|R_{ll}^{(j)}|^4\right) ^{\frac{1}{2}}\right) \frac{C}{nv}\mathrm {Im}\;\!m_n^{(j)}(z). \end{aligned}$$

(6.17)

This inequality and Lemma 7.5 together imply

$$\begin{aligned}&\mathbf{E}^{\frac{1}{2}}\frac{|\eta _j|^2}{|z+m^{(j)}(z)|^2|z+s(z)+m_n^{(j)}(z)|^2}\le \frac{C}{n\sqrt{|z^2-4|}}\mathbf{E}^{\frac{1}{2}}\frac{1}{|z+m_n^{(j)}(z)|^2}\nonumber \\&\quad + \frac{C}{n\sqrt{|z^2-4|}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathbf{E}| R_{ll}^{(j)}|^4\right) ^{\frac{1}{4}}\mathbf{E}^{\frac{1}{4}}\frac{1}{|z+m_n^{(j)}(z)|^4}\nonumber \\&\quad + \frac{C}{n\sqrt{v}|z^2-4|^{\frac{1}{4}}}\left( 1+\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j} \mathbf{E}|R_{ll}^{(j)}|^4\right) ^{\frac{1}{4}}\right) \mathbf{E}^{\frac{1}{4}}\frac{1}{|z+m_n^{(j)}(z)|^4} \end{aligned}$$

(6.18)

Inequality (6.18) and Corollary 5.14 together imply

$$\begin{aligned} |\mathfrak {T}_{22}|\le \frac{C}{n\sqrt{|z^2-4|}}+\frac{C}{n\sqrt{v}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

Applying Lemma 7.13 for $z\in \mathbb {G}$, we get

$$\begin{aligned} |\mathfrak {T}_{22}|\le \frac{C}{n\sqrt{v}|z^2-4|^{\frac{1}{4}}}\le \frac{C}{nv^{\frac{3}{4}}}. \end{aligned}$$

(6.19)

By Hölder’s inequality, we have

$$\begin{aligned} |\mathfrak {T}_{23}|&\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}\frac{|\eta _j|^2}{|z+ m^{(j)}(z)|^2|z+s(z)+m_n^{(j)}(z)|^2}\\&\quad \times \mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j4}|^4}{|z+s(z)+m_n(z)|^4}\mathbf{E}^{\frac{1}{4}}|R_{jj}|^4. \end{aligned}$$

Using now Lemmas 7.22, 7.13 and Corollary 5.14, we may write, for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathfrak {T}_{23}|\le \frac{C}{n\sqrt{v}|z^2-4|^{\frac{1}{4}}}\le \frac{C}{nv^{\frac{3}{4}}}. \end{aligned}$$

(6.20)

We continue now with $\mathfrak {T}_{21}$. We represent it in the form

$$\begin{aligned} \mathfrak {T}_{21}=H_1+H_2, \end{aligned}$$

(6.21)

where

$$\begin{aligned} H_1&=-\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{{\varepsilon }_{j2}^2}{(z+m^{(j)}(z))^2(z+s(z)+m_n(z))},\\ H_2&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{{\varepsilon }_{j2}^2(R_{jj}+\frac{1}{z+m_n^{(j)}})}{(z+m^{(j)}(z))(z+s(z)+m_n(z))}. \end{aligned}$$

Furthermore, using the representation

$$\begin{aligned} R_{jj}=-\frac{1}{z+m_n^{(j)}(z)}+\frac{1}{z+m_n^{(j)}(z)}(\varepsilon _{j1}+ \varepsilon _{j2}+\varepsilon _{j3})R_{jj} \end{aligned}$$

(6.22)

(compare with (3.2)), we bound $H_2$ in the following way

$$\begin{aligned} |H_2|\le H_{21}+H_{22}+H_{23}, \end{aligned}$$

where

$$\begin{aligned} H_{21}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{4|{\varepsilon }_{j1}|^3|R_{jj}|}{|z+m^{(j)}(z)|^2|z+s(z)+m_n(z)|},\\ H_{22}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{2|{\varepsilon }_{j2}|^3|R_{jj}|}{|z+m^{(j)}(z)|^2|z+s(z)+m_n(z)|},\\ H_{23}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{2|{\varepsilon }_{j3}|^3|R_{jj}|}{|z+m^{(j)}(z)|^2|z+s(z)+m_n(z)|}. \end{aligned}$$

Using inequality (7.42) in the Appendix and Hölder inequality, we get, for $\nu =1,2,3$

$$\begin{aligned} H_{2\nu }&\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{4|{\varepsilon }_{j\nu }|^3|R_{jj}|}{|z+m^{(j)}(z)|^2|z+s(z)+m_n^{(j)}(z)|}\nonumber \\&\quad +\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{4|{\varepsilon }_{j\nu }|^3|R_{jj}||\varepsilon _{j4}|}{|z+m^{(j)}(z)|^2|z+s(z)+m_n^{(j)}(z)||z+s(z)+m_n(z)|}\nonumber \\&\le \frac{C}{n}\sum _{j=1}^n\mathbf{E}^{\frac{3}{4}}\frac{|{\varepsilon }_{j\nu }|^4}{|z+m^{(j)}(z)|^{\frac{8}{3}}|z+s(z)+m_n^{(j)}(z)|^{\frac{4}{3}}} \mathbf{E}^{\frac{1}{4}}|R_{jj}|^4. \end{aligned}$$

(6.23)

Applying Corollary 7.17 with $\beta =\frac{4}{3}$ and $\alpha =\frac{8}{3}$, we obtain, for $z\in \mathbb {G}$, and for $\nu =1,2,3$

$$\begin{aligned} \mathbf{E}^{\frac{3}{4}}\frac{|{\varepsilon }_{j\nu }|^4}{|z+m^{(j)}(z)|^{\frac{8}{3}}|z+s(z)+ m_n^{(j)}(z)|^{\frac{4}{3}}}\le \frac{C}{(nv)^{\frac{3}{2}}}. \end{aligned}$$

This yields together with Corollary 5.14 and inequality (6.23)

$$\begin{aligned} H_2\le \frac{C}{(nv)^{\frac{3}{2}}}. \end{aligned}$$

(6.24)

Consider now $H_1$. Using the equality

$$\begin{aligned} \frac{1}{z+m_n(z)+s(z)}=\frac{1}{z+2s(z)}-\frac{\Lambda _n(z)}{(z+2s(z))(z+m_n(z)+s(z))} \end{aligned}$$

and

$$\begin{aligned} \Lambda _n=\Lambda _n^{(j)}+\varepsilon _{j4}, \end{aligned}$$

(6.25)

we represent it in the form

$$\begin{aligned} H_1=H_{11}+H_{12}+H_{13}, \end{aligned}$$

(6.26)

where

$$\begin{aligned} H_{11}= & {} -\frac{1}{(z+s(z))^2}\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j2}^2}{z+s(z)+m_n(z)}\\= & {} -s^2(z)\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j2}^2}{z+s(z)+m_n(z)},\\ H_{12}= & {} -\frac{1}{(z+s(z))}\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j2}^2\Lambda _n^{(j)} }{(z+m_n^{(j)}(z))^2(z+s(z)+m_n(z))},\\ H_{13}= & {} -\frac{1}{(z+s(z))^2}\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j2}^2 \Lambda _n^{(j)}}{(z+m_n^{(j)}(z))(z+s(z)+m_n(z))}. \end{aligned}$$

In order to apply conditional independence, we write

$$\begin{aligned} H_{11}=H_{111}+H_{112}, \end{aligned}$$

where

$$\begin{aligned} H_{111}&=-s^2(z)\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j2}^2}{z+m_n^{(j)}(z)+s(z)},\\ H_{112}&=s^2(z)\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j2}^2\varepsilon _{j4}}{(z+s(z)+m_n(z))(z+m_n^{(j)}(z)+s(z))}. \end{aligned}$$

It is straightforward to check that

$$\begin{aligned} \mathbf{E}\{\varepsilon _{j2}^2|\mathfrak {M}^{(j)}\}=\frac{1}{n^2}\mathrm{Tr}\,(\mathbf {R}^{(j)})^2-\frac{1}{n^2}\sum _{l\in \mathbb {T}_j}(R^{(j)}_{ll})^2. \end{aligned}$$

Using equality (6.3) for $m_n'(z)$ and the corresponding relation for ${m_n^{(j)}}'(z)$, we may write

$$\begin{aligned} H_{111}=L_1+L_2+L_3+L_4, \end{aligned}$$

where

$$\begin{aligned} L_1= & {} -s^2(z)\frac{1}{n}\mathbf{E}\frac{m_n'(z)}{z+m_n(z)+s(z)},\nonumber \\ L_2= & {} s^2(z)\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\frac{1}{n^2}\sum _{l\in \mathbb {T}_j} (R^{(j)}_{ll})^2}{z+m_n^{(j)}(z)+s(z)},\end{aligned}$$

(6.27)

$$\begin{aligned} L_3= & {} s^2(z)\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\frac{1}{n}((m_n^{(j)}(z))'-m_n'(z))}{z+m_n^{(j)}(z)+s(z)},\\ L_4= & {} s^2(z)\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\frac{1}{n}((m_n^{(j)}(z))'-m_n'(z))\varepsilon _{j4}}{(z+m_n(z)+s(z))(z+m_n^{(j)}(z)+s(z))}.\nonumber \end{aligned}$$

(6.28)

Using Lemmas 7.5, 7.18, 7.26, and Corollary 5.14, it is straightforward to check that

$$\begin{aligned} |L_2|\le & {} \frac{C}{n\sqrt{|z^2-4|}},\\ |L_3|\le & {} \frac{C}{n^2v^2\sqrt{|z^2-4|}},\\ |L_4|\le & {} \frac{C}{n^3v^3{|z^2-4|}}. \end{aligned}$$

Applying inequality (7.42), we may write

$$\begin{aligned} |H_{12}|\le \frac{C}{n}\sum _{j=1}^n \mathbf{E}\frac{|\varepsilon _{j2}|^2|\Lambda _n^{(j)}|}{|z+m_n^{(j)}(z)||z+m_n^{(j)}(z)+s(z)|}. \end{aligned}$$

Conditioning on $\mathfrak {M}^{(j)}$ and applying Lemma 7.15, Lemma 7.5, inequality (7.8), Corollary 5.14 and equality (6.25), we get

$$\begin{aligned} |H_{12}|\le \frac{C}{{nv}}\frac{1}{n}\sum _{j=1}^n \mathbf{E}\frac{|\Lambda _n^{(j)}|}{|z+m_n^{(j)}(z)|}\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2+\frac{C}{n^2v^2}. \end{aligned}$$

By Lemma 7.24, we get

$$\begin{aligned} |H_{12}|\le \frac{C}{n^2v^2}. \end{aligned}$$

(6.29)

Similar we get

$$\begin{aligned} |H_{13}|\le \frac{C}{n^2v^2}. \end{aligned}$$

(6.30)

We rewrite now the equations (6.2) and (6.4) as follows, using the remainder term $\mathfrak {T}_3$, which is bounded by means of inequalities (6.14), (6.19), (6.20), (6.24).

$$\begin{aligned} \mathbf{E}\Lambda _n(z)=\mathbf{E}m_n(z)-s(z)=\frac{(1-s^2(z))}{n}\mathbf{E}\frac{m_n'(z)}{z+m_n(z)+s(z)}+\mathfrak {T}_3, \end{aligned}$$

(6.31)

where

$$\begin{aligned} |\mathfrak {T}_3|\le \frac{C}{nv^{\frac{3}{4}}}+\frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

Note that

$$\begin{aligned} 1-s^2(z)=s(z)\sqrt{z^2-4}. \end{aligned}$$

In (6.31) we estimate now the remaining quantity

$$\begin{aligned} \mathfrak {T}_4=-\frac{s(z)\sqrt{z^2-4}}{n}\mathbf{E}\frac{m_n'(z)}{z+m_n(z)+s(z)}. \end{aligned}$$

6.2 Estimation of $\mathfrak {T}_4$

Using that $\Lambda _n=m_n(z)-s(z)$ we rewrite $\mathfrak {T}_4$ as

$$\begin{aligned} \mathfrak {T}_4=\mathfrak {T}_{41}+\mathfrak {T}_{42}+\mathfrak {T}_{43}, \end{aligned}$$

where

$$\begin{aligned} \mathfrak {T}_{41}&=-\frac{s(z)s'(z)}{n},\\ \mathfrak {T}_{42}&=\frac{s(z)\sqrt{z^2-4}}{n}\mathbf{E}\frac{m_n'(z)-s'(z)}{z+m_n(z)+s(z)},\\ \mathfrak {T}_{43}&=\frac{s(z)}{n}\mathbf{E}\frac{(m_n'(z)-s'(z))\Lambda _n}{z+m_n(z)+s(z)}. \end{aligned}$$

6.2.1 Estimation of $\mathfrak {T}_{42}$

First we investigate $m_n'(z)$. The following equality holds

$$\begin{aligned} m_n'(z)=\frac{1}{n}\mathrm{Tr}\,\mathbf{R}^2=\sum _{j=1}^n \varepsilon _{j4}R_{jj}= s^2(z)\sum _{j=1}^n \varepsilon _{j4}R^{-1}_{jj}+D_1, \end{aligned}$$

(6.32)

where

$$\begin{aligned} D_1=\sum _{j=1}^n \varepsilon _{j4}(1+R^{-1}_{jj}s(z))(R_{jj}-s(z)) \end{aligned}$$

(6.33)

Using equality (7.34), we may write

$$\begin{aligned} m_n'(z)=\frac{s^2(z)}{n}\sum _{j=1}^n \left( 1+\frac{1}{n}\sum _{l,k\in \mathbb {T}_j}X_{jl}X_{jk}[(R^{(j)})^2]_{lk}\right) +D_1. \end{aligned}$$

Denote by

$$\begin{aligned} \beta _{j1}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j}[(R^{(j)})^2]_{ll}-\frac{1}{n} \sum _{l=1}^n[(R)^2]_{ll} =\frac{1}{n}\sum _{l\in \mathbb {T}_j}[(R^{(j)})^2]_{ll}-m_n'(z)\nonumber \\&=\frac{1}{n}\frac{d}{dz}(\mathrm{Tr}\,\mathbf {R}-\mathrm{Tr}\,\mathbf {R}^{(j)}),\nonumber \\ \beta _{j2}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j}(X^2_{jl}-1)[(R^{(j)})^2]_{ll},\nonumber \\ \beta _{j3}&=\frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}X_{jl}X_{jk}[(R^{(j)})^2]_{lk}. \end{aligned}$$

(6.34)

Using these notation we may write

$$\begin{aligned} m_n'(z)=s^2(z)(1+m_n'(z))+\frac{s^2(z)}{n}\sum _{j=1}^n (\beta _{j1}+\beta _{j2}+\beta _{j3})+D_1. \end{aligned}$$

Solving this equation with respect to $m_n'(z)$ we obtain

$$\begin{aligned} m_n'(z)=\frac{s^2(z)}{1-s^2(z)}+\frac{1}{1-s^2(z)}(D_1+D_2), \end{aligned}$$

(6.35)

where

$$\begin{aligned} D_2=\frac{s^2(z)}{n}\sum _{j=1}^n (\beta _{j1}+\beta _{j2}+\beta _{j3}). \end{aligned}$$

Note that for the semi-circular law

$$\begin{aligned} \frac{s^2(z)}{1-s^2(z)}=\frac{s^2(z)}{1+\frac{s(z)}{z+s(z)}}=-\frac{s(z)}{z+2s(z)}= s'(z). \end{aligned}$$

Applying this relation we rewrite equality (6.35) as

$$\begin{aligned} m_n'(z)-s'(z)=\frac{1}{s(z)(z+2s(z))}(D_1+D_2). \end{aligned}$$

(6.36)

Using the last equality, we may represent $\mathfrak {T}_{42}$ now as follows

$$\begin{aligned} \mathfrak {T}_{42}=\mathfrak {T}_{421}+\mathfrak {T}_{422}, \end{aligned}$$

where

$$\begin{aligned} \mathfrak {T}_{421}&=\frac{1}{n}\mathbf{E}\frac{D_1}{z+m_n(z)+s(z)},\\ \mathfrak {T}_{422}&=\frac{1}{n}\mathbf{E}\frac{D_2}{z+m_n(z)+s(z)}. \end{aligned}$$

Recall that, by (6.33),

$$\begin{aligned} \mathfrak {T}_{421}=-\frac{1}{n}\sum _{j=1}^n \mathbf{E}\frac{\varepsilon _{j4}(1+s(z)R_{jj}^{-1})(R_{jj}-s(z))}{z+m_n(z)+s(z)}. \end{aligned}$$

(6.37)

Applying the Cauchy–Schwartz inequality, we get for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathfrak {T}_{421}|\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}|R_{jj}-s(z)|^2 \mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j4}|^4}{|z+s(z)+m_n(z)|^4}(1+|s(z)|\mathbf{E}^{\frac{1}{4}}|R_{jj}|^{-4}). \end{aligned}$$

Using Corollary 7.23 and Corollary 5.14, we get

$$\begin{aligned} |\mathfrak {T}_{421}|\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}}. \end{aligned}$$

(6.38)

6.2.2 Estimation of  $\mathfrak {T}_{422}$

We represent now $\mathfrak {T}_{421}$ in the form

$$\begin{aligned} \mathfrak {T}_{422}=\mathfrak {T}_{51}+\mathfrak {T}_{52}+\mathfrak {T}_{53}, \end{aligned}$$

where

$$\begin{aligned} \mathfrak {T}_{5\nu }=\frac{1}{n^2}\sum _{j=1}^n\mathbf{E}\frac{\beta _{j\nu }}{z+ m_n(z)+s(z)},\quad \text {for}\quad \nu =1,2,3. \end{aligned}$$

At first we investigate $\mathfrak {T}_{53}$. Note that, by Lemma 7.26,

$$\begin{aligned} |\beta _{j1}|\le \frac{C}{nv^2}. \end{aligned}$$

Therefore, for $z\in \mathbb {G}$, using Lemma 7.5, inequality (7.9), and Lemma 7.13,

$$\begin{aligned} |\mathfrak {T}_{51}|\le \frac{C}{n^2v^2\sqrt{|z^2-4|}}\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

(6.39)

Furthermore, we consider the quantity $\mathfrak {T}_{5\nu }$, for $\nu =2,3$. Applying the Cauchy–Schwartz inequality and inequality (7.42) in the Appendix as well, we get

$$\begin{aligned} |\mathfrak {T}_{5\nu }|\le \frac{C}{n^2}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}\frac{|\beta _{j\nu }|^2}{|z+m_n^{(j)}(z)+s(z)|^2}. \end{aligned}$$

By Lemma 7.25 together with Lemma 7.5 in the Appendix, we obtain

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\frac{|\beta _{j\nu }|^2}{|z+m_n^{(j)}(z)+s(z)|^2}\le \frac{C}{n^{\frac{1}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

This implies that

$$\begin{aligned} |\mathfrak {T}_{5\nu }|\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

(6.40)

Inequalities (6.39) and (6.40) yield

$$\begin{aligned} |\mathfrak {T}_{422}|\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

Combining (6.38) and (6.41), we get, for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathfrak {T}_{42}|\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

(6.41)

6.2.3 Estimation of $\mathfrak {T}_{43}$

Recall that

$$\begin{aligned} \mathfrak {T}_{43}=\frac{s(z)}{n}\mathbf{E}\frac{(m_n'(z)-s'(z))\Lambda _n}{z+m_n(z)+s(z)}. \end{aligned}$$

Applying equality (6.36), we obtain

$$\begin{aligned} \mathfrak {T}_{43}=\mathfrak {T}_{431}+\mathfrak {T}_{432}, \end{aligned}$$

where

$$\begin{aligned} \mathfrak {T}_{431}&=\frac{1}{n(z+2s(z))}\mathbf{E}\frac{D_1\Lambda _n}{z+m_n(z)+s(z)},\\ \mathfrak {T}_{432}&=\frac{1}{n(z+2s(z))}\mathbf{E}\frac{D_2\Lambda _n}{z+m_n(z)+s(z)}.\nonumber \end{aligned}$$

(6.42)

Applying the Cauchy–Schwartz inequality, we get

$$\begin{aligned} |\mathfrak {T}_{431}|\le \frac{1}{n|z+2s(z)|}\mathbf{E}^{\frac{1}{2}}\frac{|D_1|^2}{|z+m_n(z)+s(z)|^2}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

By definition of $D_1$ and Lemmas 7.18 and 7.24 , we get

$$\begin{aligned} |\mathfrak {T}_{431}| \le \frac{C}{n^{3} v^2|z^2-4|}\frac{1}{n}\sum _{j=1}^n(1+|s(z)|\mathbf{E}^{\frac{1}{4}}|R_{jj}|^{-4})\mathbf{E}{^\frac{1}{4}}{|R_{jj}-s(z)|^4}. \end{aligned}$$

Applying now Corollary 5.14 and Lemma 7.22, we get

$$\begin{aligned} |\mathfrak {T}_{431}| \le \frac{4}{n^{3} v^{2}|z^2-4|^{\frac{1}{2}}}. \end{aligned}$$

For $z\in \mathbb {G}$ this yields

$$\begin{aligned} |\mathfrak {T}_{431}| \le \frac{4}{n^{\frac{3}{2}} v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

Applying again the Cauchy–Schwartz inequality, we get for $\mathfrak {T}_{432}$ accordingly

$$\begin{aligned} |\mathfrak {T}_{432}|\le \frac{C}{n|z^2-4|^{\frac{1}{2}}}\mathbf{E}^{\frac{1}{2}}|D_2|^2\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

By Lemma 7.24, we have

$$\begin{aligned} |\mathfrak {T}_{432}|\le \frac{C}{n^2v|z^2-4|^{\frac{1}{2}}}\mathbf{E}^{\frac{1}{2}}|D_2|^2. \end{aligned}$$

(6.43)

By definition of $D_2$,

$$\begin{aligned} \mathbf{E}|D_2|^2\le \frac{1}{n}\sum _{j=1}^n(\mathbf{E}|\beta _{j1}|^2+\mathbf{E}|\beta _{j2}|^2+\mathbf{E}|\beta _{j3}|^2). \end{aligned}$$

Applying Lemmas 7.25 with $\nu =2,3$, and 7.26, we get

$$\begin{aligned} \mathbf{E}|D_2|^2\le \frac{C}{n^2v^4}+\frac{C}{nv^3}. \end{aligned}$$

(6.44)

Inequalities (6.43) and (6.44) together imply, for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathfrak {T}_{432}|\le \frac{C}{n^3v^3|z^2-4|^{\frac{1}{2}}}+\frac{C}{n^{\frac{5}{2}}v^{\frac{5}{2}}|z^2-4|^{\frac{1}{2}}}\le \frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

Finally we observe that

$$\begin{aligned} s'(z)=-\frac{s(z)}{\sqrt{z^2-4}} \end{aligned}$$

and, therefore

$$\begin{aligned} |\mathfrak {T}_{41}|\le \frac{C}{n|z^2-4|^{\frac{1}{2}}}. \end{aligned}$$

For $z\in \mathbb {G}$ we may rewrite it

$$\begin{aligned} |\mathfrak {T}_{41}|\le \frac{C}{n\sqrt{v}}. \end{aligned}$$

(6.45)

Combining now relations (6.31), (6.26), (6.24), (6.39), (6.41), (6.45), we get for $z\in \mathbb {G}$,

$$\begin{aligned} |\mathbf{E}\Lambda _n|\le \frac{C}{n v^{\frac{3}{4}}}+\frac{C}{n^{\frac{3}{2}}v^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}. \end{aligned}$$

The last inequality completes the proof of Theorem 1.3.

Appendix

1.1 Rosenthal’s and Burkholder’s inequalities

In this subsection we state the Rosenthal and Burkholder inequalities, starting with Rosenthal’s inequality. Let $\xi _1,\ldots ,\xi _n$ be independent random variables with $\mathbf{E}\xi _j=0$, $\mathbf{E}\xi _j^2=1$ and for $p\ge 1$ $\mathbf{E}|\xi _j|^p\le \mu _p$ for $j=1,\ldots ,n$.

Lemma 7.1

(Rosenthal’s inequality) There exists an absolute constant $C_1$ such that

$$\begin{aligned} \mathbf{E}\left| \sum _{j=1}^na_j\xi _j\right| ^p\le C_1^pp^p\left( \left( \sum _{j=1}^{n}|a_j|^2\right) ^{\frac{p}{2}}+\mu _p\sum _{j=1}^{n}|a_j|^p\right) . \end{aligned}$$

Proof

For a proof see [19], Theorem 3] and [16], inequality (A)]. $\square$

Let $\xi _1,\ldots \xi _n$ be a martingale-difference with respect to the $\sigma$-algebras $\mathfrak {M}_j=\sigma (\xi _1,\ldots ,\xi _{j-1})$. Assume that $\mathbf{E}\xi _j^2=1$ and $\mathbf{E}|\xi _j|^p<\infty$.

Lemma 7.2

(Burkholder’s inequality) There exist an absolute constant $C_2$ such that

$$\begin{aligned} \mathbf{E}\left| \sum _{j=1}^n\xi _j\right| ^p\le C_2^pp^p\left( \mathbf{E}\left( \sum _{k=1}^n\mathbf{E}\{\xi _k^2|\mathfrak {M}_{k-1}\}\right) ^{\frac{p}{2}}+\sum _{k=1}^{n}\mathbf{E}|\xi _k|^p\right) . \end{aligned}$$

Proof

For a proof of this inequality see [5], Theorem 3.2] and [15], Theorem 4.1]. $\square$

We rewrite the Burkholder inequality for quadratic forms in independent random variables. Let $\zeta _1,\ldots ,\zeta _n$ be independent random variables such that $\mathbf{E}\zeta _j=0$, $\mathbf{E}|\zeta _j|^2=1$ and $\mathbf{E}|\zeta _j|^p\le \mu _p$. Let $a_{ij}=a_{ji}$ for all $i,j=1,\ldots n$. Consider the quadratic form

$$\begin{aligned} Q=\sum _{1\le j\ne k\le n}a_{jk}\zeta _j\zeta _k. \end{aligned}$$

Lemma 7.3

There exists an absolute constant $C_2$ such that

$$\begin{aligned} \mathbf{E}|Q|^p\le C_2^p\left( \mathbf{E}\left( \sum _{j=2}^{n}\left( \sum _{k=1}^{j-1}a_{jk}\zeta _k\right) ^2\right) ^{\frac{p}{2}}+\mu _p\sum _{j=2}^n\mathbf{E}\left| \sum _{k=1}^{j-1}a_{jk}\zeta _k\right| ^p\right) . \end{aligned}$$

Proof

Introduce the random variables, for $j=2,\ldots ,n$,

$$\begin{aligned} \xi _j=\zeta _j\sum _{k=1}^{j-1}a_{jk}\zeta _k. \end{aligned}$$

It is straightforward to check that

$$\begin{aligned} \mathbf{E}\{\xi _j|\mathfrak {M}_{j-1}\}=0, \end{aligned}$$

and that $\xi _j$ are $\mathfrak {M}_{j}$ measurable. That means that $\xi _1,\ldots ,\xi _n$ are martingale-differences. We may write

$$\begin{aligned} Q=2\sum _{j=2}^n\xi _j. \end{aligned}$$

Applying now Lemma 7.2 and using

$$\begin{aligned} \mathbf{E}\{|\xi _j|^2|\mathfrak {M}_{j-1}\}&=\left( \sum _{k=1}^{j-1}a_{jk}\zeta _k\right) ^2\mathbf{E}\zeta _j^2,\nonumber \\ \mathbf{E}|\xi _j|^p&=\mathbf{E}|\zeta _j|^p\mathbf{E}\left| \sum _{k=1}^{j-1}a_{jk}\zeta _j\right| ^p, \end{aligned}$$

(7.1)

we get the claim. Thus, Lemma 7.3 is proved. $\square$

We prove as well the following simple Lemma

Lemma 7.4

Let $t>r\ge 1$ and $a,b>0$. Any $x>0$ satisfying the inequality

$$\begin{aligned} x^t\le a+bx^r \end{aligned}$$

(7.2)

is explicitly bounded as follows

$$\begin{aligned} x^t\le \text {e}a+\left( \frac{2t-r}{t-r}\right) ^{\frac{t}{t-r}}b^{\frac{t}{t-r}}. \end{aligned}$$

(7.3)

Proof

First assume that $x\le a^{\frac{1}{t}}$. Then inequality (7.3) holds. If $x\ge a^{\frac{1}{t}}$, then according to inequality (7.2)

$$\begin{aligned} x^{t-r}\le a^{\frac{t-r}{t}}+b, \end{aligned}$$

or

$$\begin{aligned} x^t\le (a^{\frac{t-r}{t}}+b)^{\frac{t}{t-r}}. \end{aligned}$$

Using that for any $\alpha >0$ and $a>0,b>0$

$$\begin{aligned} (a+b)^{\alpha }\le \left( a+\frac{a}{\alpha }\right) ^{\alpha }+(b+ {\alpha } b)^{\alpha } \le \text {e}a^{\alpha }+(1+\alpha )^{\alpha }b^{\alpha }, \end{aligned}$$

we get the claim. $\square$

In what follows we prove several lemmas about the resolvent matrices. Recall the equation, for $j\in \mathbb {T}_{\mathbb {J}}$, (compare with (3.2))

$$\begin{aligned} R_{jj}^{(\mathbb {J})}=-\frac{1}{z+m_n^{(\mathbb {J})}(z)}+\frac{1}{z+m_n^{(\mathbb {J})}(z)} \varepsilon _j^{(\mathbb {J})}R^{(\mathbb {J})}_{jj}, \end{aligned}$$

where

$$\begin{aligned} \varepsilon _{j1}^{(\mathbb {J})}= & {} \frac{1}{\sqrt{n}}X_{jj},\quad \varepsilon _{j2}^{(\mathbb {J})}=-\frac{1}{n}\sum _{k\ne l\in \mathbb {T}_{\mathbb {J},j}}X_{jk}X_{jl}R^{(\mathbb {J},j)}_{kl},\\ \varepsilon _{j3}^{(\mathbb {J})}= & {} -\frac{1}{n}\sum _{k\in \mathbb {T}_{j}}(X_{jk}^2-1)R^{(\mathbb {J},j)}_{kk},\quad \varepsilon _{j4}^{(\mathbb {J})}=m_n^{(\mathbb {J})}(z)-m_n^{(\mathbb {J},j)}(z). \end{aligned}$$

Summing these equations for $j\in \mathbb {T}_{\mathbb {J}}$, we get

$$\begin{aligned} m_n^{(\mathbb {J})}(z)=-\frac{n-|\mathbb {J}|}{n(z+m_n^{(\mathbb {J}))}(z))}+\frac{T_n^{(\mathbb {J})}}{z+m_n^{(\mathbb {J})}(z)}, \end{aligned}$$

(7.4)

where

$$\begin{aligned} T_n^{(\mathbb {J})}=\frac{1}{n}\sum _{j=1}^n\varepsilon _j^{(\mathbb {J})} R_{jj}^{(\mathbb {J})}. \end{aligned}$$

Note that

$$\begin{aligned} \frac{1}{z+m_n^{(\mathbb {J})}(z)}=\frac{1}{z+s(z)}-\frac{m_n^{(\mathbb {J})} (z)-s(z)}{(s(z)+z)(z+m_n^{(\mathbb {J})}(z))}= -s(z)+\frac{s(z)\Lambda _n^{(\mathbb {J})}(z)}{z+m_n^{(\mathbb {J})}(z)}, \end{aligned}$$

(7.5)

where

$$\begin{aligned} \Lambda _n^{(\mathbb {J})}=\Lambda _n^{(\mathbb {J})}(z)=m_n^{(\mathbb {J})}(z)-s(z). \end{aligned}$$

Equalities (7.4) and (7.5) together imply

$$\begin{aligned} \Lambda _n^{(\mathbb {J})}=-\frac{s(z)\Lambda _n^{(\mathbb {J})}}{z+ m_n^{(\mathbb {J})}(z)}+\frac{T_n^{(\mathbb {J})}}{z+m_n^{(\mathbb {J})}(z)}+ \frac{|\mathbb {J}|}{n(z+m_n^{(\mathbb {J})}(z))}. \end{aligned}$$

(7.6)

Solving this with respect to $\Lambda _n^{(\mathbb {J})}$, we get

$$\begin{aligned} \Lambda _n^{(\mathbb {J})}=\frac{T_n^{(\mathbb {J})}}{z+m_n^{(\mathbb {J})}(z)+ s(z)}+\frac{|\mathbb {J}|}{n(z+m_n^{(\mathbb {J})}(z)+s(z))}. \end{aligned}$$

(7.7)

Lemma 7.5

Assuming the conditions of Theorem 1.1, there exists an absolute constant $c_0>0$ such that for $\mathbb {J}\subset \mathbb {T}$,

$$\begin{aligned} |z+m_n^{(\mathbb {J})}(z)+s(z)|\ge \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z), \end{aligned}$$

(7.8)

moreover, for any $z\in \mathbb {G}$

$$\begin{aligned} |z+m_n^{(\mathbb {J})}(z)+s(z)|\ge c_0\sqrt{|z^2-4|}. \end{aligned}$$

(7.9)

Proof

Firstly note that $\mathrm {Im}\;\!(z+s(z))\ge 0$ and $\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z)\ge 0$. Therefore

$$\begin{aligned} |z+m_n^{(\mathbb {J})}(z)+s(z)|\ge \mathrm {Im}\;\!(z+m_n^{(\mathbb {J})}(z)+s(z))\ge \mathrm {Im}\;\!m_n^{(\mathbb {J})}(z). \end{aligned}$$

Furthermore,

$$\begin{aligned} |z+m_n^{(\mathbb {J})}(z)+s(z)|\ge \mathrm {Im}\;\!(z+s(z))=\frac{1}{2}\mathrm {Im}\;\!(z+\sqrt{z^2-4})\ge \frac{1}{2}\mathrm {Im}\;\!\sqrt{z^2-4}. \end{aligned}$$

Note that for $z\in \mathbb {G}$

$$\begin{aligned} \mathrm {Re}\;\!(z^2-4)<0. \end{aligned}$$

(7.10)

Therefore,

$$\begin{aligned} \mathrm {Im}\;\!\sqrt{z^2-4}\ge \frac{\sqrt{2}}{2}\sqrt{|z^2-4|}. \end{aligned}$$

Thus Lemma 7.5 is proved. $\square$

Lemma 7.6

For any $z=u+iv$ with $v>0$ and for any $\mathbb {J}\subset \mathbb {T}$, we have

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{kl}|^2\le v^{-1}\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z). \end{aligned}$$

(7.11)

For any $l\in \mathbb {T}_{\mathbb {J}}$

$$\begin{aligned} \sum _{k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{kl}|^2\le v^{-1}\mathrm {Im}\;\!R^{(\mathbb {J})}_{ll}. \end{aligned}$$

(7.12)

and

$$\begin{aligned} \sum _{k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{kl}|^2\le v^{-3}\mathrm {Im}\;\!R^{(\mathbb {J})}_{ll}. \end{aligned}$$

(7.13)

Moreover, for any $\mathbb {J}\subset T$ and for any $l\in \mathbb {T}_{\mathbb {J}}$ we have

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|^2\le v^{-3}\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z), \end{aligned}$$

(7.14)

and, for any $p\ge 1$

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|^p\le v^{-p}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\mathrm {Im}\;\!^pR^{(\mathbb {J})}_{ll}. \end{aligned}$$

(7.15)

Finally,

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^2\le v^{-3}\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z), \end{aligned}$$

(7.16)

and

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^{2p}\le v^{-3p}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\mathrm {Im}\;\!^pR^{(\mathbb {J})}_{ll}, \end{aligned}$$

(7.17)

We have as well

$$\begin{aligned} \frac{1}{n^2}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^{2p}\le v^{-2p}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\mathrm {Im}\;\!^pR^{(\mathbb {J})}_{ll}\right) ^2. \end{aligned}$$

(7.18)

Proof

For $l\in \mathbb {T}_{\mathbb {J}}$ let us denote by $\lambda ^{(\mathbb {J})}_l$ for $l\in \mathbb {T}_{\mathbb {J}}$ eigenvalues of matrix $\mathbf {W}^{(\mathbb {J})}$. Then we may write (compare (7.20))

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{kl}|^2\le \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\frac{1}{|\lambda ^{(\mathbb {J})}_l-z|^2}. \end{aligned}$$

(7.19)

Note that, for any $x\in \mathbb {R}^1$

$$\begin{aligned} \mathrm {Im}\;\!\frac{1}{x-z}=\frac{v}{|x-z|^2}. \end{aligned}$$

We may write

$$\begin{aligned} \frac{1}{|\lambda ^{(\mathbb {J})}_l-z|^2}=v^{-1}\mathrm {Im}\;\!\frac{1}{\lambda ^{(\mathbb {J})}_l-z} \end{aligned}$$

and

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{kl}|^2\le v^{-1}\mathrm {Im}\;\!\left( \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\frac{1}{\lambda ^{(\mathbb {J})}_l-z}\right) =v^{-1}\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z). \end{aligned}$$

So, inequality (7.11) is proved. Let denote now by $\mathbf {u}_l^{(\mathbb {J})}=(u^{(\mathbb {J})}_{lk})_{k\in \mathbb {T}_{\mathbb {J}}}$ the eigenvector of matrix $\mathbf {W}^{(\mathbb {J})}$ corresponding to the eigenvalue $\lambda ^{(\mathbb {J})}_l$. Using this notation we may write

$$\begin{aligned} R^{(\mathbb {J})}_{lk}=\sum _{q\in \mathbb {T}_{\mathbb {J}}}\frac{1}{ \lambda _q^{(\mathbb {J})}-z}u^{(\mathbb {J})}_{lq}u^{(\mathbb {J})}_{kq}. \end{aligned}$$

(7.20)

It is straightforward to check the following inequality

$$\begin{aligned} \sum _{k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{kl}|^2&\le \sum _{q\in \mathbb {T}_{\mathbb {J}}}\frac{1}{|\lambda ^{(\mathbb {J})}_q-z|^2}|u^{(\mathbb {J})}_{lq}|^2\nonumber \\&=v^{-1}\mathrm {Im}\;\!\left( \sum _{q\in \mathbb {T}_{\mathbb {J}}}\frac{1}{\lambda ^{(\mathbb {J})}_q-z}|u^{(\mathbb {J})}_{lq}|^2\right) =v^{-1}\mathrm {Im}\;\!R_{ll}^{(\mathbb {J})}. \end{aligned}$$

(7.21)

Thus, inequality (7.12) is proved. Similar we get

$$\begin{aligned} \sum _{k\in \mathbb {T}_{\mathbb {J}}}|[(R^{(\mathbb {J})})^2]_{kl}|^2\le \sum _{q\in \mathbb {T}_{\mathbb {J}}}\frac{1}{|\lambda ^{(\mathbb {J})}_q-z|^4}|u^{(\mathbb {J})}_{lq}|^2\le v^{-3}\mathrm {Im}\;\!R^{(\mathbb {J})}_{ll}. \end{aligned}$$

(7.22)

This proves inequality (7.13). To prove inequality (7.14) we observe that

$$\begin{aligned} |[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|\le \sum _{k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{lk}|^2. \end{aligned}$$

(7.23)

This inequality implies

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|^2\le \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}} \left( \sum _{k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{lk}|^2\right) ^2. \end{aligned}$$

Applying now inequality (7.12), we get

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|^2\le v^{-2}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\mathrm {Im}\;\!^2R^{(\mathbb {J})}_{ll}. \end{aligned}$$

This leads (using $|R^{(\mathbb {J})}_{ll}|\le v^{-1}$ ) to the following bound

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|^2\le v^{-3}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\mathrm {Im}\;\!R^{(\mathbb {J})}_{ll}= v^{-3}\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z). \end{aligned}$$

Thus inequality (7.14) is proved. Furthermore, applying inequality (7.23), we may write

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll} |^4\le \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}} \left( \sum _{k\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{lk}|^2\right) ^4. \end{aligned}$$

Applying (7.12), this inequality yields

$$\begin{aligned} \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{ll}|^4\le v^{-4}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}} \mathrm {Im}\;\!^4R^{(\mathbb {J})}_{ll}. \end{aligned}$$

The last inequality proves inequality (7.15). Note that

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^2&\le \frac{1}{n}\mathrm{Tr}\,|\mathbf {R}^{(\mathbb {J})}|^4=\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}} \frac{1}{|\lambda ^{(\mathbb {J})}_l-z|^4}\\&\le v^{-3}\mathrm {Im}\;\!\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}\frac{1}{\lambda ^{(\mathbb {J})}_l-z}=v^{-3}\mathrm {Im}\;\!m_n^{(\mathbb {J})}(z). \end{aligned}$$

Thus, inequality (7.16) is proved. To finish we note that

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^4\le \frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}} \left( \sum _{k\in \mathbb {T}_{\mathbb {J}}}|[(R^{(\mathbb {J})})^2]_{lk}|^2\right) ^2. \end{aligned}$$

Applying inequality (7.13), we get

$$\begin{aligned} \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^4\le v^{-6}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}} (\mathrm {Im}\;\!R^{(\mathbb {J})}_{ll})^2. \end{aligned}$$

To prove inequality (7.18), we note

$$\begin{aligned} |[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^2\le \left( \sum _{q\in \mathbb {T}_{\mathbb {J}}}| R^{(\mathbb {J})}_{lq}|^2\right) \left( \sum _{q\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{kq}|^2\right) . \end{aligned}$$

This inequality implies

$$\begin{aligned} \frac{1}{n^2}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}|[(\mathbf {R}^{(\mathbb {J})})^2]_{lk}|^{2p}\le \left( \frac{1}{n}\sum _{l,k\in \mathbb {T}_{\mathbb {J}}}\left( \sum _{q\in \mathbb {T}_{\mathbb {J}}}|R^{(\mathbb {J})}_{lq}|^2\right) ^p\right) ^2. \end{aligned}$$

Applying inequality (7.11), we get the claim. Thus, Lemma 7.6 is proved. $\square$

Lemma 7.7

For any $s\ge 1$, and for any $z=u+iv$ and for any $\mathbb {J}\subset \mathbb {T}$, and $j\in \mathbb {T}_{\mathbb {J}}$,

$$\begin{aligned} |R_{jj}^{(\mathbb {J})}(u+iv/s)|\le s|R_{jj}^{(\mathbb {J})}(u+iv)|. \end{aligned}$$

Proof

See [17], Lemma 3.4]. For the readers convenience we include the short argument here. Note that, for any $j\in \mathbb {T}_{\mathbb {J}}$,

$$\begin{aligned} \left| \frac{d}{dv}\log R^{(\mathbb {J})}_{jj}(u+iv)\right| \le \frac{1}{|R_{jj}^{(\mathbb {J})}(u+iv)|}|\frac{d}{dv} R_{jj}^{(\mathbb {J})}(u+iv)|. \end{aligned}$$

Furthermore,

$$\begin{aligned} \frac{d}{dv} R_{jj}^{(\mathbb {J})}(u+iv)=[(\mathbf {R}^{(\mathbb {J})})^2]_{jj}(u+iv) \end{aligned}$$

and

$$\begin{aligned} |[(\mathbf {R}^{(\mathbb {J})})^2]_{jj}(u+iv)|\le v^{-1}\mathrm {Im}\;\!R^{(\mathbb {J})}_{jj}. \end{aligned}$$

From here it follows that

$$\begin{aligned} \left| \frac{d}{dv}\log R_{jj}^{(\mathbb {J})}(u+iv)\right| \le v^{-1}. \end{aligned}$$

We may write now

$$\begin{aligned} \left| \log R_{jj}^{(\mathbb {J})}(u+iv)-\log R_{jj}^{(\mathbb {J})}(u+iv/s)\right| \le \int _{v/s}^v\frac{du}{u}=\log s. \end{aligned}$$

The last inequality yields the claim. Thus Lemma 7.7 follows. $\square$

Lemma 7.8

Assuming the conditions of Theorem 1.1, we get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j1}|^2\le \frac{C}{n}. \end{aligned}$$

Proof

The proof follows immediately from the definition of $\varepsilon _{j1}$ and conditions of Theorem 1.1. $\square$

1.1.1 Some auxiliary bounds for resolvent matrices for $\mathrm {Im}\;\!z=4$

We need the bound for the $\varepsilon _{j\nu }$, and $\eta _{j}$ for $V=4$.

Lemma 7.9

Assuming the conditions of Theorem 1.1, we get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}|^2\le \frac{C}{n}. \end{aligned}$$

Proof

Conditioning on $\mathfrak {M}^{(j)}$, we get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}|^2\le n^{-2}\sum _{k,l\in \mathbb {T}_j}\mathbf{E}|R^{(j)}_{kl}|^2. \end{aligned}$$

Applying now Lemma 7.6, inequality (7.12), we get with $\mathrm {Im}\;\!R^{(\mathbb {J})}\le \frac{1}{4}$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j2}|^2\le \frac{1}{4}n^{-2}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!R_{ll}^{(j)}\le \frac{1}{16n}. \end{aligned}$$

Thus Lemma 7.9 is proved. $\square$

Lemma 7.10

Assuming the conditions of Theorem 1.1, we get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j3}|^2\le \frac{C}{n}. \end{aligned}$$

Proof

Conditioning on $\mathfrak {M}^{(j)}$, we obtain as above

$$\begin{aligned} \mathbf{E}|\varepsilon _{j3}|^2\le \frac{\mu _4}{n^2}\mathbf{E}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2\le \frac{\mu _4}{16n}. \end{aligned}$$

Thus Lemma 7.10 is proved. $\square$

Lemma 7.11

Assuming the conditions of Theorem 1.1, we get

$$\begin{aligned} \mathbf{E}|\eta _{j}|^2\le \frac{C}{n}. \end{aligned}$$

Proof

The proof is similar to proof of Lemma 7.9. We need to use that $|[(R^{(j)})^2]_{kl}|\le V^{-2}=\frac{1}{16}$ and $\sum _{l\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\le V^{-4}$. $\square$

Lemma 7.12

Assuming the conditions of Theorem 1.1, we get

$$\begin{aligned} \mathbf{E}|\varepsilon _{j4}|^q\le \frac{C^q}{n^{ q}}. \end{aligned}$$

k,

Proof

The result follows immediately from the bound

$$\begin{aligned} |\varepsilon _{j4}|\le \frac{1}{nv}, \nonumber \text {a.s.} \end{aligned}$$

See for instance [10], Lemma 3.3]. $\square$

1.2 Some auxiliary bounds for resolvent matrices for $z\in \mathbb {G}$

Introduce now the region

$$\begin{aligned} \mathbb {G}&:=\{z=u+iv\in \mathbb {C}^+:\,u\in \mathbb {J}_{\varepsilon }, v\ge v_0/\sqrt{\gamma }\},\quad \text {where} \quad v_0= A_0 n^{-1},\\ \mathbb {J}_{\varepsilon }&=[-2+\varepsilon ,2-\varepsilon ], \quad \varepsilon :=c_1n^{-\frac{2}{3}}, \quad \gamma =\gamma (u)= \min \{2-u, 2+u\}.\nonumber \end{aligned}$$

(7.24)

In the next lemma we state some useful inequalities for the region $\mathbb {G}$.

Lemma 7.13

For any $z\in \mathbb {G}$ we have

$$\begin{aligned} |z^2-4|\ge 2\max \{\gamma ,v\},\qquad {nv}\sqrt{|z^2-4|}\ge 2A_0. \end{aligned}$$

Proof

We observe that

$$\begin{aligned} |z^2-4|=|z-2||z+2|\ge 2\sqrt{\gamma ^2+v^2}. \end{aligned}$$

This inequality proves the Lemma. $\square$

Lemma 7.14

Assuming the conditions of Theorem 1.1, there exists an absolute constant $C>0$ such that for any $j=1,\ldots ,n$,

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j1}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C\mu _4}{n^2}. \end{aligned}$$

(7.25)

Proof

The result follows immediately from the definition of $\varepsilon _{j1}$. $\square$

Lemma 7.15

Assuming the conditions of Theorem 1.1, there exists an absolute constant $C>0$ such that for any $j=1,\ldots ,n$,

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j2}|^2\big |\mathfrak {M}^{(j)}\}\le \frac{C}{nv}\mathrm {Im}\;\!m_n^{(j)}(z), \end{aligned}$$

(7.26)

and

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j2}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C\mu _4^2}{n^2v^2}\mathrm {Im}\;\!^2m_n^{(j)}(z). \end{aligned}$$

(7.27)

Proof

Note that r.v.’s $X_{jl}$, for $l\in \mathbb {T}_j$ are independent of $\mathfrak {M}^{(j)}$ and that for $l,k\in \mathbb {T}_j$ $R^{(j)}_{lk}$ are measurable with respect to $\mathfrak {M}^{(j)}$. This implies that $\varepsilon _{j2}$ is a quadratic form with coefficients $R^{(j)}_{lk}$ independent of $X_{jl}$. Thus its variance and fourth moment are easily available.

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j2}|^2\big |\mathfrak {M}^{(j)}\}=\frac{1}{n^2}\sum _{l\ne k\in \mathbb {T}_j}|R^{(j)}_{lk}|^2\le \frac{1}{n^2}\mathrm{Tr}\,\mathbb |\mathbf {R}^{(j)}|^2. \end{aligned}$$

Here we use the notation $|\mathbf {A}|^2=\mathbf {A}\mathbf {A}^*$ for any matrix $\mathbf {A}$. Applying Lemma 7.6, inequality (7.11), we get equality (7.26).

Furthermore, direct calculations show that

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j2}|^4\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^2}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|R^{(j)}_{lk}|^2\right) ^2 +\frac{C\mu _4^2}{n^2}\frac{1}{n^2}\sum _{l\in \mathbb {T}_{j}}|R^{(j)}_{lk}|^4\\&\le \frac{C\mu _4^2}{n^2}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|R^{(j)}_{lk}|^2\right) ^2 \le \frac{C\mu _4^2}{n^2v^2}(\mathrm {Im}\;\!m_n^{(j)}(z))^2. \end{aligned}$$

Here again we used Lemma 7.6, inequality (7.11). Thus Lemma 7.15 is proved. $\square$

Lemma 7.16

Assuming the conditions of Theorem 1.1, there exists an absolute constant $C>0$ such that for any $j=1,\ldots ,n$,

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j3}|^2\big |\mathfrak {M}^{(j)}\}\le \frac{C\mu _4}{n}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2, \end{aligned}$$

(7.28)

and

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j3}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C}{n^2}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2\right) ^2 +\frac{C\mu _4}{n^2}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^4. \end{aligned}$$

(7.29)

Proof

The first inequality is easy. To prove the second, we apply Rosenthal’s inequality. We obtain

$$\begin{aligned} \mathbf{E}\{|\varepsilon _{j3}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C\mu _4}{n^2}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2\right) ^2 +\frac{C\mu _8}{n^3}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^4. \end{aligned}$$

Using $|X_{jl}|\le Cn^{\frac{1}{4}}$ we get $\mu _8\le Cn\mu _4$ and the claim. Thus Lemma 7.16 is proved. $\square$

Corollary 7.17

Assuming the conditions of Theorem 1.1, there exists an absolute constant $C>0$, depending on $\mu _4$ and $D_0$ only, such that for any $j=1,\ldots ,n$, $\nu =1,2,3$ $z\in \mathbb {G}$, and $0\le \alpha \le \frac{1}{2}A_1(nv)^{\frac{1}{4}}$ and $\beta \ge 1$

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j\nu }|^2}{|z+m_n^{(j)}(z)+s(z)|^{\beta }|z+ m_n^{(j)}(z)|^{\alpha }}\le \frac{C}{nv|z^2-4|^{\frac{\beta -1}{2}}} \end{aligned}$$

(7.30)

and for $\beta \ge 2$

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^{\beta }|z+m_n^{(j)}(z)|^{\alpha }}\le \frac{C}{n^2v^2|z^2-4|^{\frac{\beta -2}{2}}}. \end{aligned}$$

(7.31)

We have as well, for $\nu =2,3$, and $\beta \ge 4$

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j\nu }|^8}{|z+m_n^{(j)}(z)+s(z)|^{\beta }|z+m_n^{(j)}(z)|^{\alpha }}\le \frac{C}{n^4v^4|z^2-4|^{\frac{\beta -4}{2}}} \end{aligned}$$

(7.32)

Proof

For $\nu =1$, by Lemma 7.5, we have

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j\nu }|^2}{|z+m_n^{(j)}(z)+s(z)||z+m_n^{(j)}(z)|^{\alpha }}\le \frac{1}{n|z^2-4|^{\frac{\beta }{2}}}\mathbf{E}|X_{jj}|^4\mathbf{E}\frac{1}{|z+m_n^{(j)}(z)|^{\alpha }}. \end{aligned}$$

Applying now Corollary 5.14 and Lemma 7.13, we get the claim. The proof of the second inequality for $\nu =1$ is similar. For $\nu =2$ we apply Lemmas 7.15 and 7.5, inequality (7.26) and obtain, using inequality (7.8),

$$\begin{aligned}&\mathbf{E}\frac{|\varepsilon _{j2}|^2}{|z+m_n^{(j)}(z)+s(z)||z+m_n^{(j)}(z)|^{\alpha }}\\&\qquad \le \frac{1}{nv|z^2-4|^{\frac{\beta -1}{2}}}\mathbf{E}\frac{\mathrm {Im}\;\!m_n^{(j)}(z)}{|z+m_n^{(j)}(z)+s(z)||z+m_n^{(j)}(z)|^{\alpha }}\\&\qquad \le \frac{C}{nv}\mathbf{E}\frac{1}{|z+m_n^{(j)}(z)|^{\alpha }}. \end{aligned}$$

Similar, using Lemma 7.15, inequality (7.27), and inequality (7.8) we get

$$\begin{aligned}&\mathbf{E}\frac{|\varepsilon _{j2}|^4}{|z+m_n^{(j)}(z)+s(z)|^{\beta }|z+m_n^{(j)}(z)|^{\alpha }}\\&\quad \le \frac{C}{n^2v^2|z^2-4|^{\frac{\beta -2}{2}}}\mathbf{E}\frac{\mathrm {Im}\;\!^2 m_n^{(j)}(z)}{|z+m_n^{(j)}(z)+s(z)|^2 |z+m_n^{(j)}(z)|^{\alpha }}\\&\quad \le \frac{C}{n^2v^2|z^2-4|^{\frac{\beta -2}{2}}}\mathbf{E}\frac{1}{|z+m_n^{(j)}(z)|^{\alpha }}. \end{aligned}$$

Applying Corollary 5.14, we get the claim. For $\nu =3$, we apply Lemma 7.16, inequalities (7.28) and (7.29) and Lemma 7.5. We get

$$\begin{aligned}&\mathbf{E}\frac{|\varepsilon _{j3}|^2}{|z+m_n^{(j)}(z)+s(z)||z+m_n^{(j)}(z)|^{\alpha }}\nonumber \\&\quad \le \frac{C}{n\sqrt{|z^2-4}}\mathbf{E}\frac{1}{|z+m_n^{(j)}(z)|^{\alpha }}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2\right) , \end{aligned}$$

(7.33)

and

$$\begin{aligned}&\mathbf{E}\frac{|\varepsilon _{j3}|^4}{|z+m_n^{(j)}(z)+s(z)|^2|z+m_n^{(j)}(z)|^{\alpha }}\\&\quad \le \frac{C}{n^2{|z^2-4|}}\mathbf{E}\frac{1}{|z+m_n^{(j)}(z)|^{\alpha }}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^4\right) . \end{aligned}$$

Using now the Cauchy–Schwartz inequality and Corollary 5.14, we get the claim. $\square$

Lemma 7.18

Assuming the conditions of Theorem 1.1, there exists an absolute constant $C>0$ such that for any $j=1,\ldots ,n$,

$$\begin{aligned} |\varepsilon _{j4}|\le \frac{C}{nv}\quad \text {a.s.} \end{aligned}$$

Proof

This inequality follows from

$$\begin{aligned} \mathrm{Tr}\,\mathbf {R}-\mathrm{Tr}\,\mathbf {R}^{(j)}=\left( 1+\frac{1}{n}\sum _{l,k\in \mathbb {T}_j}X_{jl}X_{jk}[(R^{(j)})^2]_{kl}\right) R_{jj}=R^{-1}_{jj}\frac{dR_{jj}}{dz}, \end{aligned}$$

(7.34)

which may be obtained using the Schur complement formula. See, for instance [10], Lemma 3.3]. $\square$

Corollary 7.19

Assuming the conditions of Theorem 1.1, there exists an absolute constant $C>0$ such that for any $j=1,\ldots ,n$,

$$\begin{aligned} \mathbf{E}|\varepsilon _{j}|^4\le \frac{C}{n^2v^2}. \end{aligned}$$

(7.35)

Proof

By definition of $\varepsilon _j$ (see (3.2)), we have

$$\begin{aligned} \mathbf{E}|\varepsilon _j|^4\le 4^3(\mathbf{E}|\varepsilon _{j1}|^4+\cdots +\mathbf{E}|\varepsilon _{j4}|^4). \end{aligned}$$

Applying now Lemmas 7.14, 7.15 (inequality (7.27)), 7.16 (inequality (7.29)) and 7.18 and taking expectation where needed (7.35) follows. Thus the Corollary is proved.   $\square$

Introduce quantities

$$\begin{aligned} \eta _{j1}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j} [(R^{(j)})^2]_{ll}=\frac{1}{n}\mathrm{Tr}\,(\mathbf {R}^{(j)})^2,\nonumber \\ \eta _{j2}&=\frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j} X_{jl}X_{jk}[(\mathbf {R}^{(j)})^2]_{lk},\nonumber \\ \eta _{j3}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j} (X_{jl}^2-1)[(\mathbf {R}^{(j)})^2]_{ll}. \end{aligned}$$

(7.36)

Lemma 7.20

Assuming the conditions of Theorem 1.1, we have, for $z\in \mathbb {G}$, for any $j=1,\ldots ,n$

$$\begin{aligned} \mathbf{E}\{|\eta _{j3}|^4\big |\mathfrak {M}^{(j)}\}&\le Cn^{-2}v^{-6}\mathrm {Im}\;\!^2m_n^{(j)}(z) +Cn^{-2}v^{-4}\mu _4\frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^4R^{(j)}_{ll},\end{aligned}$$

(7.37)

$$\begin{aligned} \mathbf{E}\{|\eta _{j3}|^8\big |\mathfrak {M}^{(j)}\}&\le \frac{C\mu _4^4}{n^4v^{12}}\mathrm {Im}\;\!^4m_n^{(j)}(z) +\frac{C\mu _4^4}{n^4v^{8}}\frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^8R^{(j)}_{ll}. \end{aligned}$$

(7.38)

Proof

Direct calculation shows

$$\begin{aligned} \mathbf{E}\{|\eta _{j3}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C\mu _4}{n^2}\left( \sum _{l\in \mathbb {T}_j}|[(R^{(j)})^2]_{ll}|^2\right) ^2+ \frac{C\mu _8}{n^3}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|[(R^{(j)})^2]_{ll}|^4. \end{aligned}$$

Applying Lemma 7.6, the inequality (7.15) and the inequality $\mu _8\le Cn\mu _4$, we get

$$\begin{aligned} \mathbf{E}\{|\eta _{j3}|^4\big |\mathfrak {M}^{(j)}\}\le Cn^{-2}v^{-6}\mathrm {Im}\;\!^2m_n^{(j)}(z)+Cn^{-2}\mu _4\frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^4R^{(j)}_{ll}. \end{aligned}$$

Thus, inequality (7.37) is proved. Consider now the $8$th moment of $\eta _{j3}$. Applying Rosenthal’s inequality, we get

$$\begin{aligned} \mathbf{E}\{|\eta _{j3}|^8\big |\mathfrak {M}^{(j)}\}\le \frac{C\mu _4^4}{n^4}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}|[(R^{(j)})^2]_{ll}|^2\right) ^4+ \frac{C\mu _{16}}{n^7}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|[(R^{(j)})^2]_{ll}|^8. \end{aligned}$$

Using now Lemma 7.6, inequalities (7.14) and (7.15), and that $\mu _{16}\le Cn^3\mu _4$, we obtain

$$\begin{aligned} \mathbf{E}\{|\eta _{j3}|^8\big |\mathfrak {M}^{(j)}\} \le \frac{C\mu _4^4}{n^4v^{12}}\mathrm {Im}\;\!^4m_n^{(j)}(z)+\frac{C\mu _4^4}{n^4v^{8}}\frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^8R^{(j)}_{ll}. \end{aligned}$$

Thus, inequality (7.38) is proved. $\square$

Lemma 7.21

Assuming the conditions of Theorem 1.1, we have, for $z\in \mathbb {G}$, for any $j=1,\ldots ,n$

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^4\big |\mathfrak {M}^{(j)}\}&\le \frac{C\mathrm {Im}\;\!^2m_n^{(j)}(z)}{n^2v^{6}}+ \frac{C\mu _4^2}{n^3v^6}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2,\end{aligned}$$

(7.39)

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^8\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^4v^{12}}\left( \mathrm {Im}\;\!m_n^{(j)}(z)\right) ^4+ \frac{C\mu _4^2}{n^4v^{8}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^4R^{(j)}_{ll}\right) ^2\nonumber \\&\quad +\frac{C\mu _4^2}{n^4v^{8}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^3 R^{(j)}_{ll}\right) ^2 \left( \mathrm {Im}\;\!m_n^{(j)}(z)\right) +\frac{C\mu _4^4}{n^6v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j} \left( \mathrm {Im}\;\!R^{(j)}_{ll}\right) ^2\right) ^2\nonumber \\&\quad +\frac{C\mu _4^2}{n^6v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}\left( \mathrm {Im}\;\!R^{(j)}_{ll}\right) ^2\right) \left( \mathrm {Im}\;\!m_n^{(j)}(z)\right) ^2. \end{aligned}$$

(7.40)

Proof

Direct calculation shows that

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C}{n^2}\left( \frac{1}{n} \sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) ^2+ \frac{C\mu _4^2}{n^4}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^4. \end{aligned}$$

Applying inequalities (7.16) and (7.17) of Lemma 7.6, we get

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^4\big |\mathfrak {M}^{(j)}\}\le \frac{C\mathrm {Im}\;\!^2 m_n^{(j)}(z)}{n^2v^{6}}+ \frac{C\mu _4^2}{n^3v^6}\frac{1}{n}\sum _{l\in \mathbb {T}_j}|R^{(j)}_{ll}|^2. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^8\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^4}\left( \frac{1}{n} \sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) ^4+ \frac{C\mu _8^2}{n^8}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^8\\&\quad +\frac{C\mu _6^2}{n^6}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl} |^6\right) \left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) \\&\quad +\frac{C\mu _4^4}{n^6}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^4\right) ^2\\&\quad +\frac{C\mu _4^2}{n^6}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^4\right) \left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) ^2. \end{aligned}$$

Note that

$$\begin{aligned} \mu _6\le C\sqrt{n}\mu _4,\quad \mu _8\le Cn\mu _4. \end{aligned}$$

Using this relations, we get

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^8\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^4}\left( \frac{1}{n} \sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) ^4+ \frac{C\mu _4^2}{n^6}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}\Big |^8\\&\quad + \frac{C\mu _4^2}{n^5}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^6\right) \left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) \\&\quad + \frac{C\mu _4^4}{n^6}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^4\right) ^2 \\&\quad + \frac{C\mu _4^2}{n^6}\left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^4\right) \left( \frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2\right) ^2. \end{aligned}$$

Applying now Lemma 7.6, we obtain

$$\begin{aligned} \mathbf{E}\{|\eta _{j2}|^8\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^4v^{12}}(\mathrm {Im}\;\!m_n^{(j)}(z))^4+ \frac{C\mu _4^2}{n^4v^{8}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^4\right) ^2 \\&\quad + \frac{C\mu _4^2}{n^5v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}\mathrm {Im}\;\!^3R^{(j)}_{ll}\right) (\mathrm {Im}\;\!m_n^{(j)}(z))\!+\! \frac{C\mu _4^4}{n^6v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^2\right) ^2 \\&\quad + \frac{C\mu _4^2}{n^6}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^2\right) (\mathrm {Im}\;\!m_n^{(j)}(z))^2. \end{aligned}$$

Thus, the Lemma is proved. $\square$

Lemma 7.22

Assuming the conditions of Theorem 1.1, we have

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j4}|^4}{|z+s(z)+m_n(z)|^4}\le \frac{C}{n^4v^4}. \end{aligned}$$

Proof

Using the representations (7.34)–(7.36) we have

$$\begin{aligned} \varepsilon _{j4}=\frac{1}{n}\left( 1+\frac{1}{n}\sum _{l, k\in \mathbb {T}_j}X_{jl}X_{jk}[(R^{(j)})^2]_{lk}\right) R_{jj}=\frac{1}{n}(1+\eta _{j1} +\eta _{j2}+\eta _{j3}) R_{jj}. \end{aligned}$$

(7.41)

Applying the Cauchy–Schwartz inequality, we get

$$\begin{aligned}&\mathbf{E}|\frac{\varepsilon _{j4}}{z+s(z)+m_n(z)}|^4\\&\quad \le \frac{C}{n^4}\left( 1+\mathbf{E}^{\frac{1}{2}}\left( \frac{|\frac{1}{n}\sum _{l,k\in \mathbb {T}_j} X_{jl}X_{jk}[(R^{(j)})^2]_{lk}|}{|z+s(z)+m_n(z)|}\right) ^8\right) \mathbf{E}^{\frac{1}{2}}|R_{jj}|^8. \end{aligned}$$

Using Corollary 5.14, we may write

$$\begin{aligned}&\mathbf{E}|\frac{\varepsilon _{j4}}{z+s(z)+m_n(z)}|^4\\&\quad \le \frac{C}{n^4}\left( 1+\mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j1}}{z+m_n(z)+s(z)}\right| ^8 +\mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j2}}{z+m_n(z)+s(z)}\right| ^8\right. \\&\qquad \left. +\mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j3}}{z+m_n(z)+s(z)}\right| ^8\right) . \end{aligned}$$

Observe that,

$$\begin{aligned} \frac{1}{|z+s(z)+m_n(z)|}\le \frac{1}{|z+s(z)+m_n^{(j)}(z)|}(1+ \frac{|\varepsilon _{j4}|}{|z+s(z)+m_n(z)|} ). \end{aligned}$$

Therefore, by Lemmas 7.18, 7.5 and 7.13, for $z\in \mathbb {G}$,

$$\begin{aligned} \frac{1}{|z+s(z)+m_n(z)|}\le \frac{C}{|z+s(z)+m_n^{(j)}(z)|}. \end{aligned}$$

(7.42)

Using (7.42) we get by definition of $\eta _{j1}$, Lemma 7.6, and inequality (7.11)

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j1}}{z+m_n(z)+s(z)}\right| ^8\le C\mathbf{E}^{\frac{1}{2}}\frac{v^{-8}\mathrm {Im}\;\!^8 m_n^{(j)}(z)}{|z+m_n^{(j)}(z)+s(z)|^8}\le v^{-4}. \end{aligned}$$

Furthermore, applying inequality (7.42) again, we obtain

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j2}}{z+m_n(z)+s(z)}\right| ^8\le \mathbf{E}^{\frac{1}{2}} \left| \frac{\eta _{j2}}{z+m_n^{(j)}(z)+s(z)}\right| ^8. \end{aligned}$$

Conditioning with respect to $\mathfrak {M}^{(j)}$ and applying Lemma 7.20, we obtain

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j2}}{z+m_n(z)+s(z)}\right| ^8&\le \mathbf{E}^{\frac{1}{2}}\frac{C}{|z+m_n^{(j)}(z)+s(z)|^8}\left( \frac{1}{n^4v^{12}}(\mathrm {Im}\;\!m_n^{(j)}(z))^4\right. \\&\quad \left. +\,\frac{\mu _4^4}{n^4v^8}\frac{1}{n}\sum _{l\in \mathbb {T}_{\mathbb {J}}}(\mathrm {Im}\;\!R^{(j)}_{ll})^8\right) . \end{aligned}$$

Using Lemma 7.5, inequality (7.8), together with Corollary 5.14 we get

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j2}}{z+m_n(z)+s(z)}\right| ^8\le \frac{C}{n^2v^6|z^2-4|}+\frac{C\mu _{4}^2}{n^{2}v^4|z^2-4|^2}. \end{aligned}$$

Applying inequality (7.42) and conditioning with respect to $\mathfrak {M}^{(j)}$ and applying Lemma 7.21, we get

$$\begin{aligned}&\mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j3}}{z+m_n(z)+s(z)}\right| ^8\le \mathbf{E}^{\frac{1}{2}}\frac{1}{|z+s(z)+m_n^{(j)}(z)|^8}\left( \frac{C}{n^4v^{12}}(\mathrm {Im}\;\!m_n^{(j)}(z))^4\right. \nonumber \\&\quad +\frac{C\mu _4^2}{n^4v^{8}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^4\right) ^2+ \frac{C\mu _4^2}{n^5v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^3\right) (\mathrm {Im}\;\!m_n^{(j)}(z))\nonumber \\&\quad \left. +\frac{C\mu _4^4}{n^6v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^2\right) ^2+ \frac{C\mu _4^2}{n^6v^{12}}\left( \frac{1}{n}\sum _{l\in \mathbb {T}_j}(\mathrm {Im}\;\!R^{(j)}_{ll})^2\right) (\mathrm {Im}\;\!m_n^{(j)}(z))^2\right) . \end{aligned}$$

(7.43)

Using that $|z+m_n^{(j)}(z)+s(z)|\ge \mathrm {Im}\;\!m_n^{(j)}(z)$ together with Lemma 7.6, we arrive at

$$\begin{aligned} \mathbf{E}^{\frac{1}{2}}\left| \frac{\eta _{j3}}{z+m_n(z)+s(z)}\right| ^8&\le \frac{C}{n^{2}v^{6}|z^2-4|}+\frac{C}{n^{2}v^{4}|z^2-4|^2} \\&\quad + \frac{C}{n^{\frac{5}{2}}v^{6}|z^2-4|^{\frac{7}{4}}} + \frac{C}{n^{3}v^{6}|z^2-4|^{2}}+ \frac{C}{n^{3}v^{6}|z^2-4|^{\frac{3}{2}}}. \end{aligned}$$

Summarizing we may write now, for $z\in \mathbb {G}$,

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j4}|^4}{|z+s(z)+m_n(z)|^4}&\le \frac{C}{n^4v^4}+\frac{C}{n^{6}v^{6}|z^2-4|}+\frac{C}{n^{6}v^{4}|z^2-4|^2}\\&\quad + \frac{C}{n^{\frac{13}{2}}v^{6}|z^2-4|^{\frac{7}{4}}} + \frac{C}{n^{7}v^{6}|z^2-4|^{2}}+ \frac{C}{n^{7}v^{6}|z^2-4|^{\frac{3}{2}}}. \end{aligned}$$

For $z\in \mathbb {G}$, see (7.24) and Lemma 7.13, this inequality may be simplified by means of the following bounds (with $v_0= A_0n^{-1}$)

$$\begin{aligned} n^{\frac{5}{2}}v^2|z^2-4|^{\frac{7}{4}}\ge n^{\frac{5}{2}}v_0^2\gamma ^{-1+\frac{7}{4}}\ge C\sqrt{n}\gamma ^{\frac{3}{4}}&\ge C,\nonumber \\ n^3v^2|z^2-4|\ge C,\quad n^3v^2|z^2-4|^{\frac{3}{2}}&\ge C,\nonumber \\ n^2|z^2-4|^2&\ge C. \end{aligned}$$

(7.44)

Using these relation, we obtain

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j4}|^4}{|z+s(z)+m_n(z)|^4}\le \frac{C}{n^4v^4}. \end{aligned}$$

Thus Lemma 7.22 is proved. $\square$

Corollary 7.23

Assuming the conditions of Theorem 1.1, we have

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j4}|^2}{|z+s(z)+m_n(z)|^2}\le \frac{C}{n^2v^{2}}. \end{aligned}$$

Proof

The result follows immediately from Lemma 7.22 and Jensen’s inequality. $\square$

Lemma 7.24

Assuming the conditions of Theorem 1.1, we have, for $z\in \mathbb {G}$,

$$\begin{aligned} \mathbf{E}|\Lambda _n|^2\le \frac{C}{n^2v^2}. \end{aligned}$$

(7.45)

Proof

We write

$$\begin{aligned} \mathbf{E}|\Lambda _n|^2=\mathbf{E}\Lambda _n\overline{\Lambda }_n=\mathbf{E}\frac{T_n}{z+m_n(z)+s(z)} \overline{\Lambda }_n=\sum _{\nu =1}^4\mathbf{E}\frac{T_{n\nu }}{z+m_n(z)+s(z)} \overline{\Lambda }_n, \end{aligned}$$

where

$$\begin{aligned} T_{n\nu }:=\frac{1}{n}\sum _{j=1}^n\varepsilon _{j\nu }R_{jj}, \quad \text {for}\quad \nu =1,\ldots ,4. \end{aligned}$$

First we observe that by (7.34)

$$\begin{aligned} |T_{n4}|=\frac{1}{n}|m_n'(z)|\le \frac{1}{nv}\mathrm {Im}\;\!m_n(z). \end{aligned}$$

Hence $|z+m_n(z)+s(z)|\ge \mathrm {Im}\;\!m_n(z)$ and Jensen’s inequality yields

$$\begin{aligned} |\mathbf{E}\frac{T_{n4}}{z+m_n(z)+s(z)}\overline{\Lambda }_n|\le \frac{1}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.46)

Furthermore, we represent $T_{n1}$ as follows

$$\begin{aligned} T_{n1}=T_{n11}+T_{n12}, \end{aligned}$$

where

$$\begin{aligned} T_{n11}&=-\frac{1}{n}\sum _{j=1}^n\varepsilon _{j1}\frac{1}{z+m_n(z)},\\ T_{n12}&=\frac{1}{n}\sum _{j=1}^n\varepsilon _{j1}\left( R_{jj}+\frac{1}{z+m_n(z)}\right) . \end{aligned}$$

Using these notations we may write

$$\begin{aligned} V_1:=\mathbf{E}\frac{T_{n11}}{z+m_n(z)+s(z)}\overline{\Lambda }_n=-\mathbf{E}\frac{\left( \frac{1}{n}\sum _{j=1}^n\varepsilon _{j1}\right) }{(z+m_n(z))(z+s(z)+m_n(z))}\overline{\Lambda }_n. \end{aligned}$$

Applying the Cauchy–Schwartz inequality twice and using the definition of $\varepsilon _{j1}$ (see (3.2)), we get by Lemma 7.5

$$\begin{aligned} |V_1|\le \frac{1}{|z^2-4|^{\frac{1}{2}}}\mathbf{E}^{\frac{1}{4}}\left| \frac{1}{n\sqrt{n}}\sum _{j=1}^nX_{jj}\right| ^4\mathbf{E}^{\frac{1}{4}}|\frac{1}{z+m_n(z)|^4}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

By Rosenthal’s inequality, we have, for $z\in \mathbb {G}$

$$\begin{aligned} |V_1|\le \frac{C}{n|z^2-4|^{\frac{1}{2}}}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\le \frac{1}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.47)

Using (3.2) we rewrite $T_{n12}$, obtaining

$$\begin{aligned} V_2:=\mathbf{E}\frac{T_{n12}}{z+m_n(z)+s(z)}\overline{\Lambda }_n=\frac{1}{n\sqrt{n}}\sum _{j=1}^n\mathbf{E}\frac{X_{jj}\varepsilon _jR_{jj}}{(z+m_n(z))(z+m_n(z)+s(z))}\overline{\Lambda }_n. \end{aligned}$$

By the Cauchy–Schwartz inequality, using the definition of $\varepsilon _j$ (see representation (3.2)), we obtain

$$\begin{aligned} |V_2|\le \frac{1}{\sqrt{n}}\sum _{\nu =1}^4\mathbf{E}^{\frac{1}{2}}\frac{|\frac{1}{n}\sum _{j=1}^n\varepsilon _{j\nu }X_{jj}R_{jj}|^2}{|z+m_n(z)+s(z)|^2|z+m_n(z)|^2} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2=:\sum _{\nu =1}^4V_{2\nu }. \end{aligned}$$

(7.48)

For $\nu =1$, we have

$$\begin{aligned} V_{21}\le \frac{1}{ n}\mathbf{E}^{\frac{1}{2}}\frac{\left| \frac{1}{n}\sum _{j=1}^nX_{jj}^2R_{jj}\right| ^2}{|z+m_n(z)+s(z)|^2|z+m_n(z)|^2}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

Applying the Cauchy–Schwartz inequality twice and Lemma 7.5, we arrive at

$$\begin{aligned} V_{21}\!\le \! \frac{1}{ n\sqrt{|z^2-4|}}\mathbf{E}^{\frac{1}{4}}\left( \frac{1}{n}\sum _{j=1}^nX_{jj}^4\right) ^2\mathbf{E}^{\frac{1}{8}}\left( \frac{1}{n}\sum _{j=1}^n|R_{jj}|^2\right) ^4 \mathbf{E}^{\frac{1}{8}}\frac{1}{|z+m_n(z)|^8}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.49)

Observe that

$$\begin{aligned} \mathbf{E}\left( \frac{1}{n}\sum _{j=1}^nX_{jj}^4\right) ^2&=\left( \frac{1}{n}\sum _{j=1}^n\mathbf{E}X_{jj}^4\right) ^2+\mathbf{E}\left( \frac{1}{n}\sum _{j=1}^n(X_{jj}^4-\mathbf{E}X_{jj}^4)\right) ^2\nonumber \\&\le \mu _4^2+\frac{2}{n^2}\sum _{j=1}^n\mathbf{E}|X_{jj}|^8\le (\mu _4 +D_0^4)\mu _4\le C. \end{aligned}$$

(7.50)

The last inequality, inequality (7.49), Corollary 5.14 and Lemma 7.13 together imply

$$\begin{aligned} V_{21}\le \frac{C}{n\sqrt{|z^2-4|}}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.51)

Furthermore, for $\nu =4$, by Lemma 7.18 we have

$$\begin{aligned} V_{24}\le \frac{1}{nv\sqrt{n}}\mathbf{E}^{\frac{1}{2}}\frac{\frac{1}{n}\sum _{j=1}^n|X_{jj}|^2|R_{jj}|^2}{|z+m_n(z)+s(z)|^2|z+m_n(z)|^2} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

Applying the Cauchy–Schwartz inequality and Lemma 7.5, we get

$$\begin{aligned} V_{24}\le \frac{1}{nv\sqrt{n}\sqrt{|z^2-4|}}\mathbf{E}^{\frac{1}{4}}\left( \frac{1}{n}\sum _{j=1}^nX_{jj}^4\right) ^2\mathbf{E}^{\frac{1}{8}}\left( \frac{1}{n}\sum _{j=1}^n|R_{jj}|^2\right) ^4 \mathbf{E}^{\frac{1}{8}}\frac{1}{|z+m_n(z)|^8}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

Similar to inequality (7.51), applying Lemma 7.13, inequality (7.50) and Corollary 5.14, we get

$$\begin{aligned} V_{24}\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.52)

By Hölder’s inequality, we have for $\nu =2,3$,

$$\begin{aligned} V_{2\nu }\le \frac{1}{\sqrt{n}}\frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4|X_{jj}|^4}{|z+m_n(z)+s(z)|^4}\mathbf{E}^{\frac{1}{8}}\frac{1}{|z+m_n(z)|^8} \mathbf{E}^{\frac{1}{8}}|R_{jj}|^8 \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

Note that for $\nu =2,3$, r.v. $X_{jj}$ doesn’t depend on $\varepsilon _{j\nu }$ and on $\sigma$-algebra $\mathfrak {M}^{(j)}$. Using inequality (7.42) for $z\in \mathbb {G}$, we get

$$\begin{aligned} \mathbf{E}\frac{|\varepsilon _{j\nu }|^4|X_{jj}|^4}{|z+m_n(z)+s(z)|^4}\le C\mathbf{E}\frac{|\varepsilon _{j\nu }|^4|X_{jj}|^4}{|z+m_n^{(j)}(z)+s(z)|^4}\le C\mu _4\mathbf{E}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^4}. \end{aligned}$$

Applying now Lemmas 7.15, 7.16 and 7.13, arrive at

$$\begin{aligned} V_{2\nu }\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2,\quad \text {for} \quad \nu =2,3. \end{aligned}$$

(7.53)

Inequalities (7.51), (7.52), (7.53) together imply

$$\begin{aligned} V_2\le \frac{C}{nv}\mathbf{E}|\Lambda _n|^2. \end{aligned}$$

(7.54)

Consider now the quantity

$$\begin{aligned} Y_{\nu }:=\mathbf{E}\frac{T_{n\nu }}{z+m_n(z)+s(z)}\overline{\Lambda }_n, \end{aligned}$$

for $\nu =2,3$. We represent it as follows

$$\begin{aligned} Y_{\nu }=Y_{\nu 1}+Y_{\nu 2}, \end{aligned}$$

where

$$\begin{aligned} Y_{\nu 1}= & {} -\displaystyle \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }\overline{\Lambda }_n}{(z+m_n^{(j)}(z))(z+m_n(z)+s(z))},\\ Y_{\nu 2}= & {} \displaystyle \frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }(R_{jj}+\frac{1}{z+m_n^{(j)}(z)})\overline{\Lambda }_n}{z+m_n(z)+s(z)}. \end{aligned}$$

By the representation (5.32), which is similar to (3.2) we have

$$\begin{aligned} Y_{\nu 2}=\sum _{\mu =1}^3\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }\varepsilon _{j\mu } \overline{\Lambda }_nR_{jj}}{(z+m_n(z)+s(z))(z+m_n^{(j)}(z))}. \end{aligned}$$

Using inequality (7.42), we may write, for $z\in \mathbb {G}$

$$\begin{aligned} |Y_{\nu 2}|\le \sum _{\mu =1}^3\frac{C}{n}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }||\varepsilon _{j\mu }||\overline{\Lambda }_n||R_{jj}|}{|z+m_n^{(j)}(z)+s(z)||z+m_n^{(j)}(z)|}. \end{aligned}$$

Applying the Cauchy–Schwartz inequality and the inequality $ab\le \frac{1}{2}(a^2+b^2)$, we get

$$\begin{aligned} |Y_{\nu 2}|&\le \sum _{\mu =1}^3\frac{C}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}\frac{| \varepsilon _{j\nu }|^2|\varepsilon _{j\mu }|^2|R_{jj}|^2}{|z+m_n^{(j)}(z)+s(z)|^2|z+m_n^{(j)}(z)|^2} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\\&\le \sum _{\mu =2}^3\frac{C}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}\frac{|\varepsilon _{j\mu }|^4|R_{jj}|^2}{|z+m_n^{(j)}(z)+s(z)|^2|z+m_n^{(j)}(z)|^2} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\\&\quad +\frac{C}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{2}}\frac{|\varepsilon _{j\nu }|^2|\varepsilon _{j1}|^2|R_{jj}|^2}{|z+m_n^{(j)}(z)+s(z)|^2|z+m_n^{(j)}(z)|^2} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\\&\le \sum _{\mu =2}^3\frac{C}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\mu }|^8}{|z+m_n^{(j)}(z)+s(z)|^4|z+m_n^{(j)}(z)|^4} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\mathbf{E}^{\frac{1}{4}}|R_{jj}|^4\\&\quad +\frac{C}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4|\varepsilon _{j1}|^4}{|z+m_n^{(j)}(z)+s(z)|^4|z+m_n^{(j)}(z)|^4} \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\mathbf{E}^{\frac{1}{4}}|R_{jj}|^4. \end{aligned}$$

Using Corollary 7.17 with $\alpha =2$, we arrive at

$$\begin{aligned} |Y_{\nu 2}|\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.55)

In order to estimate $Y_{\nu 1}$ we introduce now the quantity

$$\begin{aligned} \Lambda _n^{(j1)}=\frac{1}{n}\mathrm{Tr}\,\mathbf {R}^{(j)}-s(z)+\frac{s(z)}{n}+\frac{1}{n^2} \mathrm{Tr}\,{\mathbf {R}^{(j)}}^2s(z). \end{aligned}$$

Recall that

$$\begin{aligned} \eta _{j1}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j}[(\mathbf {R}^{(j)})^2]_{ll},\quad \eta _{j2}=\frac{1}{n}\sum _{k\ne l\in \mathbb {T}_j}X_{jk}X_{jl}[(\mathbf {R}^{(j)})^2]_{kl},\nonumber \\ \eta _{j3}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j}(X_{jl}^2-1)[(\mathbf {R}^{(j)})^2]_{ll}. \end{aligned}$$

(7.56)

Note that

$$\begin{aligned} |\eta _{j1}|=\frac{1}{n}|\mathrm{Tr}\,{\mathbf {R}^{(j)}}^2|\le v^{-1}\mathrm {Im}\;\!m_n^{(j)}(z). \end{aligned}$$

(7.57)

We use that (see Lemma 5.5 in [11])

$$\begin{aligned} \varepsilon _{j4}=\frac{1}{n}(1+\eta _{j1}+\eta _{j2}+\eta _{j3})R_{jj}. \end{aligned}$$

(7.58)

Note that

$$\begin{aligned} \delta _{nj}:= & {} \Lambda _n-\widetilde{\Lambda }_n^{(j)}=\varepsilon _{j4}-\frac{s(z)}{n}-\frac{1}{n}\eta _{j0}s(z)\\= & {} \frac{1}{n}(R_{jj}-s(z))(1+\eta _{j1}) +\frac{1}{n}(\eta _{j2}+\eta _{j3})R_{jj}. \end{aligned}$$

This yields

$$\begin{aligned} |\delta _{nj}|\le \frac{1}{n}(1+|\eta _{j1}|)|R_{jj}-s(z)|+\frac{1}{n}|\eta _{j2}+\eta _{j3}|)|R_{jj}|. \end{aligned}$$

(7.59)

We represent $Y_{\nu 1}$ in the form

$$\begin{aligned} Y_{\nu 1}=Z_{\nu 1}+Z_{\nu 2}+Z_{\nu 3}, \end{aligned}$$

where

$$\begin{aligned} Z_{\nu 1}&=-\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }{\overline{\Lambda }}_n^{(j1)}}{(z+m_n^{(j)}(z))(z+m_n^{(j)}(z)+s(z))},\\ Z_{\nu 2}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }\overline{\delta }_{nj}}{(z+m_n^{(j)}(z))(z+m_n(z)+s(z))},\\ Z_{\nu 3}&=\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }{\overline{\Lambda }}_n\varepsilon _{j4}}{(z+m_n^{(j)}(z))(z+m_n^{(j)}(z)+s(z))(z+m_n(z)+s(z))},\\ Z_{\nu 4}&=-\frac{1}{n}\sum _{j=1}^n\mathbf{E}\frac{\varepsilon _{j\nu }\varepsilon _{j4}\overline{\delta }_{nj}}{(z+m_n^{(j)}(z))(z+m_n^{(j)}(z)+s(z))(z+m_n(z)+s(z))}. \end{aligned}$$

First, note that by conditional independence

$$\begin{aligned} Z_{\nu 1}=0. \end{aligned}$$

(7.60)

Furthermore, applying Hölder’s inequality, we get

$$\begin{aligned} |Z_{\nu 3}|&\le \frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)|^4|z+m_n^{(j)}(z)+s(z)|^4}\\&\quad \times \mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j4}|^4}{|z+m_n(z)+s(z)|^4}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

Using Corollary 7.17 with $\alpha =4$ and Lemmas 7.22 and 7.5, we obtain

$$\begin{aligned} |Z_{\nu 3}|\le \frac{C}{(nv)^{\frac{3}{2}}|z^2-4|^{\frac{1}{4}}}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

For $z\in \mathbb {G}$ we may rewrite this bound using Lemma 7.13

$$\begin{aligned} |Z_{\nu 3}|\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.61)

Furthermore, note that

$$\begin{aligned} |1+\eta _{j1}|\le v^{-1}\mathrm {Im}\;\!\{z+m_n^{(j)}(z)\}\le \mathrm {Im}\;\!\{z+m_n^{(j)}(z)+s(z)\} . \end{aligned}$$

This inequality together with (7.59) implies that

$$\begin{aligned} |Z_{\nu 4}|\le \widetilde{Z}_{\nu 4}+\widehat{Z}_{\nu 4}, \end{aligned}$$

(7.62)

where

$$\begin{aligned} \widetilde{Z}_{\nu 4}&=\frac{1}{n^2v}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }\varepsilon _{j4}||R_{jj}-s(z)|}{|z+m_n^{(j)}(z)||z+m_n(z)+s(z)|}\\ \widehat{Z}_{\nu 4}&=\frac{1}{n^2}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }\varepsilon _{j4}||\eta _{j2}+\eta _{j3}||R_{jj}|}{|z+m_n^{(j)}(z)||z+m_n^{(j)}(z)+s(z)||z+m_n(z)+s(z)|}. \end{aligned}$$

By representation (3.2), we have

$$\begin{aligned} |R_{jj}-s(z)|\le |\Lambda _n||R_{jj}|+|\varepsilon _j||R_{jj}|. \end{aligned}$$

(7.63)

This implies that

$$\begin{aligned} \widetilde{Z}_{\nu 4}\le \widetilde{Z}_{\nu 41}+\widetilde{Z}_{\nu 42}, \end{aligned}$$

where

$$\begin{aligned} \widetilde{Z}_{\nu 41}&=\frac{1}{n^2v}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }\varepsilon _{j4}||\Lambda _n||R_{jj}|}{|z+m_n^{(j)}(z)||z+m_n(z)+s(z)|}\\ \widetilde{Z}_{\nu 42}&= \frac{1}{n^2v}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }\varepsilon _{j4}||\varepsilon _j||R_{jj}|}{|z+m_n^{(j)}(z)||z+m_n(z)+s(z)|}. \end{aligned}$$

Applying Hölder inequality, we get

$$\begin{aligned} \widetilde{Z}_{\nu 41}&\le \displaystyle \frac{1}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\mathbf{E}^{\frac{1}{8}}\left( \frac{1}{n}\sum _{j=1}^n| \varepsilon _{j\nu }|^4\right) ^2\mathbf{E}^{\frac{1}{16}}\left( \frac{1}{n}\sum _{j=1}^n|R_{jj}|^{16} \right) \\&\quad \times \mathbf{E}^{\frac{1}{16}}\left( \frac{1}{n}\sum _{j=1}^n\frac{1}{|z+m_n^{(j)}|^{16}}\right) \frac{1}{n}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j4}|^4}{|z+m_n(z)+s(z)|^4}. \end{aligned}$$

By Lemma 7.21, inequality (7.39) and Corollary 5.14, we have

$$\begin{aligned} \widetilde{Z}_{\nu 41}\le \frac{C}{n^2v^2}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2. \end{aligned}$$

(7.64)

Furthermore, applying Hölder inequality again, we get

$$\begin{aligned} \widetilde{Z}_{\nu 42}\le & {} \frac{1}{nv}\mathbf{E}^{\frac{1}{4}}\left( \frac{1}{n}\sum _{j=1}^n| \varepsilon _{j\nu }|^4\right) \mathbf{E}^{\frac{1}{4}}\left( \frac{1}{n}\sum _{j=1}^n\frac{| \varepsilon _{j4}|^4}{|z+m_n(z)+s(z)|^4}\right) \nonumber \\&\times \mathbf{E}^{\frac{1}{4}}\left( \frac{1}{n}\sum _{j=1}^n|\varepsilon _{j}|^4\right) \mathbf{E}^{\frac{1}{8}}\left( \frac{1}{n}\sum _{j=1}^n|R_{jj}|^8\right) \mathbf{E}^{\frac{1}{8}}\left( \frac{1}{n}\sum _{j=1}^n\frac{1}{|z+m_n^{(j)}|^8}\right) .\qquad \quad \end{aligned}$$

(7.65)

The last inequality, Corollaries 5.14, 7.17, Lemma 7.22 together imply

$$\begin{aligned} \widetilde{Z}_{\nu 42}\le \frac{C}{n^2v^2}. \end{aligned}$$

(7.66)

Inequalities (7.64) and (7.66) together imply

$$\begin{aligned} |Z_{\nu 4}|\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2+\frac{C}{n^2v^2}. \end{aligned}$$

(7.67)

To bound $Z_{\nu 2}$ we first apply inequality (7.42) and obtain

$$\begin{aligned} |Z_{\nu 2}|\le \frac{C}{n}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }||\delta _{nj}|}{|z+m_n^{(j)}(z)||z+m_n^{(j)}(z)+s(z)|}. \end{aligned}$$

Furthermore, similar to bound of $Z_{\nu 4}$—inequality (7.62)—we may write

$$\begin{aligned} |Z_{\nu 2}|\le \widetilde{Z}_{\nu 2}+\widehat{Z}_{\nu 2}, \end{aligned}$$

where

$$\begin{aligned} \widetilde{Z}_{\nu 2}&=\frac{C}{n^2}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }||R_{jj}-s(z)|}{|z+m_n^{(j)}(z)|},\\ \widehat{Z}_{\nu 2}&= \frac{C}{n^2}\sum _{j=1}^n\mathbf{E}\frac{|\varepsilon _{j\nu }||\eta _{j2}+\eta _{j3}||R_{jj}|}{|z+m_n^{(j)}(z)||z+m_n^{(j)}(z)+s(z)|} . \end{aligned}$$

Applying inequality (7.63) and Cauchy–Schwartz inequality, we get

$$\begin{aligned} \widetilde{Z}_{\nu 2}\le & {} \displaystyle \mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2\frac{C}{n^2}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}| \varepsilon _{j\nu }|^4\mathbf{E}^{\frac{1}{8}}\frac{1}{|z+m_n^{(j)}(z)|^8}\mathbf{E}^{\frac{1}{8}}|R_{jj}|^8\nonumber \\&+\displaystyle \frac{C}{n^2}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}{|\varepsilon _{j\nu }|^4} \mathbf{E}^{\frac{1}{4}}{|\varepsilon _{j}|^4}\mathbf{E}^{\frac{1}{8}}\frac{1}{|z+m_n^{(j)}(z)|^8}\mathbf{E}^{\frac{1}{8}}|R_{jj}|^8. \end{aligned}$$

(7.68)

Lemmas 7.15, 7.16, 7.22, inequality (7.39) and Corollary 5.14 together imply

$$\begin{aligned} \widetilde{Z}_{\nu 2}\le \frac{C}{nv}\mathbf{E}|^{\frac{1}{2}}|\Lambda _n|^2+\frac{C}{n^2v^2}. \end{aligned}$$

(7.69)

Applying now Hölder’s inequality, we get

$$\begin{aligned} |\widehat{Z}_{\nu 2}|\le & {} \frac{C}{n^2}\sum _{j=1}^n\mathbf{E}^{\frac{1}{4}}\frac{|\varepsilon _{j\nu }|^4}{|z+m_n^{(j)}(z)+s(z)|^2} \mathbf{E}^{\frac{1}{4}}\frac{ |\eta _{j2}+\eta _{j3}|^4}{|z+m_n^{(j)}(z)+s(z)|^2}\\&\times \mathbf{E}^{\frac{1}{4}}|R_{jj}|^4\mathbf{E}^{\frac{1}{4}} \frac{1}{|z+m_n^{(j)}(z)|^4} . \end{aligned}$$

The last inequality together with Lemmas 7.22, 7.20, 7.21 and Corollary 5.14, 7.17 implies

$$\begin{aligned} |\widehat{Z}_{\nu 2}|\le \frac{C}{n^2v^2}. \end{aligned}$$

(7.70)

Combining now inequalities (7.46), (7.47), (7.54), (7.55), (7.60), (7.61), (7.67), (7.69), (7.70), we get

$$\begin{aligned} \mathbf{E}|\Lambda _n|^2\le \frac{C}{nv}\mathbf{E}^{\frac{1}{2}}|\Lambda _n|^2+\frac{C}{n^2v^2}. \end{aligned}$$

(7.71)

Applying Lemma 7.4 with $t=2$, $r=1$ completes the proof of Lemma 7.24. $\square$

We relabel $\eta _{j2},\,\eta _{j3}$ and introduce the following quantity

$$\begin{aligned} \beta _{j1}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j}[(R^{(j)})^2]_{ll}- \frac{1}{n}\sum _{l=1}^n[(R)^2]_{ll},\\ \beta _{j2}&=\frac{1}{n}\sum _{l\ne k\in \mathbb {T}_j}X_{jl}X_{jk}[(R^{(j)})^2]_{lk}=\eta _{j2},\\ \beta _{j3}&=\frac{1}{n}\sum _{l\in \mathbb {T}_j}(X^2_{jl}-1)[(R^{(j)})^2]_{ll}=\eta _{j3}. \end{aligned}$$

Lemma 7.25

Assuming the conditions of Theorem 1.1, we have, for $\nu =2,3$,

$$\begin{aligned} \mathbf{E}\{|\beta _{j\nu }|^2|\mathfrak {M}^{(j)}\}\le \frac{C}{nv^3}\mathrm {Im}\;\!m_n^{(j)}(z). \end{aligned}$$

Proof

We recall that by $C$ we denote the generic constant depending on $\mu _4$ and $D_0$ only. By definition of $\beta _{j\nu }$ for $\nu =2,3$, conditioning on $\mathfrak {M}^{(j)}$, we get

$$\begin{aligned} \mathbf{E}\{|\beta _{j2}|^2\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^2}\sum _{l\ne k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2 \le \frac{C}{n^2}\sum _{l, k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2,\\ \mathbf{E}\{|\beta _{j3}|^2\big |\mathfrak {M}^{(j)}\}&\le \frac{C}{n^2}\sum _{l\in \mathbb {T}_j}|[(R^{(j)})^2]_{ll}|^2\le \frac{C}{n^2}\sum _{l, k\in \mathbb {T}_j}|[(R^{(j)})^2]_{kl}|^2. \end{aligned}$$

Applying Lemma 7.6, we get the claim. Thus Lemma 7.25 is proved. $\square$

Lemma 7.26

Assuming the conditions of Theorem 1.1, we have, for $j=1,\ldots ,n$,

$$\begin{aligned} \mathbf{E}|\beta _{j1}|\le \frac{C}{nv^2}. \end{aligned}$$

Proof

Let $\mathcal {F}_n^{(j)}(x)$ denote empirical spectral distribution function of matrix $\mathbf {W}^{(j)}$. According to interlacing eigenvalues Theorem (see [18], Theorem 4.38]) we have

$$\begin{aligned} \sup _x|\mathcal {F}_n(x)-\mathcal {F}_n^{(j)}(x)|\le \frac{C}{n}. \end{aligned}$$

Furthermore, we represent

$$\begin{aligned} \beta _{j1}=\int _{-\infty }^{\infty }\frac{1}{(x-z)^2}d(\mathcal {F}_n(x)- \mathcal {F}_n^{(j)}(x))+\frac{1}{n(n-1)}\mathrm{Tr}\,(\mathbf {R}^{(j)})^2. \end{aligned}$$

Integrating by parts, we get the claim.

Thus Lemma 7.26 is proved. $\square$